{"id":28398,"date":"2021-07-05T12:56:49","date_gmt":"2021-07-05T07:26:49","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=28398"},"modified":"2021-07-05T12:56:49","modified_gmt":"2021-07-05T07:26:49","slug":"ncert-solutions-for-class-8-maths-chapter-11-intext-questions","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-8-maths-chapter-11-intext-questions\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions"},"content":{"rendered":"

These NCERT Solutions for Class 8 Maths<\/a> Chapter 11 Mensuration InText Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions<\/h2>\n

NCERT Intext Question Page No. 170<\/span>
\nQuestion 1.
\n(a) Match the following figures with their respective areas in the box.
\n\"NCERT<\/p>\n

(b) Write the perimeter of each shape.
\n\"NCERT
\nAnswer:
\n\"NCERT<\/p>\n

(b)
\n(i) \"NCERT
\nThe given figure is a rectangle in which
\nLength = 14cm
\nBreadth = 7cm
\n\u2235Perimeter of a rectangle = 2 \u00d7 [Length + Breadth]
\n\u2234 Perimeter of the given figure = 2 \u00d7 [14cm + 7cm]<\/p>\n

(ii) \"NCERT
\nThe figure is a square housing its side as 7cm. y
\n\u2235 Perimeter of a square = 4 \u00d7 side
\n\u2234 Perimeter of the given figure = 4 \u00d7 7cm = 28cm<\/p>\n

\"NCERT<\/p>\n

NCERT Intext Question Page No. 172<\/span>
\nQuestion 1.
\nNazma\u2019s sister also has a trapezium shaped plot. Divide it into three parts as shown (Fig 11.4). Showthat the area of trapezium
\n\"NCERT<\/p>\n

Answer:
\nArea of \u0394PWZ
\n= \\(\\frac { 1 }{ 2 }\\) \u00d7 Base \u00d7 Altitiude = \\(\\frac { 1 }{ 2 }\\) \u00d7 c \u00d7 h = \\(\\frac { 1 }{ 2 }\\)ch
\nArea of the rectangle PQYZ = b \u00d7 h = bh
\nArea of \u0394QXY = \\(\\frac { 1 }{ 2 }\\) \u00d7 d \u00d7 h = \\(\\frac { 1 }{ 2 }\\)dh
\n\u2234 Area of the trapezium XYZW
\n\"NCERT<\/p>\n

Question 2.
\nIf h = 10 cm, c = 6 cm, b = 12 cm, d = 4 cm, find the values of each of its parts separetely and add to find the area WXYZ. Verify it by putting the values of h, a and b in the expression \\(\\frac{h(a+b)}{2}\\) .
\nAnswer:
\nArea of \u0394PWZ = \\(\\frac { 1 }{ 2 }\\)ch = \\(\\frac { 1 }{ 2 }\\) x 6 x 10cm2<\/sup>
\n= 30cm2
\n<\/sup>Area of \u0394QXY = \\(\\frac { 1 }{ 2 }\\) dh = \\(\\frac { 1 }{ 2 }\\) x 4 x 10cm2<\/sup>
\n= 20 cm2<\/sup>
\nArea of rectangle PQYZ = Length x Breadth = b x h = 12 x 10cm2<\/sup> = 120cm2<\/sup>
\nArea of trapezium WXYZ = Area of \u0394PWZ + Area of \u0394QXY+Area of rectangle PQYZ
\n= 30cm2<\/sup> + 20cm2<\/sup> + 120cm2<\/sup> = 170cm2<\/sup> Also, area of the trapezium WXYZ
\n= \\(\\frac{h(a+b)}{2}\\)
\n= \\(\\frac{10(22+12)}{2}\\) cm2<\/sup> [a = c + b + d = 6cm + 12cm + 4cm = 22cm]
\n= \\(\\frac{10 \\times 34}{2}\\) cm2<\/sup> = 170 cm2<\/sup>
\nHence, the area of trapezium is verified.<\/p>\n

\"NCERT<\/p>\n

NCERT Intext Question Page No. 172<\/span>
\nQuestion 1.
\nWe know that parallelogram is also a quadrilateral. Let us also split such a quadrilateral into two triangles, find their areas and hence that of the parallelogram. Does this agree with the formula that you know already? (Fig 11.12)
\n\"NCERT
\nAnswer:
\nThe diagonal BD of quadrilateral ABCD is joined and it divides the quadrilateral into two triangles.
\n\"NCERT
\nNow,
\nArea of quadrilateral ABCD
\n= Area of \u0394ABD + Area of \u0394BCD
\n= \\(\\frac { 1 }{ 2 }\\)b \u00d7 h)+ (\\(\\frac { 1 }{ 2 }\\) b \u00d7 h) = 2(\\(\\frac { 1 }{ 2 }\\) b \u00d7 h) = bh
\nIn fact ABCD is a parallelogram.
\n\u2234 Area of a parallelogram ABCD = b \u00d7 h
\nArea of a parallelogram = Base \u00d7 Height
\nWe know that a parallelogram can also be a trapezium. We already know that
\nArea of trapezium ABCD
\n= \\(\\frac { 1 }{ 2 }\\) \u00d7 (Sum of parallel sides) \u00d7 (Perpendicular distance between parallel sides)
\n= \\(\\frac { 1 }{ 2 }\\) \u00d7 (b + b) \u00d7 h = \\(\\frac { 1 }{ 2 }\\) \u00d7 (2b) \u00d7 h = bh
\nor Area of parallelogram ABCD = bh
\nYes, the above relation agrees with formula that we know already.<\/p>\n

