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

(i)  base = 16 \ cm , height = 7.5 \ cm

(ii) base = 5.6 \ m , height = 3.5 \ m

(iii) base = 6 .4 \ m , height = 8 \ dm

(iv) base = 9.5 \ cm , height = 6 \ mm

Answer:

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

(ii) 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) 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: 10 \ dm = 1 m

(iv) 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: 10 \ mm = 1 cm

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Question 2: Find the height of the triangle whose:

(i) area =  28.9 \ m^2 , base  = 8.5 \ m

(ii) area = 56 \ dm^2 , base  =  2.8 \ m

Answer:

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

(ii) 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) area = 4.2 \ m^2 , height = 2.4 \ m

(ii) area = 2.4 \ dm^2 , height = 80 \ cm

Answer:

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

(ii) 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 13 \ cm, 20 \ cm and 21 \ cm . Also find the altitude of the triangle corresponding to the largest side.

Answer:

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

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

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

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

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 50 \ cm, 48 \ cm and 14 \ cm . Find the height of the triangle corresponding to the side measuring 48 \ cm .

Answer:

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

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

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

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

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 17 \ m, 19 \ m and 32 \ m . Find the altitude of the triangle corresponding to the smallest side.

Answer:

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

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

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

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

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 30 \ cm and the third side is 48 \ cm long.

Answer:

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

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

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

Therefore Area of triangle = \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 5: 3 and its area is 43.2 \ m^2 . Find the base and the height of the triangle.

Answer:

Let the Base = 5x and Height = 3x

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

\Rightarrow 43.2 = 7.5x^2

\Rightarrow x = 2.4

Therefore Base = 12 \ m and 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) 12 \ cm (ii) 10 \ m (iii) 6.4 \ m (Take \sqrt{3} = 1.73 in each case)

Answer:

(i) a= 12 \ cm, b = 12 \ cm, c = 12 \ cm

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

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

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

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

(ii) a= 10 \ m, b = 10 \ m, c = 10 \ m

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

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

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

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

(iii) a= 6.4 \ m, b = 6.4 \ m, c = 6.4 \ m

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

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

Therefore Area of triangle = \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

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 26 \ cm long and one of the sides containing the right-angle measures 10 \ cm .

Answer:

height = \sqrt{26^2 - 10^2} = \sqrt{576} = 24 \ cm

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 240 \ cm^2 and one of its legs is 16 \ cm long. Find the length of the other leg.

Answer:

Let us say, height = 16 \ cm

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 3 : 4 and its area is 1014 \ cm^2 . Find its hypotenuse.

Answer:

Let the Base = 3x and Height = 4x

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

\Rightarrow 1014 = 6x^2

\Rightarrow x = 13

Therefore Hypotenuse = \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 13 : 14: 15 and its perimeter is 84 \ cm . Find the area of the triangle.

Answer:

Let the sides be 13x, 14x , and 15x

Therefore 84 = 13x + 14x + 15x

\Rightarrow x = 2

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

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

Answer:

Let the side of the isosceles triangle be x, x, 12

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

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

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

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

Therefore Area of triangle = \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 80 paisa per square meter is Rs.\ 176.40 . If the height of the board measures 24.5 \ m , find its base.

Answer:

Area of the triangular base = \frac{176.40}{0.80} = 220.5 \ m^2

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

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821Question 16: Calculate the area of the quadrilateral ABCD in which AB = BD = AD = 10 \ cm, \angle BCD = 90^o and CD = 8 \ cm . (Take \sqrt{3} = 1.732 )

Answer:

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

For \triangle BCD

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

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

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

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Question 17: Calculate the area of the quadrilateral PQRS shown in the adjoining figure, it being given that PR = 8 \ cm, RQ =17 \ cm \angle RPQ = 90^o, RS = 6 \ cm and \angle PRS = 90^o 823

Answer:

In \triangle RPQ

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

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

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

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

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822Question 18: Find the area of the quadrilateral ABCD whose diagonal AC is 25 \ cm long and the lengths of the perpendiculars from the opposite vertices B and D on AC are BE = 3.6 \ cm and DF = 2.4 \ cm .

Answer:

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

Area of \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 ABCD = 30+45 = 75 \ cm^2

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

Answer:824

For \triangle ABD

a= 30 \ cm, b = 28 \ cm, c = 50 \ cm

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

S = \frac{1}{2} (a+b+c) = \frac{1}{2} (30+28+50) = 54

Therefore Area of \triangle ABD = \sqrt{54(54-30)(54-28)(54-50)} = \sqrt{54 \times 24 \times 26 \times 4} = 367.13 \ cm^2

For \triangle BCD

a= 50 \ cm, b = 78 \ cm, c = 112 \ cm

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

S = \frac{1}{2} (a+b+c) = \frac{1}{2} (50+78+112) = 120

Therefore Area of \triangle ABD = \sqrt{120(120-50)(120-78)(120-112)} = \sqrt{120 \times 70 \times 42 \times 8} = 1680 \ cm^2

Hence the area of quadrilateral ABCD = 367.13+1680 = 2047.13 \ cm^2

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