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Rbse Solutions for Class 9 Maths Chapter 10 Exercise 10.1 | HERON’S FORMULA

Aman Singh07 mins
Rbse Solutions for Class 9 Maths Chapter 10 Exercise 10.1 | HERON’S FORMULA

Rbse Solutions for Class 9 Maths Chapter 10 Exercise 10.1

Master Class 9 Heron’s Formula with step-by-step NCERT solutions for Exercise 10.1. Learn to calculate triangle area using side lengths and the semi-perimeter $s$. Essential for exams!

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?

Traffic Signal Board (Equilateral Triangle)

Part 1: Area using Heron’s formula (side ‘a’)

  1. Sides: $a, a, a$
  2. Semi-Perimeter ($s$):$$s = \frac{a+a+a}{2} = \frac{3a}{2}$$
  3. Area ($A$):$$A = \sqrt{s(s-a)(s-a)(s-a)}$$$$A = \sqrt{\frac{3a}{2} \left(\frac{3a}{2} – a\right) \left(\frac{3a}{2} – a\right) \left(\frac{3a}{2} – a\right)}$$$$A = \sqrt{\frac{3a}{2} \left(\frac{a}{2}\right) \left(\frac{a}{2}\right) \left(\frac{a}{2}\right)}$$$$A = \sqrt{\frac{3a^4}{16}} = \frac{\sqrt{3} \cdot \sqrt{a^4}}{\sqrt{16}}$$$$\mathbf{A = \frac{\sqrt{3}}{4} a^2 \text{ square units}}$$

Part 2: Area if Perimeter is $180 \text{ cm}$

  1. Find side ($a$):$$\text{Perimeter} = 3a$$$$180 \text{ cm} = 3a \implies a = 60 \text{ cm}$$
  2. Semi-Perimeter ($s$):$$s = \frac{180}{2} = 90 \text{ cm}$$
  3. Area ($A$): Using the derived formula or Heron’s:
    • $A = \frac{\sqrt{3}}{4} (60)^2 = \frac{\sqrt{3}}{4} (3600)$$$\mathbf{A = 900\sqrt{3} \text{ cm}^2}$$

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 (see Fig. 10.6). 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?

Flyover Advertisement Wall

Sides: $a = 122 \text{ m}, b = 22 \text{ m}, c = 120 \text{ m}$

Rate: $₹5000 \text{ per } m^2 \text{ per year}$

Time: $3 \text{ months} = \frac{3}{12} = \frac{1}{4} \text{ year}$

Step 1: Find the Area of the Wall

  1. Semi-Perimeter ($s$):$$s = \frac{122 + 22 + 120}{2} = \frac{264}{2} = 132 \text{ m}$$
  2. Area ($A$):$$A = \sqrt{132(132-122)(132-22)(132-120)}$$$$A = \sqrt{132 \times 10 \times 110 \times 12}$$$$A = \sqrt{(12 \times 11) \times 10 \times (11 \times 10) \times 12}$$$$A = \sqrt{10^2 \times 11^2 \times 12^2} = 10 \times 11 \times 12$$$$\text{Area} = \mathbf{1320 \text{ m}^2}$$

Step 2: Calculate the Rent Paid

$$\text{Rent} = \text{Area} \times \text{Rate per } m^2 \text{ per year} \times \text{Time in years}$$

$$\text{Rent} = 1320 \times 5000 \times \frac{1}{4}$$

$$\text{Rent} = 1320 \times 1250$$

$$\mathbf{\text{Rent} = ₹1,650,000}$$

    3.There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEPTHE PARK GREEN AND CLEAN” (see Fig. 10.7 ). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

    Sides: $a = 15 \text{ m}, b = 11 \text{ m}, c = 6 \text{ m}$

    1. Semi-Perimeter ($s$):$$s = \frac{15 + 11 + 6}{2} = \frac{32}{2} = 16 \text{ m}$$
    2. Area ($A$):$$A = \sqrt{16(16-15)(16-11)(16-6)}$$$$A = \sqrt{16 \times 1 \times 5 \times 10}$$$$A = \sqrt{16 \times 50} = \sqrt{16 \times 25 \times 2}$$$$A = 4 \times 5 \sqrt{2}$$$$\mathbf{\text{Area painted} = 20\sqrt{2} \text{ m}^2}$$

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

    Two sides: $a = 18 \text{ cm}, b = 10 \text{ cm}$

    Perimeter: $P = 42 \text{ cm}$

    1. Find the third side ($c$):$$c = P – (a+b) = 42 – (18+10)$$$$c = 42 – 28 = 14 \text{ cm}$$
    2. Semi-Perimeter ($s$):$$s = \frac{42}{2} = 21 \text{ cm}$$
    3. Area ($A$):$$A = \sqrt{21(21-18)(21-10)(21-14)}$$$$A = \sqrt{21 \times 3 \times 11 \times 7}$$$$A = \sqrt{(3 \times 7) \times 3 \times 11 \times 7}$$$$A = \sqrt{3^2 \times 7^2 \times 11} = 3 \times 7 \sqrt{11}$$$$\mathbf{\text{Area} = 21\sqrt{11} \text{ cm}^2}$$

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

    Sides in Ratio $12:17:25$

    Ratio: $12:17:25$

    Perimeter: $P = 540 \text{ cm}$

    1. Find the sides:Let the sides be $12x, 17x, 25x$.$$12x + 17x + 25x = 540$$$$54x = 540 \implies x = 10$$
      • $a = 12 \times 10 = 120 \text{ cm}$
      • $b = 17 \times 10 = 170 \text{ cm}$
      • $c = 25 \times 10 = 250 \text{ cm}$
    2. Semi-Perimeter ($s$):$$s = \frac{540}{2} = 270 \text{ cm}$$
    3. Area ($A$):$$A = \sqrt{270(270-120)(270-170)(270-250)}$$$$A = \sqrt{270 \times 150 \times 100 \times 20}$$$$A = \sqrt{(27 \times 10) \times (15 \times 10) \times 100 \times (2 \times 10)}$$$$A = \sqrt{(27 \times 15 \times 2) \times (10 \times 10 \times 100 \times 10)}$$$$A = \sqrt{810 \times 100000} = \sqrt{81 \times 10 \times 100000} = \sqrt{81 \times 1000000}$$$$A = \sqrt{81} \times \sqrt{1000000} = 9 \times 1000$$$$\mathbf{\text{Area} = 9000 \text{ cm}^2}$$

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

    Perimeter: $P = 30 \text{ cm}$

    Equal sides: $a = 12 \text{ cm}, b = 12 \text{ cm}$

    1. Find the third side ($c$):$$c = P – (a+b) = 30 – (12+12)$$$$c = 30 – 24 = 6 \text{ cm}$$
    2. Semi-Perimeter ($s$):$$s = \frac{30}{2} = 15 \text{ cm}$$
    3. Area ($A$):$$A = \sqrt{15(15-12)(15-12)(15-6)}$$$$A = \sqrt{15 \times 3 \times 3 \times 9}$$$$A = \sqrt{(3 \times 5) \times 3^2 \times 3^2}$$$$A = \sqrt{3^4 \times 5} = 3^2 \sqrt{5}$$$$\mathbf{\text{Area} = 9\sqrt{5} \text{ cm}^2}$$

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