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

For an equilateral triangle, all sides are equal. Let the side length be 'a'.

The perimeter of the triangle is given by P = a + a + a = 3a.

The semi-perimeter (s) is half the perimeter: $s\; =\; \backslash frac\{3a\}\{2\}$.

Using Heron's formula for the area of a triangle, Area = $\backslash sqrt\{s(s-a)(s-b)(s-c)\}$.

Since it's an equilateral triangle, a = b = c. So, the formula becomes:

$$Area\; =\; \backslash sqrt\{s(s-a)(s-a)(s-a)\}\; =\; \backslash sqrt\{s(s-a)^3\}$$

Substitute $s\; =\; \backslash frac\{3a\}\{2\}$:

$$Area\; =\; \backslash sqrt\{\backslash frac\{3a\}\{2\}\backslash left(\backslash frac\{3a\}\{2\}-a\backslash right)^3\}\; =\; \backslash sqrt\{\backslash frac\{3a\}\{2\}\backslash left(\backslash frac\{a\}\{2\}\backslash right)^3\}\; =\; \backslash sqrt\{\backslash frac\{3a\}\{2\}\; \backslash times\; \backslash frac\{a^3\}\{8\}\}\; =\; \backslash sqrt\{\backslash frac\{3a^4\}\{16\}\}$$

$$Area\; =\; \backslash frac\{a^2\}\{4\}\backslash sqrt\{3\}\; \backslash text\{\; sq\; units\}$$

This is the general formula for the area of an equilateral triangle using Heron's formula.

Now, we are given that the perimeter is 180 cm.

Perimeter = 3a = 180 cm.

Therefore, the length of each side is $a\; =\; \backslash frac\{180\}\{3\}\; =\; 60$ cm.

Using the derived area formula for an equilateral triangle:

$$Area\; =\; \backslash frac\{a^2\}\{4\}\backslash sqrt\{3\}\; =\; \backslash frac\{(60)^2\}\{4\}\backslash sqrt\{3\}\; =\; \backslash frac\{3600\}\{4\}\backslash sqrt\{3\}\; =\; 900\backslash sqrt\{3\}\; \backslash text\{\; cm\}^2$$

Answer: The area of the signal board is $\backslash frac\{a^2\}\{4\}\backslash sqrt\{3\}$ sq units. If the perimeter is 180 cm, the area is $900\backslash sqrt\{3\}$ cm².

First, we need to calculate the area of the triangular wall using Heron's formula. The sides of the triangle are a = 122 m, b = 120 m, and c = 22 m.

Calculate the semi-perimeter (s):

$s\; =\; \backslash frac\{a+b+c\}\{2\}\; =\; \backslash frac\{122\; +\; 120\; +\; 22\}\{2\}\; =\; \backslash frac\{264\}\{2\}\; =\; 132$ m.

Now, apply Heron's formula for the area:

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

$$Area\; =\; \backslash sqrt\{132(132-122)(132-120)(132-22)\}$$

$$Area\; =\; \backslash sqrt\{132\; \backslash times\; 10\; \backslash times\; 12\; \backslash times\; 110\}$$

To simplify the calculation, we can factorize the numbers:

$$Area\; =\; \backslash sqrt\{(12\; \backslash times\; 11)\; \backslash times\; 10\; \backslash times\; 12\; \backslash times\; (10\; \backslash times\; 11)\}\; =\; \backslash sqrt\{12^2\; \backslash times\; 10^2\; \backslash times\; 11^2\}\; =\; 12\; \backslash times\; 10\; \backslash times\; 11\; =\; 1320\; \backslash text\{\; m\}^2$$

The area of the triangular wall is 1320 m².

The earning rate is Rs 5000 per m² per year.

Rent per m² per month = $\backslash frac\{5000\}\{12\}$ Rs.

The company hired the wall for 3 months. So, the total rent paid is:

$$Rent\; =\; \backslash left(\; \backslash frac\{5000\}\{12\}\; \backslash right)\; \backslash times\; 1320\; \backslash times\; 3$$

$$Rent\; =\; 5000\; \backslash times\; 1320\; \backslash times\; \backslash frac\{3\}\{12\}\; =\; 5000\; \backslash times\; 1320\; \backslash times\; \backslash frac\{1\}\{4\}$$

$$Rent\; =\; 5000\; \backslash times\; 330\; =\; 1,650,000$$ Rs.

Answer: The company paid Rs 16,50,000 as rent.

The side wall is a triangle with sides a = 15 m, b = 11 m, and c = 6 m. We need to find the area of this triangle using Heron's formula.

First, calculate the semi-perimeter (s):

$s\; =\; \backslash frac\{a+b+c\}\{2\}\; =\; \backslash frac\{15+11+6\}\{2\}\; =\; \backslash frac\{32\}\{2\}\; =\; 16$ m.

Now, apply Heron's formula:

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

$$Area\; =\; \backslash sqrt\{16(16-15)(16-11)(16-6)\}$$

$$Area\; =\; \backslash sqrt\{16\; \backslash times\; 1\; \backslash times\; 5\; \backslash times\; 10\}$$

$$Area\; =\; \backslash sqrt\{16\; \backslash times\; 50\}\; =\; \backslash sqrt\{16\; \backslash times\; 25\; \backslash times\; 2\}\; =\; 4\; \backslash times\; 5\; \backslash sqrt\{2\}\; =\; 20\backslash sqrt\{2\}\; \backslash text\{\; m\}^2$$

Answer: The area painted in colour is $20\backslash sqrt\{2\}$ m².

We are given two sides of the triangle, a = 18 cm and b = 10 cm, and the perimeter P = 42 cm.

First, we need to find the length of the third side (c). The perimeter is the sum of all sides: P = a + b + c.

$42\; =\; 18\; +\; 10\; +\; c$

$42\; =\; 28\; +\; c$

$c\; =\; 42\; -\; 28\; =\; 14$ cm.

Now we have all three sides: a = 18 cm, b = 10 cm, and c = 14 cm.

Next, calculate the semi-perimeter (s):

$s\; =\; \backslash frac\{a+b+c\}\{2\}\; =\; \backslash frac\{42\}\{2\}\; =\; 21$ cm.

Apply Heron's formula to find the area:

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

$$Area\; =\; \backslash sqrt\{21(21-18)(21-10)(21-14)\}$$

$$Area\; =\; \backslash sqrt\{21\; \backslash times\; 3\; \backslash times\; 11\; \backslash times\; 7\}$$

To simplify, factorize the numbers:

$$Area\; =\; \backslash sqrt\{(3\; \backslash times\; 7)\; \backslash times\; 3\; \backslash times\; 11\; \backslash times\; 7\}\; =\; \backslash sqrt\{3^2\; \backslash times\; 7^2\; \backslash times\; 11\}$$

$$Area\; =\; 3\; \backslash times\; 7\; \backslash sqrt\{11\}\; =\; 21\backslash sqrt\{11\}\; \backslash text\{\; cm\}^2$$

Answer: The area of the triangle is $21\backslash sqrt\{11\}$ cm².