Class 8: Perimeter and Area of Plane Figures – Exercise 36(A) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find the area, of the triangle, having:

i) $\displaystyle \text{Base } = 16 \ cm$ , $\displaystyle \text{Height } = 7.5 \ cm$ ii) $\displaystyle \text{Base } = 5.6 \ m$ , $\displaystyle \text{Height } = 3.5 \ m$

iii) $\displaystyle \text{Base } = 6 .4 \ m$ , $\displaystyle \text{Height } = 8 \ dm$ iv) $\displaystyle \text{Base } = 9.5 \ cm$ , $\displaystyle \text{Height } = 6 \ mm$

Answer:

i) $\displaystyle \text{Area of a triangle } = \frac{1}{2} \times base \times height = \frac{1}{2} \times 16 \times 7.5 = 60 \ cm^2$

ii) $\displaystyle \text{Area of a triangle } = \frac{1}{2} \times base \times height = \frac{1}{2} \times 5.6 \times 3.5 = 9.8 \ m^2$

iii) $\displaystyle \text{Area of a triangle } = \frac{1}{2} \times base \times height = \frac{1}{2} \times 6.4 \times 0.8 = 2.56 \ m^2$

Note: $\displaystyle 10 \ dm = 1 m$

iv) $\displaystyle \text{Area of a triangle } = \frac{1}{2} \times base \times height = \frac{1}{2} \times 9.5 \times 0.6 = 2.85 \ cm^2$

Note: $\displaystyle 10 \ mm = 1 cm$

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Question 2: Find the height of the triangle whose:

i) $\displaystyle \text{Area } = 28.9 \ m^2$ , $\displaystyle \text{Base } = 8.5 \ m$ ii) $\displaystyle \text{Area } = 56 \ dm^2$ , $\displaystyle \text{Base } = 2.8 \ m$

Answer:

i) $\displaystyle \text{Height } = \frac{2 \times Area}{Base} = \frac{2 \times 28.9}{8.5} = 6.8 \ m$

ii) $\displaystyle \text{Height } = \frac{2 \times Area}{Base} = \frac{2 \times 56}{28} = 4 \ dm$

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Question 3: Find the base of the triangle whose:

i) $\displaystyle \text{Area } = 4.2 \ m^2$ , $\displaystyle \text{Height } = 2.4 \ m$ ii) $\displaystyle \text{Area } = 2.4 \ dm^2$ , $\displaystyle \text{Height } = 80 \ cm$

Answer:

i) $\displaystyle \text{Base } = \frac{2 \times Area}{Height} = \frac{2 \times 4.2}{2.4} = 3.5 \ m$

ii) $\displaystyle \text{Base } = \frac{2 \times Area}{Height} = \frac{2 \times 2.4}{8} = 0.6 \ m$

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Question 4: Find the area of the triangle whose sides are $\displaystyle 13 \ cm, 20 \ cm$ and $\displaystyle 21 \ cm$ . Also find the altitude of the triangle corresponding to the largest side.

Answer:

$\displaystyle a= 13 \ cm, b = 20 \ cm, c = 21 \ cm$

Area of triangle $\displaystyle = \sqrt{s(s-a)(s-b)(s-c)}$

$\displaystyle S = \frac{1}{2} (a+b+c) = \frac{1}{2} (13+20+21) = 27$

Therefore Area of triangle $\displaystyle = \sqrt{27(27-13)(27-20)(27-21)} = \sqrt{27 \times 14 \times 7 \times 6} = 126 \ cm^2$

$\displaystyle \text{Height } = \frac{2 \times Area}{Base} = \frac{2 \times 126}{21} = 12 \ cm$

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Question 5: Find the area of the triangle whose sides are $\displaystyle 50 \ cm, 48 \ cm$ and $\displaystyle 14 \ cm$ . Find the height of the triangle corresponding to the side measuring $\displaystyle 48 \ cm$ .

