MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.2

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 11 Constructions Ex 11.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.2

In each of the following, give also the justification of the construction:

Question 1.
Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Solution:
Steps of Construction :
I. Draw a circle of radius 6 cm. Let its centre be O.
II. Take a point P such that OP = 10 cm. Join OP.
MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.2 1
III. Bisect OP and let M be its midpoint.
IV. Taking M as centre and MP or MO as radius draw a circle.
Let the new circle intersects the given circle at A and B. Join PA and PB.
Thus, PA and PB are the required tangents. By measurement, we have : PA = PB = 8 cm.
Justification:
Join OA and OB
Since PO is a diameter.
∴ ∠OAP = 90° = ∠OBP [Angles in a semicircle]
Also, OA and OB are radii of the same circle.
⇒ PA and PB are tangents to the circle.

Question 2.
Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Solution:
Steps of Construction :
I. Draw two concentric circles with centre O and radii 4 cm and 6 cm.
II. Take any point P on outer circle.
III. Join PO and bisect it and let the midpoint of PO is represented by M.
IV. Taking M as centre and OM or MP as radius, draw a circle such that this circle intersects the circle (of radius 4 cm) at A and B.
V. Join AP.
Thus, PA is the required tangent.
By measurement, we have PA = 4.5 cm
MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.2 2
Justification:
Join OA. As PO is diameter
∴ ∠PAO = 90° [Angle in a semi-circle]
⇒ PA⊥OA
∵ OA is a radius of the inner circle.
∴ PA has to be a tangent to the inner circle.
Verification:
In right ∆PAO, PO = 6 cm, OA = 4 cm.
MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.2 3
Hence both lengths are approximately equal.

MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.2

Question 3.
Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Solution:
Steps of Construction :
I. Draw a circle of radius 3 cm with centre O and draw a diameter.
II. Extend its diameter on both sides and cut OP = OQ = 7 cm
III. Bisect PO such that M be its mid-point.
IV. Taking M as centre and MO as radius, draw a circle. Let it intersect the given circle at A and B.
V. Join PA and PB.
Thus, PA and PB are the two required tangents from P.
VI. Now bisect OQ such that N is its mid point.
VII. Taking N as centre and NO as radius, draw a circle. Let it intersect the given circle at C and D.
VIII. Join QC and QD.
Thus, QC and QD are the required tangents from Question
MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.2 4
Justification:
Join OA to get ∠OAP = 90°
[Angle in a semi-circle]
⇒ PA⊥OA
⇒ PA is a tangent.
Similarly, PB⊥OB
⇒ PB is a tangent.
Now, join OC to get ∠QCO = 90°
[Angle in a semi-circle]
⇒ QC⊥OC ⇒ QC is a tangent.
Similarly, QD⊥OD
⇒ QD is a tangent.

Question 4.
Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.
Solution:
Steps of Construction :
I. With centre O and radius = 5 cm, draw a circle.
II. Taking a point A on the circle draw ∠AOB = 120°.
MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.2 5
III. Draw a perpendicular on OA at A.
IV. Draw another perpendicular on OB at B.
V. Let the two perpendiculars meet at C.
Thus CA and CB are the two required tangents to the given circle which are inclined to each other at 60°.
Justification:
In a quadrilateral OACB, using angle sum
property, we have
120° + 90° + 90° + ∠ACB = 360°
⇒ 300° + ∠ACB = 360°
⇒ ∠ACB = 360° – 300° = 60°.

Question 5.
Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Solution:
Steps of Construction:
I. Draw a line segment AB = 8 cm
II. Draw a circle with centre A and radius 4 cm, draw another circle with centre B and radius 3 cm.
III. Bisect the line segment AB. Let its mid point be M.
IV. With centre as M and MA (or MB) as radius, draw a circle such that it intersects the two circles at points P, Q, R and S.
V. Join BP and BQ.
Thus, BP and BQ are the required two tangents from B to the circle with centre A.
VI. Join RA and SA.
Thus, RA and SA are the required two tangents from A to the circle with centre B.
MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.2 6
Justification:
Let us join A and P.
∵ ∠APB = 90° [Angle in a semi-circle]
∴ BP⊥AP
But AP is radius of the circle with centre A.
⇒ BP has to be a tangent to the circle with centre A.
Similarly, BQ has to be tangent to the circle with centre A.
Also AR and AS are tangents to the circle with centre B.

MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.2

Question 6.
Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Solution:
Steps of construction :
I. Draw ∆ABC such that AB = 6 cm, BC = 8 cm and ∠B = 90°.
II. Draw BD⊥AC. Now bisect BC and let its midpoint be O.
So O is centre of the circle passing through B, C and D.
MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.2 7
III. Join AO
IV. Bisect AO. Let M be the mid-point of AO.
V. Taking M as centre and MA as radius, draw a circle intersecting the given circle at B and E.
VI. Join AB and AE. Thus, AB and AE are the required two tangents to the given circle from A.
Justification:
Join OE, then ∠AEO = 90° [Angle in a semi circle]
∴ AE⊥OE.
But OE is a radius of the given circle.
⇒ AE has to be a tangent to the circle.
Similarly, AB is also a tangent to the given circle.

Question 7.
Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
Solution:
Steps of Construction:
I. Draw the given circle using a bangle.
II. Take two non parallel chords PQ and RS of this circle.
MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.2 8
III. Draw the perpendicular bisectors of PQ and RS such that they intersect at O. Therefore, O is the centre of the given circle.
IV. Take a point N outside this circle.
V. Join ON and bisect it. Let M be the mid-point of ON.
VI. Taking M as centre and OM as radius, draw a circle. Let it intersect the given circle at A and B.
VII. Join NA and NB. Thus, NA and NB are the required two tangents.
Justification:
Join OA and OB.
Since ∠OAN = 90° [Angle in a semi circle]
∴ NA⊥OA.
Also NA is a radius.
∴ NA has to be a tangent to the given circle.
Similarly, NB is also a tangent to the given circle.