NCERT Intext Question Page No. 175<\/span>
\nQuestion 1.
\nA parallelogram is divided into two congruent triangles by drawing a diagonal across it. Can we divide a trapezium into two congruent triangles?
\nAnswer:
\nNo, a trapezium cannot be divided into two congruent triangles.<\/p>\n

\"NCERT<\/p>\n

NCERT Intext Question Page No. 176<\/span>
\nQuestion 1.
\nDivide the following polygons (Fig 11.17) into parts (triangles and trapezium) to find out its area.
\n\"NCERT
\nAnswer:
\n(a) We draw perpendiculars from opposite vertices on FI, i.e. GL \u22a5 FI, HM \u22a5 FI and EN\u22a5 FI
\nArea of the polygon EFGHI
\n= ar (\u0394GFL) + ar (trapezium GLMH) + ar (\u0394HMI) + ar (\u0394NEI) + ar (\u0394EFN)
\n\"NCERT<\/p>\n

(b) NQ is a diagonal. Draw OA \u22a5 NQ, MB \u22a5 NQ, PC \u22a5NQ and RD \u22a5 NQ
\n\"NCERT
\n\u2234 Area of Polygon OPQRMN
\n= ar (\u0394OAN) + ar (trap. CPOA) + ar (\u0394PCQ) + ar (\u0394RDQ) + ar (trap. MBDR) + ar (\u0394MBN)
\n\"NCERT<\/p>\n

Question 2.
\nPolygon ABCDE is divided into parts as shown below (Fig 11.18). Find its area if AD = 8 cm, AH = 6 cm, AG = 4 cm, AF = 3 cm and perpendiculars BF = 2 cm, CH = 3 cm, EG = 2.5 cm.
\n\"NCERT
\nArea of polygon ABCDE = area of \u0394AFB + ….
\nArea of \u0394AFB = \\(\\frac { 1 }{ 2 }\\) \u00d7 AF \u00d7 BF
\n= \\(\\frac { 1 }{ 2 }\\) \u00d7 3 \u00d7 2 = …
\nArea of trapezium FBCH = FH \u00d7 \\(\\frac{(\\mathrm{BF}+\\mathrm{CH})}{2}\\)
\n= 3 \u00d7 \\(\\frac{(2+3)}{2}\\) [FH = AH – AF]
\nArea of \u0394CHD = \\(\\frac { 1 }{ 2 }\\) \u00d7 HD \u00d7 CH = ….
\nArea of \u0394ADE = \\(\\frac { 1 }{ 2 }\\) \u00d7 AD \u00d7 GE = ….
\nSo, the area of polygon ABCDE = ….
\nAnswer:
\nArea of polygon ABCDE = Area of \u0394AFB + Area of \u0394ADE
\nArea of \u0394AFB = \\(\\frac { 1 }{ 2 }\\) \u00d7 AF \u00d7 BF
\n= \\(\\frac { 1 }{ 2 }\\) \u00d7 3 \u00d7 2 = 3cm2<\/sup>
\nArea of trapezium FBCH = FH \u00d7 \\(\\frac{[\\mathrm{BF}+\\mathrm{CH}]}{2}\\)
\n= 3 \u00d7 \\(\\frac{(2+3)}{2}\\))cm2<\/sup> [\u2235 FH = AH – AF]
\n3 \u00d7 \\(\\frac{5}{2}\\) cm2<\/sup> = \\(\\frac{15}{2}\\) cm2<\/sup> = 7.5 cm2<\/sup>
\nArea of \u0394CHD= \\(\\frac { 1 }{ 2 }\\) \u00d7 HD \u00d7 CH
\n= \\(\\frac { 1 }{ 2 }\\) \u00d7 (AD – AH) \u00d7 CH [\u2235HD = AD – AH]
\n= \\(\\frac { 1 }{ 2 }\\) \u00d7 (8 – 6) \u00d7 3cm2 = 3cm2
\nArea of \u0394ADE = \\(\\frac { 1 }{ 2 }\\) \u00d7 AD \u00d7 GE
\n= \\(\\frac { 1 }{ 2 }\\) \u00d7 8 \u00d7 2.5cm2<\/sup> = 4 \u00d7 2.5cm2<\/sup> = 10cm2<\/sup>
\nSo, the area of polygon ABCD = 3cm2<\/sup> + 7.5cm2<\/sup> + 3cm2<\/sup> + 10cm2<\/sup> = 23.5cm2<\/sup><\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions NCERT Intext Question Page No. 170 Question 1. (a) Match the following figures with their respective areas in the box. (b) Write …<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions<\/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":{"spay_email":""},"categories":[7],"tags":[],"yoast_head":"\nNCERT Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-8-maths-chapter-11-intext-questions\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions and Answers are prepared by our highly skilled subject experts. 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NCERT Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions NCERT Intext Question Page No. 170 Question 1. (a) Match the following figures with their respective areas in the box. 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