\nA 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path.
\nSolution:
\nLength of rectangular park (l) = 125 m
\nand breadth (b) = 65 m
\nWidth of path around it = 3 m
\n
![\"ML](\"https:\/\/i2.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-Solutions-for-ICSE-Maths-Chapter-16-Perimeter-and-Area-Check-Your-Progress-Q1.1.png?resize=206%2C159&ssl=1\")
\nOuter length (L) = 125 + 2 \u00d7 3 = 125 + 6 = 131 m
\nand breadth = 65 + 2 \u00d7 3 = 65 + 6 = 71 m
\nArea of path = L \u00d7 B – l \u00d7 b
\n= 131 \u00d7 71 – 125 \u00d7 65
\n= 9301 – 8125 = 1176 m2<\/sup><\/p>\n
Question 2. Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Question 11. Question 12. ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 16 Perimeter and Area Check Your Progress Question 1. A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path. Solution: Length of rectangular park (l) = 125 m and breadth …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/43151"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=43151"}],"version-history":[{"count":1,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/43151\/revisions"}],"predecessor-version":[{"id":159032,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/43151\/revisions\/159032"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=43151"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=43151"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=43151"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nIn the given figure, all adjacent line segments are at right angles. Find:
\n(i) the area of the shaded region
\n(ii) the area of the unshaded region.
\n![\"ML](\"https:\/\/i1.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-Solutions-for-ICSE-Maths-Chapter-16-Perimeter-and-Area-Check-Your-Progress-Q2.1.png?resize=262%2C174&ssl=1\")
\nSolution:
\nIn the given figure,
\nLength (l) = 22 m
\nand breadth (b) = 14 m
\nWidth of length wire region = 3 m
\nand breadth wire = 2 m
\nArea of shaded portion = 22 \u00d7 3 + 14 \u00d7 2 – 3 \u00d7 2 m2<\/sup>
\n= 66 + 28 – 6
\n= 88 m2<\/sup>
\nTotal area = l \u00d7 b = 22 \u00d7 14 = 308 m2<\/sup>
\nArea of unshaded region = 308 – 88 = 220 m2<\/sup><\/p>\n
\nFind the area of a triangle whose:
\n(i) base = 2 m, height = 1.5 m
\n(ii) base = 3.4 m and height = 90 cm
\nSolution:
\n(i) Base of a triangle (b) = 2 m
\nHeight (h) = 1.5 m
\nArea = \\(\\frac { 1 }{ 2 }\\) \u00d7 b \u00d7 h
\n= \\(\\frac { 1 }{ 2 }\\) \u00d7 2 \u00d7 1.5 = 1.5 m2<\/sup>
\n(ii) Base of the triangle (b) = 3.4 m
\nand height (h) = 90 cm = \\(\\frac { 90 }{ 100 }\\) = \\(\\frac { 9 }{ 10 }\\) m
\nArea = \\(\\frac { 1 }{ 2 }\\) \u00d7 b \u00d7 h
\n= \\(\\frac { 1 }{ 2 }\\) \u00d7 3.4 \u00d7 \\(\\frac { 9 }{ 10 }\\)
\n= \\(\\frac { 30.6 }{ 20 }\\) = 1.53 m2<\/sup><\/p>\n
\nIn the given figure, PQRS is a parallelogram. QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm, PS = 8 cm and QM = 7.6 cm, find:
\n(i) the area of the parallelogram PQRS
\n(ii) the length of QN.
\n![\"ML](\"https:\/\/i1.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-Solutions-for-ICSE-Maths-Chapter-16-Perimeter-and-Area-Check-Your-Progress-Q4.1.png?resize=283%2C159&ssl=1\")
\nSolution:
\nIn the given figure,
\nPQRS is a parallelogram in which QM \u22a5 SR and QN \u22a5 PQ
\nSR = 12 cm, PS = 8 cm, QM = 7.6 cm
\n(i) Area of ||gm ABCD = b \u00d7 h
\n= SR \u00d7 QM
\n= 12 \u00d7 7.6 cm2<\/sup>
\n= 91.2 cm2<\/sup>
\nBase PS = 8 cm
\nArea of ||gm = 91.2 cm2<\/sup>
\nHeigh QN = \\(\\frac { Area }{ Base }\\) = \\(\\frac { 91.2 }{ 8 }\\) = 11.4 cm<\/p>\n
\nFrom the given figure, find
\n(i) the area of \u0394ABC
\n(ii) length of BC
\n(iii) the length of altitude from A to BC.