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.1

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.1

Question 1.
Check whether the following are quadratic equations:
(i) (x + 1 )2 = 2(x – 3)
(ii) x2 – 2x = (-2)(3 – x)
(iii) (x – 2)(x + 1) = (x – 1 )(x + 3)
(iv) (x – 3)(2x + 1) = x(x + 5)
(v) (2x – 1 )(x – 3) = (x + 5)(x – 1)
(vi) x2 + 3x + 1 = (x – 2)2
(vii) (x + 2)3 = 2x(x2 – 1)
(viii) x3 – 4x2 – x + 1 = (x – 2)3
Solution:
(i) We have, (x + 1)2 = 2(x – 3)
⇒ x2 + 2x + 1 = 2x – 6
⇒ x2 + 2x + 1 – 2x + 6 = 0
⇒ x2 + 7 = 0
Since, x2 + 7 is a quadratic polynomial.
∴ (x + 1)2 = 2(x – 3) is a quadratic equation.

(ii) We have, x2 – 2x = (-2) (3 – x)
⇒ x2 – 2x = – 6 + 2x
⇒ x2 – 2x – 2x + 6 = 0
⇒ x2 – 4x + 6 = 0
Since, x2 – 4x + 6 is a quadratic polynomial
∴ x2 – 2x = (-2)(3 – x) is a quadratic equation.

(iii) We have, (x – 2)(x + 1) = (x – 1) (x + 3)
⇒ x2 – x2 = x2 + 2x3
⇒ x2 – x – 2 – x2 – 2x + 3 = 0
⇒ – 3x +1 = 0
Since, – 3x + 1 is a linear polynomial.
∴ (x – 2)(x + 1) = (x – 1)(x + 3) is not a quadratic equation.

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.1

(iv) We have, (x – 3) (2x + 1) = x(x + 5)
⇒ 2x2 + x – 6x – 3 = x2 + 5x
⇒ 2x2 – 5x – 3 – x2 – 5x = 0
⇒ x2 – 10x – 3 = 0
Since, x2 – 10x – 3 is a quadratic polynomial.
∴ (x – 3)(2x + 1) = x(x + 5) is a quadratic equation.

(v) We have, (2x – 1) (x – 3) = (x + 5) (x – 1)
⇒ 2x2 – 6x – x + 3 = x2 – x + 5x – 5
⇒ 2x2 – 7x + 3 = x2 + 4x – 5
⇒ 2x2 – 7x + 3 – x2 – 4x + 5 = 0
⇒ x2 – 11x + 8 = 0
Since, x2 – 11x + 8 is a quadratic polynomial
∴ (2x – 1)(x – 3) = (x + 5)(x -1) is a quadratic equation.

(vi) We have, x2 + 3x + 1 = (x – 2)2
⇒ x2 + 3x + 1 = x2 – 4x + 4
⇒ x2 + 3x + 1 – x2 + 4x – 4 = 0
⇒ 7x – 3 = 0
Since, 7x – 3 is a linear polynomial.
x2 + 3x + 1 = (x – 2)2 is not a quadratic equation.

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.1

(vii) We have, (x + 2)3 = 2x(x2 – 1)
⇒ x3 + 3x2(2) + 3x(2)2 + (2)3 = 2x3 – 2x
⇒ x3 + 6x2 + 12x + 8 = 2x3 – 2x
⇒ x3 + 6x2 + 12x + 8 – 2x3 + 2x = 0
⇒ -x3 + 6x2 + 14x + 8 = 0
Since, -x3 + 6x2 + 14x + 8 is a polynomial of degree 3.
∴ (x + 2)3= 2x(x2 – 1) is not a quadratic equation.

(viii) We have, x3 – 4x2 – x + 1 = (x – 2)3
⇒ x3 – 4x2 – x + 1 = x3 + 3x2(-2) + 3x(-2)2 + (-2)3
⇒ x3 – 4x2 – x + 1 = x3 – 6x2 + 12x – 8
⇒ x3 – 4x2 – x + 1 – x3 + 6x2 – 12x + 8 = 0 ⇒ 2x2 – 13x + 9 = 0
Since, 2x2 – 13x + 9 is a quadratic polynomial.
∴ x3 – 4x2 – x + 1 = (x – 2)3 is a quadratic equation.

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.1

Question 2.
Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/hr less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Solution:
(i) Let the breadth = x metres
∵ Length = 2(Breadth) + 1
∴ Length = (2x + 1)metres
Since, Length × Breadth = Area
∴ (2x + 1) × x = 528 ⇒ 2x2 + x = 528
⇒ 2x2 + x – 528 = 0
Thus, the required quadratic equation is 2x2 + x – 528 = 0

(ii) Let the two consecutive positive integers be x and (x + 1).
∵ Product of two consecutive positive integers = 306
∴ x(x + 1) = 306 ⇒ x2 + x = 306
⇒ x2 + x – 306 = 0
Thus, the required quadratic equation is x2 + x – 306 = 0

(iii) Let the present age of Rohan be x years
∴ Mother’s age = (x + 26) years
After 3 years,
Rohan’s age = (x + 3) years
Mother’s age = [(x + 26) + 3] years
= (x + 29) years
According to the condition,
(x + 3) × (x + 29) = 360
⇒ x2 + 29x + 3x + 87 = 360
⇒ x2 + 29x + 3x + 87 – 360 = 0
⇒ x2 + 32x – 273 = 0
Thus, the required quadratic equation is x2 + 32x – 273 = 0

(iv) Let the speed of the train = u km/hr Distance covered = 480 km Distance
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 1
In second case,
Speed = (u – 8) km/hour
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 2
⇒ 480u – 480(u – 8) = 3u(u – 8)
⇒ 480u – 480u + 3840 = 3u2 – 24u
⇒ 3840 – 3u2 + 24u = 0
⇒ u2 – 8u – 1280 = 0
Thus, the required quadratic equation is u2 – 8u – 1280 = 0

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3
Question 1.
Find the sum of the following APs:
(i) 2,7,12, ….. to 10 terms
(ii) -37, -33, -29, …. to 12 terms
(iii) 0.6, 1.7, 2.8, …. to 100 terms
(iv) \(\frac{1}{15}, \frac{1}{12}, \frac{1}{10}\), ……
Solution:
(i) Here, a = 2, d = 7 – 2 = 5, n = 10
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 1
Thus, the sum of first 10 terms is 245.

(ii) Here, a = -37, d = -33 – (-37) = 4, n = 12
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 2
Thus, the sum of first 12 terms is -180.

