Bhagya

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5

Question 1.
A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm³.
Solution:
Since, diameter of the cylinder = 10 cm 10
∴ Radius of the cylinder (r) = \(\frac{10}{2}\) cm = 5cm
⇒ Length of wire in one round = 2πr
= 2 × 3.14 × 5 cm = 31.4 cm
∵ Diameter of wire = 3 mm = \(\frac{3}{10}\) cm
∴ The thickness of cylinder covered in one round = \(\frac{3}{10}\) cm
⇒ Number of rounds (turns) of the wire to cover 12 cm = \(\frac{12}{3 / 10}=12 \times \frac{10}{3}\) = 40
∴ Length of wire required to cover the whole surface = Length of wire required to complete 40 rounds
= l = 40 × 31.4 cm = 1256 cm
Now, radius of the wire = \(\frac{3}{2}\) mm = \(\frac{3}{20}\) cm
∴ Volume of wire = πr²l
= 3.14 × \(\frac{3}{20} \times \frac{3}{20}\) × 1256 cm³
∵ Density of wire = 8.88 g/cm³
∴ Mass of the wire = [Volume of the wire] × density

Question 2.
A right triangle, whose sides are 3cm and 4cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate).
Solution:
Let us consider the right ABAC, right angled at A such that AB = 3 cm, AC = 4 cm
∴ Hypotenuse BC = \(\sqrt{3^{2}+4^{2}}\) = 5cm
Obviously, we have obtained two cones on the same base AA’ such that radius = DA or DA’.
Now,

Question 3.
A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm³ of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?
Solution:
Dimensions of the cistern are 150 cm, 120 cm and 100 cm.
∴ Volume of the cistern = 150 × 120 × 110 cm³ = 1980000 cm³
Volume of water contained in the cistern = 129600 cm³
∴ Free space (volume) which is not filled with water = (1980000 – 129600) cm³
= 1850400 cm³
Now, volume of one brick
= (22.5 × 7.5 × 6.5) cm³ = 1096.875 cm³
∴ Volume of water absorbed by one brick
= \(\frac{1}{17}\) × 1096.875 cm³
Let n bricks can be put in the cistern.
∴ Volume of water absorbed by n bricks
= \(\frac{n}{17}\) × 1096.875 cm³
∴ Volume occupied by n bricks = [free space in the cistern + volume of water absorbed by n bricks]
⇒ [n × 1096.875] = [1850400 + \(\frac{n}{17}\)(1096.875)]
⇒ 1096.875 n – \(\frac{n}{17}\)(1096.875) = 1850400
⇒ (n – \(\frac{n}{17}\)) × 1096.875 = 1850400
⇒ \(\frac{16}{17} n=\frac{1850400}{1096.875} \Rightarrow n=\frac{1850400}{1096.875} \times \frac{17}{16}\)
= 1792.4102 ≈ 1792
Thus, 1792 bricks can be put in the cistern.

Question 4.
In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280 km², show that the total rainfall was approximately equivalent to the addition to the norma water of three rivers each 1072 km long, 75 m wide and 3 m deep.
Solution:
Volume of three rivers = 3 {(Surface area of a river) × Depth}

Since 0.7236 km³ ≠ 9.728 km³
∴ The additional water in the three rivers is not equivalent to the rainfall.

Question 5.
An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see figure).

Solution:
We have,
For the cylindrical part:
Diameter = 8 cm ⇒ Radius (r) = 4 cm
Height = 10 cm

Question 6.
Derive the formula for the curved surface area and total surface area of the frustum of a cone.
Solution:

Since, ∆OC₁Q ~ ∆OC₂S [By AA similarity]

Now, the total surface area of the frustum = (curved surface area) + (base surface area) + (top surface area)
= πl(r₁ + r₂) + πr₂² + πr₁² = π [(r₁ + r₂)l + r₁² + r₂²]

Question 7.
Derive the formula for the volume of the frustum of a cone.
Solution:
We have,
[Volume of the frustum RPQS] = [Volume of right circular cone OPQ] – [Volume of right circular cone ORS]

Since, ∆OC₁Q ~ ∆OC₂S [By AA similarity]

From (1) and (2), we have
Volume of the frustum RPQS

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.3

Question 1.
Evaluate:
(i) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
(ii) \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\)
(iii) cos48° – sin42°
(iv) cosec31°- sec59°
Solution:

(iii) cos 48° – sin 42°
cos 48° = cos (90° – 42°) = sin 42°
[∵ cos (90° – A) = sin A]
∴ cos 48° – sin 42° = sin 42° – sin 42° = 0
(iv) cosec 31° – sec 59°
cosec 31° = cosec (90° – 59°) = sec 59° [ ∵ cosec (90° – A) = sec A]
∴ cosec 31° – sec 59° = sec 59° – sec 59° = 0

Question 2.
Show that:
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Solution;
(i) L.H.S. = tan 48° tan 23° tan 42° tan 67°
= tan (90° – 42°) tan 23° tan 42° tan (90° – 23°)
= cot 42° tan 23° tan 42° cot 23° [ ∵ tan (90° – A) = cot A]

(ii) L.H.S. = cos 38° cos 52° – sin 38° sin 52°
= cos 38° cos (90° – 38°) – sin38° sin(90° – 38°)
= cos 38° sin 38° – sin 38° cos 38° [ ∵ sin(90° – A) = cosA and cos(90° – A)= sinA]
= 0 = R.H.S.
⇒ L.H.S. = R.H.S.

Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution:
tan 2A = cot(A – 18°)
cot(90 – 2A) = cot (A – 18)
90 – 2A = A – 18
90 + 18 = A + 2A
3A= 108
A = \(\frac {108}{3}\)
A = 36°.

Question 4.
If tan A = cotB, prove that A + B = 90°.
Solution:
tan A = cot B and cot B = tan (90° – B) [∵ tan (90° – θ) = cot θ]
∴ A = 90° – B ⇒ A + B = 90°

Question 5.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
sec 4A= cosec (A – 20)
cosec (90 – 4A) = cosec (A – 20)
90 – 4A = A – 20
90 + 20 =A + 4A
110 = 5A
A = \(\frac{110}{5}\)
A = 22°.

Question 6.
If A, 6 and C are interior angles of a triangle ABC, then show that \(\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}\)
Solution:
Since, sum of the angles of ∆ABC is 180° i.e.,
A + B + C = 180°
∴ B + C = 180° – A
Dividing both sides by 2, we get

Question 7.
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution:
sin 67° + cos 75° = sin (90 – 23) + cos (90 – 15)
[sin(90 – θ) = cosθ
cos(90 – θ) = sinθ]
= cos 23 + sin 15

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 12 प्रायिकता Ex 12.1 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Book Solutions Chapter 12 प्रायिकता Ex 12.1

ग्राफीय विधि से निम्न रैखिक प्रोग्रामन समस्याओं को हल कीजिए।
प्रश्न 1.
निम्न अवरोधों के अन्तर्गत Z = 3x + 4y का अधिकतमीकरण कीजिए-
x + y ≤ 4, x ≥ 0, y ≥ 0
हल:
अधिकतम
Z = 3x +4y
x + y ≤ 4
x ≥ 0, y ≥ 0

पहले सभी समस्याओं को समीकरण के रूप में लिखने पर
x + y = 4
x = 0   ….(ii)
y = 0   …(iii)
अब ग्राफ बनाने पर सुसंगत क्षेत्र OAB प्राप्त होता है।
Z के मान की गणना प्रत्येक कोणीय बिन्दु पर करने पर

अतः Z का अधिकतम मान 16 बिन्दु (0, 4) पर है।

प्रश्न 2.
निम्न अवरोधों के अन्तर्गत Z = – 3x + 4y का न्यूनतमीकरण कीजिए-
x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0
हल:
सर्वप्रथम सभी असमीकरणों को समीकरण के रूप में लिखने पर
x + 2y =  8 ……(i)
3x + 2y = 12 ……(ii)
x = 0, y = 0 ……(iii)
अब आलेख बनाने पर सुसंगत क्षेत्र OABC प्राप्त होता है।
समी०
(i) व
(ii) को हल करने पर
x = 2, y = 3 प्राप्त होता है।
∵ रेखा
(i) व
(ii) बिन्दु (2, 3) पर मिलती हैं।

अतः बिन्दु (4,0) पर Z का मान न्यूनतम है।

प्रश्न 3:
निम्न अवरोधों के अन्तर्गत Z = 5x + 3y का अधिकतमीकरण कीजिए
3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0
हल:
सर्वप्रथम असमीकरणों को समीकरणों के रूप में लिखने पर,
3x + 5y =15 ….(i)
5x + 2y=10 …(ii)
x = 0 …(iii)
y = 0 …(iv)
अब समीकरणों का ग्राफ बनाने पर सुसंगत क्षेत्र OABC प्राप्त होता है।

अब Z का मान प्रत्येक कोणीय बिन्दु पर ज्ञात करने पर

अतः बिन्दु Z का अधिकतम मान \(\frac{245}{19}\) है।
\(\left(\frac{20}{19}, \frac{45}{19}\right)\)

प्रश्न 4.
निम्न अवरोधों के अन्तर्गत z = 3x + 5y का न्यूनतमीकरण कीजिए-
x + 3y ≥ 3, x + y ≥ 2, x ≥ 0, y ≥ 0
हल:
सर्वप्रथम सभी असमीकरणों को समीकरणों के रूप में लिखने पर,
x + 3y =3 …(i)
x + y=2 …(ii)
x = 0 …(iii)
y = 0 …(iv)
अब ग्राफ बनाने पर सुसंगत क्षेत्र X ABCY प्राप्त होता हो।

