GSEB Class 9 Maths Solutions Chapter 12 Herons Formula Exercise 12.1

Gujarat Board Textbook Solutions Class 9 Maths Chapter 12 Heron's Formula Ex 12.7

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?
Answer: According to the question, the semi-perimeter of triangle ABC is given by:
\( s = \frac {a + a + a}{2} = \frac {3a}{2} \)
Now, we calculate the area using Heron's formula:
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
\( = \sqrt{\frac {3a}{2}(\frac {3a}{2}-a)(\frac {3a}{2}-a)(\frac {3a}{2}-a)} \)
\( = \sqrt{\frac {3a}{2} \times \frac {a}{2} \times \frac {a}{2} \times \frac {a}{2}} \)
\( = \sqrt{3 \times \frac {a^4}{16}} \)
\( = \frac{\sqrt{3}}{4} a^2 \) sq. units
Next, the perimeter of this triangle is 180 cm.
\( 3a = 180 \)
\( a = \frac {180}{3} \)
\( a = 60 \) cm
Now, we find the area of the signal board using the side length:
Area \( = \frac{\sqrt{3}}{4} (60)^2 \)
\( = \frac{\sqrt{3}}{4} \times 3600 \)
\( = 900\sqrt{3} \) cm\(^2\)
Alternatively, we can calculate the semi-perimeter:
\( s = \frac {3a}{2} = \frac {3}{2} \times 60 = 90 \) cm
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
\( = \sqrt{90(90-60)(90-60)(90-60)} \)
\( = \sqrt{90 \times 30 \times 30 \times 30} \)
\( = \sqrt{3 \times 30 \times 30 \times 30 \times 30} \)
\( = 900\sqrt{3} \) cm\(^2\)

In simple words: First, we use the side 'a' to find the area formula for an equilateral triangle. Then, we use the given perimeter to find the value of 'a'. Finally, we put 'a' into the area formula to get the exact area of the signal board.

Exam Tip: Remember that Heron's formula applies to all triangles, but for an equilateral triangle, a simpler formula \( \frac{\sqrt{3}}{4} a^2 \) can be used directly if the side length is known, making calculations quicker.

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 Rs. 5000 per m\(^2\) per year. A company hired one of its walls for 3 months. How much rent did it pay?
Answer: Given the sides of the triangular wall are \( a = 122 \) m, \( b = 22 \) m, and \( c = 120 \) m.
First, we calculate the semi-perimeter (s):
\( s = \frac {a + b + c}{2} \)
\( s = \frac {122 + 22 + 120}{2} = \frac {264}{2} = 132 \) m
Now, we find the area of the wall using Heron's formula:
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
\( = \sqrt{132(132-122)(132-22)(132-120)} \)
\( = \sqrt{132 \times 10 \times 110 \times 12} \)
\( = \sqrt{(11 \times 12) \times 10 \times (11 \times 10) \times 12} \)
\( = \sqrt{11 \times 12 \times 10 \times 11 \times 10 \times 12} \)
\( = \sqrt{11^2 \times 12^2 \times 10^2} \)
\( = 11 \times 12 \times 10 = 1320 \) m\(^2\)
The company hired the wall for 3 months. The rate of earning is Rs. 5000 per m\(^2\) per year.
To calculate the rent for 3 months, we convert the yearly rate to a monthly rate:
Yearly rate = Rs. 5000 per m\(^2\)
Monthly rate \( = \frac {5000}{12} \) per m\(^2\)
Rent for 3 months \( = (\text{Monthly rate}) \times (\text{Number of months}) \times (\text{Area}) \)
Rent paid \( = \frac {5000}{12} \times 3 \times 1320 \)
\( = 5000 \times \frac {3}{12} \times 1320 \)
\( = 5000 \times \frac {1}{4} \times 1320 \)
\( = 5000 \times 330 \)
\( = \text{Rs. } 16,50,000 \)

In simple words: First, we use Heron's formula to find the wall's area. Then, we calculate how much money is earned per square meter for 3 months. Finally, we multiply this rate by the total area to find the total rent paid.

Exam Tip: Pay close attention to unit conversions, especially when dealing with time (e.g., years to months) and money, as this is a common source of errors in such problems.

Question 3. There is a slide in a park. One of its side walls has been painted in some color 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 color.
Answer: Given the sides of the slide wall are \( a = 15 \) m, \( b = 11 \) m, and \( c = 6 \) m.
First, we calculate the semi-perimeter (s):
\( s = \frac {a + b + c}{2} \)
\( s = \frac {15 + 11 + 6}{2} = \frac {32}{2} = 16 \) m
Now, we find the area painted in color using Heron's formula:
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
\( = \sqrt{16(16-15)(16-11)(16-6)} \)
\( = \sqrt{16 \times 1 \times 5 \times 10} \)
\( = \sqrt{16 \times 50} \)
\( = \sqrt{4 \times 4 \times 5 \times 5 \times 2} \)
\( = 4 \times 5 \times \sqrt{2} \)
\( = 20\sqrt{2} \) m\(^2\)

In simple words: We first find the semi-perimeter using the given side lengths. Then, we use Heron's formula to calculate the exact area of the triangular wall that was painted.