(iii) Here, a = 0.6, d = 1.7 – 0.6 = 1.1, n = 100
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 3
Thus, the required sum of first 100 terms is 5505.

(iv)
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 4
Thus, the required sum of first 11 terms is \(\frac{33}{20}\)

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Question 2.
Find the sums given below:
(i) 7+10\(\frac{1}{2}\) + 14 + …… + 84
(ii) 34 + 32 + 30+ …….. + 10
(iii) -5 + (-8) + (-11) + ……. + (-230)
Solution:
(i) Here,
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 5
Let n be the number of terms.
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 6

(ii) Here, a = 34, d = 32 – 34 = -2, l = 10
Let n be the number of terms
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 7

(iii) Here, a = -5,d = -8- (-5) = -3,1 = -230
Let n be the number of terms.
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 8

Question 3.
In an AP:
(i) given a = 5, d = 3, an = 50,find n and Sn.
(ii) given a = 7, a13 = 35, find d and S13.
(iii) given a12 = 37, d = 3, find a and S12.
(iv) given a3 = 15, S10 = 125, find d and a10.
(v) given d = 5, S9 = 75, find a and a9.
(vi) given a = 2, d = 8, Sn = 90, find n and an.
(vii) given a = 8, an = 62, Sn = 210, find n and d.
(viii) given an = 4, d = 2,Sn = -14, find n and a.
(ix) given a = 3, n = 8, S = 192, find d.
(x) given l = 28, S = 144, and there are total 9 terms. Find a.
Solution:
(i) Here, a = 5, d = 3 and an = 50 = l
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 9
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 10
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 11
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 12
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 13
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 14
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 15
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 16
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 17

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Question 4.
How many terms of the AP : 9,17, 25, ….. must be taken to give a sum of 636?
Solution:
Here, a = 9, d = 17 – 9 = 8, Sn = 636
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 18

Question 5.
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 19

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Question 6.
The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution:
We have, first term (a) = 17, last term (l) = 350 = Tn and common difference (d) = 9
Let the number of terms be n.
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 20

Question 7.
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution:
Here, n = 22, T22 = 149 = l, d = 7
Let the first term of the AP be a.
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 21

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Question 8.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 22

Question 9.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Solution:
Here, we have S7 = 49 and S17 = 289
Let the first term of the AP be a and d be the common difference, then
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 23
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 24

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Question 10.
Show that a1, a2,…… an,………, form an AP where an is defined as below:
(i) an = 3 + 4n
(ii) an = 9 – 5n
Also find the sum of the first 15 terms in each case.
Solution:
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 25
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 26

Question 11.
If the sum of the first n terms of an AP is 4n – n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Solution:
We have, Sn = 4n – n2
S1 = 4(1) – (1)2 = 4 – 1 = 3 ⇒ First term (a) = 3
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 27

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Question 12.
Find the sum of the first 40 positive integers divisible by 6.
Solution:
The first 40 positive integers divisible by 6 are 6,12,18, ……. , (6 × 40)
And, these numbers are in AP, such that a = 6,
d = 12 – 6 = 6 and n = 6 × 40 = 240 = l
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 28
= 20[12 + 39 × 6] = 20[12 + 234]
= 20 × 246 = 4920

Question 13.
Find the sum of the first 15 multiples of 8.
Solution:
The first 15 multiples of 8 are 8, 16, 24, 32, ….., 120.
These numbers are in AP, where, a = 8 and l = 120
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 29
Thus, the sum of first 15 multiples of 8 is 960.

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Question 14.
Find the sum of the odd numbers between 0 and 50.
Solution:
Odd numbers between 0 and 50 are 1, 3, 5, 7, …… , 49
These numbers are in AP such that a = 1 and l = 49
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 30
Thus, the sum of odd numbers between 0 and 50 is 625.

Question 15.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows : ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
Solution:
Here, penalty for delay on 1st day = ₹ 200
2nd day = ₹ 250
3rd day = ₹ 300
……………………………………….
……………………………………….
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 31

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Question 16.
A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.
Solution:
Sum of all the prizes = ₹ 700
Let the first prize be a
2nd prize = (a – 20)
3rd prize = (a- 40)
4th prize = (a – 60)
The above prizes form an AP
Now, we have, first term = a
Common difference = d = (a – 20) – a = -20
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 32

Question 17.
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students ?
Solution:
Number of classes = 12
∵ Each class has 3 sections.
∴ Number of trees planted by class I = 1 × 3 = 3
Number of trees planted by class II =2 × 3 = 6
Number of trees planted by class III = 3 × 3 = 9
Number of trees planted by class IV = 4 × 3 = 12
………………………………………………………………….
Number of trees planted by class XII = 12 × 3 = 36
The numbers 3, 6, 9,12, ……. , 36 form an AP
Here, a = 3, d = 6 – 3 = 3 and n = 12
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 33

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Question 18.
A spiral is made up of successive semi-circles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, …… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semi-circles? (Take \(\pi=\frac{22}{7}\))
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 34
Solution:
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 35
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 36

Question 19.
200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see figure). In how many rows are the 200 logs placed and how many logs are in the top row?
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 37
Solution:
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 38
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 39

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Question 20.
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see figure).
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 40
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint :To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)]
Solution:
Here, number of potatoes = 10
The to and fro distance of the bucket:
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 41

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3
Question 1.
Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) 2x2 – 7x + 3 = 0
(ii) 2x2 + x – 4 = 0
(iii) 4x2 + 4\(\sqrt{3}\)x + 3 = 0
(iv) 2x2 + x + 4 = 0
Solution:
(i) We have, 2x2 – 7x + 3 = 0
Dividing both sides by 2, we get
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 1

(ii) We have 2x2 + x – 4 = 0
Divide both sides by 2, we get
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 2

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 3
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 4

(iv) We have, 2x2 + x + 4 = 0
Dividing both sides by 2, we get
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 5
Since, the square of a number cannot be negative.
∴ There is no real value of x satisfying the given equation.