\n![\"ML](\"https:\/\/i0.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-Solutions-for-ICSE-Maths-Chapter-16-Perimeter-and-Area-Check-Your-Progress-Q5.1.png?resize=163%2C170&ssl=1\")
\nSolution:
\nIn the given figure,
\nABC is a right angled triangle in which
\nAB = 3 cm , AC = 4 cm
\nAD \u22a5 BC
\n(i) Area \u0394ABC = \\(\\frac { 1 }{ 2 }\\) \u00d7 Base \u00d7 Height
\n= \\(\\frac { 1 }{ 2 }\\) \u00d7 3 \u00d7 4 = 6 cm2<\/sup>
\n(ii) BC2<\/sup> = AB2<\/sup> + AC2<\/sup> (Pythagoras Theorem)
\n= 32<\/sup> + 42<\/sup> = 9 + 16 = 25 = (5)2<\/sup>
\nBC = 5 cm
\n(iii) Now length of altitude AD
\n![\"ML](\"https:\/\/i2.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-Solutions-for-ICSE-Maths-Chapter-16-Perimeter-and-Area-Check-Your-Progress-Q5.2.png?resize=159%2C119&ssl=1\")
<\/p>\n
\nIn the given figure, the area of the right-angled triangle is 54 cm2<\/sup>. If one of its legs is 12 cm long, find its perimeter.
\n![\"ML](\"https:\/\/i0.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-Solutions-for-ICSE-Maths-Chapter-16-Perimeter-and-Area-Check-Your-Progress-Q6.1.png?resize=140%2C180&ssl=1\")
\nSolution:
\nIn the given figure,
\n![\"ML](\"https:\/\/i1.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-Solutions-for-ICSE-Maths-Chapter-16-Perimeter-and-Area-Check-Your-Progress-Q6.2.png?resize=144%2C208&ssl=1\")
\nArea of right-angled triangle = 54 cm2<\/sup>
\nLength of one leg AB = 12 cm
\n![\"ML](\"https:\/\/i1.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-Solutions-for-ICSE-Maths-Chapter-16-Perimeter-and-Area-Check-Your-Progress-Q6.3.png?resize=225%2C113&ssl=1\")
\nNow AC2<\/sup> = AB2<\/sup> + BC2<\/sup> (Pythagoras Theorem)
\n= 122<\/sup> + 92<\/sup> = 144 + 81 = 225 = (15)2<\/sup>
\nAC = 15 cm
\nNow perimeter = AB + BC + AC = 12 + 9 + 15 = 36 cm<\/p>\n
\nIf the area of a circle is 78.5 cm2<\/sup>, find its circumference. (Take \u03c0 = 3.14)
\nSolution:
\nArea of a circle = 78.5 cm2<\/sup>
\n![\"ML](\"https:\/\/i2.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-Solutions-for-ICSE-Maths-Chapter-16-Perimeter-and-Area-Check-Your-Progress-Q7.1.png?resize=259%2C103&ssl=1\")
\nCircumference = 2\u03c0r = 2 \u00d7 3.14 \u00d7 5 = 31.4 cm<\/p>\n
\nFind the circumference of the circle whose area is 16 times the area of the circle with diameter 7 cm.
\nSolution:
\nDiameter of first circle = 7 cm
\nRadius (r) = \\(\\frac { 7 }{ 2 }\\) cm
\n![\"ML](\"https:\/\/i2.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-Solutions-for-ICSE-Maths-Chapter-16-Perimeter-and-Area-Check-Your-Progress-Q8.1.png?resize=320%2C210&ssl=1\")
\n![\"ML](\"https:\/\/i1.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-Solutions-for-ICSE-Maths-Chapter-16-Perimeter-and-Area-Check-Your-Progress-Q8.2.png?resize=295%2C122&ssl=1\")
<\/p>\n
\nFrom square cardboard, a circle of the biggest area was cut out. If the area of the circle is 154 cm2<\/sup>, calculate the original area of the cardboard.
