Class 7

MP Board Class 7th Maths Solutions Chapter 7 त्रिभुजों की सर्वांगसमता Ex 7.2

प्रश्न 1.
निम्न में आप कौन-से सर्वांगसम प्रतिबन्धों का प्रयोग करेंगे ?
(a) दिया है : AC = DF AB = DE, BC = EF

इसलिए, ∆ABC ≅ ∆DEF
(b) दिया है : ZX = RP RQ = ZY
∠PRQ = ∠XZY
इसलिए, ∆PQR = ∆XYZ
(c) दिया है: ∠MLN = ∠FGH
∠NML = ∠GFH
ML = FG
इसलिए, ∆LMN ≅ ∆GFH
(d) दिया है: EB = DB
AE = BC
∠A = ∠C
इसलिए, ∆ABE ≅ ∆CDB
उत्तर:
(a) S.S.S. सर्वांगसमता प्रतिबन्ध द्वारा,
∆ABC ≅ ∆DEE
(b) S.A.S. सर्वांगसमता प्रतिबन्ध द्वारा,
∆PQR ≅ ∆XYZ.
(c) A.S.A. सर्वांगसमता प्रतिबन्ध द्वारा,
∆LMN ≅ ∆GFH.
(d) R.H.S. सर्वांगसमता प्रतिबन्ध द्वारा,
∆ABE ≅ ∆CDB.

प्रश्न 2.
आप ∆ART ≅ ∆PEN दर्शाना चाहते हैं।
(a) यदि आप S.S.S. सर्वांगसमता प्रतिबन्ध का प्रयोग करें तो आपको दर्शाने की आवश्यकता है:
(i) AR =
(ii) RT =
(iii) AT =
(b) यदि यह दिया गया है कि ∠T = ∠N और आपको S.A.S. प्रतिबन्ध का प्रयोग करना है, तो आपको आवश्यकता होगी:
(i) RT = और (ii) PN =

(c) यदि यह दिया गया है कि AT = PN और आपको A.S.A. प्रतिबन्ध का प्रयोग करना है, तो आपको आवश्यकता होगी:
(i) ? =
(ii) ? =
हल:
(a) ∆ART ≅ ∆PEN को S.S.S. सर्वांगसमता प्रतिबन्ध द्वारा दर्शाने के लिए दर्शाना होगा –
(i) AR = PE
(ii) RT = EN
(iii) AT = PN
(b) ∴ ∠T = ∠N
∴ (i) RT = EN
(ii) PN = AT
(c) यदि AT = PN और A.S.A. सर्वांगसमता के लिए आवश्यकता होगी –
(i) ∠RAT = ∠EPN
(ii) ∠ATR = ∠PNE

प्रश्न 3.
आपको ∆AMP ≅ ∆AMQ दर्शाना है। निम्न चरणों में, रिक्त कारणों को भरिए:

उत्तर:
(i) दिया है
(ii) दिया है
(iii) उभयनिष्ठ
(iv) S.A.S. सर्वांगसमता प्रतिबन्ध।

प्रश्न 4.
∆ABC में ∠L = 30°, ∠B = 40° और ∠C = 110°, ∆PQR में, ∠P = 30° ∠Q = 40° और ∠R = 110°. एक विद्यार्थी कहता है कि A.A.A. सर्वांगसमता प्रतिबन्ध से ∆ABC ≅ ∆PQR है।
क्या यह कथन सत्य है ? क्यों या क्यों नहीं ?
हल:
यहाँ ∆MBC के तीनों कोण ∆PQR के तीनों कोणों के बराबर हैं। तो यह आवश्यक नहीं कि त्रिभुज सर्वांगसम हों क्योंकि यदि ∆ABC में, भुजा BC = 3.0-सेमी तथा ∆POR में, भुजा QR = 4.0 सेमी हो, तो इस दशा में त्रिभुज के संगत कोण तो बराबर हैं परन्तु यह सर्वांगसम नहीं हैं। क्योंकि BC ≠ QR अतः विद्यार्थी की A.A.A. सर्वांगसमता का प्रतिबन्ध तर्कसंगत नहीं है।

प्रश्न 5.
संलग्न आकृति में दो त्रिभुज ART तथा OWN सर्वांगसम हैं जिनके संगत भागों को अंकित किया गया है। हम लिख सकते हैं ∆RAT = ?

हल:
हम लिख सकते हैं ∆RAT ≅ ∆WON
(∴ O ↔ A, N ↔ T, W ↔ R)

प्रश्न 6.
कथनों को पूरा कीजिए –

∆BCA ≅ ? ∆QRS ≅ ?
उत्तर:
∆BCA ≅ ∆ABTA, ∆QRS = ∆TPQ

प्रश्न 7.
एक वर्गांकित शीट पर, बराबर क्षेत्रफलों वाले दो त्रिभुजों को इस प्रकार बनाइए कि
(i) त्रिभुज सर्वांगसम हों
(ii) त्रिभुज सर्वांगसम न हों। आप उनके परिमाप के बारे में क्या कह सकते हैं?
हल:

(i) चित्र 7.19 (1) में,
∆ ABC का क्षेत्रफल = ∆EDC का क्षेत्रफल = \(\frac { 1 }{ 2 } \) × 3 × 4 = 6 cm²
∆ ABC का परिमाप = 3 + 4 + 5 = 12 cm
∆ EDE का परिमाप = 3 + 4 + 5 = 12 cm
∆ ABC का परिमाप = ∆EDC का परिमाप,
अतः चित्र 7.19 में, ∆ABC ≅ ∆EDC है।
(ii) चित्र 7.19 (ii) में,
∆ PQR का क्षेत्रफल = \(\frac { 1 }{ 2 } \) × PQ × PR
= \(\frac { 1 }{ 2 } \) × 3 × 4 = 6 cm²
तथा ∆ PSR का क्षेत्रफल = \(\frac { 1 }{ 2 } \) × ST × PR
\(\frac { 1 }{ 2 } \) × 3 × 4 = 6 cm²

∴ ∆ POR का क्षेत्रफल = ∆ PSR का क्षेत्रफल
अब, ∆ PQR का परिमाप = 3 + 4 + 5 = 12 cm
तथा ∆ PRS का परिमाप = 4 + 35 + 4 = 11’5 cm
∆ POR का परिमाप ≠ ∆PRS का परिमाप
अत: चित्र 7.19 (ii) में ∆POR व ∆PRS सर्वांगसम नहीं हैं क्योंकि इनके क्षेत्रफल तो समान हैं परन्तु परिमाप समान नहीं

प्रश्न 8.
संलग्न आकृति में एक सर्वांगसम भागों का एक अतिरिक्त युग्म बताइए जिससे ∆ABC और ∆PQR सर्वांगसम हो जाएँ। आपने किस प्रतिबन्ध का प्रयोग किया ?

हल:
यहाँ, ∆ABC ≅ ∆PQR
∴ ∠B = ∠Q IR ∠C = ∠R
∴ सर्वांगसम भागों का अतिरिक्त युग्म –
BC = QR
उत्तर हमने यहाँ A.S.A. सर्वांगसम प्रतिबन्ध का प्रयोग किया है।

प्रश्न 9.
चर्चा कीजिए, क्यों?
∆ABC ≅ ∆FED.

