{"id":27086,"date":"2021-06-29T13:49:50","date_gmt":"2021-06-29T08:19:50","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=27086"},"modified":"2022-03-02T10:47:24","modified_gmt":"2022-03-02T05:17:24","slug":"ncert-solutions-for-class-7-maths-chapter-11-ex-11-2","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-7-maths-chapter-11-ex-11-2\/","title":{"rendered":"NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2"},"content":{"rendered":"

These NCERT Solutions for Class 7 Maths<\/a> Chapter 11 Perimeter and Area Ex 11.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.2<\/h2>\n

Question 1.
\nFind the area of each of the following parallelograms:
\n\"NCERT
\n\"NCERT
\nAnswer:
\nHere, Base(b) = 7cm
\nHeight (h) = 4 cm
\n\u2234 Area of the parallelogram
\n= b \u00d7 h sq. units
\n= 7 \u00d7 4 cm2<\/sup>
\n= 28 cm2<\/sup><\/p>\n

\"NCERT<\/p>\n

(b) Here Base (b) = 5 cm
\nHeight (h) = 3 cm
\n\u2234 Area of the parallelogram
\n= b x h sq. units
\n= 5 x 3 cm2<\/sup> =15 cm2<\/sup><\/p>\n

(c) Here Base (b) = 2.5 cm
\nHeight (h) = 3.5 cm
\n\u2234 Area of the parallelogram
\n= b x h sq. units
\n= 2.5 x 3.5 cm2<\/sup>
\n= 8.75 cm2<\/sup><\/p>\n

(d) Here Base (b) = 5 cm
\nHeight (h) = 4.8 cm
\n\u2234 Area of the parallelogram
\n= b x h sq. units
\n= 5 x 4.8 cm2<\/sup>
\n= 24 cm2<\/sup><\/p>\n

\"NCERT<\/p>\n

(e) Here Base (b) = 2 cm
\nHeight (h) =4.4 cm
\n\u2234 Area of the parallelogram
\n= b x h sq. units
\n= 2 x 4.4 cm2<\/sup>
\n= 8.8 cm2<\/sup><\/p>\n

Question 2.
\nFind the area of each of the following triangles:
\n\"NCERT
\nAnswer:
\n(a) Here, Base (b) = 4cm
\nHeight (h) = 3 cm
\nArea of the triangle
\n= \\(\\frac { 1 }{ 2 }\\) x b x h sq.m
\n= \\(\\frac { 1 }{ 2 }\\) x 4 x 3 cm2<\/sup>
\n= 6 cm2<\/sup><\/p>\n

(b) Here base (b) = 5 cm
\nHeight (h) = 3.2 cm
\nArea of the triangle
\n= \\(\\frac { 1 }{ 2 }\\) x b x h sq. units
\n= \\(\\frac { 1 }{ 2 }\\) x 5 x 3.2 cm2<\/sup>
\n= 8 cm2<\/sup><\/p>\n

\"NCERT<\/p>\n

(c) Here base (b) = 3 cm
\nHeight = 4 cm
\nArea of the triangle = \\(\\frac { 1 }{ 2 }\\) x b x h sq. units
\n= \\(\\frac { 1 }{ 2 }\\) x 3 x 4 cm2<\/sup>
\n= 6 cm2<\/sup><\/p>\n

(d) Here base (b) = 3 cm
\nHeight (h) = 2m
\nArea of the triangle
\n= \\(\\frac { 1 }{ 2 }\\) x b x h sq units
\n= \\(\\frac { 1 }{ 2 }\\) x 3 x 2 cm2<\/sup> = 3 cm2<\/sup><\/p>\n

Question 3.
\nFind the missing values:<\/p>\n\n\n\n\n\n\n\n
Base<\/td>\nHeight<\/td>\nArea of the Parallelogram<\/td>\n<\/tr>\n
(a) 20 cm<\/td>\n<\/td>\n246 cm;<\/sup><\/td>\n<\/tr>\n
(b)<\/td>\n15 cm<\/td>\n154.5 cm2<\/sup><\/td>\n<\/tr>\n
(c)<\/td>\n8.4 cm<\/td>\n48.72 cm2<\/sup><\/td>\n<\/tr>\n
(d) 15.6 cm<\/td>\n<\/td>\n16.38 cm2<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:
\n(a) Here, base of the parallelogram
\n(b) = 20 cm
\nLet the height \u2018h\u2019
\nArea of the Parallelogram = 246 cm2<\/sup>
\nb x h = 246
\n20 x h = 246
\nh = \\(\\frac{246}{20}\\)
\n\\(\\frac{123}{10}\\) = 12.3 cm
\n\u2234 The missing value height = 12.3 cm<\/p>\n

