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

The traffic signal board is an equilateral triangle. Let the side length of the triangle be denoted by 'a'.

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

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

According to Heron's formula, the area of a triangle is given by $Area\; =\; \backslash sqrt\{s(s-a)(s-b)(s-c)\}$.

Since the triangle is equilateral, all sides are equal: a = b = c.

Substituting these into Heron's formula:

$$Area\; =\; \backslash sqrt\{\backslash frac\{3a\}\{2\}\backslash left(\backslash frac\{3a\}\{2\}-a\backslash right)\backslash left(\backslash frac\{3a\}\{2\}-a\backslash right)\backslash left(\backslash frac\{3a\}\{2\}-a\backslash right)\}$$

$$Area\; =\; \backslash sqrt\{\backslash frac\{3a\}\{2\}\backslash left(\backslash frac\{a\}\{2\}\backslash right)\backslash left(\backslash frac\{a\}\{2\}\backslash right)\backslash left(\backslash frac\{a\}\{2\}\backslash right)\}$$

$$Area\; =\; \backslash sqrt\{\backslash frac\{3a^4\}\{16\}\}\; =\; \backslash frac\{a^2\}\{4\}\backslash sqrt\{3\}$$

So, the area of the signal board is $\backslash frac\{a^2\}\{4\}\backslash sqrt\{3\}$ square units.

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

Perimeter = 180 cm

Since it's an equilateral triangle, each side is $a\; =\; \backslash frac\{180\}\{3\}\; =\; 60$ cm.

Using the formula derived above, the area of the signal board is:

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

Answer: The area of the signal board is $900\backslash sqrt\{3\}$ cm².

The side walls are triangular in shape with side lengths a = 122 m, b = 120 m, and c = 22 m.

First, we need to find the area of the triangular wall using Heron's formula.

Calculate the semi-perimeter (s):

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

Now, apply Heron's formula for the area:

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

$$Area\; =\; \backslash sqrt\{132(132-122)(132-120)(132-22)\}\; \backslash text\{\; m\}^2$$

$$Area\; =\; \backslash sqrt\{132\; \backslash times\; 10\; \backslash times\; 12\; \backslash times\; 110\}\; \backslash text\{\; m\}^2$$

To simplify the square root:

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

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

The area of the triangular wall is 1320 m².

The earning from advertisements is Rs 5000 per m² per year.

Rent for 1 m² for 1 year = Rs 5000.

Rent for 1 m² for 1 month = $\backslash frac\{5000\}\{12\}$ Rs.

The company hired the wall for 3 months. So, the rent for 3 months for 1 m² is $\backslash frac\{5000\}\{12\}\; \backslash times\; 3$ Rs.

The total rent for the complete wall (1320 m²) for 3 months is:

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

$$Rent\; =\; \backslash frac\{5000\}\{4\}\; \backslash times\; 1320\; =\; 1250\; \backslash times\; 1320\; =\; 1,650,000$$

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

The side wall is a triangle with side lengths a = 15 m, b = 11 m, and c = 6 m.

We need to find the area painted in colour, which is the area of the triangle.

First, calculate the semi-perimeter (s):

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

Now, apply Heron's formula:

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

$$Area\; =\; \backslash sqrt\{16(16-15)(16-11)(16-6)\}\; \backslash text\{\; m\}^2$$

$$Area\; =\; \backslash sqrt\{16\; \backslash times\; 1\; \backslash times\; 5\; \backslash times\; 10\}\; \backslash text\{\; m\}^2$$

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

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

Let the sides of the triangle be a, b, and c. We are given:

a = 18 cm

b = 10 cm

Perimeter = 42 cm

The perimeter of a triangle is the sum of its sides: $a\; +\; b\; +\; c\; =\; 42$.

Substitute the given values:

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

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

Solve for c:

$c\; =\; 42\; -\; 28\; =\; 14\; \backslash text\{\; cm\}$

Now we have all three sides of the triangle: 18 cm, 10 cm, and 14 cm.

Next, calculate the semi-perimeter (s):

$s\; =\; \backslash frac\{Perimeter\}\{2\}\; =\; \backslash frac\{42\}\{2\}\; =\; 21\; \backslash text\{\; cm\}$

Apply Heron's formula to find the area:

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

$$Area\; =\; \backslash sqrt\{21(21-18)(21-10)(21-14)\}\; \backslash text\{\; cm\}^2$$

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

To simplify the square root, find the prime factors:

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

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

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