हल:
∠B = ∠E = 90°,
∠A = ∠F (दिया हुआ है)
∴ ∠C = ∠D (तीसरा कोण)
BC = DE (दिया हुआ है)
अत: ASA सर्वांगसम प्रतिबन्ध से ∆ ABC ≅ ∆ FED परिणाम प्राप्त होगा।

पाठ्य-पुस्तक पृष्ठ संख्या # 163

ज्ञानवर्धक क्रियाकलाप

प्रश्न 1.
अलग-अलग माप के वर्गों के कट-आउट सोचिए। अध्यारोपण विधि का प्रयोग वर्गों की सर्वांगसमता के लिए प्रतिबन्ध ज्ञात करने के लिए कीजिए। कैसे “सर्वांगसम भागों” की संकल्पना सर्वांगसम के अंतर्गत उपयोग होती है ? क्या यहाँ संगत भुजाएँ हैं ? क्या यहाँ संगत विकर्ण हैं ?
हल:
हम जानते हैं कि समतल आकृतियाँ सर्वांगसम होती हैं। जब आकृतियों के आकार समान होते हैं तो वे एक-दूसरे की ठीक-ठीक पूरा ढक लेती हैं। सभी वर्ग समान आकृति के होते हैं लेकिन वर्ग का आकार उनकी भुजाओं की लम्बाई पर निर्भर करता है।

ABCD व PQRS दो वर्ग हैं। वर्ग ABCD के कट-आउट को वर्ग PQRS के ऊपर इस प्रकार रखते हैं कि शीर्ष A, वर्ग PQRS के शीर्ष P पर और भुजा AB भुजा PQ पर आए।

स्पष्ट है कि ABCD वर्ग PQRS को पूर्णतया ढक लेता है।

यदि AB = PQ तो दो वर्ग सर्वागसम होंगे यदि उनकी भुजाओं की लम्बाइयाँ समान हों।

अत: वर्ग ABCD ≅ वर्ग PORS यदि AB = PQ

हम एक वर्ग की किसी भी भुजा को दूसरे वर्ग की किसी भुजा के संगत ले सकते हैं। दूसरी संगत भुजाओं के युग्म इसी प्रकार बदल जाएँगे। यह बात विकर्णों के लिए भी सत्य है।

प्रश्न 2.
यदि आप वृत्त लेते हैं तो क्या होता है ? दो वृत्तों की सर्वांगसमता के लिए प्रतिबन्ध क्या है ? क्या, आप फिर अध्यारोपण विधि का प्रयोग कर सकते हैं ? पता लगाइए।
हल:
सभी वृत्तों की समान आकृति होती है और वृत्त का आकार वृत्त की त्रिज्या पर निर्भर करता है। यहाँ दो वृत्त C₁ व C₂ हैं। इनमें से किसी एक वृत्त का कट-आउट (माना वृत्त C₂ का) वृत्त C₁ पर रखते हैं। वृत्त C₂ वृत्त C₁ को पूरी तरह ठीक-ठीक ढल लेता है। यदि दोनों वृत्तों की त्रिज्याएँ समान होंगी तो दोनों वृत्त सर्वांगसम होंगे।

वृत्त C₁ वृत्त C₂ जबकि C₁ वृत्त की त्रिज्या = C₂ वृत्त की त्रिज्या।

प्रश्न 3.
इस संकल्पना को बढ़ाकर तल की दूसरी आकृतियाँ जैसे समषद्भुज इत्यादि के लिए प्रयत्न कीजिए।
हल:
हम जानते हैं कि समतल आकृतियाँ सर्वांगसम होती हैं यदि वे एक-दूसरे को पूर्णतया ढक लेती हैं। सभी समषट्भुज समान आकृति के होते हैं और इनका आकार समषट्भुज की भुजा की लम्बाई पर निर्भर करता है। दो समषट्भुज ABCDEF व PQRSTU लेते हैं। इनके कट-आउट लेते हैं जिनमें से प्रत्येक की सभी भुजाएँ समान हों।

अब PQRSTU के कट-आउट को ABCDEF पर इस प्रकार रखते हैं कि PQRSTU का बिन्दु P बिन्दु A पर आए तथा भुजा PQ भुजा AB पर आए। यदि PQ = AB तो समषट्भुज PQRSTU, समषट्भुज ABCDEF को पूर्णतया ठीक-ठीक ढक लेता है। अत: दो समषट्भुज सर्वांगसम होते हैं यदि इनकी भुजाओं की लम्बाई समान हो।

अत: समषट्भुज ABCDEF = समषट्भुज PQRSTU.

प्रश्न 4.
एक त्रिभुज की दो सर्वांगसम प्रतिलिपियाँ लीजिए। कागज को मोड़कर पता लगाइए कि क्या उनके शीर्ष लम्ब बराबर हैं ? क्या उनकी माध्यिकाएँ समान हैं ? आप उनके परिमाप तथा क्षेत्रफल के बारे में क्या कह सकते हैं ?
हल:
माना ∆ABC ≅ ∆DEF
कागज को मोड़कर प्रत्येक त्रिभुज के शीर्ष बनाए। हम देखते हैं कि
AL = DP BM = EQ और CN = FR
अर्थात् संगत शीर्ष लम्ब समान हैं।

इसी प्रकार हम देख सकते हैं कि सर्वांगसम त्रिभुजों में संगत माध्यिकाएँ समान होती हैं और इनके परिमाप व क्षेत्रफल समान होते हैं।

MP Board Class 7th Maths Solutions

MP Board Class 7th Social Science Solutions Chapter 22 Aurangzeb and the Decline of the Mughal Empire

MP Board Class 7th Social Science Chapter 22 Text Book Questions

Choose the correct alternatives:

Question 1.
Aurangzeb ruled for:
(a) 30 years
(b) 40 years
(c) 50 years
(d) 60 years
Answer:
(b) 40 years

Question 2.
The Jats lived near:
(a) Delhi and Agra
(b) Agra and Mathura
(c) Mathura and Bharatpur
(d) Arga and Jhansi
Answer:
(b) Agra and Mathura

Question 3.
The third battle of Panipat was fought in:
(a) 1526 AD
(b) 1556 AD
(c) 1560 AD
(d) 1761 AD
Answer:
(d) 1761 AD

Question 4.
The Indian goods which were in great demand in the European markets were:
(a) Clothes and spices
(b) Clothes and silver
(c) Spices and gems
(d) Spices and horses
Answer:
(a) Clothes and spices

Fill in the blanks:

1. ……………. and …………. were die Bundela rulers who revolted against Aurangzeb.
2. Gurudwara Sheshganj was built at the Martyrdom place of ……………
3. Nadh  Shah invaded India in ………….
4. Due to Aurangzeb’s policy of …………….. the Mughal administration weakend.

Answer:

1. Compestral, Chhatrasal
2. Guru Teg Bahadur
3. 1739 AD
4. Deccan

MP Board Class 7th Social Science Chapter 22 Short Answer Type Questions

Question 1.
Which taxes were imposed on the Hindus by Aurangzeb?
Answer:
Aurangzeb imposed the Jazia tax on the Hindus.

