RS Aggarwal Class 7 Mathematics Twentieth Chapter Mensuration Exercise 20D Solution
EXERCISE 20D
(1) Find the area of the triangle in which
(i) Base = 42 cm and height = 25 cm,
Solution: Area = (1/2 × 42 × 25) cm2 = 525 cm2
(ii) Base = 16.8 m and height = 75 cm.
Solution: Here, 75 cm = 0.75 m
Area = (1/2 × 16.8 × 0.75) m2 = 6.3 m2
(iii) Base = 8 dm and height = 35 cm.
Solution: Here, 8 dm = 80 cm
Area = (1/2 × 80 × 35) cm2 = 1400 cm2.
(2) Find the height of a triangle having an area of 72 cm2 and base 16 cm.
Solution: Let height be x cm.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Find-the-height-of-a-triangle-having-an-area-of-72-cm2-and-base-16-cm..png)
Hence, height of the triangle is 9 cm.
(3) Find the height of a triangular region having an area of 224 m2 and base 28 m.
Solution: Let the height be x m.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Find-the-height-of-a-triangle-having-an-area-of-72-cm2-and-base-16-cm.-1.png)
(4) Find the base of a triangle whose area is 90 cm2 and height 12 cm.
Solution: Let the base be x cm.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Class-7-math-ex-20D.png)
Hence, the base of the triangle is 15 cm.
(5) The base of a triangular field is three times its height. If one of the cultivating the field at Rs 1080 per hectare is Rs 14580, find its base and height.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Class-7-math-ex-20D-1.png)
(6) The area of right triangular region is 129.5 cm2. If one of the sides containing the right angle is 14.8 cm, find the other one.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Class-7-math-ex-20D-2.png)
(7) Find the area of a right triangle whose base is 1.2 m and hypotenuse 3.7 m.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Class-7-math-ex-20D-4.png)
(8) The legs of a right triangle are in the ratio 3 : 4 and its area is 1014 cm2. Find the lengths of its legs.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Class-7-math-ex-20D-3.png)
(9) One side of a right-angled triangular scarf is 80 cm and its longest side is 1 m. Find its cost at the rate of Rs 250 per m2.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Class-7-math-ex-20D-5.png)
(10) Find the area of an equilateral triangle each of whose sides measures 9i) 18 cm, (ii) 20 cm.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Class-7-math-ex-20D-6.png)
(11) Area of an equilateral triangle is (16 × √3) cm2. Find the length of each side of the triangle.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Class-7-math-ex-20D-7.png)
(12) Find the length of the height of an equilateral triangle of side 24 cm.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Class-7-math-ex-20D-8.png)
(13) Find the area of the triangle in which
(i) a = 13 m, b = 14 m, c = 15m.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Class-7-math-ex-20D-9.png)
(ii) a = 52 cm, b = 56 cm, c = 60 cm.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Class-7-math-ex-20D-10.png)
(iii) a = 91 m, b = 98 m, c = 105 m.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Class-7-math-ex-20D-11.png)
(14) The lengths of the sides of a triangle are 33 cm, 44 cm and 55 cm respectively. Find the area of the triangle and hence find the height corresponding to the side measuring 44 cm.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Class-7-math-ex-20D-12.png)
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Class-7-math-ex-20D-13.png)
(15) The sides of a triangle are in the ratio 13 : 14 : 15 and its perimeter is 84 cm. Find the area of the triangle.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Class-7-math-ex-20D-14.png)
(16) The sides of a triangle are 42 cm, 34 cm and 20 cm. Calculate its area and the length of the height on the longest side.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Class-7-math-ex-20D-15.png)
(17) The base of an isosceles triangle is 48 cm and one of its equal sides is 30 cm. Find its area of triangle.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Class-7-math-ex-20D-16.png)
(18) The base of an isosceles triangle is 12 cm and its perimeter is 32 cm. Find its area.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Class-7-math-ex-20D-17.png)
(19) A diagonal of a quadrilateral is 26 cm and perpendiculars drawn to it from the opposite vertices are 12.8 cm and 11.2 cm. Find the area of the quadrilateral.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Class-7-math-ex-20D-18.png)
(20) In a quadrilateral ABCD, AB = 28 cm, BC = 26 cm, CD = 50 cm, DA = 40 cm and diagonal AC = 30 cm. Find the area of the quadrilateral.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Class-7-math-ex-20D-19.png)
(21) In the given figure, ABCD is a rectangle with length = 36 m breadth = 24 m, In ∆ADE, EF ⊥ AD and EF = 15 m. Calculate the area of the shaded region.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Class-7-math-ex-20D-20-300x240.png)
(22) In the given figure, ABCD is rectangle in which AB = 40 cm and BC = 25 cm. If P, Q, R,S be the midpoints of AB, BC, CD and Da respectively, find the area of the shaded region.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Class-7-math-Ex-20D.jpg)
(23) In the following figures, find the area of the shaded region.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/23-In-the-following-figures-find-the-area-of-the-shaded-region-i.png)
![](https://www.netexplanations.com/wp-content/uploads/2018/11/23-In-the-following-figures-find-the-area-of-the-shaded-region-ii.png)
(24) Find the area of quadrilateral ABCD in which diagonal BD = 24 cm. AL ⊥ BD and CM ⊥ BD such that AL = 5 cm and CM = 8 cm.
![](https://www.netexplanations.com/wp-content/uploads/2018/11/Find-the-area-of-quadrilateral-ABCD-in-which-diagonal-BD-24-cm.-AL-⊥-BD-and-CM-⊥-BD-such-that-AL-5-cm-and-CM-8-cm..png)
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