(b) Here, height (h) = 15 cm
\nLet the base of the parallelogram be \u2018b\u2019
\nArea of a parallelogram = 154.5 cm2<\/sup>
\nb x h = 154.5 cm2<\/sup>
\nb = \\(\\frac{154.5}{15}\\) = 12.3
\n= \\(\\frac{1545}{15 \\times 10}\\) = 12.3cm
\n= \\(\\frac{103}{10}\\)10.3
\n\u2234 The missing value base = 10.3 cm.<\/p>\n

\"NCERT<\/p>\n

(c) Here, height (h) = 8.4 cm
\nLet the base of the parallelogram be \u2018b’
\nArea of the parallelogram = 48.72 cm2<\/sup>
\nb \u00d7 h = 48.72
\nb \u00d7 8.4 =48.72
\nb = \\(\\frac{48.72}{8.4}=\\frac{48.72}{8.4}\\)
\n= 5.8 cm
\nThus, the missing value base = 5.8 cm.<\/p>\n

(d) Here, base (b) = 15.6 cm
\nLet the height of the parallelogram
\nbe \u2018h\u2019
\nArea of the parallelogram = 16.38 cm2<\/sup>
\nb \u00d7 h = 16.38
\n15.6 \u00d7 h = 16.38
\nh = \\(\\frac{16.38}{15.6}\\)
\n= \\(\\frac{163.8}{156}\\)
\n= 1.05 cm
\nThus, the missing value (height)
\n= 1.05 cm.<\/p>\n

Question 4.
\nFind the missing values:<\/p>\n\n\n\n\n\n\n
Base<\/td>\nHeight<\/td>\nArea of Triangle<\/td>\n<\/tr>\n
15 cm<\/td>\n<\/td>\n87cm2<\/sup><\/td>\n<\/tr>\n
<\/td>\n3.14 mm<\/td>\n1256mm2<\/sup><\/td>\n<\/tr>\n
22 cm<\/td>\n<\/td>\n170.5cm2<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:
\n(i) Let the height of the triangle be \u2018h\u2019
\nHere base (b) = 15 cm
\nArea of the triangle = 87 cm2<\/sup>
\n\\(\\frac { 1 }{ 2 }\\) \u00d7 b \u00d7 h =87 2
\n\\(\\frac { 1 }{ 2 }\\) \u00d7 15 \u00d7 h =87
\nh = \\(\\frac{87 \\times 2}{15}\\)
\n= \\(\\frac{29 \\times 2}{5}\\)
\n= \\(\\frac{58}{5}\\)
\n= 11.6 cm
\n\u2234 The missing value height = 11.6 cm<\/p>\n

(ii) Here Height = 31.4 mm
\nLet the base be \u201cb\u201d
\nArea of a triangle = 1256 mm2<\/sup>
\n\\(\\frac { 1 }{ 2 }\\) \u00d7 b \u00d7 h = 1256 2
\n\\(\\frac { 1 }{ 2 }\\) \u00d7 b \u00d7 31.4 = 1256
\nb = \\(\\frac{1256 \\times 2}{31.4}\\)
\nb = \\(\\frac{1256 \\times 2 \\times 10}{314}\\)
\n= 4 \u00d7 2 \u00d7 10 mm
\n= 80 mm.
\n\u2234 The missing value base = 80 mm.<\/p>\n

\"NCERT<\/p>\n

(iii) Let the height of the triangle be \u2018h\u2019
\nbase (b) = 22 cm.
\nArea of the triangle = 170.5 cm2<\/sup>
\n\\(\\frac { 1 }{ 2 }\\) \u00d7 b \u00d7 h = 170.5
\n\\(\\frac { 1 }{ 2 }\\) \u00d7 22 \u00d7 h = 170.5
\n11 \u00d7 h = 170.5
\nh = \\(\\frac{170.5}{11}\\)
\n= 15.5 cm
\n\u2234 The missing value height = 15.5 cm.<\/p>\n

