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

Answer:

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?

Answer:

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 .

Answer:

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 .

Answer:

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.

Answer:

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.

Answer:2019-01-09_21-20-52

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.

Answer:

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.

Answer:2019-01-09_21-18-29

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?

Answer:2019-01-09_21-17-57

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 .

Answer:2019-01-09_21-17-32

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.

Answer:2019-01-09_21-15-58

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 .

Answer:2019-01-09_21-15-16

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 .

Answer:2019-01-09_21-15-36

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 .

Answer:2019-01-09_21-13-56

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.

Answer:2019-01-09_21-14-26

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.

Answer:

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.

Answer:2019-01-09_21-13-12

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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