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

Question 2.
Find the roots of the following quadratic equations, using the quadratic formula:
(i) 2x2 – 7x + 3 = 0
(ii) 2x2 + x – 4 = 0
(iii) 4x2 + 4\(\sqrt{3}\)x + 3 = 0
(iv) 2x2 + x + 4 = 0
Solution:
(i) Comparing the given equation with ax2 + bx + c = 0, we get a = 2, b = -7, c = 3
∴ b2 – 4ac = (-7)2 – 4(2)(3) = 49 – 24 = 25 > 0
Since b2 – 4ac > 0, therefore the given equation has real roots, which are given by
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 6
Taking negative sign, x = \(\frac{7-5}{4}=\frac{2}{4}=\frac{1}{2}\)
Thus, the roots of the given equation are 3 and \(\frac{1}{2}\).

(ii) Comparing the given equation with ax2 + bx + c = 0, we get a = 2, b = 1, c = – 4
b2 – 4ac = (1)2 – 4(2)(-4) = 1 + 32 = 33 > 0
Since b2 – 4ac > 0, therefore the given equation has real roots, which are given by
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 7

(iii) Comparing the given equation with
ax2 + bx + c = 0, we get
a = 4, b= 473, c = 3
b2 – 4ac = (473)2 – 4(4)(3)
= (16 × 3) – 48 = 48 – 48 = 0
Since b2 – 4ac = 0, therefore the given equation has real and equal roots, which are given by
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 8

(iv) Comparing the given equation with ax2 + bx + c = 0, we get a = 2, b = 1, c = 4
b2 – 4ac = (1 )2 – 4(2)(4) = 1 – 32 = -31 < 0 Since
b2 – 4ac < 0, therefore the given equation does not have real roots.

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

Question 3.
Find the roots of the following equations:
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 9
Solution:
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 10
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 11
Taking negative sign, x = \(\frac{3-1}{2}\)
Thus, the required roots of the given equation are 2 and 1.

Question 4.
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is \(\frac{1}{3}\). Find his present age.
Solution:
Let the present age of Rehman be x years.
3 years ago, Rehman’s age = (x – 3) years
5 years later, Rehman’s age = (x + 5) years
According to the condition,
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 12
⇒ 3(2x + 2) = x2 + 2x – 15
⇒ 6x + 6 = x2 + 2x – 15
⇒ x2 + 2x – 6x – 15 – 6 = 0
⇒ x2 – 4x – 21 = 0 …(1)
Comparing equation (1) with ax2 + bx + c = 0,
we get a = 1, b = -4, c = -21
b2 – 4ac = (-4)2 – 4(1)(-21) = 16 + 84 = 100
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 13
Since, age cannot be negative.
∴ x ≠ -3 ⇒ x = 7
So, the present age of Rehman = 7 years

Question 5.
In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
Solution:
Let Shefali’s marks in Mathematics = x
∴ Marks in English = 30 – x
According to the condition,
(x + 2) × [(30 – x) – 3] = 210
⇒ (x + 2) × (30 – x – 3) = 210
⇒ (x + 2)(- x + 27) = 210
⇒ -x2 + 25x + 54 = 210
⇒ -x2 + 25x + 54 – 210 = 0
⇒ -x2 + 25x – 156 = 0
⇒ x2 – 25x + 156 = 0 …(1)
Comparing equation (1) with ax2 + bx + c = 0, we get a = 1,b = -25, c = 156
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 14
When x = 12, then 30 – x = 30 – 12 = 18 Thus, marks in Mathematics = 13, marks in English = 17 or
Marks in Mathematics = 12, Marks in English = 18.

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

Question 6.
The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Solution:
Let the shorter side i.e., breadth = x metres
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 15
∴ The longer side i.e., length = (x + 30) metres and diagonal = (x + 60) metres
In a rectangle,
(diagonal)2 = (breadth)2 + (length)2
⇒ (x + 60)2 = x2 + (x + 30)2
⇒ x2 + 120 x + 3600 = x2 + x2 + 60x + 900
⇒ x2 + 120x + 3600 = 2x2 + 60x + 900
⇒ 2x2 – x2 + 60x – 120x + 900 – 3600 = 0
⇒ x2 – 60x – 2700 = 0 …(1)
Comparing equation (1) with ax2 + bx + c = 0, we get a = 1, b = -60, c = -2700
∴ b2 – 4ac = (-60)2 – 4(1)(-2700)
⇒ b2 – 4ac = 3600 + 10800
⇒ b2 – 4ac = 14400
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 16
Since breadth cannot be negative,
x ≠ -30 ⇒ x = 90
∴ x + 30 = 90 + 30 = 120
Thus, the shorter side is 90 m and the longer side is 120 m.

Question 7.
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Solution:
Let the larger number be x.
Since, (smaller number)2 = 8(larger number)
⇒ (smaller number)2 = 8
⇒ smaller number = \(\sqrt{8 x}\)
According to the condition,
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 17
Thus, the smaller number = 12 or -12
Thus, the two numbers are 18 and 12 or 18 and -12.

Question 8.
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 18
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 19
Thus, speed of the train is 40 km/hr.

Question 9.
Two water taps together can fill a tank in \(9 \frac{3}{8}\) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Solution:
Let the smaller tap fills the tank in x hours
∴ The larger tap fills the tank in (x – 10) hours.
Amount of water flowing through both the taps in one hour
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 20
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 21
[ ∵ Time cannot be negative]
x = 25 ⇒ x – 10 = 25 – 10 = 15
Thus, time taken to fill the tank by the smaller tap alone is 25 hours and by the larger tap alone is 15 hours.

