RowQ
The Vault
RowQ
The Vault
CBSE Class 9 Maths · 11 questions · 26 marks
Perimeter measures the walk around a shape and area measures the paint needed to cover it, and this chapter finally lets you find the area of a triangle even when no height is given. Heron's formula does that using the three sides alone. Once you can handle a triangle, any straight-sided field can be chopped into triangles and measured.
The area of a triangle whose sides are 13 cm, 14 cm and 15 cm is:
Answer
The area is 84 cm². Semi-perimeter s = (13 + 14 + 15)/2 = 42/2 = 21 cm. By Heron's formula, area = √(21(21 - 13)(21 - 14)(21 - 15)) = √(21 × 8 × 7 × 6). 21 × 8 = 168 and 7 × 6 = 42, so the product is 168 × 42 = 7056. Area = √7056 = 84 cm².
An isosceles triangle has two equal sides of 10 cm and a base of 12 cm. Its area is:
Answer
The area is 48 cm². The height to the base splits the base into two halves of 6 cm each. Height = √(10² - 6²) = √(100 - 36) = √64 = 8 cm. Area = (1/2) × base × height = (1/2) × 12 × 8 = 48 cm².
The sides of a triangle are in the ratio 3 : 4 : 5 and its perimeter is 54 cm. Its area is:
Answer
The area is 121.5 cm². Let the sides be 3k, 4k and 5k. Then 3k + 4k + 5k = 54, so 12k = 54 and k = 4.5. The sides are 13.5 cm, 18 cm and 22.5 cm. Since 13.5² + 18² = 182.25 + 324 = 506.25 = 22.5², the triangle is right-angled with legs 13.5 cm and 18 cm. Area = (1/2)(13.5)(18) = 121.5 cm².
The area of an equilateral triangle is 25√3 cm². The length of its side is:
Answer
The side is 10 cm. For an equilateral triangle of side a, area = (√3/4)a². (√3/4)a² = 25√3 Dividing both sides by √3 gives a²/4 = 25, so a² = 100 and a = 10 cm.
Assertion (A): A triangle with sides 8 cm, 15 cm and 17 cm has an area of 60 cm². Reason (R): The area of a triangle whose three side lengths are known can be found from Heron's formula, area = √(s(s - a)(s - b)(s - c)) with s the semi-perimeter.
Answer
Both A and R are true and R is the correct explanation of A. Reason R states Heron's formula correctly. Applying it: s = (8 + 15 + 17)/2 = 40/2 = 20 cm. Area = √(20(20 - 8)(20 - 15)(20 - 17)) = √(20 × 12 × 5 × 3) = √3600 = 60 cm². So Assertion A is true and it is obtained precisely by using the formula in R, making R the correct explanation.
Find the area of a triangle whose sides measure 9 cm, 12 cm and 15 cm using Heron's formula.
Answer
Semi-perimeter s = (9 + 12 + 15)/2 = 36/2 = 18 cm. Area = √(18(18 - 9)(18 - 12)(18 - 15)) = √(18 × 9 × 6 × 3). 18 × 9 = 162 and 6 × 3 = 18, so the product is 162 × 18 = 2916. Area = √2916 = 54 cm².
A triangular signboard has sides 40 cm, 24 cm and 32 cm. Find its area, and the cost of painting one face of it at ₹0.50 per cm².
Answer
Semi-perimeter s = (40 + 24 + 32)/2 = 96/2 = 48 cm. Area = √(48(48 - 40)(48 - 24)(48 - 32)) = √(48 × 8 × 24 × 16). 48 × 8 = 384 and 24 × 16 = 384, so the product is 384 × 384. Area = √(384²) = 384 cm². Cost = 384 × 0.50 = ₹192.
A triangle has sides 5 cm, 12 cm and 13 cm. Find the length of the altitude drawn to the longest side.
Answer
Since 5² + 12² = 25 + 144 = 169 = 13², the triangle is right-angled with legs 5 cm and 12 cm. Area = (1/2)(5)(12) = 30 cm². Taking the longest side 13 cm as the base, area = (1/2) × 13 × h. 30 = (1/2)(13)h, so h = 60/13 ≈ 4.62 cm.
A field is in the shape of quadrilateral ABCD in which ∠DAB = 90°, AB = 9 m, AD = 40 m, BC = 28 m and CD = 15 m. Find the total area of the field.
Answer
Step 1: Join the diagonal BD, splitting the field into triangle ABD and triangle BCD. Step 2: Triangle ABD is right-angled at A, so its area = (1/2) × AB × AD = (1/2)(9)(40) = 180 m². Step 3: Find BD using Pythagoras in triangle ABD. BD² = 9² + 40² = 81 + 1600 = 1681, so BD = 41 m. Step 4: For triangle BCD the sides are 28 m, 15 m and 41 m. s = (28 + 15 + 41)/2 = 84/2 = 42 m. Area = √(42(42 - 28)(42 - 15)(42 - 41)) = √(42 × 14 × 27 × 1). 42 × 14 = 588 and 588 × 27 = 15876. Area = √15876 = 126 m². Step 5: Total area = 180 + 126 = 306 m².
(a) Using Heron's formula, show that the area of an equilateral triangle of side a is (√3/4)a². (b) A rhombus-shaped park has a perimeter of 400 m and one of its diagonals measures 120 m. Find the area of the park.
Answer
(a) For an equilateral triangle all sides equal a, so s = (a + a + a)/2 = 3a/2. Then s - a = 3a/2 - a = a/2, and the same for the other two brackets. Area = √((3a/2)(a/2)(a/2)(a/2)) = √(3a⁴/16) = (a²/4)√3 = (√3/4)a². Proved. (b) A rhombus has four equal sides, so each side = 400/4 = 100 m. The diagonal of 120 m divides the rhombus into two congruent triangles with sides 100 m, 100 m and 120 m. For one such triangle, s = (100 + 100 + 120)/2 = 320/2 = 160 m. Area = √(160(160 - 100)(160 - 100)(160 - 120)) = √(160 × 60 × 60 × 40). 160 × 40 = 6400 and 60 × 60 = 3600, so the product is 6400 × 3600 = 23040000. Area of one triangle = √23040000 = 4800 m². Area of the park = 2 × 4800 = 9600 m².
Read the following and answer the questions that follow: A school's craft club is stitching triangular bunting flags for Sports Day. Every flag is a triangle with two equal sides of 20 cm and a base of 24 cm, and the club plans to make 50 such flags. (a) Find the semi-perimeter of one flag. (b) Using Heron's formula, find the area of cloth in one flag. (c) Find the total cloth needed for 50 flags, in square metres. (d) Find the height of the flag measured from the 24 cm base.
Answer
(a) s = (20 + 20 + 24)/2 = 64/2 = 32 cm. (b) Area = √(32(32 - 20)(32 - 20)(32 - 24)) = √(32 × 12 × 12 × 8). 32 × 8 = 256 and 12 × 12 = 144, so the product is 256 × 144 = 36864. Area = √36864 = 192 cm². (c) Total cloth = 50 × 192 = 9600 cm². Since 1 m² = 10000 cm², total = 9600/10000 = 0.96 m². (d) Using area = (1/2) × base × height with the 24 cm base: 192 = (1/2)(24)h, so 192 = 12h and h = 16 cm.
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