Class 9 Maths: Heron’s Formula – Extra Questions with Answers

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

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

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

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

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

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: $6\sqrt{2}$

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

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

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

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