Question 10.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
Solution:
Let the average speed of the passenger train be x km/h.
∴ Average speed of the express train = (x + 11) km/h
Total distance covered = 132 km
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 22
Comparing equation (1) with ax2 + bx + c = 0, we get a = 1, b = 11, c = -1452
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 23

Question 11.
Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
Solution:
Let the side of the smaller square be x m.
⇒ Perimeter of the smaller square = 4x m
∴ Perimeter of the larger square = (4x + 24) m
⇒ Side of the larger square
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 24
Area of the smaller square = x2 m2
Area of the larger square = (x + 6)2 m2
According to the condition,
x2 + (x + 6)2 = 468
⇒ x2 + x2 + 12x + 36 = 468
⇒ 2x2 + 12x – 432 = 0
⇒ x2 + 6x – 216 = 0 …(1)
[Dividing both sides by 2] Comparing equation (1) with ax2 + bx + c = 0, we get
a = 1, b = 6, c = -216
b2 – 4ac = (6)2 – 4(1)(-216) = 36 + 864 = 900
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 26
But the length of a square cannot be negative,
∴ x ≠ -18 ⇒ x = 12
Length of the smaller square = 12 m
and the length of the larger square = x + 6
= 12+ 6 = 18 m

MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 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 13.2

Unless stated otherwise, take π = \(\frac{22}{7}\)

Question 1.
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.
Solution:
Here, r = 1 cm and h = 1 cm.
Volume of the conical part = \(\frac{1}{3}\) πr2h and volume of the hemispherical part = \(\frac{2}{3}\) πr3h
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 1
∴ Volume of the solid shape
= \(\frac{1}{3}\) πr2h + \(\frac{2}{3}\) πr3h = \(\frac{1}{3}\) πr2h[h + 2r]
= \(\frac{1}{3}\) π(1)2 [1 + 2(1)] cm3
= (\(\frac{1}{3}\) π × 1 × 3) cm3 = π cm3

MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2

Question 2.
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Solution:
Here, diameter = 3 cm
⇒ Radius (r) = \(\frac{3}{2}\) cm
Total height = 12 cm
Height of a cone (h1) = 2 cm
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 2
∴ Height of both cones = 2 × 2 = 4 cm
⇒ Height of the cylinder (h2) = (12 – 4) cm = 8 cm.
Now, volume of the cylindrical part = πr2h2
Volume of both conical parts = 2[\(\frac{1}{3}\) πr2h1]
∴ Volume of the whole model
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 3

Question 3.
A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see figure).
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 4
Solution:
Since a gulab jamun is like a cylinder with hemispherical ends.
Total height of the gulab jamun = 5 cm.
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 5
Diameter = 2.8 cm
⇒ Radius (r) = 1.4 cm
∴ Length of the cylindrical part (h)
= 5 cm – (1.4 + 1.4) cm = 5 cm – 2.8 cm = 2.2 cm
Now, volume of the cylindrical part = πr2h and volume of both the hemispherical ends
= 2(\(\frac{2}{3}\) πr3) = \(\frac{4}{3}\) πr3
∴ Volume of a gulab jamun
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 6
Volume of 45 gulab jamuns
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 7

MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2

Question 4.
A pen stand made up of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see figure).
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 8
Solution:
Dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm.
∴ Volume of the cuboid = 15 × 10 × \(\frac{35}{10}\) cm3
= 525 cm3
Since, each depression is conical in shape with base radius (r) = 0.5 cm and depth (h) = 1.4 cm
∴ Volume of each depression
= \(\frac{1}{3} \pi r^{2} h=\frac{1}{3} \times \frac{22}{7} \times\left(\frac{5}{10}\right)^{2} \times \frac{14}{10} \mathrm{cm}^{3}=\frac{11}{30} \mathrm{cm}^{3}\)
Since, there are 4 depressions
∴ Total volume of 4 depressions = \(\frac{44}{30}\) cm3
Now, volume of the wood in entire stand = [Volume of the wooden cuboid] – [Volume of 4 depressions]
= 525cm3 – \(\frac{44}{30}\) cm3
= \(\frac{15750-44}{30} \mathrm{cm}^{3}=\frac{15706}{30} \mathrm{cm}^{3}\)
= 523.53 cm3

MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2

Question 5.
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:
Height of the conical vessel (h) = 8 cm
Base radius (R) = 5 cm
Volume of water in conical vessel = \(\frac{1}{3}\) πR2h
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 9
Thus, the required number of lead shots = 100

Question 6.
A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass. (Use π = 3.14)
Solution:
Height of the big cylinder (h) = 220 cm
Base radius (r) = \(\frac{24}{2}\) cm = 12cm
∴ Volume of the big cylinder
= πr2h = (π(12)2 × 220) cm3
Also, height of smaller cylinder (h1) = 60 cm
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 10
Base radius (r1) = 8 cm
∴ Volume of the smaller cylinder πr12h1
= (π(8)2 × 60) cm3
∴ Volume of iron pole = [Volume of big cylinder] + [Volume of the smaller cylinder]
= (π × 220 × 122 × π × 60 × 82) cm3
= 3.14[220 × 12 × 12 + 60 × 8 × 8] cm3
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 11

Question 7.
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Solution:
Height of the conical part = 120 cm
Base radius of the conical part = 60 cm
Volume of the conical part
= \(\frac{1}{3}\) πr2h = [\(\frac{1}{3} \times \frac{22}{7}\) × (60)2 × 120] cm3
Radius of the hemispherical part = 60 cm.
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 12

MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2

Question 8.
A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Solution:
Volume of the cylindrical part = πr2h
= (3.14 × 12 × 8) cm3 = \(\frac{314}{100}\) × 8 cm3
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 13
∴ Total volume of the glass vessel = [Volume of cylindrical part] + [Volume of spherical part]
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 14
⇒ Volume of water in the vessel = 346.51 cm3
Since the child finds the volume as 345 cm3
∴ The child’s answer is not correct.
The correct answer is 346.51 cm3.

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

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:
MP Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.3 1

(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]
MP Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.3 2
(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.

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

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°.

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

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
MP Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.3 3

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

MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2

Question 1.
Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3.
Solution:
Let the required point be P(x, y).
Here the end points are (-1, 7) and (4, – 3)
Ratio = 2 : 3 = m1 : m2
∴ x = \(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}\)
= \(\frac{(2 \times 4)+3 \times(-1)}{2+3}=\frac{8-3}{5}=\frac{5}{5}=1\)
And y = \(\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}\)
= \(\frac{2 \times(-3)+(3 \times 7)}{2+3}=\frac{-6+21}{5}=\frac{15}{5}=3\)
Thus, the required point is (1, 3)

Question 2.
Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
Solution:
Let the given points be A(4, -1) and B(-2, -3)
MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2 1
Let the points P and Q trisects AB.
i.e., AP = PQ = QB
i.e., P divides AB in the ratio of 1 : 2 and Q divides AB in the ratio of 2 : 1
Let the coordinates of P be (x, y)
MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2 2

MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2

Question 3.
To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the figure. Niharika runs 1/4th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag ?.
MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2 3
Solution:
Let us consider ‘A’ as origin, then
AB is the x-axis.
AD is the y-axis.
Now, the position of green flag-post is (2, \(\frac{100}{4}\)) or (2, 25)
And the position of the red flag-post is (8, \(\frac{100}{5}\)) or (8, 20)
Distance between both the flags
MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2 4
Let the mid-point of the line segment joining the two flags be M(x, y).
MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2 5
or x = 5 and y = 22.5
Thus, the blue flag is on the 5th line at a distance 22-5 m above AB.