सारणी से Z का न्यूनतम मान बिन्दु B\(\left(\frac{3}{2}, \frac{1}{2}\right)\) पर 7 है।

प्रश्न 5.
निम्न अवरोधों के अन्तर्गत Z = 3x +2y का अधिकतमीकरण कीजिए-
x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0
हल:
सर्वप्रथम असमीकरणों को समीकरणों के रूप में लिखने पर
x + 2y = 10 …(i)
3x + y = 15 …(ii)
x = 0, y = 0 …(iii)
अब ग्राफ बनाने पर सुसंगत क्षेत्र OABC प्राप्त होता है।
समीकरण (i) व (ii) को हल करने पर,
x =4, y=3
ये रेखाएँ बिन्दु B(4,3) पर प्रतिच्छेदित करती हैं।

सारणी से बिन्दु (4,3) पर Z का अधिकतम मान 18 है।

प्रश्न 6.
निम्न अवरोधों के अन्तर्गत Z = x+2y का न्यूनतमीकरण कीजिए-
2x+y ≥ 3, x+2y ≥ 6, x,y ≥ 0
हल:
सर्वप्रथम असमीकरणों को समीकरणों के रूप में लिखने पर
2x + y =3 ….(i)
x + 2y =6 …(ii)
x = 0, y = 0 …(iii)

ग्राफ बनाने पर सुसंगत क्षेत्र XABY प्राप्त होता है।
z का मान प्रत्येक कोणीय बिन्दु पर ज्ञात करने पर

यहाँ z का प्रत्येक मान 6 है।।
अतः बिन्दुओं (6,0) और (0,3) को मिलाने वाली रेखा खण्ड पर स्थित सभी बिन्दुओं पर Z का न्यूनतम मान 6 है।

दिखाइए कि z का न्यूनतम मान दो बिन्दुओं से अधिक बिन्दुओं पर घटित होता है।

प्रश्न 7.
निम्न अवरोधों के अन्तर्गत Z = 5x + 10y का न्यूनतमीकरण तथा अधिकतमीकरण कीजिए-
x + 2y ≤ 120; x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0
हल:
दिया है : उद्देश्य फलन : Z = 5x + 10y
अवरोध : x + 2y ≤ 120, x + y ≥ 60
x – 2y ≥ 0, x, y ≥ 0

(1) x +2y ≤ 120 का आरेख,
रेखा x + 2y =120, बिन्दु A(120, 0) और बिन्दु B(0, 60) से होकर जाती है।
∴ x + 2y =120 का आरेख रेखा AB है।
x + 2y ≤ 120 में x = 0, y = 0 रखने पर,
0 ≤ 120, जो सत्य है।
∴ x +2y ≤ 120 के क्षेत्र में बिन्दु रेखा AB पर और उसके नीचे मूल बिन्दु की ओर स्थित है।
(2) x + y ≥ 60 का आरेख
रेखा x + y = 60, बिन्दु P(60, 0), B(0, 60) से होकर जाती है।
∴ x + y = 60 का आरेख रेखा PB है।
x + y ≥ 60 में x = 0, y = 0 रखने पर, 0 ≥ 60 जो सत्य नहीं है।
⇒ x +y ≥ 60 क्षेत्र के बिन्दु रेखा PB पर और उसके ऊपर होते हैं।
(3) x – 2y ≥ 0 का आरेख
रेखा x – 2y = 0 मूल बिन्दु 0 और Q(120, 60) से होकर जाती है।
∴ x – 2y ≥ 0 का आरेख रेखा OQ है।
x – 2y ≥ 0 में x =1, y = 0 रखने पर 1 ≥ 0 जो सत्य है।
⇒ (1, 0) इस क्षेत्र में स्थित है। x – 2y ≤ 0 क्षेत्र के बिन्दु रेखा OQ पर और इसके नीचे (1, 0) की ओर हैं।
(4) x ≥ 0 क्षेत्र के बिन्दु y- अक्ष पर और y- अक्ष के दायीं ओर है।
(5) y ≥ 0 क्षेत्र के बिन्दु x- अक्ष पर और इसके ऊपर हैं।
इस समस्या का सुसंगत क्षेत्र PSRA है।
जबकि बिन्दु S(40, 20) PB: x + y = 60 और OQ: x – 2y = 0 का प्रतिच्छेद बिन्दु है।
और R(60, 30), AB: x + 2y =120 और x – 2y = 0 का प्रतिच्छेद बिन्दु है।
उद्देश्य फलन : Z = 5x + 10y
बिन्दु A(120, 0) पर,
Z = 5 x 120 + 10 x 0 = 600
बिन्दु R(60, 30) पर,
Z = 5 x 60 + 10 x 30
= 300 + 300 = 600
बिन्दु S(40, 20) पर,
Z = 5 x 40 + 10x 20
= 200 + 200 = 400
बिन्दु P(60, 0) पर,
Z = 5 x 60 + 10 x 0
= 300 + 0 = 300
⇒ Z का न्यूनतम मान P(60, 0) पर 300 है।
और Z का अधिकतम मान RA के सभी बिन्दुओं पर 600 है।

प्रश्न 8.
निम्न अवरोधों के अन्तर्गत Z = x + 2y का न्यूनतमीकरण तथा अधिकतमीकरण कीजिए-
x + 2y ≥ 100, 2x – y ≤ 0, 2x +y ≤ 200, x, y ≥ 0
हल:
सर्वप्रथम असमीकरणों को समीकरणों के रूप में लिखने पर
x + 2y = 100
2x – y = 0
2x + y = 200
x = 0 y=0
ग्राफ बनाने पर सुसंगत क्षेत्र BEDC प्राप्त होता है।
समी० 2x + y = 200 तथा 2x – y = 0 को हल करने पर x = 50, y = 100 प्राप्त होता है।
⇒ D(50,100)
पुनः समी० x + 2y = 100 तथा 2x – y = 0 को हल करने पर, x = 20, y = 40 प्राप्त होता है।

अतः Z का न्यूनतम मान 100 है तथा अधिकतम मान बिन्दु (0, 200) पर 400 है।

प्रश्न 9.
निम्न अवरोधों के अन्तर्गत Z = – x + 2y का अधिकतमीकरण कीजिए-
x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0
हल:
दिया है : उद्देश्य फलन :
Z = – x + 2y
(1) x + y ≥ 5 का आरेख
रेखा x + y =5, बिन्दु A(5, 0) और B(0, 5) से होकर जाती है।
∴ x + y =5 का आरेख रेखा AB है।
x + y ≥ 5 में x =0, y=0 रखने पर,
0 ≥ 5 जो सत्य नहीं है।
∴ x + y ≥ 5 क्षेत्र के बिन्दु रेखा AB पर और उसके ऊपर है।

(2) x + 2y ≥ 6 का आरेख
रेखा x + 2y = 6, बिन्दु C (6, 0) और D (0, 3) से होकर जाती है।
∴ x + 2y = 6 रेखा का आरेख रेखा CD है।
⇒ x + 2y ≥ 6 में x =0, y = 0 रखने पर, 0 ≤ 6 जो सत्य नहीं है।
∴ x + 2y ≥ 6 का क्षेत्र के बिन्दु CD पर या उसके ऊपर है।
(3) x ≥ 3 क्षेत्र के बिन्दु रेखा PQ: x =3 पर या उसके दायीं ओर है।
(4) y ≥ 0 क्षेत्र के बिन्दु x- अक्ष पर और उसके ऊपर होते हैं। समस्या का सुसंगत क्षेत्र PQRCX है।
बिन्दु रेखा PQ =3 और AB: x + y =5 का प्रतिच्छेदन बिन्दु Q के निर्देशांक (3, 2) है।
बिन्दु R रेखा CD: x + 2y = 6 और AB: x + y =5 का प्रतिच्छेदन बिन्दु (4, 1) है।
उद्देश्य फलन : Z = – x + 2y
अब, बिन्दु Q (3, 2) पर,
Z = – 3 + 2 x 2 = – 3 + 4 =1
बिन्दु R(4, 1) पर,
Z = – 4 + 2 x 1 = – 4 + 2 = – 2
बिन्दु C(6, 0) पर,
Z = – 6 + 0 = – 6
⇒ z का अधिकतम मान 1 है परन्तु सुसंगत क्षेत्र अपरिबद्ध है तो – x + 2y > 1 क्षेत्र पर विचार करें। .
– x + 2y > 1 तथा सुसंगत क्षेत्र में अनेकों बिन्दु उभयनिष्ठ है।
अतः Zका कोई अधिकतम मान नहीं है।

प्रश्न 10.
निम्न अवरोधों के अन्तर्गत Z = x + y का अधिकतमीकरण कीजिए-
x – y ≤ – 1, – x + y ≤ 0, x, y ≥ 0
हल:
(i) x – y ≤ -1 का क्षेत्र
रेखा x – y = – 1 बिन्दु A(-1,0), B(0, 1) से होकर जाती है, जो AB आरेख है।
x – y ≤ – 1 में x =0, y = 0 रखने पर,
0 ≤ -1 जो सत्य नहीं है।
⇒ x – y ≤ – 1 के क्षेत्र बिन्दु रेखा AB पर और उसके ऊपर है।
(ii) – x + y ≤ का क्षेत्र
रेखा – x + y = 0, मूल बिन्दु O और C(1, 1) से होकर जाती है।
– x + y ≤ 0 में x = 1, y = 0 रखने पर, -1 ≤ 0 जो सत्य है।
⇒ – x + y ≤ 0 के क्षेत्र बिन्दु OC पर या उसके नीचे (1,0) ओर हैं।

