Heron's Formula MCQ Test — Class 9 CBSE

Solution:

The standard formula for the area of a triangle is 1/2 × base × height. This requires knowing the height corresponding to a specific base.

Heron's Formula, Area = √[s(s-a)(s-b)(s-c)], allows calculating the area solely from the lengths of the three sides (a, b, c) and the semi-perimeter (s).

Therefore, it is most useful when the height is not readily available or difficult to calculate, but all three side lengths are known.

Solution:

The perimeter of a triangle is the sum of its three sides: P = a + b + c.

Given side lengths a = 7 cm, b = 8 cm, c = 9 cm, the perimeter P = 7 + 8 + 9 = 24 cm.

The semi-perimeter (s) is half of the perimeter: s = P / 2. — s = (7 + 8 + 9) / 2 = 24 / 2

Therefore, s = 12 cm.

Solution:

According to the triangle inequality theorem, the sum of any two sides of a triangle must be greater than the third side (e.g., a + b > c).

If a + b > c, then a + b + c > 2c, which means 2s > 2c, or s > c. Similarly, s > a and s > b.

This implies that (s-a), (s-b), and (s-c) must all be positive values. If any of these terms were zero or negative, it would mean the sides cannot form a valid triangle.

The area calculated using Heron's formula, being a physical quantity, must always be positive.

Solution:

Given sides a=5 cm, b=12 cm, c=13 cm.

Step 1: Calculate semi-perimeter (s). s = (a + b + c) / 2 = (5 + 12 + 13) / 2 = 30 / 2 = 15 cm. Ravi's s = 15 cm is correct.

Step 2: Calculate (s-a), (s-b), (s-c). — s-a = 15-5 = 10 cm s-b = 15-12 = 3 cm s-c = 15-13 = 2 cm

Ravi's calculations for (s-a), (s-b), (s-c) are correct.

Step 3: Apply Heron's Formula: Area = √[s(s-a)(s-b)(s-c)]. — Area = √(15 × 10 × 3 × 2) = √(30 × 30) = √(900) = 30 cm².

Ravi's multiplication and square root calculation are also correct. Therefore, he made no mistake.

Solution:

Let the sides be a = 10 cm, b = 17 cm, c = 21 cm.

Step 1: Calculate the semi-perimeter (s). — s = (a + b + c) / 2 = (10 + 17 + 21) / 2 = 48 / 2 = 24 cm.

Step 2: Calculate (s-a), (s-b), (s-c). — s-a = 24 - 10 = 14 cm s-b = 24 - 17 = 7 cm s-c = 24 - 21 = 3 cm

Step 3: Apply Heron's Formula: Area = √[s(s-a)(s-b)(s-c)]. — Area = √(24 × 14 × 7 × 3) = √(2 × 12 × 2 × 7 × 7 × 3) = √(2 × 2 × 6 × 2 × 7 × 7 × 3) = √(2^4 × 3^2 × 7^2) = 2^2 × 3 × 7 = 4 × 21 = 84 cm².

Solution:

The triangle inequality theorem states that for a triangle with sides a, b, c, the following must be true: a + b > c, a + c > b, and b + c > a.

Let's check each option:

A) 6, 8, 10: 6+8=14 > 10 (True). 6+10=16 > 8 (True). 8+10=18 > 6 (True). This can form a triangle.

B) 3, 4, 5: 3+4=7 > 5 (True). 3+5=8 > 4 (True). 4+5=9 > 3 (True). This can form a triangle (it's a right-angled triangle).

C) 7, 7, 14: 7+7=14. This is NOT greater than 14. It is equal. Therefore, these sides cannot form a triangle (they would form a straight line).

D) 5, 12, 13: 5+12=17 > 13 (True). 5+13=18 > 12 (True). 12+13=25 > 5 (True). This can form a triangle (it's a right-angled triangle).

Solution:

Let the sides of the triangular park be a = 40 m, b = 24 m, c = 32 m.

Step 1: Calculate the semi-perimeter (s). — s = (a + b + c) / 2 = (40 + 24 + 32) / 2 = 96 / 2 = 48 m.

Step 2: Calculate (s-a), (s-b), (s-c). — s-a = 48 - 40 = 8 m s-b = 48 - 24 = 24 m s-c = 48 - 32 = 16 m

Step 3: Apply Heron's Formula: Area = √[s(s-a)(s-b)(s-c)]. — Area = √(48 × 8 × 24 × 16) = √( (3 × 16) × 8 × (3 × 8) × 16 ) = √( 3^2 × 16^2 × 8^2 )

Area = 3 × 16 × 8 = 384 m².

Solution:

For a right-angled triangle, the two shorter sides (legs) are perpendicular to each other, thus one can be considered the base and the other the corresponding height.

Given sides 6 cm, 8 cm, 10 cm, the hypotenuse is 10 cm. The base and height are 6 cm and 8 cm.

Using standard formula: Area = 1/2 × base × height = 1/2 × 6 × 8 = 24 cm².

Using Heron's Formula: s = (6+8+10)/2 = 12 cm. Area = √[12(12-6)(12-8)(12-10)] = √[12 × 6 × 4 × 2] = √[576] = 24 cm².

Both formulas give the same result. However, for a right-angled triangle, the standard formula is generally simpler and quicker to apply since the height is directly given by one of the sides.

Solution:

Let the sides of the triangular field be a = 3x, b = 5x, and c = 7x.

The perimeter of the field is given as 300 m. So, a + b + c = 300.

Substitute the expressions for the sides: 3x + 5x + 7x = 300.

Combine like terms: 15x = 300.

Solve for x: x = 300 / 15 = 20 m.

The shortest side is 3x. So, shortest side = 3 × 20 = 60 m.

Solution:

Heron's Formula is a general formula for finding the area of *any* triangle when all three side lengths are known. It is not restricted to only equilateral triangles; it works for scalene, isosceles, and right-angled triangles as well.

Calculating semi-perimeter (s) as (a+b+c)/2 is correct.

The expression s(s-a)(s-b)(s-c) must indeed be non-negative (specifically positive for a non-degenerate triangle) because you cannot take the square root of a negative number in real-number geometry.

Area, being a physical measurement, must always be a positive value.