\nSolution:
\nFrom square cardboard, the biggest circle is cut out
\n![\"ML](\"https:\/\/i0.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-Solutions-for-ICSE-Maths-Chapter-16-Perimeter-and-Area-Check-Your-Progress-Q9.1.png?resize=281%2C331&ssl=1\")
\nSide of square = diameter of the circle = 2 \u00d7 7 = 14 cm
\nArea of square cardboard = (Side)2<\/sup> = (14)2<\/sup> = 196 cm2<\/sup><\/p>\n
\nA road 3.5 m wide surrounds a circular park whose circumference is 88 m. Find the cost of paving the road at the rate of \u20b9 60 per square meter.
\nSolution:
\nCircumference of a circular park = 88 m
\n![\"ML](\"https:\/\/i0.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-Solutions-for-ICSE-Maths-Chapter-16-Perimeter-and-Area-Check-Your-Progress-Q10.1.png?resize=214%2C115&ssl=1\")
\nWidth of road surounded it = 3.5 m
\nOuter radius (R) = 14 + 3.5 = 17.5 m
\nArea of road = \u03c0R2<\/sup> – \u03c0r2<\/sup> = \u03c0(R2<\/sup> – r2<\/sup>)
\n= \\(\\frac { 1 }{ 2 }\\) \u00d7 (17.52<\/sup> – 142<\/sup>)
\n= \\(\\frac { 1 }{ 2 }\\) \u00d7 31.5 \u00d7 3.5
\n= 346.5 m2<\/sup>
\nCost of paving the road = \u20b9 60 per m2<\/sup>
\nTotal cost = \u20b9 60 \u00d7 346.5 = \u20b9 20790<\/p>\n
\nIn the given figure, ABCD is a square of side 14 cm. Find the area of the shaded region.
\nTake \u03c0 = \\(\\frac { 22 }{ 7 }\\)
\n![\"ML](\"https:\/\/i1.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-Solutions-for-ICSE-Maths-Chapter-16-Perimeter-and-Area-Check-Your-Progress-Q11.1.png?resize=148%2C166&ssl=1\")
\nSolution:
\nIn the given figure, ABCD is a square whose side is 14 cm.
\nArea of square = (Side)2<\/sup> = (14)2<\/sup> cm2<\/sup> = 14 \u00d7 14 = 196 cm2<\/sup>
\nRadius of each circle in it = \\(\\frac { 7 }{ 2 }\\) cm
\nArea of 4 circles = 4 \u00d7 \u03c0r2<\/sup>
\n= 4 \u00d7 \\(\\frac { 22 }{ 7 }\\) \u00d7 \\(\\frac { 7 }{ 2 }\\) \u00d7 \\(\\frac { 7 }{ 2 }\\)
\n= 154 cm2<\/sup>
\nArea of shaded portion =196 – 154 = 42 cm2<\/sup><\/p>\n
\nThe boundary of a shaded region in the given figure consists of three semicircles, the smaller being equal. If the diameter of the larger one is 28 cm, find
\n(i) the length of the boundary
\n(ii) the area of the shaded region.
\n![\"ML](\"https:\/\/i0.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-Solutions-for-ICSE-Maths-Chapter-16-Perimeter-and-Area-Check-Your-Progress-Q12.1.png?resize=144%2C115&ssl=1\")
\nSolution:
\nIn the given figure,
\nThere are a bigger semicircle and two small semicircles
\nDiameter of bigger semicircle = 28 cm
\nRadius (R) = \\(\\frac { 28 }{ 2 }\\) = 14 cm
\nand radius of each of smaller semicircles = \\(\\frac { 14 }{ 2 }\\) = 7 cm
\nNow the area of shaded portion
\n= Area of bigger semicircle + Area of one smaller on semicircle
\n– Area of another smaller semicircle
\nBoth small semicircles have the same area
\n![\"ML](\"https:\/\/i1.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-Solutions-for-ICSE-Maths-Chapter-16-Perimeter-and-Area-Check-Your-Progress-Q12.2.png?resize=329%2C141&ssl=1\")
<\/p>\nML Aggarwal Class 7 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"