Question 2.
Who do you know about the Satnamis?
Answer:
Satnamis were the followers of truth. They used to dress like the purohits orpriests. They revolted due to the religious oppression of Aurangzeb. Their main center was Namaul and Mewar. However Aurangzeb suppressed their revolts ruthlessly.

Question 3.
Due to which policy of Aurangzeb his successors remained weak?
Answer:
Due to Aurangzeb’s policy of expansion his successors remained weak. The Mughal Empire was very vast and it was difficult to manage such a vast empire. The incompetent successors were not able to protect it.

Question 4.
What were the outcomes of the invasions of Nadir shah and Ahmad Shah Abdali?
Answer:
As a result of the invasions of Nadir Shah and Ahmad Shah Abdali the Mughal Empire became weak and remained restricted to Delhi and surrounding area.

MP Board Class 7th Social Science Chapter 22 Long Answer Type Questions

Question 1.
Describe the causes of the decline of the Mughal Empire.
Answer:
After the death of Aurangzeb in 1707, the Mughal Empire broke up. Its fall had already begun during Aurangzeb’s reign. But during the eighteenth century the fall was nearly complete.

The factors that led to the down-break of the Mughal empire were as follows:
1. Weak and incapable successors:
The successors of Aurangzeb were very weak and incapable. They had no capacity to rule over such a vast empire. They remained puppets in the hands of their subedars and other officers.

2. Wars of succession:
After the death of Aurangzeb war of succession broke out. They reduced the strength of the Kingdom.

3. The Suspicious nature of Aurangzeb and his rigid religious policy:
Aurabgzeb’s nature was very suspicious. Due this nature he could not provide administrative and military training to his sons. As a result they became incapable of managing the empire. His religious policy was based on intolerance. This made the Sikhs, Jats, Satnamis, Rajputs and Marathas his great enemies. They revolted against him.

4. Aurangzeb’s Deccan Policy:
Aurangzeb spent the last 25 years of his life fighting battles in the south. This made the Mughal administration in the north very weak.

5. Luxurious life styles of the Mughal rulers and Aristocrats:
Resulted in the decline of the Mughal Empire.

6. Foreign invasion:
Nadirshah, the ruler of Iran invaded the Mughal Empire in 1739. He looted the wealth about seventy crores of rupees. After his assassination in 1747 AD his general Ahmadshah Abdali, became the ruler and fought third battle of Panipat with Maratha in which Maratha were defeated. Because of these invasions Mughal Empire became weak and remained restricted to Delhi and surrounding areas.

7. Lack of Naval Power:
Neglect of navy proved fatal for the Mughal Empire.

Project Work:
On an outline map of India mark the extent of the Mughal Empire. Also mark the areas of Bundelas, Jats and Satnamis.
Answer:

MP Board Class 7th Social Science Solutions

MP Board Class 7th Maths Solutions Chapter 12 Algebraic Expressions Ex 12.4

Question 1.
Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic
watches or calculators,

If the number of digits formed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern. How many segments are required to form 5, 10, 100 digits of the kind

Solution:
(a) It is given that the number of segments required to form n digits of the kind
is (5n + 1).
Number of segments required to form 5 digits = (5 × 5 + 1) = 25 + 1 = 26
Number of segments required to form 10 digits = (5 × 10 + 1) = 50 + 1 = 51
Number of segments required to form 100 digits = (5 × 100 + 1) = 500 + 1 = 501

(b) It is given that the number of segments required to form n digits of the kind
[ is (3n +1).
Number of segments required to form 5 digits = (3 × 5 + 1) = 15 + 1 = 16
Number of segments required to form 10 digits = (3 × 10 + 1) = 30 + 1 = 31
Number of segments required to form 100 digits = (3 × 100 + 1) = 300 + 1 = 301

(c) It is given that the number of segments required to form n digits of the kind
is (5n + 2).
Number of segments required to form 5 digits = (5 × 5 + 2) = 25 + 2 = 27
Number of segments required to form 10 digits = (5 × 10 + 2) = 50 + 2 = 52
Number of segments required to form 100 digits = (5 × 100 + 2) = 500 + 2 = 502

Question 2.
Use the given algebraic expression to complete the table of number patterns.

Solution:
(i) Number pattern for expression 2n – 1
Put n = 1, 2, 3,…. and so on, we get

(ii) For expression 3 n + 2, 5^th, 10^th term and 100th term of the pattern are 3 × 5 + 2 = 17, 3 × 10 + 2 = 32 and 3 × 100 + 2 = 302 respectively.
(iii) For expression 4n +1, 5^th, 10^th and 100^th term of the pattern are 4 × 5 +1 = 21, 4 × 10 + 1 = 41 and 4 × 100 + 1 = 401 respectively.
(iv) For expression 7n + 20, 5^th, 10^th and 100^th term of the pattern are 7 × 5 + 1 = 36, 7 × 10 + 20 = 90 and 7 × 100 + 20 = 720 respectively.
(v) For expression n² + 1, 5^th and 10th term of the pattern are 5² + 1 = 26 and 10² + 1 = 101 respectively.

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 8 Comparing Quantities Ex 8.1

Question 1.
Find the ratio of:
(a) ₹ 5 to 50paise
(b) 15 kg to 210 g
(c) 9 m to 27 cm
(d) 30 days to 36 hours
Solution:
(a) ₹ 5 to 50 paise
1 rupee = 100 paise
∴ 5 rupee = 500 paise
Hence, the required ratio is

(b) 15 kg to 210 g
1kg = 1000g
15 kg = 15000 g
Hence, the required ratio is

(c) 9 m to 27 cm
1m = 100 cm
9 m = 900 cm
Hence, the required ratio is

(d) 30 days to 36 hours
1 day = 24 hours
30 days = 24 × 30 = 720 hrs
Hence, the required ratio is

Question 2.
In a computer lab, there are 3 computers for every 6 students. How many computers will
be needed for 24 students?
Solution:
For 6 students, number of computers required = 3
For 1 student, number of computers required = \(\frac{3}{6}\)
For 24 students, number of computers required = \(24 \times \frac{3}{6}=12\)
Hence, 12 computers are required for 24 students.

Question 3.
Population of Rajasthan = 570 lakhs and population of UP = 1660 lakhs.
Area of Rajasthan = 3 lakhs km² and Area of UP = 2 lakhs km²
(i) How many people are there per km² in both these states ?
(ii) Which states is less populated?
Solution:
(i) Population of Rajasthan in 3 lakhs km²
area = 570 lakhs
Population of rajasthan in 1 km² area
\(=\frac{570}{3}=190\) lakhs
Population of U.P. in 2 km² = 1660 lakhs
Population of U.P. in 1 km² = \(\frac{1660}{2}\)
= 830 lakhs

(ii) It can be observed that population per km² area is lesser for Rajasthan. Therefore, Rajasthan is less populated.