(iii) x ≥ 0 क्षेत्र के बिन्दु y- अक्ष पर और -अक्ष के दायीं ओर हैं।
(iv) y ≥ 0 क्षेत्र के बिन्दु x- अक्ष पर और x- अक्ष के ऊपर स्थित हैं।
इस समस्या का कोई सुसंगत क्षेत्र नहीं है।
अतः Z का अधिकतम मान नहीं है।

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3

Question 1.
Find the area of the triangle whose vertices are:
(i) (2, 3), (-1, 0), (2, -4)
(ii) (-5, -1), (3, -5), (5, 2)
Solution:
(i) Let the vertices of the triangle be A(2, 3), B(-1, 0) and C(2, – 4)
Here, x₁ = 2, y₁ = 3
x₂ = -1, y₂ = 0
x₃ = 2, y₃ = -4
∵ Area of a triangle
= \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
∴ Area of the ∆ABC
= \(\frac{1}{2}\) [2{0 – (-4)} + (-1){-4 – (3)} + 2{3 – 0}]
= \(\frac{1}{2}\) [2(0 + 4) + (-1)(-4 – 3) + 2(3)]
= \(\frac{1}{2}\) [8 + 7 + 6] = \(\frac{1}{2}\) [21] = \(\frac{21}{2}\) sq. units

(ii) Let the vertices of the triangle be A(-5, -1), B(3, -5) and C(5, 2)
Here, x₁ = -5, y₁ = -1
x₂ = 3, y₂ = -5
x₃ = 5, y₃ = 2
∵ Area of a triangle
= \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
∴ Area of the ∆ABC
= \(\frac{1}{2}\) [-5{-5 – 2} + 3{2 – (-1)} + 5{-1 – (-5)}]
= \(\frac{1}{2}\) [-5{-7} + 3{2 + 1} + 5{-1 + 5}]
= \(\frac{1}{2}\) [35 + 3(3) + 5(4)]
= \(\frac{1}{2}\) [35 + 9 + 20] = \(\frac{1}{2}\) × 64 = 32 sq. units

Question 2.
In each of the following find the value of’k’, for which the points are collinear.
(i) (7, -2), (5, 1), (3, k)
(ii) (8, 1), (k, -4), (2, -5)
Solution:
The given three points will be collinear if the triangle formed by them has zero area.
(i) Let A(7, -2), B(5, 1) and C(3, k) be the vertices of a triangle.
∴ The given points will be collinear, if ar (∆ABC) = 0
or \(\frac{1}{2}\) [7(1 – k) + 5(k + 2) + 3(-2 – 1)] = 0
⇒ 7 – 7k + 5k + 10 + (-6) – 3 = 0
⇒ 17 – 9 + 5k – 7k = 0
⇒ 8 – 2k = 0 ⇒ 2k = 8 ⇒ k = \(\frac{8}{2}\) = 4
The required value of k = 4.
(ii) \(\frac{1}{2}\) [8(- 4 + 5) + k(- 5 -1) + 2(1 + 4)] = 0
⇒ 8 – 6k + 10 = 0 ⇒ 6k = 18 ⇒ k = 3.

Question 3.
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Solution:
Let the vertices of the triangle be A(0, -1), B(2, 1) and C(0, 3).
Let D, E and F be the mid-points of the sides BC, CA and AB respectively.
∴ Coordinates of D are


ar(∆DEF) : ar(∆ABC) = 1 : 4.

Question 4.
Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).
Solution:
Let A(- 4, – 2), B(- 3, – 5), C(3, – 2) and D(2, 3) be the vertices of the quadrilateral.
Let us join diagonal BD.

Now, ar(∆ABD)

Question 5.
You have studied in class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ABC whose vertices are A(A, -6), B(3, -2) and C(5, 2).
Solution:
Here, the vertices of the triangle are A(4, -6), B(3, -2) and C(5, 2).
Let D be the midpoint of BC.
∴ The coordinates of the point D are
\(\left(\frac{3+5}{2}, \frac{-2+2}{2}\right)\) or ( 4, 0)
AD divides the triangle ABC into two parts i.e., ∆ABD and ∆ACD.

ar(∆ADC) = \(\frac{1}{2}\) [4(0 – 2) + 4(2 + 6) + 5(-6 – 0)]
= \(\frac{1}{2}\) [-8 + 32 – 30] = \(\frac{1}{2}\) [-6] = -3
= 3 sq. units (numerically) ………… (2)
From (1) and (2),
ar(∆ABD) = ar(∆ADC)
Thus, a median divides the triangle into two triangles of equal areas.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 15 Probability Ex 15.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 15 Probability Ex 15.1

Question 1.
Complete the following statements:
(i) Probability of an event E + Probability of the event ‘not E’ = _________
(ii) The probability of an event that cannot happen is ______.Such an event is called _________.
(iii) The probability of an event that is certain to happen is ______ Such an event is called ______
(iv) The sum of the probabilities of all the elementary events of an experiment is ______.
(v) The probability of an event is greater than or equal to _______ and less than or equal to ______.
Solution:
(i) 1 : Probability of an event E + Probability of the event ‘not E’ = 1.
(ii) 0, impossible: The probability of an event that cannot happen is 0. Such an event is called impossible event.
(iii) 1, certain: The probability of an event that is certain to happen is 1. Such an event is called a sure or certain event.
(iv) 1: The sum of the probabilities of all the elementary events of an experiment is 1.
(v) 0, 1: The probability of an event is greater than or equal to 0 and less than or equal to 1.

Question 2.
Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
Solution:
(i) It depends on various factors such as whether the car will start or not. So, the probability of car will start does not equal to the probability of car will not start.
∴ The outcomes are not equally likely.
(ii) It depends on the player’s ability. So, probability that the player shot the ball is not the same as the probability that the player misses the shot.
(iii) The outcomes are equally likely as the probability of answer either right or wrong is \(\frac{1}{2}\)
(iv) The outcomes are equally likely as the probability of ‘newly born baby to be either bay or girls’ is \(\frac{1}{2}\) .

Question 3.
Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Solution:
Since on tossing a coin, the outcomes ‘head’ and ‘tail’ are equally likely, the result of tossing a coin is completely unpredictable and so it is a fairway.

Question 4.
Which of the following cannot be the probability of an event?
(A) \(\frac{2}{3}\)
(B) -1.5
(C) 15%
(D) 0.7
Solution:
Since, the probability of an event cannot be negative.
∴ -1.5 cannot be the probability of an event.

Question 5.
If P(E) = 0.05, what is the probability of ‘not E’?
Solution:
∵ P(E) + P(not E) = 1
∴ 0.05 + P(not E) = 1 ⇒ P(not E) = 0.95
Thus, probability of ‘not E’ = 0.95.

Question 6.
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
Solution:
(i) Since there are only lemon flavoured candies in the bag.
∴ Taking out orange flavoured candy is not possible.
⇒ Probability of taking out an orange flavoured candy = 0.

(ii) Probability of taking out a lemon flavoured candy = 1.

Question 7.
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Solution:
Let the probability of 2 students having same birthday = P(SB)
And the probability of 2 students not having the same birthday = P(NSB)
∴ P(SB) + P(NSB) = 1
⇒ P(SB) + 0.992 = 1 ⇒ P(SB) = 1 – 0.992 = 0.008

Question 8.
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is
(i) red?
(ii) not red?
Solution:
Total number of balls = 3 + 5 = 8
∴ umber of possible outcomes = 8
(i) ∵ There are 3 red balls.
∴ Number of favourable outcomes = 3
∴ P (red) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}\)
= \(\frac{3}{8}\)
(ii) Probability of the ball drawn which is not red = 1 – P(red) = \(1-\frac{3}{8}=\frac{8-3}{8}=\frac{5}{8}\)

Question 9.
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) red?
(ii) white?
(iii) not green?
Solution:
Total number of marbles = 5 + 8 + 4 = 17
∴ Number of all possible outcomes = 17
(i) ∵ Number of red marbles = 5
∴ Number of favourable outcomes = 5
∴ Probability of red marbles, P(red) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}=\frac{5}{17}\)

(ii) Number of white marbles = 8
∴ Probability of white marbles, P(white) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}=\frac{8}{17}\)
_ Number of favourable outcomes _ 8 Number of all possible outcomes 17

(iii) Number of green marbles = 4 Number of marbles which are not green
= 17-4 = 13
i.e., Favourable outcomes = 13
∴ Probability of marbles ‘not green’, P(not greeen)
\(\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}=\frac{13}{17}\)

Question 10.
A piggy bank contains hundred 50p coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin
(i) will be a 50p coin?
(ii) will not be a ₹ 5 coin?
Solution:
Number of coins 50 p = 100, ₹ 1 = 50 ₹ 2 = 20, ₹ 5 = 10
Total number of coins = 100 + 50 + 20 +10 = 180
∴ Total possible outcomes = 180

(i) For a 50 p coin:
Favourable outcomes = 100

(ii) For not a ₹ 5 coin:
Y Number of ₹ 5 coins = 10
∴ Number of ‘not ₹ 5’ coins = 180 – 10 = 170
⇒ Favourable outcomes = 170

Question 11.
Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?
Solution:
Number of male fishes = 5
Number of female fishes = 8
∴ Total number of fishes = 5 + 8 = 13
⇒ Total number of outcomes = 13
For a male fish:
Number of favourable outcomes = 5

Question 12.
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see figure), and these are equally likely outcomes. What is the probability that it will point at
(i) 8?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?

