Question 1: Find the area of a triangle with base $16 \ cm$ and height $7 \ cm$.

Area of a triangle $=$ $\frac{1}{2}$ $\times Base \times Height =$ $\frac{1}{2}$ $\times 16 \times 7 = 56 \ cm^2$

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Question 2: The area of a triangle is $48 \ cm^2$. Its base is $12 \ cm$. What is its altitude?

Let the height of the triangle $= h$

Area of triangle $= 48 \ cm^2$

$\therefore 48 =$ $\frac{1}{2}$ $\times 12 \times h$

$\Rightarrow h =$ $\frac{48}{6}$ $= 8 \ cm$

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Question 3: Find the area of an equilateral triangle whose perimeter is $12 \ m$.

Perimeter $= 12 \ m$

Therefore side $=$ $\frac{12}{3}$ $=4 \ m$ (since triangle is equilateral triangle)

Therefore $h = \sqrt{4^2 - 2^2}= \sqrt{16-4} = 2\sqrt{3}$

Therefore Area of  triangle $=$ $\frac{1}{2}$ $\times 4 \times 2\sqrt{3} = 4\sqrt{3} \ m^2$

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Question 4: Find the area of a right-angled triangle if the radius of its circumcircle is $3 \ cm$ and altitude drawn to the hypotenuse is $2 \ cm$.

Radius $= 3 \ cm$

Therefore Diameter $=$ Base $= 6 \ cm$ (remember, the angle subtended by the diameter on any point on the circumference is a right angle)

Therefore Area of triangle $=$ $\frac{1}{2}$ $\times 6 \times 2 = 6 \ cm^2$

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Question 5: Find the area of a right-angled triangle if the diameter of its circumcircle is $10 \ cm$ and altitude drawn to the hypotenuse is $4.5 \ cm$ long.

Diameter $=$ Base $= 10 \ cm$

Therefore Area of triangle $=$ $\frac{1}{2}$ $\times 10 \times 4.5 = 22.5 \ cm^2$

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Question 6: The perimeter of a right triangle is $60 \ cm$. Its hypotenuse is $26 \ cm$. Find the other two sides and the area of the triangle.

Perimeter of triangle $= 60 \ cm$

Given hypotenuse $= 26$. Let one side be $x$, then the other side would be $(34-x)$

Applying Pythagoras theorem we get

$26^2 = (34-x)^2 + x^2$

$\Rightarrow 676 = 1156 + x^2 - 68x + x^2$

$\Rightarrow 2x^2 - 68x + 480 = 0$

$\Rightarrow x^2 - 34x + 240 = 0$

$\Rightarrow (x-24)(x-10) = 0$

$\Rightarrow x = 24 or 10$

When $x = 24$, the other side is $(34-24) = 10$. Similarly, when $x = 10 \ cm$ the other side $= (34-10) = 24 \ cm$

Therefore the sides are $10 \ cm, 24 \ cm$, and $26 \ cm$.

Therefore the area of the triangle $=$ $\frac{1}{2}$ $\times 10 \times 24 = 120 \ cm^2$

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Question 7: The cost of turfing a triangular field at the rate of $Rs. \ 45$ per $100 \ m^2$ is $Rs. \ 900$. Find the height, if double the base of the triangle is $5$ times the height.

Let the base Base $=x$

$\Rightarrow Height =$ $\frac{2}{5}$ $x$

The rate of turfing $= Rs. \ 45$ per $100 \ m^2$

Total Cost of turfing  $= 900 \ Rs.$

Therefore $\frac{1}{2} \times x \times \frac{2}{5}$ $x \times$ $\frac{45}{100}$ $= 900$

$\Rightarrow x^2 = 100$

$\Rightarrow x = 100$

Therefore Height $=$ $\frac{2}{5}$ $\times 100 = 40 \ m$

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Question 8: The base of a triangular field is three times its height. If the cost of cultivating the field at $Rs. \ 36$ per hectare is $Rs \ 486$, find its base and height.

Let the Height $= h$

Therefore the Base $= 3h$

Therefore Area of triangle $=$ $\frac{1}{2}$ $\times 3h \times h =$ $\frac{3}{2}$ $h^2$

1 hectare $= 10000 \ m^2$

Therefore cost of cultivation $=$ $\frac{36}{10000}$ $Rs./m^2$

Therefore $\frac{3}{2} h^2 \times$ $\frac{36}{10000}$ $latex = 486$

$\Rightarrow h^2 = 90000$

$\Rightarrow h = 300 \ m$

Hence the Base $= 3 \times 300 = 900 \ m$

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Question 9: An isosceles right triangle has area $200 \ cm^2$. what is the length of its hypotenuse?

Let the Base and Height of the isosceles right triangle $= x$

Therefore $\frac{1}{2}$ $\times x \times = 200$

$\Rightarrow x^2 = 400$

$\Rightarrow x = 20$

Therefore hypotenuse $= \sqrt{20^2 + 20^2} = 20\sqrt{2} \ cm$ or $28.28 \ cm$

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Question 10: Find the base of an isosceles triangle whose area is $6 \ cm^2$ and the length of one of its equal sides is $13 \ cm$.

