Important Formulas, Rules & Key Points

• Heron's Formula for Area of Triangle: $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$
• Semi-perimeter ($s$): $s = \frac{a + b + c}{2}$, where $a$, $b$, and $c$ are the side lengths of the triangle.
• Equilateral Triangle Area: $\text{Area} = \frac{\sqrt{3}}{4} a^2$
• Always ensure that all side lengths are in the same unit of measurement before calculation.

Question 1:

A traffic signal board, indicating 'SCHOOL AHEAD', is an equilateral triangle with side 'a'. Find the area of the signal board, using Heron's formula. If its perimeter is 180 cm, what will be the area of the signal board?

Solution:

1. For an equilateral triangle with side $a$:

Semi-perimeter ($s$) = $\frac{a + a + a}{2} = \frac{3a}{2}$

2. Applying Heron's Formula:

$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{3a}{2} \left(\frac{3a}{2} - a\right) \left(\frac{3a}{2} - a\right) \left(\frac{3a}{2} - a\right)}$

$\text{Area} = \sqrt{\frac{3a}{2} \cdot \frac{a}{2} \cdot \frac{a}{2} \cdot \frac{a}{2}} = \frac{\sqrt{3}}{4} a^2$

3. Given Perimeter = $180\text{ cm}$:

$3a = 180 \implies a = 60\text{ cm}$

4. Calculating actual area:

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

Question 2:

The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m. The advertisements yield an earning of ₹ 5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?

Solution:

1. Let the sides be $a = 122\text{ m}$, $b = 22\text{ m}$, $c = 120\text{ m}$.

Semi-perimeter ($s$) = $\frac{122 + 22 + 120}{2} = \frac{264}{2} = 132\text{ m}$.

2. Applying Heron's Formula:

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

$\text{Area} = \sqrt{132 \times 10 \times 110 \times 12} = \sqrt{(12 \times 11) \times 10 \times (11 \times 10) \times 12}$

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

3. Calculation of Rent:

Rent for 1 m² per year = ₹ 5000

Rent for 1320 m² for 3 months ($\frac{3}{12}$ year) = $1320 \times 5000 \times \frac{3}{12} = ₹ 16,50,000$

Question 3:

There is a slide in a park. One of its side walls has been painted in some colour with a message "KEEP THE PARK GREEN AND CLEAN". If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Solution:

1. Let sides be $a = 15\text{ m}$, $b = 11\text{ m}$, $c = 6\text{ m}$.

Semi-perimeter ($s$) = $\frac{15 + 11 + 6}{2} = \frac{32}{2} = 16\text{ m}$.

2. Applying Heron's Formula:

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

$\text{Area} = \sqrt{16 \times 1 \times 5 \times 10} = \sqrt{16 \times 50} = \sqrt{16 \times 25 \times 2} = 4 \times 5 \sqrt{2} = 20\sqrt{2}\text{ m}^2$

Question 4:

Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.

Solution:

1. Given $a = 18\text{ cm}$, $b = 10\text{ cm}$, Perimeter = $42\text{ cm}$.

Third side $c = 42 - (18 + 10) = 42 - 28 = 14\text{ cm}$.

2. Semi-perimeter ($s$) = $\frac{42}{2} = 21\text{ cm}$.

3. Applying Heron's Formula:

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

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

Question 5:

Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.

Solution:

1. Let sides be $12x$, $17x$, and $25x$.

Perimeter = $12x + 17x + 25x = 54x = 540 \implies x = 10$.

Hence, $a = 120\text{ cm}$, $b = 170\text{ cm}$, $c = 250\text{ cm}$.

2. Semi-perimeter ($s$) = $\frac{540}{2} = 270\text{ cm}$.

3. Applying Heron's Formula:

$\text{Area} = \sqrt{270(270 - 120)(270 - 170)(270 - 250)}$

$\text{Area} = \sqrt{270 \times 150 \times 100 \times 20} = \sqrt{81000000} = 9000\text{ cm}^2$

Question 6:

An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

Solution:

1. Equal sides $a = 12\text{ cm}$, $b = 12\text{ cm}$. Perimeter = $30\text{ cm}$.

Third side $c = 30 - (12 + 12) = 6\text{ cm}$.

2. Semi-perimeter ($s$) = $\frac{30}{2} = 15\text{ cm}$.

3. Applying Heron's Formula:

$\text{Area} = \sqrt{15(15 - 12)(15 - 12)(15 - 6)} = \sqrt{15 \times 3 \times 3 \times 9} = 3 \times 3 \sqrt{15} = 9\sqrt{15}\text{ cm}^2$

Aptitude Corner: Quick Calculation Tips

Factorization Technique: Instead of multiplying large numbers under the square root, split every term into prime factors or perfect squares. This saves calculation time and minimizes mistakes!

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