Solution:
Total number marked = 8
∴ Total number of possible outcomes = 8
(i) When pointer points at 8:
Number of favourable outcomes = 1

(ii) When pointer points at an odd number:
∵ Odd numbers are 1, 3, 5 and 7
∴ Total odd numbers from 1 to 8 = 4
⇒ Number of favourable outcomes = 4

(iii) When pointer points at a number greater than 2:
∵ The numbers 3, 4, 5, 6, 7 and 8 are greater than 2
∴ Total numbers greater than 2 = 6
⇒ Number of favourable outcomes = 6

(iv) When pointer points at a number less than 9:
∵ The numbers 1, 2, 3, 4, 5, 6, 7 and 8 are less than 9.
∴ Total numbers less than 9 = 8
∴ Number of favourable outcomes = 8

Question 13.
A die is thrown once. Find the probability of getting:
(i) a prime number;
(ii) a number lying between 2 and 6;
(iii) an odd number.
Solution:
Since, numbers on a die are 1, 2, 3, 4, 5 and 6.
∴ Total number of possible outcomes = 6
(i) Since 2, 3 and 5 are prime number.
∴ Favourable outcomes = 3

(ii) Since the numbers between 2 and 6 are 3, 4 and 5
∴ Favourable outcomes = 3

(iii) Since 1, 3 and 5 are odd numbers.
⇒ Favourable outcomes = 3

Question 14.
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds
Solution:
Number of cards in deck = 52
∴ Total number of possible outcomes = 52
(i) ∵ Number of red colour kings = 2
[∵ King of diamond and heart is red]
Number of favourable outcomes = 2

(ii) For a face card:
∵ 4 kings, 4 queens and 4 jacks are face cards
∴ Number of face cards = 12
⇒ Number of favourable outcomes = 12

(iii) Since, cards of diamond and heart are red
∴ There are 2 kings, 2 queens, 2 jacks i.e., 6 cards are red face cards.
∴ Number of favorable outcomes = 6

(iv) Since, there is only 1 jack of hearts.
∴ Number of favourable outcomes = 1

(v) There are 13 spades in a pack of 52 cards.
∴ Number of favourable outcomes = 13

(vi) ∵ There is only one queen of diamond.
∴ Number of favourable outcomes = 1

Question 15.
Five cards-the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Solution:
We have five cards.
∴ Total number of possible outcomes = 5
(i) ∵ Number of queen = 1
∴ Number of favourable outcomes = 1

(ii) The queen is drawn and put aside.
∴ Only 5 – 1 = 4 cards are left.
∴ Total number of possible outcomes = 4
(a) ∵ There is only one ace.
∴ Number of favourable outcomes = 1

(b) Since, the only queen has been put aside already.
∴ Number of favourable outcomes = 0

Question 16.
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Solution:
We have number of good pens = 132 and number of defective pens = 12
∴ Total number of pens = 132 + 12 = 144 = Total possible outcomes
There are 132 good pens.
∴ Number of favourable outcomes = 132

Question 17.
(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Solution:
Since, there are 20 bulbs in the lot.
Total number of possible outcomes = 20
(i) ∵ Number of defective bulbs = 4
∴ Favourable outcomes = 4

(ii) ∵ The bulb drawn above is not included in the lot.
∴ Number of remaining bulbs = 20 – 1 = 19.
⇒ Total number of possible outcomes = 19.
∵ Number of bulbs which are not defective = 19 – 4 = 15
⇒ Number of favourable outcomes = 15

Question 18.
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by 5.
Solution:
We have total number of discs = 90
∴ Total number of possible outcomes = 90
(i) Since the two-digit numbers are 10, 11, 12, ………, 90.
∴ Number of two-digit numbers = 90 – 9 = 81
∴ Number of favourable outcomes = 81

(ii) Perfect square from 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64 and 81.
∴ Number of perfect squares = 9
∴ Number of favourable outcomes = 9

(iii) Numbers divisible by 5 from 1 to 90 are 5, 10,15, 20, 25, 30, 35,40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90
i. e., There are 18 numbers from (1 to 90) which are divisible by 5.
∴ Numbers of favourable outcomes = 18

Question 19.
A child has a die whose six faces show the letters as given below:

The die is thrown once. What is the probability of getting (i) A? (ii) D?
Solution:
Since there are six faces of the given die and these faces are marked with letters

∴ Total number of letters = 6
∴ Total number of possible outcomes = 6
(i) ∵ Number of faces having the letter A = 2
∴ Number of favourable outcomes = 2

(ii) ∵ Number of faces having the letter D = 1
∴ Number of favourable outcomes = 1

Question 20.
20. Suppose you drop a die at random on the rectangular region shown in figure. What is the probability that it will land inside the circle with diameter 1 m?

Solution:
Here, area of the rectangle = 3m × 2m = 6 m²
And, the area of the circle = πr²

Question 21.
A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) She will buy it? (ii) She will not buy it?
Solution:
Total number of ball pens = 144
⇒ Total number of possible outcomes = 144
(i) Since there are 20 defective pens.
∴ Number of good pens = 144 – 20 = 124
⇒ Number of favourable outcomes = 124

(ii) Probability that Nuri will not buy it = 1 – [Probability that she will buy it]
= \(1-\frac{31}{36}=\frac{36-31}{36}=\frac{5}{36}\)

Question 22.
Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. An event is defined as the sum of the two numbers appearing on the top of the dice.
(i) Complete the following table

(ii) A student argues that’there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability \(\frac{1}{11}\) Do you agree with this argument? Justify your answer.
Solution:
∵ The two dice are thrown together.
∴ Following are the possible outcomes :
(1, 1) ; (1, 2); (1, 3); (1, 4); (1, 5); (1, 6).
(2, 1) ; (2, 2); (2, 3); (2, 4); (2, 5); (2, 6).
(3, 1); (3, 2); (3, 3); (3, 4); (3, 5); (3, 6).
(4, 1) ; (4, 2); (4, 3); (4, 4); (4, 5); (4, 6).
(5, 1) ; (5, 2); (5, 3); (5, 4); (5, 5); (5, 6).
(6, 1) ; (6,.2); (6, 3); (6, 4); (6, 5); (6, 6).
∴ Total number of possible outcomes is 6 × 6 = 36
(i) (a) The sum on two dice is 3 for (1, 2) and (2, 1)
∴ Number of favourable outcomes = 2
⇒ P(3) = \(\frac{2}{36}\)

(b) The sum on two dice is 4 for (1, 3), (2, 2) and (3, 1).
∴ Number of favourable outcomes = 3
⇒ P(4) = \(\frac{3}{36}\)

(c) The sum on two dice is 5 for (1, 4), (2, 3), (3, 2) and (4,1)
∴ Number of favourable outcomes = 4
⇒ P(5) = \(\frac{5}{36}\)

(d) The sum on two dice is 6 for (1, 5), (2, 4), (3, 3), (4, 2) and (5,1)
∴ Number of favourable outcomes = 5
⇒ P(6) = \(\frac{5}{36}\)

(e) The sum on two dice is 7 for (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6,1)
∴ Number of favourable outcomes = 6
⇒ P(7) = \(\frac{62}{36}\)

(f) The sum on two dice is 9 for (3, 6), (4, 5), (5, 4) and (6, 3)
∴ Number of favourable outcomes = 4
⇒ P(9) = \(\frac{4}{36}\)

(g) The sum on two dice is 10 for (4, 6), (5, 5), (6,4)
∴ Number of favourable outcomes = 3
⇒ P(10) = \(\frac{3}{36}\)

(h) The sum on two dice is 11 for (5, 6) and (6,5)
∴ Number of favourable outcomes = 2
⇒ P(11) = \(\frac{2}{36}\)

Thus, the complete table is as follows:

(ii) No. The number of all possible outcomes is 36 not 11.
∴ The argument is not correct.

Question 23.
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Solution:
Let T denotes the tail and H denotes the head.
∴ All the possible outcomes are:
{H H H, H H T, H T T, T T T, T T H, T H T, T H H, H T H)
∴ Number of all possible outcomes = 8
Let the event that Hanif will lose the game denoted by E.
∴ Favourable events are: {HHT, HTH, THH, THT, TTH, HTT}
⇒ Number of favourable outcomes = 6
∴ P(E) = \(\frac{6}{8}=\frac{3}{4}\)

Question 24.
A die is thrown twice. What is the probability that
(i) 5 will not come up either time?
(ii) 5 will come up at least once?
[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]
Solution:
Since, throwing a die twice or throwing two dice simultaneously is the same.
∴ All possible outcomes are:
(1, 1); (1, 2); (1, 3); (1, 4); (1, 5); (1, 6).
(2, 1); (2, 2); (2, 3); (2, 4); (2, 5); (2, 6).
(3, 1) ; (3, 2); (3, 3); (3, 4); (3, 5); (3, 6).
(4, 1); (4, 2); (4, 3); (4, 4); (4, 5); (4, 6).
(5, 1) ; (5, 2); (5, 3); (5, 4); (5, 5); (5, 6).
(6, 1) ; (6, 2); (6, 3); (6, 4); (6, 5); (6, 6).
∴ All possible outcomes = 36
(i) Let E be the event that 5 does not come up either time.
∴ Numebr of favourable outcomes = [36 – (5 + 6)] = 25
∴ P(E) = \(\frac{25}{36}\)
(ii) Let N be the event that 5 will come up at least once, then number of favourable outcomes = 5 + 6 = 11
∴ P(N) = \(\frac{11}{36}\)

Question 25.
Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(i) If two coins are tossed simultaneously there are three possible outcomes – two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is \(\frac{1}{3}\).
(ii) If a die is thrown, there are two possible outcomes – an odd number or an even number. Therefore, the probability of getting an odd number is \(\frac{1}{3}\)
Solution:
(i) Given argument is not correct. Because, if two coins are tossed simultaneously then four outcomes are possible (HH, HT, TH, TT). So total outcomes is 4.
∴ The required probability = \(\frac{1}{4}\).
(ii) Given argument is correct.
Since, total numebr of possible outcomes = 6
Odd numbers = 3 and even numbers = 3
So, favourable outcomes = 3 (in both the cases even or odd).
∴ Probability = \(\frac{3}{6}=\frac{1}{2}\)

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 10 Circles Ex 10.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 10 Circles Ex 10.2

In questions 1 to 3, choose the correct option and give justification.