Answer:

$\displaystyle a= 50 \ cm, b = 48 \ cm, c = 14 \ cm$

Area of triangle $\displaystyle = \sqrt{s(s-a)(s-b)(s-c)}$

$\displaystyle S = \frac{1}{2} (a+b+c) = \frac{1}{2} (50+48+14) = 56$

Therefore Area of triangle $\displaystyle = \sqrt{56(56-50)(56-48)(56-14)} = \sqrt{56 \times 6 \times 8 \times 42} = 336 \ cm^2$

$\displaystyle \text{Height } = \frac{2 \times Area}{Base} = \frac{2 \times 336}{48} = 14 \ cm$

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Question 6: Find the area of a triangular field whose sides are $\displaystyle 17 \ m, 19 \ m$ and $\displaystyle 32 \ m$ . Find the altitude of the triangle corresponding to the smallest side.

Answer:

$\displaystyle a= 17 \ m, b = 19 \ m, c = 32 \ m$

Area of triangle $\displaystyle = \sqrt{s(s-a)(s-b)(s-c)}$

$\displaystyle S = \frac{1}{2} (a+b+c) = \frac{1}{2} (17+19+32) = 34$

Therefore Area of triangle $\displaystyle = \sqrt{34(34-17)(34-19)(34-32)} = \sqrt{34 \times 17 \times 15 \times 2} = 131.68 \ m^2$

$\displaystyle \text{Height } = \frac{2 \times Area}{Base} = \frac{2 \times 131.68}{17} = 15.49 \ cm$

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Question 7: Find the area of an isosceles triangle in which each of the equal sides measures $\displaystyle 30 \ cm$ and the third side is $\displaystyle 48 \ cm$ long.

Answer:

$\displaystyle a= 30 \ cm, b = 30 \ cm, c = 48 \ cm$

Area of triangle $\displaystyle = \sqrt{s(s-a)(s-b)(s-c)}$

$\displaystyle S = \frac{1}{2} (a+b+c) = \frac{1}{2} (30+30+48) = 54$

Therefore Area of triangle $\displaystyle = \sqrt{54(54-30)(54-30)(54-48)} = \sqrt{54 \times 24 \times 24 \times 6} = 432 \ cm^2$

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Question 8: The base and the height of a triangle are in the ratio $\displaystyle 5: 3$ and its area is $\displaystyle 43.2 \ m^2$ . Find the base and the height of the triangle.

Answer:

Let the $\displaystyle \text{Base } = 5x$ and $\displaystyle \text{Height } = 3x$

$\displaystyle \text{Area of a triangle } = \frac{1}{2} \times base \times height = \frac{1}{2} \times 5x \times 3x$

$\displaystyle \Rightarrow 43.2 = 7.5x^2$

$\displaystyle \Rightarrow x = 2.4$

Therefore $\displaystyle \text{Base } = 12 \ m$ and $\displaystyle \text{Height } = 7.2 \ m$

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Question 9: Find. the area and the height of an equilateral triangle whose each side measures: (i) $\displaystyle 12 \ cm$ (ii) $\displaystyle 10 \ m$ (iii) $\displaystyle 6.4 \ m$ (Take $\displaystyle \sqrt{3} = 1.73$ in each case)

Answer:

i) $\displaystyle a= 12 \ cm, b = 12 \ cm, c = 12 \ cm$

Area of triangle $\displaystyle = \sqrt{s(s-a)(s-b)(s-c)}$

$\displaystyle S = \frac{1}{2} (a+b+c) = \frac{1}{2} (12+12+12) = 18$

Therefore Area of triangle $\displaystyle = \sqrt{18(18-12)(18-12)(18-12)} = \sqrt{18 \times 6 \times 6 \times 6} = 62.28 \ cm^2$