Area $= 6 \ cm^2$

Therefore $h = \sqrt{13^2 - a^2}$

Hence $\frac{1}{2}$ $\times (2a) \times \sqrt{13^2 - a^2} = 60$

$\Rightarrow a \sqrt{13^2 - a^2} = 60$

$\Rightarrow a^2(13^2 - a^2) = 3600$

$\Rightarrow (a^2 - 144)(a^2 - 25) = 0$

$\Rightarrow a^2 = 144$ or  $a = 12$

and $a^2 = 25$ or $a = 5$

Hence $a$ is either $12$ or $5$

Therefore Base is either $24 \ cm$ or $10 \ cm$

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Question 11: The perimeter of an isosceles triangle is $40 \ cm$. The base is two-third of the sum of equal sides. Find the length of each side.

Perimeter $= 40 \ cm$

Let the two equal sides be $x$

Given: $b =$ $\frac{2}{3}$ $(2x) =$ $\frac{4}{3}$ $x$

Therefore $x +$ $\frac{4}{3}$ $x + x = 40$

$\Rightarrow$ $\frac{10x}{3}$ $= 4$

$\Rightarrow x = 12$

Therefore sides are $12 \ cm, 12 \ cm$, and $16 \ cm$

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Question 12: Find the area of an isosceles triangle whose equal sides are $12 \ cm$ each and the perimeter is $30 \ cm$.

$b = 30 - 12 - 12 = 6 \ cm$

Therefore $h = \sqrt{12^2 - 3^2} = \sqrt{135}$

Therefore area $= \frac{1}{2} \times 6 \times \sqrt{135} = 3\sqrt{135} \ cm^2 = 34.86 \ cm^2$

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Question 13: Find the area of an isosceles triangle whose base is $6 \ cm$ and perimeter is $16 \ cm$.

Perimeter $= 16 \ cm$

Therefore $2x + 6 = 16 \Rightarrow x = 5 \ cm$

Therefore $h = \sqrt{5^2 - 3^2} = \sqrt{25-9} = 4 \ cm$

Therefore Area $=$ $\frac{1}{2}$ $\times 6 \times 4 = 12 \ cm^2$

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Question 14: The sides of a right triangle containing the right angle $5x \ cm$ and $(3x-1) \ cm$. Calculate the length of the hypotenuse of the triangle, if its area is $60 \ cm^2$.

Area $= 60 \ cm^2$

Therefore    $\frac{1}{2} \times (3x-1) \times 5x = 60$

$\Rightarrow (3x-1)x = 24$

$\Rightarrow 3x^2 - x - 24 = 0$

$\Rightarrow x = 3 \ or \ - 2.67$ (this is not possible as $x$ is positive)

Therefore sides are $15, 8$

Hence Hypotenuse $= \sqrt{15^2 + 8^2} = \sqrt{289} = 17 \ cm$

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Question 15: $ABC$ is a right triangle right angled at $B$. lf $AB = (2x +1) \ cm$, $BC = (x + 1) \ cm$ and area of $\triangle ABC$ is 60 \ cm^2 \$, find its perimeter.

Area $= 60 \ cm^2$

Therefore    $\frac{1}{2} \times (x+1) \times (2x+1) = 60$

$\Rightarrow (x+1)(2x+1)= 120$

$\Rightarrow 2x^2 + 3x - 119 = 0$

$\Rightarrow x = 7 \ or \ -8.5$  (this is not possible as $x$ is positive)

Therefore sides are $15 \ cm$ and $8 \ cm$

Hence Hypotenuse $= \sqrt{15^2 + 8^2} = \sqrt{289} = 17 \ cm$

and Perimeter $= 15 + 8 + 17 = 40 \ cm$

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Question 16: In the adjoining figure, $ABC$ is an equilateral triangle with each side of length $10 \ cm$. $D$ is a point inside $\triangle ABC$ such that $\angle BDC=90^o$ and $CD =6 \ cm$. Find the area of the shaded region.

Height of equilateral triangle   $h_1 = \sqrt{10^2 - 5^2} = \sqrt{75} = 5\sqrt{3} \ cm$

Therefore are of equilateral triangle  $=$ $\frac{1}{2}$ $\times 10 \times 5\sqrt{3} = 25\sqrt{3} \ cm^2$

Height of right angle triangle $h_2 = \sqrt{10^2 - 6^2} = \sqrt{64} = 8 \ cm$

Therefore are of right triangle  $=$ $\frac{1}{2}$ $\times 6 \times 8 = 24 \ cm^2$

Therefore shaded area $= 25 \sqrt{3} - 24 = 19.3 \ cm^2$

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Question 17: If the difference between the two sides of a right-angled triangle is $2 \ cm$ and the area of the triangle is $24 \ cm^2$ , find the perimeter of the triangle.

Area $= 24 \ cm^2$

Therefore    $\frac{1}{2}$ $\times x \times (x+2) = 24$

$\Rightarrow x(x+2) = 48$

$\Rightarrow x^2 + 2x - 48 = 0$

$\Rightarrow x = 6$ or $-8$ (this is not possible as $x$ is positive)

Therefore sides are $6 \ cm$ and $8 \ cm$

Therefore Hypotenuse $= \sqrt{8^2 + 6^2} = \sqrt{100} = 10 \ cm$

Hence Perimeter $= 6 + 8 + 10 = 24 \ cm$

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