Question 1.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
Solution:
(A): ∵ QT is a tangent to the circle at T and OT is radius

∴ OT⊥QT
Also, OQ = 25 cm and QT = 24 cm
∴ Using Pythagoras theorem, we get
OQ² = QT² + OT²
⇒ OT² = OQ² – QT² = 25² – 24² = 49
⇒ OT = 7
Thus, the required radius is 7 cm.

Question 2.
In figure, if TP and TQ are the two tangents to a circle with centre 0 so that ∠POQ =110°, then ∠PTQ is equal to
(A) 60°
(B) 70°
(C) 80°
(D) 90°

Solution:
(B): TQ and TP are tangents to a circle with centre O and ∠POQ = 110°
∴ OP⊥PT and OQ⊥QT
⇒ ∠OPT = 90° and ∠OQT = 90°
Now, in the quadrilateral TPOQ, we get
∠PTQ + 90° + 110° + 90° = 360° [Angle sum property of a quadrilateral]
⇒ ∠PTQ + 290° = 360°
⇒ ∠PTQ = 360° – 290° = 70°

Question 3.
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to
(A) 50°
(B) 60°
(C) 70°
(D) 80°
Solution:
(A) : Since, O is the centre of the circle and two tangents from P to the circle are PA and PB.
∴ OA⊥AP and OB⊥BP
⇒ ∠OAP = ∠OBP = 90°

Now, in quadrilateral PAOB, we have
∠BPA + ∠PAO + ∠AOB + ∠OBP = 360°
⇒ 80° + 90° + ∠AOB + 90° = 360°
⇒ 260° + ∠AOB = 360°
⇒ ∠AOB = 360° – 260° ⇒ ∠AOB = 100°
In right ∆OAP and right ∆OBP, we have
OP = OP [Common]
∠OAP = ∠OBP [Each 90°]
OA = OB [Radii of the same circle]
∴ ∆OAP ≅ ∆OBP [By RHS congruency]
⇒ ∠POA = ∠POB [By CPCT]
∴ ∠POA = \(\frac{1}{2}\) ∠AOB = \(\frac{1}{2}\) × 100° = 50°

Question 4.
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Solution:
In the figure, PQ is diameter of the given circle and O is its centre.
Let tangents AB and CD be drawn at the end points of the diameter PQ.
Since, the tangents at a point to a circle is perpendicular to the radius through the point.

∴ PQ⊥AB
⇒ ∠APQ = 90°
And PQ⊥CD
⇒ ∠PQD = 90° ⇒ ∠APQ = ∠PQD
But they form a pair of alternate angles.
∴ AB || CD
Hence, the two tangents are parallel.

Question 5.
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Solution:
In the figure, the centre of the circle is O and tangent AB touches the circle at P. If possible, let PQ be perpendicular to AB such that it is not passing through O.
Join OP.
Since, tangent at a point to a circle is perpendicular to the radius through that point.

∴ OP⊥AB
⇒ ∠OPB = 90° ……….. (1)
But by construction, PQ⊥AB
⇒ ∠QPB = 90° ………….. (2)
From (1) and (2),
∠QPB = ∠OPB
which is possible only when O and Q coincide. Thus, the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Question 6.
The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Solution:
∵ The tangent to a circle is perpendicular to the radius through the point of contact.
∴ ∠OTA = 90°
Now, in the right ∆OTA, we have
OA² = OT² + AT² [Pythagoras theorem]

⇒ OT² = 5² – 4²
⇒ OT² = (5 – 4)(5 + 4)
⇒ OT² = 1 × 9 = 9 = 3²
⇒ OT = 3
Thus, the radius of the circle is 3 cm.

Question 7.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Solution:
In the figure, O is the common centre, of the given concentric circles.
AB is a chord of the bigger circle such that it is a tangent to the smaller circle at P.
Since, OP is the radius of the smaller circle.
∴ OP⊥AB ⇒ ∠APO = 90°
Also, radius perpendicular to a chord bisects the chord.

∴ OP bisects AB
⇒ AP = \(\frac{1}{2}\) AB
Now, in right ∆APO,
OA² = AP² + OP²
⇒ 5² = AP² + 3² ⇒ AP² = 5² – 3²
⇒ AP² = 4² ⇒ AP = 4 cm
⇒ \(\frac{1}{2}\) AB = 4 ⇒ AB = 2 × 4 = 8 cm
Hence, the required length of the chord AB is 8 cm.

Question 8.
A quadrilateral ABCD is drawn to circumscribe a circle (see figure).

Prove that AB + CD = AD + BC
Solution:
Since, the sides of quadrilateral ABCD, i.e., AB, BC, CD and DA touches the circle at P, Q, R and S respectively, and the lengths of two tangents to a circle from an external point are equal.
∴ AP = AS, BP = BQ,
DR = DS and CR = CQ
Adding them, we get
(AP + BP) + (CR + RD) = (BQ + QQ) + (DS + SA)
⇒ AB + CD = BC + DA

Question 9.
In figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B.

Prove that ∠AOB = 90°.
Solution:
∵ The tangents drawn to a circle from an external point are equal.
∴ AP = AC ……… (1)
Join OC.
In ∆PAO and ∆CAO, we have
AO = AO [Common]
OP = OC [Radii of the same circle]
AP = AC [From (1)]
⇒ ∆PAO ≅ ∆CAO [SSS congruency]
∴ ∠PAO = ∠CAO
⇒ ∠PAC = 2∠CAO …………. (2)
Similarly, ∠CBQ = 2∠CBO ……………… (3)

Again, we know that sum of internal angles on the same side of a transversal is 180°.
∴ ∠PAC + ∠CBQ = 180°
2∠CAO + 2∠CBO = 180° [From (2) and (3)]
⇒ ∠CAO + ∠CBO = \(\frac{180^{\circ}}{2}\) = 90° …………… (4)
Also, in ∆AOB,
∠BAO + ∠OBA + ∠AOB = 180° [Sum of angles of a triangle]
⇒ ∠CAO + ∠CBO + ∠AOB = 180°
⇒ 90° + ∠AOB = 180° [From (4)]
⇒ ∠AOB = 180° – 90°
⇒ ∠AOB = 90°

Question 10.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Solution:
Let PA and PB be two tangents drawn from an external point P to a circle with centre O.

Now, in right ∆OAP and right ∆OBP, we have
PA = PB [Tangents to circle from an external point]
OA = OB [Radii of the same circle]
OP = OP [Common]
⇒ ∆OAP ≅ ∆OBP [By SSS congruency]
∴ ∠OPA = ∠OPB [By CPCT]
and ∠AOP = ∠BOP
⇒ ∠APB = 2∠OPA and ∠AOB = 2∠AOP
In right ∆OAP,
∠AOP + ∠OPA + ∠PAO = 180°
⇒ ∠AOP = 180° – 90° – ∠OPA
⇒ ∠AOP = 90° – ∠OPA
⇒ 2∠AOP = 180° – 2∠OPA
⇒ ∠AOB = 180° – ∠APB
⇒ ∠AOB + ∠APB = 180°

Question 11.
Prove that the parallelogram circumscribing a circle is a rhombus.
Solution:
We have ABCD, a parallelogram which circumscribes a circle (i.e., its sides touch the circle) with centre O.
Since, tangents to a circle from an external point are equal in length

∴ AP = AS
BP = BQ
CR = CQ
DR = DS
On adding, we get
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
But AB = CD [Opposite sides of parallelogram]
and BC = AD
∴ AB + CD = AD + BC ⇒ 2AB = 2BC
⇒ AB = BC
Similarly, AB = DA and DA = CD
Thus, AB = BC = CD = DA
Hence, ABCD is a rhombus.

Question 12.
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC.

Solution:
Here ∆ABC circumscribes the circle with centre O. Also, radius = 4 cm
Let AC and AB touches the circle at E and F, respectively and join OE and OF.
∵ The sides BC, CA and AB touches the circle at D, E and F respectively.
∴ BF = BD = 8 cm
[ ∵ Tangents to a circle from an external point are equal]
CD = CE = 6 cm
AF = AE = x cm (say)

∴ The sides of the ∆ABC are 14 cm, (x + 6) cm and (x + 8) cm
Perimeter of ∆ABC
= [14 + (x + 6) + (x + 8)] cm
= [14 + 6 + 8 + 2x] cm
= (28 + 2x) cm
⇒ Semi perimeter of ∆ABC,
s = \(\frac{1}{2}\) [28 + 2x] cm = (14 + x) cm
∴ s – a = (14 + x) – (8 + x) = 6
s – b = (14 + x) – (14) = x
s – c = (14 + x) – (6 + x) = 8
where, a = AB, b = BC, c = AC

Squaring both sides, we get
(14 + x)² = (14 + x)3x
⇒ 196 + x² + 28x = 42x + 3x²
⇒ 2x² + 14x – 196 = 0
⇒ x² + 7x – 98 = 0
⇒ (x – 7)(x + 14) = 0
⇒ x – 7 = 0 or x + 14 = 0
⇒ x = 7 or x = -14
But x = -14 is rejected.
∴ x = 7
Thus, AB = 8 + 7 = 15 cm, BC = 8 + 6 = 14 cm and CA = 6 + 7 = 13 cm

Question 13.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution:
We have a circle with centre O. A quadrilateral ABCD is such that the sides AB, BC, CD and DA touches the circle at P, Q, R and S respectively.
Join OP, OQ, OR and OS.
We know that two tangents drawn from an external point to a circle subtend equal angles at the centre.