Question 5.
\nPQRS is a parallelogram. QM is the height from Q to SR and QN is the height from Q to PS.
\nIf SR = 12 cm and QM = 7.6 cm. Find:
\n(a) the area of the parallelogram PQRS
\n(b) QN, if PS = 8 cm
\n\"NCERT
\nAnswer:
\nHere, Base (SR) = 12 cm
\nCorresponding height (QM) = 7.6 cm
\n(a) Area of the parallelogram
\n= b \u00d7 h sq units
\n= 12 \u00d7 7.6 cm2<\/sup>
\n= 91.2 cm2<\/sup><\/p>\n

\"NCERT<\/p>\n

(b) Base of the parallelogram (PS)
\n= 8 cm
\nArea of the parallelogram = 91.2 cm2<\/sup>
\nb \u00d7 h = 91.2
\n8 \u00d7 h = 91.2
\nh = \\(\\frac{91.2}{8}\\) = 11.4 cm
\nThe height QN =11.4 cm.<\/p>\n

Question 6.
\nDL and BM are the heights on sides AB and AD respectively of parallelogram ABCD. If the area of the parallelogram is 1470 cm2<\/sup>, AB = 35 cm and AD = 49 cm, find the length of BM and DL.
\n\"NCERT
\nAnswer:
\nArea of the parallelogram ABCD
\n= Base x\u00d7 height
\n= AD \u00d7 BM
\nArea of a parallelogram
\n= 1470cm2<\/sup>
\n\u2234 AD \u00d7 BM = 1470cm2<\/sup>
\n49 \u00d7 BM = 1470
\nBM = \\(\\frac{1470}{49}\\)
\nBM = \\(\\frac{210}{7}\\) = 30 cm
\nArea of the parallelogram ABCD = 1470cm2<\/sup>
\nAB \u00d7 DL = 1470
\n35 \u00d7 DL = 1470
\nDL = \\(\\frac{1470}{35}\\)
\n= \\(\\frac{210}{7}\\) = 42 cm
\n\u2234 Length of BM = 30 cm
\nLength of DL = 42 cm]<\/p>\n

Question 7.
\n\u0394ABC is right angled at A. AD is perpendicular to BC. If AB = 5 cm, BC =13 cm and AC = 12 cm, find the area of \u0394ABC. Also, find the length of AD.
\nAnswer:
\nArea of AABC = \\(\\frac { 1 }{ 2 }\\) \u00d7 b \u00d7 h sq. units 2 H
\n(base = 5 cm and height =12 cm)
\n\"NCERT
\n= \\(\\frac { 1 }{ 2 }\\) \u00d7 5 \u00d7 12 cm2 <\/sup>= 5 \u00d7 6 cm2<\/sup> = 30 cm2<\/sup>
\nArea of the AABC = \\(\\frac { 1 }{ 2 }\\) x base x height
\n30 = \\(\\frac { 1 }{ 2 }\\) \u00d7 13 \u00d7 AD
\nAD = \\(\\frac{30 \\times 2}{13}\\) cm = \\(\\frac{60}{13}\\) cm
\nHence, length of AD = \\(\\frac{60}{13}\\) cm = 4.6 cm.<\/p>\n

\"NCERT<\/p>\n

Question 8.
\n\u2206ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm. The height AD from A to BC, is 6 cm. Find the area of \u2206ABC. What will be the height from C to AB i.e., CE?
\nAnswer:
\nHere base BC = 9 cm
\nCorresponding height AD = 6 cm
\n\u2234 Area of \u2206ABC
\n= \\(\\frac { 1 }{ 2 }\\) \u00d7 base \u00d7 height
\n= \\(\\frac { 1 }{ 2 }\\) \u00d7 BC \u00d7 AD
\n= \\(\\frac { 1 }{ 2 }\\) \u00d7 9 \u00d7 6 cm2<\/sup> = 27 cm2<\/sup>
\n\"NCERT
\nLet the height from C to AB be ‘h’
\n= \\(\\frac { 1 }{ 2 }\\) \u00d7 AB \u00d7 h = 27
\n= \\(\\frac { 1 }{ 2 }\\) \u00d7 7.5 \u00d7 h = 27
\nh = \\(\\frac{27 \\times 3}{7.5}\\)
\n= \\(\\frac{27 \\times 2 \\times 10}{75}\\)
\n= 7.2 cm
\n\u2234 The height CE = 7.2 cm.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.2 Question 1. Find the area of each of the following parallelograms: Answer: Here, Base(b) = 7cm Height …<\/p>\n

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NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.2 Question 1. 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