Exam Tip: Always double-check your calculations when simplifying square roots, looking for perfect square factors to make the number outside the root as large as possible.

Question 4. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Answer: Given two sides of the triangle are \( a = 18 \) cm, \( b = 10 \) cm, and the perimeter is 42 cm.
We know that the perimeter of a triangle is the sum of its three sides:
\( a + b + c = 42 \)
\( 18 + 10 + c = 42 \)
\( 28 + c = 42 \)
\( c = 42 - 28 \)
\( c = 14 \) cm
Now, we calculate the semi-perimeter (s):
\( s = \frac {\text{perimeter}}{2} = \frac {42}{2} = 21 \) cm
Now, we find the area of the triangle using Heron's formula:
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
\( = \sqrt{21(21-18)(21-10)(21-14)} \)
\( = \sqrt{21 \times 3 \times 11 \times 7} \)
\( = \sqrt{(3 \times 7) \times 3 \times 11 \times 7} \)
\( = \sqrt{3^2 \times 7^2 \times 11} \)
\( = 3 \times 7 \times \sqrt{11} \)
\( = 21\sqrt{11} \) cm\(^2\)
In simple words: First, we use the given two sides and the perimeter to find the length of the third side. Then, we calculate the semi-perimeter. Finally, we apply Heron's formula to get the triangle's area.

Exam Tip: Remember to always find the third side first when the perimeter and two sides are given, as this is a necessary step before applying Heron's formula.

Question 5. Sides of a triangle are in the ratio of 12: 17 : 25 and its perimeter is 540 cm. Find its area.
Answer: Let the sides of the triangle be \( 12x \) cm, \( 17x \) cm, and \( 25x \) cm.
The perimeter is given as 540 cm.
Perimeter \( = 12x + 17x + 25x = 54x \) cm
So, \( 54x = 540 \)
\( x = \frac {540}{54} \)
\( x = 10 \)
Now, we find the actual lengths of the sides:
\( a = 12 \times 10 = 120 \) cm
\( b = 17 \times 10 = 170 \) cm
\( c = 25 \times 10 = 250 \) cm
Next, we calculate the semi-perimeter (s):
\( s = \frac {a + b + c}{2} = \frac {120 + 170 + 250}{2} = \frac {540}{2} = 270 \) cm
Now, we find the area of the triangle using Heron's formula:
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
\( = \sqrt{270(270-120)(270-170)(270-250)} \)
\( = \sqrt{270 \times 150 \times 100 \times 20} \)
To simplify the square root, we can factorize the numbers:
\( = \sqrt{(9 \times 30) \times (5 \times 30) \times (5 \times 20) \times 20} \)
\( = \sqrt{9 \times 30^2 \times 5^2 \times 20^2} \)
\( = 3 \times 30 \times 5 \times 20 \)
\( = 9000 \) cm\(^2\)
In simple words: We first use the ratio and perimeter to find the actual lengths of the triangle's sides. Then, we calculate the semi-perimeter. Finally, we apply Heron's formula to determine the triangle's total area.

Exam Tip: When given side ratios and a perimeter, always use a variable (like 'x') to find the exact side lengths before proceeding with area calculations. This prevents errors that might occur from misinterpreting the ratio.

Question 6. A triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Answer: Given that the perimeter of the triangle is 30 cm, and each of the equal sides is 12 cm.
This means it is an isosceles triangle with two sides equal.
Let \( a = 12 \) cm, \( b = 12 \) cm. Let the third side be \( c \).
Perimeter \( = a + b + c = 30 \)
\( 12 + 12 + c = 30 \)
\( 24 + c = 30 \)
\( c = 30 - 24 \)
\( c = 6 \) cm
Now, we calculate the semi-perimeter (s):
\( s = \frac {\text{perimeter}}{2} = \frac {30}{2} = 15 \) cm
Next, we find the area of the triangle using Heron's formula:
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
\( = \sqrt{15(15-12)(15-12)(15-6)} \)
\( = \sqrt{15 \times 3 \times 3 \times 9} \)
\( = \sqrt{(3 \times 5) \times 3 \times 3 \times (3 \times 3)} \)
\( = \sqrt{3^2 \times 3^2 \times 3 \times 5} \)
\( = 3 \times 3 \times \sqrt{3 \times 5} \)
\( = 9\sqrt{15} \) cm\(^2\)

In simple words: First, we use the perimeter and the lengths of the two equal sides to find the length of the third side. Then, we calculate the semi-perimeter. Finally, we use Heron's formula to find the triangle's area.

Exam Tip: For isosceles triangles, you can also draw an altitude to the unequal side and use Pythagoras theorem to find its height, then apply the formula \( \frac{1}{2} \times \text{base} \times \text{height} \) as an alternative method.