∴ ∠1 = ∠2
∠3 = ∠4
∠5 = ∠6 and ∠7 = ∠8
Also, the sum of all the angles around a point is 360°.
∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
∴ 2(∠1 + ∠8 + ∠5 + ∠4) = 360°
⇒ (∠1 + ∠8 + ∠5 + ∠4) = 180° …………. (1)
and 2(∠2 + ∠3 + ∠6 + ∠7) = 360°
⇒ (∠2 + ∠3 + ∠6 + ∠7) = 180° ……………. (2)
Since, ∠2 + ∠3 = ∠AOB, ∠6 + ∠7 = ∠COD, ∠1 + ∠8 = ∠AOD and ∠4 + ∠5 = ∠BOC
∴ From (1) and (2), we have
∠AOD + ∠BOC = 180°
and ∠AOB + ∠COD = 180°

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.4

Question 1.
Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, -2) and 8(3, 7).
Solution:
Let the required ratio be k : 1 and the point C divide them in the above ratio.
∴ Coordinates of C are \(\left(\frac{3 k+2}{k+1}, \frac{7 k-2}{k+1}\right)\)
Since, the point C lies on the given line 2x + y – 4 = 0.
∴ We have \(2\left(\frac{3 k+2}{k+1}\right)+\left(\frac{7 k-2}{k+1}\right)-4\) = 0
⇒ 2(3k + 2) + (7k – 2) = 4 × (k + 1)
⇒ 6k + 4 + 7k – 4k – 4 – 2 = 0
⇒ (6 + 7 – 4)k + (-2) = 0 ⇒ 9k – 2 = 0
⇒ k = \(\frac{2}{9}\)
The required ratio = k : 1 = \(\frac{2}{9}\) : 1 = 2 : 9

Question 2.
Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
Solution:
Let the given points be A(x, y), B( 1, 2) and C(7, 0) are collinear.
The points A, B and C will be collinear if area of ∆ABC = 0
⇒ \(\frac{1}{2}\) [x(2 – 0) + 1(0 – y) + 7(y – 2) = 0
or 2x – y + 7y – 14 = 0
or 2x + 6y – 14 = 0 or x + 3y – 7 = 0, which is the required relation between x and y.

Question 3.
Find the centre of a circle passing through the points (6, -6), (3, -7) and (3, 3).
Solution:
Let the points are A (6, -6), B(3, -7) and C(3, 3)
Let P(x, y) be the centre of the circle. Since, the circle is passing through A, B and C.
∴ AP = BP = CP
Taking AP = BP, we have AP² = BP²
⇒ (x – 6)² + (y + 6)² = (x – 3)² + (y + 7)²
⇒ x² – 12x + 36 + y² + 12y + 36 = x² – 6x + 9 + y² + 14y + 49
⇒ – 12x + 6x + 12y – 14y + 72 – 58 = 0
⇒ – 6x – 2y + 14 = 0
⇒ 3x + y – 7 = 0 ……………….. (1)
Taking BP = CP, we have BP² = CP²
⇒ (x – 3)² + (y + 7)² = (x – 3)² + (y – 3)²
⇒ x² – 6x + 9 + y² + 14y + 49 = x² – 6x + 9 + y² – 6y + 9
⇒ – 6x + 6x + 14y + 6y + 58 -18 = 0
⇒ 20y + 40 = 0
⇒ y = \(\frac{-40}{20}\) = -2
From (1) and (2), 3x – 2 – 7 = 0
⇒ 3x = 9 ⇒ x = 3
i.e., x = 3 and y = -2
∴ The required centre is (3, -2).

Question 4.
The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.
Solution:
Let we have a square ABCD such that A(-1, 2) and C(3, 2) are the opposite vertices. Let B(x, y) be an unknown vertex.

Since, all sides of a square are equal.
∴ AB = BC ⇒ AB² = BC2²
⇒ (x + 1)² + (y – 2)² = (x – 3)² + (y – 2)²
⇒ x² + 2x + 1 + y² – 4y + 4
⇒ x² – 6x + 9 + y² – 4y + 4
⇒ 2x + 1 = -6x + 9
⇒ 8x = 8 ⇒ x = 1  …………………. (1)
Since, each angle of a square = 90°.
∴ ∆ABC is a right angled triangle.
∴ Using Pythagoras theorem, we have AB² + BC² = AC²
⇒ (x + 1)² + (y – 2)²] + [(x – 3)² + (y – 2)²]
= [(3 + 1)² + (2 – 2)²]
⇒ [x² + 2x + 1 + y² – 4y + 4] + [x² – 6x + 9 + y² – 4y + 4]
= [4² + 0²]
⇒ 2x² + 2y² + 2x – 4y – 6x – 4y + 1 + 4 + 9 + 4 = 16
⇒ 2x² + 2y² – 4x – 8y + 2 = 0
⇒ x² + y² – 2x – 4y + 1 = 0 …………….. (2)
Substituting the value of x from (1) into (2), we have
1 + y² – 2 – 4y + 1 = 0
⇒ y² – 4y + 2 – 2 = 0
⇒ y² – y = 0
⇒ y(y – 4) = 0
⇒ y = 0 or y = 4
Hence, the required other two vertices are (1, 0) and (1, 4).

Question 5.
The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot.
(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of ∆PQR if C is the origin?
Also calculate the areas of the triangles in these cases. What do you observe?

Solution:
(i) By taking A as the origin and AD and AB as the coordinate axes. We have P(4, 6), Q( 3, 2) and R( 6, 5) as the vertices of ∆PQR.
(ii) By taking C as the origin and CB and CD as the coordinate axes, then the vertices of ∆PQR are P(-12, – 2), Q(-13, – 6) and R(- 10, – 3)
Case I: When P(4, 6), Q(3, 2) and R(6, 5) are the vertices.
∴ ar(∆PQR) = \(\frac{1}{2}\) [4(2 – 5) + 3(5 – 6) + 6(6 – 2)]
= \(\frac{1}{2}\) [-12 – 3 + 24] = \(\frac{9}{2}\) sq. units
Case II: When P(-12, -2), Q(-13, -6) and R(-10, -3) are the vertices.
∴ ar(∆PQR)
= \(\frac{1}{2}\) [-12(- 6 + 3) + (-13)(- 3 + 2) + (-10)(-2 + 6)]
= \(\frac{1}{2}\) [-12(-3) + (-13)(-1) + (-10) × (4)]
= \(\frac{1}{2}\) [36 + 13 – 40] = \(\frac{9}{2}\) sq. units
Thus, in both cases, the area of ∆PQR is the same.

Question 6.
The vertices of a ∆ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that \(\frac{A D}{A B}=\frac{A E}{A C}=\frac{1}{4}\) Calculate the area of the ∆ADE and compare it with the area of ∆ABC. [Recall “The converse of basis proportionality theorem”, and “theorem of similar triangles taking their areas and corresponding sides”]
Solution:

Question 7.
Let A(4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ABC.
(i) The median from A meets BCat D. Find the coordinates of the point D.
(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.
(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1
(iv) What do you observe?
[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2:1.]
(v) If A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of ∆ABC, find the coordinates of the centroid of the triangle.
Solution:
We have the vertices of ∆ABC as A (4, 2), B(6, 5) and C(1, 4).
(i) Since AD is a median

∴ Coordinates of D are


Also, CR : RF = 2 : 1 i.e., the point R divides CF in the ratio 2 : 1
∴ Coordinates of R are

(iv) We observe that P, Q and R represent the same point.
(v) Here, we have A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of ∆ABC. Also AD, BE and CF are its medians.
∴ D, E and F are the mid points of BC, CA and AB respectively.
We know, the centroid is a point on a median, dividing it in the ratio 2 : 1.
Considering the median AD, coordinates of

Question 8.
ABCD is a rectangle formed by the points A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1). P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.
Solution:
We have a rectangle whose vertices are A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1).
∵ P is mid-point of AB
∴ Coordinates of P are


We see that PQ = QR = RS = SP i.e., all sides of PQRS are equal.
∴ It can be a square or a rhombus.
But PR ≠ QS i.e., its diagonals are not equal.
∴ PQRS is a rhombus.

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 11 प्रायिकता Ex 11.2 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Book Solutions Chapter 11 प्रायिकता Ex 11.2

प्रश्न 1.
दर्शाइए कि दिक् कोसाइन \(\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}\) \(\frac{4}{13}, \frac{12}{13}, \frac{3}{13} ; \frac{3}{13}, \frac{-4}{13}, \frac{12}{13}\) वाली तीन रेखाएँ परस्पर लम्बवत्
हल:
माना


अतः तीनों रेखाएँ परस्पर लंब हैं।

प्रश्न 2.
दर्शाइए कि बिन्दुओं (1, – 1, 2), (3, 4, – 2) से होकर जाने वाली रेखा बिन्दुओं (0, 3, 2) और (3, 5, 6) से जाने वाली रेखा पर लंब है।
हल:
बिन्दुओं (1, – 1, 2) तथा (3, 4, – 2) से होकर जाने वाली रेखा के दिक् अनुपात 3 – 1, 4 + 1, – 2, – 2 या 2, 5, – 4
माना a₁ = – 2, b₁ = 5 तथा C₁ = – 4
अब बिन्दु (0, 3, 2) तथा (3, 5, 6) से होकर जाने वाली रेखा के दिक् अनुपात
3 – 0, 5 – 3, 6 – 2 या 3, 2, 4
अब a₁ a₂ + b₁ b₂ + c₁ c₂
= 2 x 3 + 5 x 2 + (- 4) x 4
= 6 + 10 – 16 = 0
∴ रेखाएँ परस्पर लंब हैं।