$\displaystyle \text{Height } = \frac{2 \times Area}{Base} = \frac{2 \times 62.28}{12} = 10.38 \ cm$

ii) $\displaystyle a= 10 \ m, b = 10 \ m, c = 10 \ m$

Area of triangle $\displaystyle = \sqrt{s(s-a)(s-b)(s-c)}$

$\displaystyle S = \frac{1}{2} (a+b+c) = \frac{1}{2} (10+10+10) = 15$

Therefore Area of triangle $\displaystyle = \sqrt{15(15-10)(15-10)(15-10)} = \sqrt{15 \times 5 \times 5 \times 5} = 43.25 \ m^2$

$\displaystyle \text{Height } = \frac{2 \times Area}{Base} = \frac{2 \times 43.25}{10} = 8.65 \ m$

iii) $\displaystyle a= 6.4 \ m, b = 6.4 \ m, c = 6.4 \ m$

Area of triangle $\displaystyle = \sqrt{s(s-a)(s-b)(s-c)}$

$\displaystyle S = \frac{1}{2} (a+b+c) = \frac{1}{2} (6.4+6.4+6.4) = 9.6$

Therefore Area of triangle $\displaystyle = \sqrt{9.6(9.6-6.4)(9.6-6.4)(9.6-6.4)} = \sqrt{9.6 \times 3.2 \times 3.2 \times 3.2} = 17.7152 \ m^2$

$\displaystyle \text{Height } = \frac{2 \times Area}{Base} = \frac{2 \times 17.7152}{6.4} = 5.536 \ m$

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Question 10: Find the area of a right triangle whose hypotenuse is $\displaystyle 26 \ cm$ long and one of the sides containing the right-angle measures $\displaystyle 10 \ cm$ .

Answer:

$\displaystyle \text{Height } = \sqrt{26^2 - 10^2} = \sqrt{576} = 24 \ cm$

$\displaystyle \text{Area of a triangle } = \frac{1}{2} \times base \times height = \frac{1}{2} \times 10 \times 24 = 120 \ cm^2$

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Question 11: The area of a right triangle is $\displaystyle 240 \ cm^2$ and one of its legs is $\displaystyle 16 \ cm$ long. Find the length of the other leg.

Answer:

Let us say, $\displaystyle \text{Height } = 16 \ cm$

$\displaystyle \text{Base } = \frac{2 \times Area}{Height} = \frac{2 \times 240}{16} = 30 \ cm$

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Question 12: The legs of a right triangle are in the ratio $\displaystyle 3 : 4$ and its area is $\displaystyle 1014 \ cm^2$ . Find its hypotenuse.

Answer:

Let the $\displaystyle \text{Base } = 3x$ and $\displaystyle \text{Height } = 4x$

$\displaystyle \text{Area of a triangle } = \frac{1}{2} \times base \times height = \frac{1}{2} \times 3x \times 4x$

$\displaystyle \Rightarrow 1014 = 6x^2$

$\displaystyle \Rightarrow x = 13$

Therefore Hypotenuse $\displaystyle = \sqrt{(39)^2 + (52)^2} = \sqrt{1521+2704} = 65 \ cm$

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Question 13: The sides of a triangle are in the ratio $\displaystyle 13 : 14: 15$ and its perimeter is $\displaystyle 84 \ cm$ . Find the area of the triangle.

Answer:

Let the sides be $\displaystyle 13x, 14x$ , and $\displaystyle 15x$

Therefore $\displaystyle 84 = 13x + 14x + 15x$

$\displaystyle \Rightarrow x = 2$

Hence the sides are $\displaystyle 26 \ cm, 28 \ cm$ and $\displaystyle 30 \ cm$

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Question 14: The base of an isosceles triangle is $\displaystyle 12 \ cm$ and its perimeter is $\displaystyle 32 \ cm$ . Find its area.