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 12 Algebraic Expressions Ex 12.3

Question 1.
If m = 2, find the value of:
(i) m – 2
(ii) 3m – 5
(iii) 9 – 5m
(iv) 3m² – 2m – 7
(v) \(\frac{5 m}{2}\) – 4
Solution:
(i) m – 2 = 2 – 2 = 0
(ii) 3m – 5 = (3 × 2) – 5 = 6 – 5 = 1
(iii) 9 – 5m = 9 – (5 × 2) = 9 – 10 = -1
(iv) 3m² – 2m – 7 = 3 × (2 × 2) – (2 × 2) – 7
= 12 – 4 – 7 = 1
(v) \(\frac{5 m}{2}\) – 4 = \(\left(\frac{5 \times 2}{2}\right)\) – 4 = 5 – 4 = 1

Question 2.
If p = -2, find the value of:
(i) 4p + 7
(ii) -3p² + 4p + 7
(iii) -2p³ – 3p² + 4p + 7
Solution:
(i) 4p + 7 = 4 × (-2) + 7 = -8 + 7 = -1
(ii) -3p² + 4p + 7 = -3 × (-2) × (-2) + 4 × (-2) + 7 = -12 – 8 + 7 = -13
(iii) -2p³ – 3p² + 4p + 7
= -2 × (-2) × (-2) × (-2) – 3 × (-2) × (-2) + 4 × (-2) + 7
= 16 – 12 – 8 + 7 = 3

Question 3.
Find the value of the following expressions, when x = -1:
(i) 2x – 7
(ii) -x + 2
(iii) x² + 2x + 1
(iv) 2x² – x – 2
Solution:
(i) 2x – 7 = 2 × (-1) – 7 = -2 – 7 = -9
(ii) – x + 2 = – (-1) + 2 = 1 + 2 = 3
(iii) x² + 2x + 1 = (-1) × (-1) + 2 × (-1) + 1 = 1 – 2 + 1 = 0
(iv) 2x² – x – 2 = 2 × (-1) × (-1) – (-1) – 2 = 2 + 1 – 2 = 1

Question 4.
If a = 2, b = -2, find the value of:
(i) a² + b²
(ii) a² + ab + b²
(iii) a² – b²
Solution:
(i) a² + b² = 2 × 2 + (-2) × (-2) = 4 + 4 = 8
(ii) a² + ab + b² = (2 × 2) + 2 × (-2) + (-2) × (-2) = 4 – 4 + 4 = 4
(iii) a² – b² = 2 × 2 – (-2) × (-2) = 4 – 4 = 0

Question 5.
When a = 0, b = -1, find the value of the given expressions:
(i) 2a + 2b
(ii) 2a² + b² + 1
(iii) 2a²b + 2ab² + ab
(iv) a² + ab + 2
Solution:
(i) 2a + 2b = 2 × (0) + 2 × (-1) = 0 – 2 = – 2
(ii) 2a² + b² + 1 = 2 × (0) × (0) + (-1) × (-1) + 1 = 0 + 1 + 1 = 2
(iii) 2a²b + 2ab² + ab = 2 × (0) × (0) × (-1) + 2 × (0) × (-1) × (-1) + 0 × (-1)
= 0 + 0 + 0 = 0
(iv) a² + ab + 2 = (0) × (0) + 0 × (-1) + 2
= 0 + 0 + 2 = 2

Question 6.
Simplify the expressions and find the value, if x is equal to 2.
(i) x + 7 + 4 (x – 5)
(ii) 3(x + 2) + 5x – 7
(iii) 6x + 5(x – 2)
(iv) 4(2x – 1) + 3x+ 11
Solution:
(i) x + 7 + 4(x – 5) = x + 7 + 4x – 20
= (1 + 4)x + (7 – 20) = 5x – 13
Putting x = 2 we get,
5x – 13 = (5 × 2) – 13 = 10 – 13 = -3

(ii) 3 (x + 2) + 5x – 7 = 3x + 6 + 5x – 7
= (3 + 5)x + (6 – 7) = 8x – 1
Putting x = 2 we get,
8x – 1 = (8 × 2) – 1 = 16 – 1 = 15

(iii) 6x + 5(x – 2) = 6x + 5x – 10
= (6 + 5)x – 10 = (11x – 10)
Putting x = 2 we get,
11x – 10 = (11 × 2) – 10 = 22 – 10 = 12

(iv) 4(2x – 1) + 3x + 11 = 8x – 4 + 3x + 11
= (8 + 3) × + (11 – 4) = 11x + 7
Putting x = 2 we get,
11x + 7= (11 × 2) + 7 = 22 + 7 = 29

Question 7.
Simplify these expressions and find their values if x = 3, a = -1, b = -2.
(i) 3x – 5 – x + 9
(ii) 2 – 8x + 4x + 4
(iii) 3a + 5 – 8o + 1
(iv) 10 – 3b – 4 – 5b
(v) 2a – 2b – 4 – 5 + a
Solution:
(i) 3x – 5 – x + 9 = (3 – 1) x + (-5 + 9)
= 2x + 4 = (2 × 3) + 4 [∵ x = 3]
= 6 + 4 = 10

(ii) 2 – 8x + 4x + 4 = 2 + 4 + (- 8 + 4)x
= 6 – 4x = 6 – (4 × 3) = 6 – 12 = -6 [∵ x = 3]

(iii) 3a + 5 – 8a +1 = (3 – 8)a + (5 + 1)
= -5a + 6
= -5 × (-1) + 6     [∵ a = -1]
= 5 + 6 = 11

(iv) 10 – 3b – 4 – 5b = 10 – 4 + (- 3 – 5)b
= 6 – 8b = 6 – 8 × (-2) [∵ b = -2]
= 6 + 16 = 22

(v) 2a – 2b – 4 – 5 + a = (2 + 1)a – 2b – 4 – 5
= 3a – 2b – 9
= 3 × (-1) – 2 × (-2) – 9 [∵ a = -1, b = -2]
= -3 + 4 – 9 = -8

Question 8.
(i) If z = 10, find the value of z³ – 3(z – 10).
(ii) If p = -10, find the value of p² – 2p -100.
Solution:
(i) For z = 10,
z³ – 3z + 30
= (10 × 10 × 10) – (3 × 10) + 30
= 1000 – 30 + 30 = 1000

(ii) For p = -10,
p² – 2p – 100
=(-10) × (-10) – 2 × (-10) – 100
= 100 + 20 – 100 = 20

Question 9.
What should be the value of a if the value of 2x² + x – a equals to 5, when x = 0?
Solution:
When x = 0; 2x² + x – a = 5,
∴ (2 × 0) + 0 – a = 5
⇒ 0 – a = 5
⇒ a = -5

Question 10.
Simplify the expression and find its value when a = 5 and b = -3.
2(a² + ab) + 3 – ab
Solution:
2(a² + ab) + 3 – ab = 2a² + 2ab + 3 – ab
= 2a² + (2 – 1) ab + 3 = 2a² + ab + 3
= 2 × (5 × 5) + 5 × (-3) + 3       [ ∵ a = 5, b = – 3]
= 50 – 15 + 3 = 38

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 7 Congruence of Triangles Ex 7.2

Question 1.
Which congruence criterion do you use in the following?
(a) Given: AC = DF
AB = DE
BC = EF
So, ∆ABC ≅ ∆DEF

(b) Given: ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ∆PQR ≅ ∆XYZ

(c) Given: ∠MLN = ∠FGH
∠NML = ∠HFG
ML = FG
So, ∆LMN ≅ ∆GFH

(d) Given EB = BD

AE = CB
∠A = ∠C = 90°
So, ∆ABE ≅ ∆CDB
Solution:
(a) SSS, as all three sides of AABC are equal to the corresponding sides of ∆DEF.
(b) SAS, as two sides and the angle included between these sides of ∆PQR are equal to the corresponding sides and included angle of ∆XYZ.
(c) ASA, as two angles and the side included between these angles of ∆LMN are equal to the corresponding angles and included side of ∆GFH.
(d) RHS, as in the given two right-angled triangles, one side and the hypotenuse are respectively equal.