प्रश्न 3.
दर्शाइए कि बिन्दुओं (4, 7, 8) (2, 3, 4) से होकर जाने वाली रेखा बिन्दुओं (- 1, – 2, 1), (1, 2, 5) से जाने वाली रेखा के समांतर है।
हल:
बिन्दुओं (4, 7, 9), (1, 2, 5) से जाने वाली रेखा के दिक् अनुपात 2, – 4, 3 – 7, 4 – 8 या – 2, – 4, – 4
माना a₁ = – 2, b₁ = – 4 तथा c₁= – 4
अब बिन्दुओं (- 1, – 2, 1) (1, 2, 5) से होकर जाने वाली रेखा के दिक् अनुपात 1 + 1, 2 + 2, 5 – 1 या 2, 4, 4
माना a₁ = 2, b₂ = 4, C₂ = 4

∴ रेखाएँ परस्पर समांतर हैं।

प्रश्न 4.
बिन्दु (1, 2, 3) से गुजरने वाली रेखा का समीकरण ज्ञात कीजिए जो सदिश \(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}\) के समांतर है।
हल:
दिया है

यही रेखा के समी० का कार्तीय रूप है।

प्रश्न 5.
बिन्दु जिसकी स्थिति सदिश \(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}\) से गुजरने व सदिश \(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}\) की दिशा में जाने वाली रेखा का सदिश और कार्तीय रूपों में समी० ज्ञात कीजिए।
हल:
दिया है

यही रेखा के समी०का कार्तीय रूप हैं।

प्रश्न 6.
उस रेखा का कार्तीय समीकरण ज्ञात कीजिए जो बिन्दु (- 2, 4, – 5) से जाती है और \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\) के समांतर हैं।
हल:
बिन्दु (- 2, 4, – 5 ) से होकर जाने वालों और रेखा
\(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\) के समांतर रेखा के समीकरण का
कार्तीय समीकरण
\(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+5}{6}\) हैं।

प्रश्न 7.
उस रेखा का कार्तीय समीकरण \(\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}\) है। इसका सदिश समीकरण ज्ञात 37 2 कीजिए।
हल:
दी गई रेखा का कार्तीय समीकरण
\(\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}\)
इससे स्पष्ट होता है कि रेखा बिन्दु A (+ 5, – 4, 6) से होकर जाती है तथा यह सदिश \(\vec{a}=3 \hat{i}+7 \hat{j}+2 \hat{k}\) के समांतर है। तथा A का स्थिति सदिश \(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}\)
∴ रेखा का सदिश समी० \(\vec{r}=\vec{a}+\lambda \vec{b}\)
या \(\vec{r}=(5 \hat{i}-4 \hat{j}+6 \hat{k})+l(3 \hat{i}+7 \hat{j}+\hat{2} \hat{k})\)

प्रश्न 8.
मूल बिन्दु और (5, – 2, 3) से जाने वाली रेखा का सदिश तथा कार्तीय रूपों में समी० ज्ञात कीजिए।
हल:
माना O(0, 0,0) तथा A (5, – 2, 3) दो बिन्दु है।

जोकि रेखा का कार्तीय रूप है।

प्रश्न 9.
बिन्दुओं (3, – 2, – 5) और (3, – 2, 6) से गुजरने वाली रेखा का सदिश तथा कार्तीय रूपों में समीकरण ज्ञात कीजिए।
हल:
माना P(3, – 2, – 5) और Q(3, – 2, 6) दो बिन्दु स्थिति सदिश हैं।

यही अभीष्ट कार्तीय समीकरण है।

प्रश्न 10.
निम्न रेखायुग्मों के बीच का कोण ज्ञात कीजिए–

हल:

(ii) माना \(\overline{b}_{1}=\hat{i}-\hat{j} \times 2 \hat{k}\), \(\overline{b}_{2}=3 \hat{i}-5 \hat{j}-4 \hat{k}\) तथा दोनों रेखाओं के मध्य कोण θ है इसलिए

प्रश्न 11.
निम्नलिखित रेखायुग्मों के बीच का कोण ज्ञात कीजिए

हल:
(i) पहली रेखा के दिक् अनुपात 2, 5, – 3 तथा दूसरी रेखा के दिक् अनुपात – 1, 8, 4 हैं।
माना इनके बीच का कोण θ है, तब

या θ = \(\cos ^{-1}\left(\frac{2}{3}\right)\)

प्रश्न 12.
p का मान ज्ञात कीजिए ताकि रेखायें \(\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2}\) और \(\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}\) परस्पर लम्ब हों।
हल:
समी० को व्यापक रूप में लिखने पर

प्रश्न 13.
दिखाइये कि रेखाएँ \(\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\) और \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\)
हल:
पहली व दूसरी रेखा के दिक् अनुपात 7, – 5, 1 तथा 1, 2, 3 हैं।
तब a₁ a₂ + b₁ b₂ + c₁ c₂ = 7 · 1 + 2 · – 5 + 3 · 1
= 7 – 10 + 3 = 10 – 10 = 0
अतः रेखायें परस्पर लम्ब हैं।

प्रश्न 14.
रेखाओं \(\vec{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})\) और \(\vec{r}=2 \hat{i}-\hat{j}-\hat{k}=\mu(2 \hat{i}+\hat{j}+2 \hat{k})\) के बीच की न्यूनतम दूरी ज्ञात कीजिए।
हल:
दी गई रेखाएँ-

रेखाओं के बीच की दूरी

प्रश्न 15.
रेखाओं \(\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\) और \(\)\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1} के बीच की न्यूनतम दूरी ज्ञात कीजिए।
हल:

प्रश्न 16.
रेखाएँ, जिनके सदिश समीकरण निम्नलिखित हैं, के बीच की न्यूनतम दूरी ज्ञात कीजिए।

हल:

प्रश्न 17.
रेखाएँ जिनकी सदिश समीकरण निम्नलिखित हैं, के बीच की न्यूनतम दूरी ज्ञात कीजिए।

हल:
पहली रेखा का समीकरण


इसका मान (1) में रखने पर

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 6 Triangles Ex 6.5 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5

Question 1.
Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm
Solution:
(i) The sides are : 7 an, 24 cm, 25 cm
Here, (7 cm)² = 49 cm²
(24 cm)² = 576 cm²
(25 cm)² = 625 cm²
∵ (49 + 576)cm² = 625 cm²
∴ It is a right triangle.
Hypotenuse = 25 cm.

(ii) The sides are: 3 cm, 8 cm, 6 cm
Here, (3 cm)² = 9 cm²
(8 cm)² = 64 cm²
(6 cm)² = 36 cm²
∵ (9 + 36) ≠ 64 cm²
∴ It is not a right triangle.

(iii) The sides are : 50 cm, 80 cm, 100 cm
Here, (50 cm)² = 2500 cm²
(80 cm)² = 6400 cm²
(100 cm)² = 10000 cm²
∵ (2500 + 6400) cm² ≠ 10000 cm²
∴ It is not a right triangle.

(iv) The sides are : 13 cm, 12 cm, 5 cm
Here, (13 cm)² = 169 cm²
(12 cm)² = 144 cm²
(5 cm)² = 25 cm²
∵ (144 + 25)cm² = 169 cm²
∴ It is a right triangle.
Hypotenuse = 13 cm.

Question 2.
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM.MR.
Solution:
In ∆QMP and ∆QPR,
∠QMP = ∠QPR [Each 90°]
∠Q = ∠Q [Common]
⇒ ∆QMP ~ ∆QPR …. (1) [AA similarity]
Again, in ∆PMR and ∆QPR,
∠PMR = ∠QPR [Each = 90°]
∠R = ∠R [Common]
⇒ ∆PMR ~ ∆QPR …… (2) [AA similarity]
From (1) and (2), we have

Question 3.
In the figure, ABD is a triangle, right angled at A and AC ⊥ BD.

Show that
(i) AB² = BC.BD
(ii) AC² = BC.DC
(iii) AD² = BD.CD
Solution:

Question 4.
ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².
Solution:
We have, right ∆ABC such that ∠C = 90° and AC = BC.

∴ By Pythagoras theorem, we have AB² = AC² + BC² = AC² + AC² = 2AC²
[∵ BC = AC (given)]
Thus, AB² = 2AC²

Question 5.
ABC is an isosceles triangle with AC = BC. If AB² = 2AC², prove that ABC is a right triangle.
Solution:
We have, an isosceles AABC such that BC = AC.

Also, AB² = 2AC²
∴ AB² = AC² + AC²
But AC =BC
∴ AB² = AC² + BC²
∴ Using the converse of Pythagoras theorem, ∠ACB = 90°
f.e., ∆ABC is a right angled triangle.

Question 6.
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution:
In equilateral triangle, altitude bisects the base.
⇒ AD = DB

Question 7.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Solution:
Let us have a rhombus ABCD.

∵ Diagonal of a rhombus bisect each other at right angles.
∴ OA = OC and OB = OD
Also, ∠AOB = ∠BOC [Each = 90°]
And ∠COD = ∠DOA [Each = 90°]
In right ∆AOB, we have,
AB² = OA² + OB² …… (1)
[Using Pythagoras theorem]
Similarly, in right ∆BOC,
BC² = OB² + OC² …… (2)
In right ∆COD,
CD² = OC² + OD² …… (3)
In right ∆AOD,
DA² = OD² + OA² ……. (4)
Adding (1), (2), (3) and (4)
AB² + BC² + CD² + DA²
= [OA² + OB²] + [OB² + OC²] + [OC² + OD²] + [OD² + OA²]
= 2OA² + 2OB² + 2 OC² + 2OD² = 2[OA² + OB² + OC² + OD²]
= 2[OA² + OB² + OA² + OB²]

Thus, sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Question 8.
In the figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²
(ii) AF² + BD² + CE² = AE² + CD² + BF²

Solution:
We have a point in the interior of a ∆ABC such that
OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.
(i) Let us join OA, OB and OC.