Question 4.
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Solution:
Let the given points are A(-3, 10) and B(6, -8).
Let the point P(-1, 6) divides AB in the ratio m1 : m2.
∴ Using the section formula, we have
MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2 6
⇒ -1(m1 + m2) = 6m1 – 3m2
and 6(m1 + m2) = – 8m1 + 10m2
⇒ -m1 – m2 – 6m1 + 3m2 = 0
and 6m1 + 6m2 + 8m1 – 10m2 = 0
⇒ -7m1 + 2m2 = 0 and 14m1 – 4m2 = 0 or 7m1 – 1m2 = 0
⇒ \(\frac{m_{1}}{m_{2}}=\frac{2}{7}\) and \(\frac{m_{1}}{m_{2}}=\frac{2}{7}\)
⇒ 2m2 = 7m1 and 7m1 = 2m2
⇒ m1 : m2 = 2 : 7 and m1 : m2 = 2 : 7
Thus, the required ratio is 2 : 7.

MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2

Question 5.
Find the ratio in which the line segment joining 4(1, -5) and B(-4, 5) is divided by the x-axis. Also, find the coordinates of the point of division.
Solution:
The given points are A( 1, -5) and B(-4, 5). Let the required ratio = k:l and the required point be P(x, y).
Since the point P lies on x-axis,
∴ Its y-coordinate is 0.
MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2 7

Question 6.
If (1, 2), (4, y), (x, 6) and (3,5) are the vertices of a parallelogram taken in order, find x and y.
Solution:
Let the given points are A( 1, 2), B(4, y), C(x, 6) and D(3, 5)
MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2 8
Since, the diagonals of a parallelogram bisect each other.
∴ The coordinates of P are :
MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2 9
∴ The required values of x and y are 6 and 3 respectively.

Question 7.
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and Bis (1,4).
Solution:
Let centre of the circle is 0(2, -3) and the end points of the diameter be A(x, y) and B(1, 4)
MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2 10
Since, the centre of a circle bisects the diameter.
∴ 2 = \(\frac{x+1}{2}\) ⇒ x + 1 = 4 or x = 3
And -3 = \(\frac{y+4}{2}\) ⇒ y + 4 = -6 or y = -10
Thus, the coordinates of A are (3, -10)

MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2

Question 8.
If A and B are (-2, -2) and (2, -4) respectively, find the coordinates of P such that AP = \(\frac{3}{7}\) AB and Plies on the line segment AB.
Solution:
Here, the given points are A(-2, -2) and B(2, -4)
Let the coordinates of P are (x, y)
Since, the point P divides AB such that
AP = \(\frac{3}{7}\)AB or \(\frac{A P}{A B}\) = \(\frac{3}{7}\)
⇒ AB = AP + BP
⇒ \(\frac{A P+B P}{A P}\) = \(\frac{7}{3}\)
MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2 11

Question 9.
Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.
Solution:
Here, the given points are A(-2, 2) and B(2, 8)
Let P1, P2 and P3 divide AB in four equal parts.
MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2 12

Question 10.
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.
[Hint: Area of a rhombus = \(\frac{1}{2}\) (product of its diagonals)]
Solution:
Let the vertices of the given rhombus are A(3, 0), B(4, 5), C(-1, 4) and D(-2, -1)
MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2 13

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

Question 1.
Find the roots of the following quadratic equations by factorisation:
(i) x2 – 3x – 10 = 0
(ii) 2x2 + x – 6 = 0
(iii) \(\sqrt{2} x^{2}+7 x+5 \sqrt{2}\) = 0
(iv) 2x2 – x + \(\frac{1}{8}\) = 0
(v) 100x2 – 20x + 1 =0
Solution:
(i) We have, x2 – 3x – 10 = 0
⇒ x2 – 5x + 2x – 10 = 0
⇒ x(x – 5) + 2(x – 5) = 0 (x – 5)(x + 2) = 0
Either x – 5 = 0 or x + 2 = 0
x = 5 or x = – 2
Thus, the required roots are 5 and -2.

(ii) We have, 2x2 + x – 6 = 0
⇒ 2x2 + 4x – 3x – 6 = 0
⇒ 2x(x + 2) – 3(x + 2) = 0
⇒ (x + 2)(2x – 3) = 0
Either x + 2 = 0 or 2x – 3 = 0
⇒ x = -2 or x = \(\frac{3}{2}\)
Thus, the required roots are -2 and \(\frac{3}{2}\)

(iii)
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.2 1

(iv) We have, 2x2 – x + \(\frac{1}{8}\) = 0
⇒ 16x2 – 8x + 1 = 0
⇒ 16x2 – 4x – 4x + 1 = 0
⇒ 4x(4x – 1) -1(4x – 1) = 0
⇒ 4x – 1 = 0
⇒ x = \(\frac{1}{4}\)
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.2 2

(v) We have 100x2 – 20x + 1 = 0
⇒ 100x2 – 10x – 10x + 1 = 0
⇒ 10x(10x – 1) -1(10x – 1) = 0
⇒ (10x – 1)(10x – 1) = 0
⇒ (10x – 1) = 0
⇒ x = \(\frac{1}{10}\)
Thus the required roots are \(\frac{1}{10}, \frac{1}{10}\)

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

Question 2.
Solve the problems:
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹ 750. We would like to find out the number of toys produced on that day.
Solution:
(i) Let John had x marbles and Jivanti had (45 – x) marbles.
When both of them lost 5 marbles then equation becomes (x – 5) × (45 – x – 5) = 124
⇒ (x – 5) × (40 – x) = 124
⇒ x2 – 45x + 324 = 0
⇒ x2 – 9x – 36x + 324 = 0
⇒ x(x – 9) – 36(x – 9) = 0
⇒ (x – 9)(x – 36) = 0
Either x – 9 = 0 or x – 36 = 0
Thus, x = 9 or x = 36
∴ If John had 9 marbles, then Jivanti had 45 – 9 = 36 marbles.
If John had 36 marbles, then Jivanti had 45 – 36 = 9 marbles.