Question 2.
You want to show that ∆ART ≅ ∆PEN,
(a) If you have to use SSS criterion, then you need to show
(i) AR =
(ii) RT =
(iii) AT =
(b) If it is given that ∠T = ∠N and you are to use SAS criterion, you need to have
(i) RT = and
(ii) PN =
(c) If it is given that AT = PN and you are to use ASA criterion, you need to have

Solution:
(a) (i) AR = PE
(ii) RT = EN
(iii) AT = PN

(b) (i) RT = EN
(ii) PN = AT

(c) (i) ∠ATR = ∠PNE
(ii) ∠RAT = ∠EPN

Question 3.
You have to show that ∆AMP ≅ ∆AMQ
In the following proof, supply the missing reasons.

Solution:
(i) Given
(ii) Given
(iii) Common
(iv) SAS congruence criterion, as the two sides and the angle included between these sides of ∆AMP are equal to corresponding sides and included angle of ∆AMQ.

Question 4.
In ∆ABC, ∠A = 30°, ∠B = 40° and ∠C = 110°
In ∆PQR, ∠P = 30°, ∠Q = 40° and ∠R = 110°
A student says that ∆ABC ≅ ∆PQR by AAA congruence criterion. Is he justified? Why or why not?
Solution:
No. This property represents that these triangles have their respective angles of equal measure. However, this does not give any information about their sides. The sides of these triangles may have a ratio different from 1 : 1. Therefore, AAA property does not prove that two triangles are congruent.

Question 5.
In the figure, the two triangles are congruent. The corresponding parts are marked. We can write ∆RAT ≅ ?

Solution:
It can be observed that,
∠RAT = ∠WON
∠ART = ∠OWN
AR = OW
Therefore, ∆RAT ≅ ∆WON, by ASA criterion.

Question 6.
Complete the congruence statement:

Solution:
In ∆BCA and ∆BTA,
BC = BT (given)
TA = CA (given)
BA is common.
Therefore, ∆BCA ≅ ∆BTA
(by SSS congruence criterion)
Now, in AQRS and ATPQ,
∠RQS = ∠PTQ (given)
RQ = PT (given)
∠SRQ = ∠QPT (given)
Therefore, ∆QRS = ∆TPQ
(by ASA congruence criterion)

Question 7.
In a squared sheet, draw two triangles of equal areas such that
(i) the triangles are congruent.
(ii) the triangles are not congruent. What can you say about their perimeters?
Solution:

Here, ∆ABC and ∆PQR have the same area and are congruent to each other also. Also the perimeter of both the triangles will be the same.

Here, ∆ABC and ∆PQR have the same height and same base. Thus, their areas are equal. However, these triangles are not congruent to each other. Also, the perimeter of both the triangles will not be the same.

Question 8.
Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent.

Solution:
∆ABC & ∆DEF are not congruent to each other but have five congruent parts i.e., ∠BAC = ∠EFD, ∠ABC = ∠DEF, ∠BCA = ∠EDF, AB = DF and BC = EF

Question 9.
If ∆ABC and ∆PQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

Solution:
∠ABC = ∠PQR = 90° and
∠BCA = ∠QRP (Given)
For triangles to be congruent, BC = QR
∆ABC ≅ ∆PQR (ASA congruence criterion)

Question 10.
Explain, why ∆ABC ≅ ∆FED

Solution:
Given that, ∠ABC = ∠FED = 90° and ∠BAC = ∠EFD
The two angles of ∆ABC are equal to the two respective angles of ∆FED. Also, the sum of all interior angles of a triangle is 180°. Therefore, third angle of both triangles will also be equal in measure.
∴ ∠BCA = ∠EDF
Now, in ∆ABC and ∆FED,
∠ABC = ∠FED, BC = ED and ∠BCA = ∠EDF
⇒ ∆ABC = ∆FED
(ASA congruence criterion)

MP Board Class 7th Maths Solutions

MP Board Class 7th Social Science Solutions Miscellaneous Questions 2

Choose the correct answer

Question 1.
The founder of the slave dynasty was:
(a) Mohammed Ghori
(b) Ill-tutmish
(c) Qutub-ud-din Aibak
(d) Alauddin Khilji
Answer:
(c) Qutub-ud-din Aibak

Question 2.
The Chariman of the National Human Rights Commission is appointed by:
(a) Prime Minister
(b) President
(c) Governor
(d) Chief Minister
Answer:
(b) President

Question 3.
The first battle of Panipat was fought in:
(a) 1526 AD
(b) 1556 AD
(c) 1529 AD
(d) 1527 AD
Answer:
(a) 1526 AD

Question 4.
The percentage of land area on earth is:
(a) 71%
(b) 72%
(c) 28%
(d) 29%
Answer:
(d) 29%

Question 5.
Spring tide occurs on:
(a) Fifth day
(b) full moon day
(c) 8th day
(d) 4th day
Answer:
(b) full moon day

Fill in the blanks:

1. The founder of the Khilji dynasty was ………….
2. Guru Nanak was bom in ………….. village.
3. The tenure of the Vice President is …………….. year.
4. ………….. is used to measure rainfall.
5. There are …………….number of seasons in India.

Answer:

1. Alauddin Khilji
2. Talwandi
3. Five years
4. Rain grange
5. three.

Match column ‘A’with column ‘B’

Answer:
1. (b) Issue of copper coins
2. (c) Riddles-mukris
3. (a) Iqta system
4. (e) Witty, advisor
5. (d) Waves

Answer the following questions in detail:

Question 1.
What are effects of ocean currents on human life?
Answer:
Ocean currents are streams of water flowing constantly in a definite direction at or near the surface of the ocean.
Ocean currents are caused by prevailing or planetary winds. They are also caused by die variation of temperature. The variation of saline water in the sea water causes ocean currents.

The ocean currents influence the climate of the regions such as coastal areas or islands. The warm currents raise the temperature whereas the cold currents reduce the temperature. Due to warm currents the harbors at higher latitudes remain, open through out the jar i. e. The harbors of Norway and Japan. The warm winds blowing over warm currents absorbs a lot of moisture and the nearby areas receive lots of rainfall.

Question 2.
Describe the administrative set up of Shershah.
Answer:
Shershah was an efficient administrator. The interest of the people ranked above everything. He started many reforms in military administration, land revenue etc that was followed by Akbar. Shershah divided his empire into Sarkars, which were again subdivided into parganas.

Officer in charge of sarkars and panganas were periodically transferred. He enforced equal law for justice and introduced a reformed system of currency. The silver coin known as the Rupee which lasted throughout the Mughal period and was maintained by the East India company down to 1835 AD.

Question 3.
Explain the process of the formation of Vidhan Sablia.
Answer:
The representatives of the Vidhan Sabha are elected in such a way that the seats are proportionately represented. If a person is elected from more than one constituency then he will have to resign from the rest retaining only one seat If there is a controversy about the election of any candidate, then a petition can be filed in the High court. After accepting the petition, the High Court gives its verdict It can also be against the candidate.

