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 $= 6 \ mm$

(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  $= 2.8 \ m$

(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 $= 80 \ cm$

(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.

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

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 $48 \ cm$.

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

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.

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

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.

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

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.

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)

(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 $10 \ cm$.

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.

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.

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.

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.

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.

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

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 $\angle PRS = 90^o$

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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Question 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 $DF = 2.4 \ cm$.

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 $DA = 30 \ cm$

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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