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>\nNCERT 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 \n \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 <\/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 <\/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>\nQuestion 2. \nFind the area of each of the following triangles: \n \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 <\/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>\nQuestion 3. \nFind the missing values:<\/p>\n
\n\n\nBase<\/td>\n Height<\/td>\n Area of the Parallelogram<\/td>\n<\/tr>\n \n(a) 20 cm<\/td>\n <\/td>\n 246 cm;<\/sup><\/td>\n<\/tr>\n\n(b)<\/td>\n 15 cm<\/td>\n 154.5 cm2<\/sup><\/td>\n<\/tr>\n\n(c)<\/td>\n 8.4 cm<\/td>\n 48.72 cm2<\/sup><\/td>\n<\/tr>\n\n(d) 15.6 cm<\/td>\n <\/td>\n 16.38 cm2<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nAnswer: \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 <\/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>\nQuestion 4. \nFind the missing values:<\/p>\n
\n\n\nBase<\/td>\n Height<\/td>\n Area of Triangle<\/td>\n<\/tr>\n \n15 cm<\/td>\n <\/td>\n 87cm2<\/sup><\/td>\n<\/tr>\n\n<\/td>\n 3.14 mm<\/td>\n 1256mm2<\/sup><\/td>\n<\/tr>\n\n22 cm<\/td>\n <\/td>\n 170.5cm2<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nAnswer: \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 <\/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>\nQuestion 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 \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 <\/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>\nQuestion 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 \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>\nQuestion 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 \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 <\/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 \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
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[9],"tags":[],"yoast_head":"\nNCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n