A person can appeal to the Supreme Court if the Verdict goes against him. The verdict of the Supreme Court is final.
The term of Vidhan Sabha is 5 years. In a case of emergency, it can be dissolved, Vidhan Sabha has a speaker and a Deputy Speaker. The Speaker conducts the business of the state Vidhan Sabha. The Deputy Speaker discharges duties in the absence of the Speaker.

Question 4.
What are the advantages of Ocean?
Answer:

• Rain on land – All the rain on land are due to the evaporation of sea water. Rains are useful for vegetation, animal life and human beings.
• Balance of temperature – Ocean shelp to maintain balance of temperature on land.
• Means of transport – The international trade is possible because of the oceans and seas, which joins the continents.

Question 5.
Write the main features of the architecture of the Sultanate period.
Answer:

• The caste rules were strictly followed. The women did not enjoy much freedom and the purdah system became very common.
• The practice of child marriage, polygamy and sati stystem was prevalent in the society.

MP Board Class 7th Social Science Solutions Miscellaneous Questions 2 Short Answer Type Questions

Question 1.
Explain the market control policy of Alauddin Khilji.
Answer:
Alauddin controlled prices in the market Those traders who did not sell goods at fixed prices or who weighed something less, were given severe punishment.

Question 2.
Write the advantages of tides.
Answer:

• The tides help in fishing.
• They increase the depth of the shallow harbors. This helps in bringing ships into the harbors.
• They increase the depth of estuaries.
• They keep rivers deep enough so to help ships placed in the rivers.

Question 3.
What is meant by ordinance?
Answer:
Ordinance is the order of the Chief Executive issued at a time when the Legislature is not in session. As soon as the legislature is convened, the ordinance is approved by it. If the legislature does not approve them, they are withdrawn.

Question 4.
Write about the progress made in the field of architecture during the reign of Shahjahan.
Answer:
Under Shahjahan Mughal architecture reached die climax. He built magnificent buildings, forts, mosques, tombs in Agra, Lahore, Delhi, Kabul, Kandhar, Ajmer, Ahamdabad etc. He built Moti-Masjid, Diwan-e-Aam, Diwan-e-Khas in the Fort at Agra. The most famous building of his time is the Tajmahal. It is made of white marble. It is regarded as jewel in the builder’s art.

Question 5.
Write a short note on cyclonic rains.
Answer:
When the hot and cold air meet, die hot air rises upwards and the cold air rushes to occupy the low-pressure area in the center. As a result there is circular movement which causes the whirling air in the center to rise upwards. This rising air cools down, condenses and brings rains. This type of rainfall is called cyclonic rain.

Question 6.
Explain the main features of Akbar’s Mansabdari system.
Answer:
‘Mansabdari System’ was a special feature of the Mughal rule. By ‘Mansab’ is meant a post or title. Each ‘Mansabdar’ was recognised as big or small as per the number of soldiers under him. It ranged between 10,000 to 50,000 soliders and the emperor could use their armies as per his wishes.

Question 7.
What was the effect of Bhakti movement on the life of the people?
Answer:
Bhakti movement had a great effect on the Indian society. The leaders of this movement used die local language which put great impact on the minds of die common people. This movement helped in arousing consciousness among people. It removed the false notions and bad practices. It also tried to bring the society together by bridging the gap between Hindu and Muslim.

MP Board Class 7th Social Science Solutions

MP Board Class 7th Maths Solutions Chapter 12 Algebraic Expressions Ex 12.2

Question 1.
Simplify by combining like terms:
(i) 21b – 32 +7b – 20b
(ii) -z² + 13z² – 5z + 7z³ – 15z
(iii) p – (p – q) – q – (q – p)
(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
(v) 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² – 3y²
(vi) (3y² + 5y – 4) – (8y – y² – 4)
Solution:
(i) 21b – 32 + 7b – 20b
= 21b + 7b – 20b – 32
= (21 + 7 – 20)b – 32
= 8b – 32

(ii) -z² + 13z² – 5z + 7z³ – 15z
= 7z³ + (-1 + 13) z² + (-5 -15) z
= 7z³ + 12z² – 20z

(iii) p – (p – q) – q – (q – p)
= p – p + q – q – q + p
= p – q

(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
= 3a – 2b – ab – a + b – ab + 3 ab + b – a
= (3 – 1 – 1)a + (-2 + 1 + 1)b + (-1 – 1 + 3)ab
= a + ab

(v) 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² – 3y²
= (5 + 3) x²y + (-5 + 1) x² + (- 3 – 1 – 3 )y² + 8xy²
= 8x²y – 4x² – 7y² + 8xy²

(vi) (3y² + 5y – 4) – (8y – y² – 4)
= 3y² + 5y – 4 – 8y + y² + 4
= (3 + 1) y² + (5 – 8) y + 4 – 4
= 4y² – 3y

Question 2.
Add:
(i) 3mn, -5mn, 8mn, -4mn
(ii) t – 8tz, 3tz – z, z – 1
(iii) -7mn + 5, 12mn + 2, 9mn – 8, -2mn – 3
(iv) a + b – 3, b – a + 3, a – b + 3
(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
(vii) 4x²y, -3xy², 5xy², 5x²y
(viii) 3 p²q² – 4pq + 5, -10p²q², 15 + 9pq + 7p²q²
(ix) ab – 4a, 4b – ab, 4a – 4b
(x) x² – y² – 1, y² – 1 – x², 1 – x² – y².
Solution:
(i) 3 mn + (-5 mn) + 8 mn + (-4 mn)
= (3 – 5 + 8 – 4 )mn = 2 mn

(ii) (t – 8tz) + (3tz – z) + (z – t)
= t – 8tz + 3tz – z + z – t
= (1 – 1)t + (- 8 + 3)tz + (-1 + 1)z
= -5 tz

(iii) (-7mn + 5) + (12mn + 2) + (9mn – 8) + (-2mn – 3)
=-7mn + 5 + 12mn + 2 + 9mn – 8 – 2mn – 3
= (-7 + 12 + 9 – 2)mn + (5 + 2 – 8 – 3)
= 12mn – 4

(iv) (a + b – 3) + (b – a + 3) + (a – b + 3)
= a + b – 3 + b – a + 3 + a – b + 3
= (1 -1 + 1)a + (1 + 1 – 1)b + (- 3 + 3 + 3)
= a + b + 3

(v) (14x + 10y – 12xy – 13) + (18 – 7x – 10y + 8xy) + 4xy
= 14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy
= (14 – 7)x + (10 – 10)y + (12 + 8 + 4 )xy + (-13 + 18)
= 7x + 5

(vi) (5m – 7n) + (3n – 4m + 2) + (2m – 3mn – 5)
= 5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5
= (5 – 4 + 2)m + (- 7 + 3)n – 3mn + (2 – 5)
= 3m – 4n – 3mn – 3

(vii) (4x²y) + (-3xy²) + (-5xy²) + (5x²y)
= 4x²y – 3xy² – 5xy² + 5x²y
= (4 + 5) x²y + (-3 – 5) xy²
= 9x²y – 8xy²

