Class 9 Maths: Heron’s Formula – Extra Questions with Answers
Q. 1 The area of an isosceles triangle with a base of 3 cm and sides of 5 cm is ______ .
Check Solution
Ans: D
Solution: The height of the triangle can be found using the Pythagorean theorem: h = sqrt(5^2 – (3/2)^2) = sqrt(25 – 2.25) = sqrt(22.75). The area is (1/2) * base * height = (1/2) * 3 * sqrt(22.75) = 1.5 * sqrt(22.75) ≈ 7.14.
Answer: 7.14
Q. 2 The perimeter of a field in the form of an equilateral triangle is 48 cm. Determine the area of the field.
Check Solution
Ans: C
Solution:
Side length = Perimeter / 3 = 48 cm / 3 = 16 cm
Area = (√3 / 4) * side² = (√3 / 4) * 16² = (√3 / 4) * 256 = 64√3 cm²
Answer: 64√3 cm²
Q. 3 The sides of a triangle are in the ratio 5:12:13. If the perimeter of the triangle is 90 cm, find the area of the triangle.
Check Solution
Ans: D
Solution:
Let the sides be 5x, 12x, and 13x.
The perimeter is 5x + 12x + 13x = 30x = 90 cm, so x = 3 cm.
The sides are 15 cm, 36 cm, and 39 cm.
Since 5^2 + 12^2 = 13^2 (or 15^2 + 36^2 = 39^2), this is a right triangle.
The area is (1/2) * 15 * 36 = 270 cm^2.
Answer: 270 cm^2
Q. 4 The base of an isosceles triangle measures 20 cm and its area is 120 $cm^2$. Find its perimeter.
Check Solution
Ans: D
Solution:
Let the base be $b=20$ cm and the area be $A=120$ $cm^2$.
The height $h$ is found by $A = \frac{1}{2}bh$, so $120 = \frac{1}{2}(20)h$, which gives $h = 12$ cm.
Since the triangle is isosceles, the height bisects the base.
We have two right triangles with legs of length $10$ cm and $12$ cm.
The hypotenuse (side) is $s = \sqrt{10^2 + 12^2} = \sqrt{100+144} = \sqrt{244} = 2\sqrt{61}$ cm.
The perimeter is $2s + b = 2(2\sqrt{61}) + 20 = 20 + 4\sqrt{61}$ cm.
Answer: $20+4\sqrt{61}$ cm
Q. 5 A decorative banner is made of 12 triangular pieces of fabric. Six triangles are red and six are blue. Each triangular piece has sides of 15 cm, 25 cm, and 25 cm. What is the total area of blue fabric required for the banner, in square centimeters?
Check Solution
Ans: D
Solution:
The triangle is isosceles with sides 15, 25, 25. Let the base be 15. The height can be found using the Pythagorean theorem: h^2 + (15/2)^2 = 25^2, so h = sqrt(25^2 – 7.5^2) = sqrt(625 – 56.25) = sqrt(568.75) = 23.85 cm.
Area of one triangle = 0.5 * 15 * 23.85 = 178.875 cm^2.
Area of 6 blue triangles = 6 * 178.875 = 1073.25 cm^2.
Answer: 1073.25
Q. 6 A right triangle has legs of length $x$ and $x$. If the area of the triangle is 18 cm$^2$, what is the length of the hypotenuse?
Check Solution
Ans: D
Solution:
The area of a right triangle is $\frac{1}{2} \times base \times height$. Since the legs have length $x$, the area is $\frac{1}{2}x^2 = 18$.
Then $x^2 = 36$, so $x=6$.
The hypotenuse is $\sqrt{x^2 + x^2} = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2}$.
Answer: $6\sqrt{2}$
Q. 7 A triangle and a parallelogram have the same area. The sides of the triangle are 13cm, 20cm, and 21cm. The parallelogram shares a base with the triangle, and the base of the parallelogram is 21cm. What is the height of the parallelogram?
Check Solution
Ans: C
Solution:
1. Calculate the area of the triangle using Heron’s formula.
s = (13 + 20 + 21) / 2 = 27
Area = sqrt(27 * (27-13) * (27-20) * (27-21)) = sqrt(27 * 14 * 7 * 6) = sqrt(15876) = 126 cm^2
2. Since the triangle and parallelogram have the same area, the area of the parallelogram is also 126 cm^2.
3. The area of a parallelogram is base * height. We know the base (21 cm) and the area (126 cm^2).
4. Height = Area / base = 126 cm^2 / 21 cm = 6 cm
Answer: 6 cm
Q. 8 If the area of an equilateral triangle is $9\sqrt{3}$ cm$^2$, what is the length of one side?
Check Solution
Ans: D
Solution:
Let $s$ be the side length of the equilateral triangle.
The area of an equilateral triangle is given by the formula $A = \frac{s^2\sqrt{3}}{4}$.
Given $A = 9\sqrt{3}$ cm$^2$, we have
$\frac{s^2\sqrt{3}}{4} = 9\sqrt{3}$
$s^2 = \frac{4 \cdot 9\sqrt{3}}{\sqrt{3}}$
$s^2 = 4 \cdot 9 = 36$
$s = \sqrt{36} = 6$ cm
Answer: 6 cm
Q. 9 The sides of a triangle are 15 cm, 20 cm, and 25 cm. Find the length of its shortest altitude.
Check Solution
Ans: A
Solution: The triangle is a right triangle (15^2 + 20^2 = 25^2). The area is (1/2)*15*20 = 150 cm^2. The shortest altitude is to the longest side (25 cm). Thus, 150 = (1/2)*25*h => h = 12 cm.
Answer: 12 cm
Q. 10 If the sides of a triangular field measure 50 m, 35 m, and 15 m, then find the approximate cost of levelling it at Rs 8 per $m^2$.
Check Solution
Ans: D
Solution:
Semi-perimeter, s = (50+35+15)/2 = 50 m
Area = sqrt(s(s-a)(s-b)(s-c)) = sqrt(50*0*15*35) = 0. Area is not possible for these dimensions. The provided side lengths do not form a valid triangle, as 15 + 35 = 50. Therefore, area calculation is not possible. Since this would result in a zero area and zero leveling costs, we cannot calculate the cost.
Answer: Not possible
Next Topic: Surface Areas and Volumes
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