(ii) Let the number of toys produced in a day be x.
Then cost of 1 toy = \(\frac{750}{x}\)
⇒ \(\frac{750}{x}\) = 55 – x
⇒ 750 = 55x – x2
⇒ x2 – 55x + 750 = 0
⇒ x2 – 30x – 25x + 750 = 0
⇒ x(x – 30) – 25(x – 30) = 0
⇒ (x – 30)(x – 25) = 0
Either x – 30 = 0 or x – 25 = 0
x = 30 or x = 25

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

Question 3.
Find two numbers whose sum is 27 and product is 182.
Solution:
Let one of the numbers be x.
∴ Other number = 27 – x
According to the condition,
x(27 – x) = 182
⇒ 21x – x2 = 182
⇒ x2 – 27x + 182 = 0
⇒ x2 – 13x – 14x +182 = 0
⇒ x(x – 13) – 14(x – 13) = 0
⇒ (x – 13)(x – 14) = 0
Either x – 13 = 0 or x -14 = 0
⇒ x = 13 or x = 14
Thus, the required numbers are 13 and 14.

Question 4.
Find two consecutive positive integers, sum of whose squares is 365.
Solution:
Let the two consecutive positive integers be x and (x + 1).
Since, the sum of the square of the numbers is 365.
∴ x2 + (x + 1)2 = 365
⇒ x2 + (x2 + 2x + 1) = 365
⇒ x2 + x2 + 2x + 1 = 365
⇒ 2x2 + 2x + 1 – 365 = 0
⇒ 2x2 + 2x – 364 = 0
⇒ x2 + x – 182 = 0
⇒ x2 + 14x – 13x – 182 = 0
⇒ x(x + 14) – 13(x + 14) = 0
⇒ (x + 14)(x -13) = 0
Either x + 14 = 0 or x – 13 = 0
⇒ x = -14 or x = 13
Since, x has to be a positive integer
⇒ x = -14 is rejected.
∴ x = 13 ⇒ x + 1 = 13 + 1 = 14
Thus, the required consecutive positive integers are 13 and 14.

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

Question 5.
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution:
Let the base of the given right triangle be x cm.
∴ Its height = (x – 7) cm
MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.2 3
Squaring both sides, we get
169 = x2 + (x – 7)2
⇒ 169 = x2 + x2 – 14x + 49
⇒ 2x2 – 14x + 49 – 169 = 0
⇒ 2x2 – 14x – 120 = 0
⇒ x2 – 7x- 60 = 0
⇒ x2 – 12x + 5x – 60 = 0
⇒ x(x – 12) + 5(x – 12) = 0
⇒ (x – 12)(x + 5) = 0
Either x – 12 = 0 or x + 5 = 0
⇒ x = 12 or x = -5
But the sides of a triangle can never be negative
⇒ x = -5 is rejected.
∴ x = 12
∴ Length of base = 12 cm
⇒ Length of altitude = (12 – 7)cm = 5 cm
Thus, the required base = 12 cm and altitude = 5 cm

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

Question 6.
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹ 90, find the number of articles produced and the cost of each article.
Solution:
Let the number of articles produced in a day = x
∴ Cost of production of each article = ₹ (2x + 3)
According to the condition,
Total cost = ₹ 90
⇒ x × (2x + 3) = 90
⇒ 2x2 + 3x = 90
⇒ 2x2 + 3x – 90 = 0
⇒ 2x2 – 12x + 15x – 90 = 0
⇒ 2x(x – 6) + 15(x – 6) = 0
⇒ (x – 6)(2x + 15) = 0
Either x – 6 = 0 or 2x + 15 = 0
⇒ x= 6 or x = \(\frac{-15}{2}\)
But the number of articles produced can never be negative.
⇒ x = \(\frac{-15}{2}\) is rejected
∴ Cost of production of each article = ₹ (2 × 6 + 3) = ₹ 15
Thus, the required number of articles produced is 6 and the cost of each article is ₹ 15.

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4
Question 1.
Which term of the AP : 121,117,113, ……., is its first negative term? [Hint: Find n for an < 0]
Solution:
We have the A.P. having a = 121 and d = 117 – 121 = -4
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 1
Thus, the first negative term is 32nd term.

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4

Question 2.
The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP
Solution:
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 2
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 3
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 4

Question 3.
A ladder has rungs 25 cm apart (see figure).The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are \(2 \frac{1}{2}\) m apart, what is the length of the wood required for the rungs? 250
[Hint: Number of rungs = \(\frac{250}{25}+1\)]
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 5
Solution:
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 6
Length of the 1st rung (bottom rung) = 45 cm
Length of the 11th rung (top rung) = 25 cm
Let the length of each successive rung decreases by x cm.
∴ Total length of the rungs = 45 cm + (45 – x) cm + (45 – 2x) cm + ….. + 25 cm
Here, the number 45, (45 – x), (45 – 2x), …., 25 are in an AP such that
First term (a) = 45
and last term (l) = 25
Number of terms, n = 11
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 7
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 8

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4

Question 4.
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the number of the houses preceding the house numbered x is equal to the sum of the number of the houses following it. Find this value of x. [Hint: Sx-1 = S49 – Sx]
Solution:
We have the following consecutive numbers on the houses of a row; 1, 2, 3, 4, 5, ….. , 49.
These numbers are in AP, such that a = 1, d = 2 – 1 = 1, n = 49
Let one of the houses be numbered as x.
∴ Number of houses preceding it = x – 1
Number of houses following it = 49 – x
Now, the sum of the house-numbers preceding
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 9
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 10
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 11

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4

Question 5.
A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of \(\frac{1}{4}\)m and a tread of \(\frac{1}{2}\)m (see fig.) Calculate the total volume of concrete required to build the terrace.
[Hint: Volume of concrete required to build the 1st step \(\frac{1}{4} \times \frac{1}{2} \times 50 m^{3}\)]
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 12
Solution:
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 13
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 14
\(d=\frac{25}{2}-\frac{25}{4}=\frac{25}{4}\)
Here, total number of steps n = 15
Total volume of concrete required to build 15 steps is given by the sum of their individual volumes.
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 15

MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 11 Constructions Ex 11.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1

In each of the following, give the justification of the construction also.