(viii) (3p²q² – 4pq + 5) + (-10p²q²) + (15 + 9pq +7p²q²)
= 3p²q² – 4pq + 5 – 10p²q² + 15 + 9pq + 7p²q²
= (3 – 10 + 7) p²q² + (-4 + 9)pq + (5 + 15)
= 5pq + 20

(ix) (ab – 4a) +(4 b – ab) + (4a- 4b)
= ab – 4a + 4b – ab + 4a – 4b
= (1 – 1)ab + (-4 + 4 )a + (4 – 4)b = 0

(x) (x² – y² – 1) + (y² – 1 – x²) + (1 – x² – y²)
= x² – y² – 1 + y² – 1 – x² + 1 – x² – y²
= (1 – 1 – 1)x² + (-1 + 1 – 1)y² + (-1 – 1 + 1)
= -x² – y² – 1

Question 3.
Subtract:
(i) -5y² from y²
(ii) 6xy from -12xy
(iii) (a – b) from (a + b)
(iv) a(b – 5) from b (5 – a)
(v) – m2 + 5 mn from 4mi2 – 3mn + 8
(vi) -x² + 10x – 5 from 5x – 10
(vii) 5a² – 7ab + 5b² from 3ab – 2a² – 2b²
(viii) 4pq – 5q² – 3p² from 5p² + 3q² – pq
Solution:
(i) y² – (-5y²) = y² + 5y² = 6y²
(ii) -12xy – (6xy) = -12xy – 6xy = -18xy
(iii) (a + b) – (a – b) = a + b – a + b = 2b

(iv) b(5 – a) – a(b – 5)
= 5b – ab – ab + 5a
= 5a + 5b – 2ab

(v) (4m² – 3mn + 8) – (- m² + 5mn)
= 4m² – 3mn + 8 + m² – 5mn
= (4 + 1)m² + (- 3 – 5 )mn + 8
= 5m² – 8mn + 8

(vi) (5x – 10) – (-x² + 10x – 5)
= 5x – 10 + x² – 10x + 5
= x² + (5 – 10)x + (-10 + 5)
= x² – 5x – 5

(vii) (3ab – 2a² – 2b²) – (5a² – 7ab + 5b²)
= 3ab – 2a² – 2b² – 5a² + 7ab – 5b²
= (3 + 7)ab + (- 2 – 5)a² + (- 2 – 5 )b²
= 10ab – 7a² – 7b²

(viii) (5p² + 3q² – pq) – (4pq – 5a² – 3p²)
= 5p² + 3q² – pq – 4pq + 5q² + 3p²
= (5 + 3 )p² + (3 + 5 )q² + (-1 – 4 )pq
= 8p² + 8q² – 5pq

Question 4.
(a) What should be added to x² + xy + y² to obtain 2x² + 3xy?
(b) What should be subtracted from 2a + 8b + 10 to get – 3a + 7b + 16?
Solution:
(a) Let a be the required term.
∴ a + (x² + y² + xy) = 2x² + 3xy
⇒ a = 2x² + 3xy – (x² + y² + xy)
= 2x² + 3xy – x² – y² – xy
= (2 – 1) x² – y² + (3 – 1)xy
= x² – y² + 2xy

(b) Let p be the required term.
∴ (2a + 8b + 10) -p = -3a + 7b + 16
⇒ p = 2a + 8b + 10 – (- 3a + 7b + 16)
= 2a + 8b + 10 + 3a – 7b – 16
= (2 + 3)a + (8 – 7)b + (10 – 16)
= 5a + b – 6

Question 5.
What should be taken away from 3x² – 4y² + 5xy + 20 to obtain -x² – y² + 6xy + 20 ?
Solution:
Required term
= (3x² – 4y² + 5xy + 20) – (-x² – y² + 6xy + 20)
= 3x² – 4y² + 5xy + 20 + x² + y² – 6xy – 20
= (3 + 1)x² + (- 4 + 1) y² + (5 – 6)xy + (20 – 20)
= 4x² – 3y² – xy

Question 6.
(a) From the sum of 3x – y + 11 and -y – 11, subtract 3x – y – 11.
(b) From the sum of 4 + 3x and 5 – 4x + 2x², subtract the sum of 3x² – 5x and -x² + 2x + 5.
Solution:
(a) Sum of 3x – y + 11 and – y – 11
= (3x – y + 11) + (-y – 11)
= 3x – y + 11 – y – 11
= 3x + (- 1 – 1) y + (11 – 11)
= 3x – 2y
Now, required difference
= (3x – 2y) – (3x – y – 11)
= 3x – 2y – 3x + y + 11
= (3 – 3)x + (- 2 + 1) y + 11
= -y + 11

(b) Sum of 4 + 3x and 5 – 4x + 2x²
= (4 + 3x) + (5 – 4x + 2x²)
= 4 + 3x + 5 – 4x + 2x²
= (3 – 4)x + 2x² + 4 + 5
= – x + 2x² + 9
Now, sum of 3x² – 5x and -x² + 2x + 5
= (3x² – 5x) + (-x² + 2x + 5)
= 3x² – 5x – x² + 2x + 5
= (3 – 1) x² + (- 5 + 2)x + 5
= 2x² – 3x + 5
Required difference
= (- x + 2x² + 9) – (2x² – 3x + 5)
= -x + 2x² + 9 – 2x² + 3x – 5
= (-1 + 3)x + (2 – 2) x² + (9 – 5)
= 2x + 4

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.4

Question 1.
Is it possible to have a triangle with the following sides?
(i) 2 cm, 3 cm, 5 cm
(ii) 3 cm, 6 cm, 7 cm
(iii) 6 cm, 3 cm, 2 cm
Solution:
In a triangle, the sum of the lengths of either two sides is always greater than the third side.
(i) Given that, the sides of the triangle are 2 cm, 3 cm, 5 cm.
It can be observed that,
(2 + 3) cm = 5 cm
However, 5 cm = 5 cm
Hence, this triangle is not possible.

(ii) Given that, the sides of the triangle are
3 cm, 6 cm, 7 cm.
Here, (3 + 6) cm = 9 cm > 7 cm
(6 + 7) cm = 13 cm > 3 cm
(3 + 7) cm = 10 cm > 6 cm
Hence, this triangle is possible.

(iii) Given that, the sides of the triangle are
6 cm, 3 cm, 2 cm.
Here, (6 + 3) cm = 9 cm > 2 cm
However, (3 + 2) cm = 5 cm < 6 cm Hence, this triangle is not possible.

Question 2.
Take any point 0 in the interior of a triangle PQR. Is

(i) OP + OQ > PQ?
(ii) OQ + OR > QR?
(iii) OR + OP > RP?
Solution:
If O is a point in the interior of a given triangle, then three triangles ∆OPQ, ∆OQR and ∆ORP can be constructed. In a triangle, the sum of the lengths of either two sides is always greater than the third side.
(i) Yes, as OQR is a triangle with sides OR, OQ and QR.

∴ OQ + OR> QR

(ii) Yes, as OQR is a triangle sides OR, OQ and QR.
∴ OQ + OR > QR

(iii) Yes, as A ORP is a triangle with sides OR, OP and PR.
∴ OR + OP > PR

Question 3.
AM is a median of a triangle ABC.
Is AB + BC + CA > 2AM?
(Consider the sides of ∆ABM and ∆AMC.)

