CBSE Class 9 Maths Exemplar Chapter 12 Heron's Formula NCERT Solutions

Solution:

Let the isosceles right triangle be ABC, with the right angle at B. In an isosceles right triangle, the two legs are equal in length. Let AB = BC = x.

The area of a triangle is given by the formula: Area = $\backslash frac\{1\}\{2\}\; \backslash times\; \backslash text\{Base\}\; \backslash times\; \backslash text\{Height\}$.

In this case, Area = $\backslash frac\{1\}\{2\}\; \backslash times\; x\; \backslash times\; x\; =\; \backslash frac\{x^2\}\{2\}$.

We are given that the area is 8 cm². So, we have:

$$\backslash frac\{x^2\}\{2\}\; =\; 8$$

Multiplying both sides by 2, we get:

$$x^2\; =\; 16$$

Taking the square root of both sides, we find the length of the legs:

$x\; =\; \backslash sqrt\{16\}\; =\; 4$ cm.

Now, we need to find the length of the hypotenuse (AC). Using the Pythagorean theorem ($a^2\; +\; b^2\; =\; c^2$):

$$AC^2\; =\; AB^2\; +\; BC^2$$

$$AC^2\; =\; x^2\; +\; x^2$$

$$AC^2\; =\; 4^2\; +\; 4^2$$

$$AC^2\; =\; 16\; +\; 16$$

$$AC^2\; =\; 32$$

Taking the square root to find the length of the hypotenuse:

$AC\; =\; \backslash sqrt\{32\}$ cm.

Therefore, the correct option is (A).

Solution:

Given that the perimeter of an equilateral triangle is 60 m.

Let the length of each side of the equilateral triangle be 'a' meters.

The perimeter of an equilateral triangle is the sum of its three equal sides, so:

$$a\; +\; a\; +\; a\; =\; 60$$

$$3a\; =\; 60$$

Dividing by 3, we find the length of one side:

$a\; =\; \backslash frac\{60\}\{3\}\; =\; 20$ m.

The formula for the area of an equilateral triangle is:

Area = $\backslash frac\{\backslash sqrt\{3\}\}\{4\}\; \backslash times\; (\backslash text\{Side\})^2$

Substituting the side length 'a = 20 m' into the formula:

Area = $\backslash frac\{\backslash sqrt\{3\}\}\{4\}\; \backslash times\; (20)^2$

Area = $\backslash frac\{\backslash sqrt\{3\}\}\{4\}\; \backslash times\; 400$

Area = $100\backslash sqrt\{3\}$ m².

Therefore, the area of the triangle is $100\backslash sqrt\{3\}$ m². The correct option is (D).

Solution:

We are given the lengths of the three sides of a triangle: a = 56 cm, b = 60 cm, and c = 52 cm.

To find the area of the triangle, we can use Heron's formula. First, we need to calculate the semi-perimeter (s) of the triangle, which is half of the perimeter:

$$s\; =\; \backslash frac\{a+b+c\}\{2\}$$

Substitute the given side lengths:

$$s\; =\; \backslash frac\{56\; +\; 60\; +\; 52\}\{2\}$$

$$s\; =\; \backslash frac\{168\}\{2\}$$

$s\; =\; 84$ cm.

Now, we apply Heron's formula for the area of the triangle:

Area = $\backslash sqrt\{s(s-a)(s-b)(s-c)\}$

Substitute the values of s, a, b, and c:

$$Area\; =\; \backslash sqrt\{84(84-56)(84-60)(84-52)\}$$

$$Area\; =\; \backslash sqrt\{84\; \backslash times\; 28\; \backslash times\; 24\; \backslash times\; 32\}$$

To simplify the calculation, we can find the prime factorization of each number under the square root:

$84\; =\; 2^2\; \backslash times\; 3\; \backslash times\; 7$

$28\; =\; 2^2\; \backslash times\; 7$

$24\; =\; 2^3\; \backslash times\; 3$

$32\; =\; 2^5$

Now, multiply these factors together:

$Area\; =\; \backslash sqrt\{(2^2\; \backslash times\; 3\; \backslash times\; 7)\; \backslash times\; (2^2\; \backslash times\; 7)\; \backslash times\; (2^3\; \backslash times\; 3)\; \backslash times\; (2^5)\}$

Combine the powers of the same bases:

$Area\; =\; \backslash sqrt\{2^\{(2+2+3+5)\}\; \backslash times\; 3^\{(1+1)\}\; \backslash times\; 7^\{(1+1)\}\}$

$Area\; =\; \backslash sqrt\{2^\{12\}\; \backslash times\; 3^2\; \backslash times\; 7^2\}$

Now, take the square root:

$Area\; =\; 2^\{12/2\}\; \backslash times\; 3^\{2/2\}\; \backslash times\; 7^\{2/2\}$

$Area\; =\; 2^6\; \backslash times\; 3^1\; \backslash times\; 7^1$

$Area\; =\; 64\; \backslash times\; 3\; \backslash times\; 7$

$Area\; =\; 192\; \backslash times\; 7$

$Area\; =\; 1344$ cm².

Hence, the correct option is (C).