Question 1.
Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
Solution:
Steps of Construction :
I. Draw a line segment AB = 7.6 cm.
II. Draw a ray AX making an acute angle with AB.
III. Mark 13 = (8 + 5) equal points on AX, such that AX1 = X1X2 = ……….. X12X13.
IV. Join points X13 and B.
V. From point X5, draw X5C || X13B, which meets AB at C.
Thus, C divides AB in the ratio 5 : 8 On measuring the two parts, we get AC = 2.9 cm and CB = 4.7 cm.
MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1 1
Justification:
In ∆ABX13 and ∆ACX5, we have
CX5 || BX13
∴ \(\frac{A C}{C B}=\frac{A X_{5}}{X_{5} X_{13}}=\frac{5}{8}\) [By Thales theorem]
⇒ AC : CB = 5 : 8.

Question 2.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle.
Solution:
Steps of Construction :
I. Draw a ∆ABC such that BC = 6 cm, AC = 5 cm and AB = 4 cm.
II. Draw a ray BX making an acute angle ∠CBX.
III. Mark three points [greater of 2 and 3 in \(\frac{2}{3}\)] X1, X2, X3 on BX1 such that BXj = X1X2 = X2X3.
MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1 2
IV. Join X3C.
V. Draw a line through X2 such that it is parallel to X3C and meets BC at C’.
VI. Draw a line through C parallel to CA which intersect BA at A’.
Thus, ∆A’BC’ is the required similar triangle.
Justification :
By construction, we have X3C || X2C’
MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1 3

MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1

Question 3.
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \(\frac{7}{5}\) of the corresponding sides of the first triangle.
Solution:
Steps of Construction :
I. Construct a ∆ABC such that AB = 5 cm, BC = 7 cm and AC = 6 cm.
MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1 4
II. Draw a ray BX such that ∠CBX is an acute angle.
III. Mark 7 points of X1, X2, X3, X4, X5, X6 and X7 on BX such that BX1 = X1X2 = X2X3 = X3X4 – X4X5 = X5X6 = X6X7
IV Join X5 to C.
V. Draw a line through X7 intersecting BC (produced) at C’ such that X5C || X7C’
VI. Draw a line through C’ parallel to CA to intersect BA (produced) at A’.
Thus, ∆A’BC’ is the required triangle.
Justification:
By construction, we have C’A’ || CA
∴ Using AA similarity, ∆ABC ~ ∆A’BC’
MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1 5

Question 4.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are \(1 \frac{1}{2}\) times the corresponding sides of the isosceles triangle.
Solution:
Steps of Construction :
I. Draw BC = 8 cm
II. Draw the perpendicular bisector of BC which intersects BC at D.
MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1 6
III. Mark a point A on the above perpendicular such that DA = 4 cm.
IV. Join AB and AC.
Thus, ∆ABC is the required isosceles triangle.
V. Now, draw a ray BX such that ∠CBX is an acute angle.
VI. On BX, mark three points [greater of 2 and 3 in \(\frac{2}{3}\)] X1, X2 and X3 such that BX1 = X1X2 = X2X3
VII. Join X2C.
VIII. Draw a line through X3 parallel to X2C and intersecting BC (extended) to C’.
IX. Draw a line through C’ parallel to CA intersecting BA (extended) to A’, thus, ∆A’BC’ is the required triangle.
Justification:
We have C’A’ || CA [By construction]
∴ Using AA similarity, ∆A’BC’ ~ ∆ABC
MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1 7

Question 5.
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are \(\frac{3}{4}\) of the corresponding sides of the triangle ABC.
Solution:
Steps of construction :
I. Construct a ∆ABC such that BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
II. Draw a ray BX such that ∠CBX is an acute angle.
MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1 8
Mark four points [greater of 3 and 4 in \(\frac{3}{4}\)] X1, X2, X3, X4 on BX such that 4
BX1 = X1X2 = X2X3 = X3X4
IV. Join X4C and draw a line through X3 parallel to X4C to intersect BC at C’.
V. Also draw another line through C’ and parallel to CA to intersect BA at A’.
Thus, ∆A’BC’ is the required triangle.
Justification:
In ∆BX4C we have
X4C || X3C’ [By construction]
MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1 9

MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1

Question 6.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°.Then, construct a triangle whose sides are \(\frac{4}{3}\) times the corresponding sides of ∆ABC.
Solution:
Steps of Construction :
I. Construct a AABC such that BC = 7 cm, ∠B = 45°, ∠A = 105° and ∠C = 30°
II. Draw a ray BX making an acute angle ∠CBX with BC.
MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1 10
III. On BX, mark four points [greater of 4 and 3 in \(\frac{4}{3}\) ] X1, X2, X3 and X4 such that BX1 = X1X2 = X2X3 = X3X4.
IV. Join X3C.
V. Draw a line through X4 parallel to X3C intersecting BC(extended) at C’.
VI. Draw a line through C parallel to CA intersecting the extended line segment BA at A’.
Thus, ∆A’BC’ is the required triangle. Justification:
By construction, we have
C’A’ || CA
∴ ∆ABC ~ ∆A’BC’ [AA similarity]
MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1 11

Question 7.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are \(\frac{5}{3}\) times the corresponding sides of the given triangle.
Solution:
Steps of Construction :
I. Construct the right triangle ABC such that ∠B = 90°, BC = 4 cm and BA = 3 cm.
II. Draw a ray BX such that an acute angle ∠CBX is formed.
III. Mark 5 points X1, X2, X3, X4 and X5 on BX such that BX1 = X1X2 = X2X3 = X3X4 = X4X5.
IV. Join X3C.
V. Draw a line through X5 parallel to X3C, intersecting the extended line segment BC at C’.
VI. Draw another line through C’ parallel to CA intersecting the extended line segment BA at A’.
Thus, ∆A’BC’ is the required triangle.
Justification:
By construction, we have C’A’ || CA
∴ ∆ABC ~ ∆A’BC’ [AA similarity]
MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1 12