Solution:
In a triangle, the sum of the lengths of either two sides is always greater than the third side.
In ∆ABM,
AB + BM > AM …(i)
Similarly, in ∆ACM,
AC + CM > AM …(ii)
Adding (i) and (ii), we obtain
AB + BM + MC +AC > AM + AM
AB + BC + AC > 2AM
Yes, the given expression is true.

Question 4.
ABCD is a quadrilateral.

Is AB + BC + CD + DA > AC + BD ?
Solution:
In a triangle, the sum of the lengths of either two sides is always greater than the third side.
In ∆ABC; AB + BC > CA …(i)
In ∆BCD; BC + CD > DB …(ii)
In ∆CDA; CD + DA > AC …(iii)
In ∆DAB; DA + AB > DB …(iv)
Adding (i), (ii), (iii) and (iv), we obtain
AB + BC + BC + CD + CD + DA + DA+ AB > AC + BD + AC + BD
⇒ 2AB + 2BC + 2CD + 2DA > 2AC + 2BD ⇒ 2{AB + BC + CD + DA) > 2(AC+BD)
⇒ (AB + BC + CD + DA) > (AC + BD)
Yes, the given expression is true.

Question 5.
ABCD is a quadrilateral.

Is AB + BC + CD + DA < 2(AC + BD)?
Solution:
In a triangle, the sum of the lengths of either two sides is always greater than the third side.
Let diagonals AC and BD of the quadrilateral ABCD intersects at O.

In ∆OAB, OA + OB > AB … (i)
In ∆OBC, OB + OC > BC … (ii)
In ∆OCD, OC + CD > CD … (iii)
In ∆ODA, OD + OA > DA … (iv)
OA + OB + OB + OC + OC + OD + OD + OA > AB + BC + CD + DA
⇒ 20A + 20B + 20C + 20D > AB + BC + CD + DA
⇒ 2(OA + OC) + 2(OB + OD) > AB + BC + CD + DA
⇒ 2(AC) + 2(BD) > AB + BC + CD + DA
⇒ 2(AC + BD) > AB + BC + CD + DA
Yes, the given expression is true.

Question 6.
The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?
Solution:
In a triangle, the Sum of the lengths of either two side is always greater than the third side.
Lengths of two sides of a triangle are 12 cm and 15 cm.
Let the third side be x cm.
∴ x + 12 > 15 ⇒ x > 3
x + 15 > 12 ⇒ x > – 3 but side length never be negative.
and 12 + 15 > x ⇒ 27 > x
Hence, third side can measure between 3 and 27.

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 4 Simple Equations Ex 4.4

Question 1.
Set up equations and solve them to find the unknown numbers in the following cases:
(a) Add 4 to eight times a number; you get 60.
(b) One-fifth of a number minus 4 gives 3.
(c) If I take three-fourths of a number and add 3 to it, I get 21.
(d) When I subtracted 11 from twice a number, the result was 15.
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
(g) Anwar thinks of a number. If he takes away 7 from \(\frac{5}{2}\) of the number, the result is 23.
Solution:
(a) Let the number be x.
8 times of this number = 8x
According to question;
8x + 4 = 60 ⇒ 8x = 60 – 4
(Transposing 4 to R.H.S.)
⇒ 8x = 56
Dividing both sides by 8,

(b) Let the number be x
One-fifth of this number \(\frac{x}{5}\)
According to question;

(Transporting -4 to R.H.S.)
Mutiplying both sides by 5,

(c) Let the number be x.
Three-fourth of this number = \(\frac{3 x}{4}\)
According to question;

Multiplying both sides by 4,

Dividing both sides by 3,

(d) Let the number be x.
Twice of this number = 2x
According to question;
2x – 11 = 15 ⇒ 2x = 15 + 11
(Transposing -11 to R.H.S.)
⇒ 2x = 26
Dividing both sides by 2,

(e) Let the number of notebooks be x.
Thrice the number of notebooks = 3x
According to question;
⇒ 50 – 3x = 8 ⇒ – 3x = 8 – 50
(Transposing 50 to R.H.S.)
⇒ – 3x = – 42
Dividing both sides by – 3,

(f) Let the number be x.
According to question;

Multiplying both sides by 5,

⇒ x = 40 – 19 (Transposing 19 to R.H.S.)
⇒ x = 21

(g) Let the number be x.
\(\frac{5}{2}\) of this number = \(\frac{5x}{2}\)
According to question;

Question 2.
Solve the following;
(a) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. What is the lowest score?
(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180°).
(c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?
Solution:
(a) Let the lowest score be l.
According to question;
Highest marks = 2 × Lowest marks + 7
⇒ 87 = (2 × l) + 7 × 2l + 7 = 87
⇒ 2l = 87 – 7
(Transposing 7 to R.H.S.)
⇒ 2l = 80
Dividing both sides by 2,

Therefore, the lowest score is 40.

(b) Let the base angles be equal to b.
Vertex angle = 40°
The sum of all interior angles of a triangle is 180°.
b + b + 40° = 180° ⇒ 2b + 40° = 180°
⇒ 2b = 180° – 40° = 140°
(Transposing 40° to R.H.S.)
Dividing both sides by 2,

Therefore, the base angles of the triangle are of 70°.

(c) Let Rahul’s score be x.
Therefore, Sachin’s score = 2x
According to question;
Rahul’s score + Sachin’s score = 200 – 2
2x + x = 198 ⇒ 3x = 198
Dividing both sides by 3,

∴ Rahul’s score = 66
Sachin’s score = 2 × 66 = 132

Question 3.
Solve the following:
(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?
(iii) People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted if the number of non-fruit trees planted was 77?
Solution:
(i) Let Parmit has x marbles.
5 times the number of marbles Parmit has = 5x
According to question;
5x + 7 = 37 ⇒ 5x = 37 – 7 = 30
(Transposing 7 to R.H.S.)
Dividing both sides by 5,

Therefore, Parmit has 6 marbles.

(ii) Let Laxmi’s age be x years.
According to question;
Her father’s age = 3 × Laxmi’s age + 4
⇒ 49 = (3 × x) + 4
⇒ 3x + 4 = 49 ⇒ 3x = 49 – 4
(Transposing 4 to R.H.S.)
⇒ 3x = 45
Dividing both sides by 3,

Therefore, Laxmi’s age is 15 years.

(iii) Let the number of fruit trees be x. According to question;
Number of non-fruit trees = 3 × Number of fruit trees + 2
77 = (3 × x) + 2
⇒ 3x + 2 = 77 ⇒ 3x = 77 – 2
(Transposing 2 to R.H.S.)
⇒ 3x = 75
Dividing both sides by 3,

Therefore, the number of fruit trees planted were 25.

Question 4.
Solve the following riddle:

Solution:
Let the number be x.
Seven times x = 7x
According to question; .
(7x + 50) + 40 = 300 ⇒ 7x + 90 = 300
⇒ 7x = 300 – 90 (Transposing 90 to R.H.S.)
⇒ 7x = 210
Dividing both sides by 7,

Therefore, the number is 30.

MP Board Class 7th Maths Solutions