Answer:

Let the side of the isosceles triangle be $\displaystyle x, x, 12$

Therefore $\displaystyle 32 = 2x+12 \Rightarrow x = 10$

$\displaystyle a= 10 \ cm, b = 10 \ cm, c = 12 \ cm$

Area of triangle $\displaystyle = \sqrt{s(s-a)(s-b)(s-c)}$

$\displaystyle S = \frac{1}{2} (a+b+c) = \frac{1}{2} (10+10+12) = 16$

Therefore Area of triangle $\displaystyle = \sqrt{16(16-10)(16-10)(16-12)} = \sqrt{16 \times 6 \times 6 \times 4} = 48 \ cm^2$

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Question 15: The cost of painting the top surface of a triangular board at $\displaystyle 80$ paisa per square meter is $\displaystyle Rs.\ 176.40$ . If the height of the board measures $\displaystyle 24.5 \ m$ , find its base.

Answer:

Area of the triangular $\displaystyle \text{Base } = \frac{176.40}{0.80} = 220.5 \ m^2$

$\displaystyle \text{Base } = \frac{2 \times Area}{Height} = \frac{2 \times 220.5}{24.5} = 18 \ m$

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Question 16: Calculate the area of the quadrilateral $\displaystyle ABCD$ in which $\displaystyle AB = BD = AD = 10 \ cm, \angle BCD = 90^o$ and $\displaystyle CD = 8 \ cm$ . (Take $\displaystyle \sqrt{3} = 1.732$ )

Answer:

Area of $\displaystyle \triangle ABC = \sqrt{15(15-10)(15-10)(15-10)} = \sqrt{15 \times 5 \times 5 \times 5} = 43.3 \ cm^2$

For $\displaystyle \triangle BCD$

$\displaystyle BC = \sqrt{10^2 - 8^2} = \sqrt{36} = 6$

Therefore Area of $\displaystyle \triangle BCD = \frac{1}{2} \times base \times height = \frac{1}{2} \times 8 \times 6 = 24 \ cm^2$

Therefore the area of quadrilateral $\displaystyle ABCD = 43.3 + 24 = 67.3 \ cm^2$

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Question 17: Calculate the area of the quadrilateral $\displaystyle PQRS$ shown in the adjoining figure, it being given that $\displaystyle PR = 8 \ cm, RQ =17 \ cm \angle RPQ = 90^o, RS = 6 \ cm$ and $\displaystyle \angle PRS = 90^o$

Answer:

In $\displaystyle \triangle RPQ$

$\displaystyle RP = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt {225} = 15 \ cm$

Area of $\displaystyle \triangle RPQ = \frac{1}{2} \times base \times height = \frac{1}{2} \times 8 \times 15 = 60 \ cm^2$

Area of $\displaystyle \triangle SRP = \frac{1}{2} \times base \times height = \frac{1}{2} \times 6 \times 15 = 45 \ cm^2$

Hence the area of quadrilateral $\displaystyle PQRS = 60+45 = 105 \ cm^2$

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Question 18: Find the area of the quadrilateral $\displaystyle ABCD$ whose diagonal $\displaystyle AC$ is $\displaystyle 25 \ cm$ long and the lengths of the perpendiculars from the opposite vertices $\displaystyle B$ and $\displaystyle D$ on $\displaystyle AC$ are $\displaystyle BE = 3.6 \ cm$ and $\displaystyle DF = 2.4 \ cm$ .

Answer:

Area of $\displaystyle \triangle ADC = \frac{1}{2} \times base \times height = \frac{1}{2} \times 25 \times 2.4 = 30 \ cm^2$

Area of $\displaystyle \triangle ABC = \frac{1}{2} \times base \times height = \frac{1}{2} \times 25 \times 3.6 = 45 \ cm^2$

Hence the area of quadrilateral $\displaystyle ABCD = 30+45 = 75 \ cm^2$

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Question 19: Find the area of the quadrilateral $\displaystyle ABCD$ , given in the adjoining figure in which $\displaystyle AB = 28 \ cm, BC = 7 8 \ cm, CD = 112 \ cm, BD = 50 \ cm$ and $\displaystyle DA = 30 \ cm$