In right ∆OAF, by Pythagoras theorem
OA² = OF² + AF² …(1)
Similarly, from right triangle ODB and OEC, we have
OB² = BD² + OD², …(2)
and OC² = CE² + OE² …(3)
Adding (1), (2) and (3), we get OA² + OB² + OC²
= (AF² + OF²) + (BD² + OD²) + (CE² + OE²)
⇒ OA² + OB² + OC²
= AF² + BD² + CE² + (OF² + OD² + OE²)
⇒ OA² + OB² + OC² – (OD² + OE² + OF²)
= AF² + BD² +CE²
⇒ OA² + OB² + OC² – OD² – OE² – OF²
= AF² + BD² + CE²

(ii) In right triangle OBD and triangle OCD, by Pythagoras theorem:
OB² = OD² + BD² and OC² = OD² + CD²
⇒ OB² – OC² = OD² + BD² – OD² – CD²
⇒ OB² – OC² = BD² – CD² ….. (1)
Similarly, we have
OC² – OA² = CE² – AE² …… (2)
and OA² – OB² = AF² – BF² ….. (3)
Adding (1), (2) and (3), we get (OB² – OC²) + (OC² – OA²) + (OA² – OB²) = (BD² – CD²) + (CE² – AE²) + (AF² – BF²)
⇒ 0 = BD² + CE² + AF² – (CD² + AE² + BF²)
⇒ BD² + CE² + AF² = CD² + AE² + BF²
or AF² + BD² + CE² = AE² + BF² + CD²

Question 9.
A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
Solution:
Let PQ be the ladder and PR be the wall

⇒ PQ = 10 m, PR = 8 m
Now, in the right ∆PQR, PQ² = PR² + QR²
⇒ 10² = 8² + QR²
[using Pythagoras theorem]
⇒ QR² = 10² – 8² = (10 + 8)(10 – 8)
= 18 × 2 = 36
QR = \(\sqrt{36}\) = 6m
Thus, the distance of the foot of the ladder from the base to the wall is 6 m.

Question 10.
A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Solution:
Let AB is the wire and BC is the vertical pole. The point A is the stake.

Now, in the right AABC, using Pythagoras Theorem, we have

Thus, the stake is required to be taken at \(6 \sqrt{7}\)m from the base of the pole to make the wire taut.

Question 11.
An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after \(1 \frac{1}{2}\) hours?
Solution:
Let the point A represent the airport. Plane-I fly towards North,
∴ Distance of the plane-I from the airport after \(1 \frac{1}{2}\) hours = speed × time

Question 12.
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops?
Solution:
Let the two poles AB and CD are such that the distance between their feet AC = 12m.
∵ Height of pole-1, AB = 11 m

Question 13.
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE² + BD² = AB² + DE².
Solution:
We have a right ∆ABC such that ∠C = 90°.
Also, D and E are points on CA and CB respectively.
rain We have a right ∆ABC such that ∠C = 90°.

Let us join AE and BD.
In right ∆ACB, using Pythagoras theorem
AB² = AC² + BC² …… (1)
In right ∆DCE, using Pythagoras theorem,
DE² = CD² + CE² …… (2)
Adding (1) and (2), we get
AB² + DE² = [AC² + BC²] + [CD² + CE²]
= AC² + BC² + CD² + CE² = [AC² + CE²] + [BC² + CD²] …. (3)
In right ∆ACE,
AC² + CE² = AE² …… (4)
In right ∆BCD,
BC² + CD² = BD² …….. (5)
From (3), (4) and (5), we have AB² + DE² = AE² + BD² or AE² + BD² = AB² + DE²

Question 14.
The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3CD (see figure). Prove that 2AB² = 2AC² + BC².

Solution:

Question 15.
In an equilateral triangle ABC, D is a point on side BC such that BD = \(\frac{1}{3}\)BC. Prove that 9AD² = 7AB².
Solution:
We have an equilateral ∆ABC; in which D is a point on BC such that BD = \(\frac{1}{3}\) BC.

Question 16.
In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Solution:
We have an equilateral ∆ABC, in which AD ⊥ BC.
Since, an altitude in an equilateral A, bisects the corresponding side.

∴ D is the mid-point

Question 17.
Tick the correct answer and justify : In ∆ABC, AB = \(6 \sqrt{3}\) cm, AC = 12 cm and BC = 6 cm. The angle B is
(A) 120°
(B) 60°
(C) 90°
(D) 45°
Solution:
(C): We have, AB = \(6 \sqrt{3}\) cm, AC = 12 cm, and BC = 6 cm

In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Ex 3.5 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5

Question 1.
Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
(i) x – 3y – 3 = 0; 3x – 9y – 2 = 0
(ii) 2x + y = 5; 3x + 2y = 8
(iii) 3x – 5y = 20; 6x -10y = 40
(iv) x – 3y – 7 = 0; 3x – 3y – 15 = 0
Solution:
(i) For x – 3y – 3 = 0, 3x – 9y – 2 = 0
∴ a₁ = 1, b₁ = – 3, C₁ = – 3, a₂ = 3, b₂ = – 9, C₂ = -2

∴ The given system has no solution.

(ii) 2x + y – 5 = 0, 3x + 2y – 8 = 0
∴ a₁ = 2, b₁ = 1, c₁ = -5, a₂ = 3, b₂ = 2, c₂ = -8

∴ The given system has a unique solution.
Now, using cross multiplication method, we have

(iii) For 3x – 5y – 20 = 0, 6x – 10y – 40 = 0

∴ The given system of linear equation has infinitely many solutions

(iv) For x – 3y – 7 = 0, 3x – 3y – 15 = 0

∴ The given statement has unique solution. Now, using cross multiplication method, we have

Question 2.
(i) For which values of a and b does the following pair of linear equations have an infinite number of solutions ?
2x + 3y = 7; (a – b) x + (a + b) y = 3a + b – 2
(ii) For which value of k will the following pair of linear equations have no solution ?
3x + y = 1; (2k – 1) x + (k – 1) y = 2k + 1
Solution:
(i) We have, 2x + 3y = 7 and
(a – b) x + (a + b) y = (3a + b – 2)
⇒ 2x + 3y – 7 = 0
and (a – b) x + (a + b) y -(3a + b – 2) = 0
a₁ = 2, b₁ = 3, c₁ = – 7, a₂ = (a- b) , b₂ = (a + b) , c₂ = -(3a + b – 2)
For infinite number of solutions,

From the first two equations, we get

Question 3.
Solve the following pair of linear equations by the substitution and cross-multiplication methods: 8x + 5y = 9; 3x + 2y = 4
Solution:
Method-1 [Substitution method]:
8x + 5y = 9 … (1)
3x + 2y = 4 … (2)

Question 4.
Form the pair of linear equations in the following problems and find their solutions (if they exist)by any algebraic method:
(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay ₹ 1000 as hostel charges whereas a student B, who takes food for 26 days, pays ₹ 1180 as hostel charges. Find the fixed charges and the cost of food per day.
(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.
(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Solution:
(i) Let the fixed charges = ₹ x
and charges of food per day = ₹ y
For student A : Number of days = 20
∴ Cost of food for 20 days = ₹ 20y
According to the problem,
x + 20y = 1000
⇒ x + 20y – 1000 = 0 …. (1)
For student B : Number of days = 26
Cost of food for 26 days = ₹ 26y
According to the problem,
x + 26y = 1180
⇒ x + 26y – 1180 = 0 … (2)
Solving these by cross multiplication, we get

Thus x = 400 and y = 30
∴ Fixed charges = ₹ 400
and cost of food per day = ₹ 30

(ii) Let the numerator = x
and the denominator = y

From equations (1) and (2) , we have

(iii) Let the number of correct answers = ₹ x
and the number of wrong answers = ₹ y
Case-I: Marks for all correct answers
= (3 × x) = 3x
Mark for all wrong answers = (1 × y) = y
∴ According to the condition :
3x – y = 40 ⇒ 3x – y – 40 = 0
Case-II: Mark for all correct answers
= (4 × x)=
Marks for all wrong answers = (2 × y)
= 2y
∴ According to the condition :
4x – 2y = 50
⇒ 2x – y = 25 ⇒ 2x – y – 25 = 0 … (2)
From (1)and (2) , we have a₁ = 3, b₁ = -1, c₁ = -40, a₂ = 2, b₂ = -1, c₂ = -25

Now, total number of questions = [Number of correct answers] + [Number of wrong answers]
= 15 + 5 = 20
Thus, required number of questions = 20.

(iv) Let the speed of car-I be x km/hr.
and the speed of car-II be y km/hr.
Case-I:

Distance travelled by car-I = AC
∵ AC = time × speed = 5 × x km, AC = 5x
Distance travelled by car-II, BC = 5y
Since AB = AC – BC,
100 = 5x – 5y
⇒ 5x – 5y – 100 = 0
⇒ x – y – 20 = 0 …. (1)


Thus, speed of car-I = 60 km/hr
Speed of car-II = 40 km/hr

(v) Let the length of the rectangle = x units
and the breadth of the rectangle = y units
∴ Area of rectangle = x × y = xy

Condition-II:
(Length + 3) × (Breadth + 2) = Area + 67
⇒ (x + 3) (y + 2)= xy + 67 ⇒ 2x + 3y + 6 = 67
⇒ 2x + 3y – 61 = 0….(2)
Now, using cross multiplication method in (1)and (2) , where

Thus, length of the rectangle = 17 units and breadth of the rectangle = 9 units.