Class 9: Perimeter and Areas of Plane Figures – Exercise 15.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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

Answer:

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

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

Answer:

Let the height of the triangle $\displaystyle = h$

$\displaystyle \text{ Area of triangle } = 48 \text{ cm} ^2$

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

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

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

Answer:

$\displaystyle \text{ Perimeter } = 12 \text{ m }$

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

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

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

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

Answer:

Radius $\displaystyle = 3 \text{ cm }$

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

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

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

Answer:

Diameter $\displaystyle =$ Base $\displaystyle = 10 \text{ cm }$

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

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

Answer:

$\displaystyle \text{ Perimeter of triangle } = 60 \text{ cm }$

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

Applying Pythagoras theorem we get

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

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

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

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

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

$\displaystyle \Rightarrow x = 24 or 10$

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

Therefore the sides are $\displaystyle 10 \ cm, 24 \text{ cm }$ , and $\displaystyle 26 \text{ cm }$ .

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

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

Answer:

Let the base Base $\displaystyle =x$

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

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

Total Cost of turfing $\displaystyle = 900 \ Rs.$

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

$\displaystyle \Rightarrow x^2 = 100$

$\displaystyle \Rightarrow x = 100$

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

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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 $\displaystyle Rs. \ 36$ per hectare is $\displaystyle Rs \ 486$ , find its base and height.

Answer:

Let the Height $\displaystyle = h$

Therefore the Base $\displaystyle = 3h$

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

1 hectare $\displaystyle = 10000 \ m^2$

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

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

$\displaystyle \Rightarrow h^2 = 90000$

$\displaystyle \Rightarrow h = 300 \text{ m }$

Hence the Base $\displaystyle = 3 \times 300 = 900 \text{ m }$

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Question 9: An isosceles right triangle has $\displaystyle \text{ Area } 200 \text{ cm} ^2$ . what is the length of its hypotenuse?

Answer:

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

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

$\displaystyle \Rightarrow x^2 = 400$

$\displaystyle \Rightarrow x = 20$

$\displaystyle \text{Therefore Hypotenuse } = \sqrt{20^2 + 20^2} = 20\sqrt{2} \text{ cm }$ or $\displaystyle 28.28 \text{ cm }$

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

Answer:

$\displaystyle \text{ Area } = 6 \text{ cm} ^2$

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

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

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

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

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

$\displaystyle \Rightarrow a^2 = 144$ or $\displaystyle a = 12$

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

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

Therefore Base is either $\displaystyle 24 \text{ cm }$ or $\displaystyle 10 \text{ cm }$

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

Answer:

$\displaystyle \text{ Perimeter } = 40 \text{ cm }$

Let the two equal sides be $\displaystyle x$

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

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

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

$\displaystyle \Rightarrow x = 12$

$\displaystyle \text{Therefore sides are } 12 \ cm, 12 \text{ cm }$ , and $\displaystyle 16 \text{ cm }$

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

Answer:

$\displaystyle b = 30 - 12 - 12 = 6 \text{ cm }$

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

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

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

Answer:

$\displaystyle \text{ Perimeter } = 16 \text{ cm }$

$\displaystyle \text{ Therefore } 2x + 6 = 16 \Rightarrow x = 5 \text{ cm }$

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

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

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

Answer:

$\displaystyle \text{ Area } = 60 \text{ cm} ^2$

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

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

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

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

$\displaystyle \text{Therefore sides are } 15, 8$

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

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

Answer:

$\displaystyle \text{ Area } = 60 \text{ cm} ^2$

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

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

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

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

$\displaystyle \text{Therefore sides are } 15 \text{ cm }$ and $\displaystyle 8 \text{ cm }$

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

and $\displaystyle \text{ Perimeter } = 15 + 8 + 17 = 40 \text{ cm }$

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

Answer:

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

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

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

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

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

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