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| Example 1 to 3 (Before Exercise 12.1) | Exercise 12.1 | Example 4 to 6 (Before Exercise 12.2) |
| Exercise 12.2 | ||
Chapter 12 Heron's Formula
Welcome to the solutions guide for Chapter 12: Heron’s Formula. This chapter introduces a remarkably elegant and powerful tool in geometry for determining the area of any triangle when only the lengths of its three sides are known. Named after Hero of Alexandria, a Greek engineer and mathematician, Heron's Formula provides a method to calculate area without any reference to the triangle's altitude or height, which is often difficult or impossible to determine directly in practical scenarios. This makes it an invaluable asset in various fields, including surveying, engineering, and design, where side lengths are more readily measurable than perpendicular heights. The solutions presented here focus on elucidating this formula and demonstrating its wide range of applications through carefully worked examples.
The essence of this chapter revolves around understanding and applying Heron's Formula itself. It states that for any triangle with side lengths denoted by $a$, $b$, and $c$, the area can be calculated using the following expression:
$Area = \sqrt{s(s-a)(s-b)(s-c)}$
Here, the crucial variable '$s$' represents the semi-perimeter of the triangle. The semi-perimeter is simply half the total perimeter of the triangle, calculated as:
$s = \frac{a+b+c}{2}$
The solutions meticulously break down the process of applying this formula into a clear sequence of steps, ensuring students can follow along and replicate the calculation accurately:
- First, calculate the semi-perimeter '$s$' using the lengths of the three sides $a, b, c$.
- Next, compute the difference between the semi-perimeter and each side: $(s-a)$, $(s-b)$, and $(s-c)$.
- Substitute the values of $s$, $(s-a)$, $(s-b)$, and $(s-c)$ into Heron's formula: $\sqrt{s(s-a)(s-b)(s-c)}$.
- Calculate the product of these four terms inside the square root: $s(s-a)(s-b)(s-c)$.
- Finally, determine the square root of this product to obtain the triangle's Area.
The versatility of Heron's formula is highlighted through examples involving various triangle types – scalene, isosceles, and even equilateral. While simpler, specific formulas exist for isosceles and equilateral triangles, the solutions demonstrate that Heron's formula works universally, providing a single, consistent method regardless of the triangle's specific shape, as long as the three side lengths are known.
Furthermore, a significant focus of the solutions lies in extending the application of Heron's formula beyond simple triangles to calculate the area of more complex polygons, particularly quadrilaterals. The standard technique involves strategically dividing the quadrilateral (or other polygon) into constituent triangles using one or more diagonals. If the lengths of all sides and the chosen diagonal(s) are known (or can be calculated), Heron's formula can be applied to find the area of each resulting triangle. The total area of the polygon is then simply the sum of the areas of these individual triangles. Solutions tackle practical word problems, illustrating how to find the area of diverse shapes like parks or fields, sometimes involving calculating associated costs, such as leveling a field at a certain rate per square meter (e.g., $\textsf{₹} 5$ per $m^2$). Clear calculation steps, including the handling of square roots, are provided to ensure mastery of this powerful geometric tool.
Example 1 to 3 (Before Exercise 12.1)
Example 1. Find the area of a triangle, two sides of which are 8 cm and 11 cm and the perimeter is 32 cm (see Fig. 12.6).
Answer:
Given:
Let the sides of the triangle be $a$, $b$, and $c$.
Given two sides: $a = 8$ cm and $b = 11$ cm.
Given perimeter: $P = 32$ cm.
To Find:
The area of the triangle.
Solution:
The perimeter of a triangle is the sum of its three sides.
$P = a + b + c$
Substitute the given values:
$32 = 8 + 11 + c$
$32 = 19 + c$
Now, find the third side $c$:
$c = 32 - 19$
$c = 13$ cm
The third side of the triangle is 13 cm.
Now, calculate the semi-perimeter, $s$.
$s = \frac{\text{Perimeter}}{2}$
$s = \frac{32}{2}$
$s = 16$ cm
We can find the area of the triangle using Heron's formula:
$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$
Substitute the values of $s$, $a$, $b$, and $c$:
$\text{Area} = \sqrt{16(16-8)(16-11)(16-13)}$
$\text{Area} = \sqrt{16(8)(5)(3)}$
$\text{Area} = \sqrt{16 \times 8 \times 15}$
$\text{Area} = \sqrt{128 \times 15}$
$\text{Area} = \sqrt{1920}$
To simplify the square root, find the prime factors of 1920:
$\begin{array}{c|cc} 2 & 1920 \\ \hline 2 & 960 \\ \hline 2 & 480 \\ \hline 2 & 240 \\ \hline 2 & 120 \\ \hline 2 & 60 \\ \hline 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$
$1920 = 2^2 \times 2^2 \times 2^2 \times 2 \times 3 \times 5$
$1920 = 16 \times 120 = 16 \times 4 \times 30 = 64 \times 30$
$\text{Area} = \sqrt{64 \times 30}$
$\text{Area} = \sqrt{64} \times \sqrt{30}$
$\text{Area} = 8\sqrt{30}$
The area of the triangle is $8\sqrt{30}$ cm$^2$.
The area of the triangle is $8\sqrt{30}$ cm$^2$.
Example 2. A triangular park ABC has sides 120m, 80m and 50m (see Fig. 12.7). A gardener Dhania has to put a fence all around it and also plant grass inside. How much area does she need to plant? Find the cost of fencing it with barbed wire at the rate of ₹ 20 per metre leaving a space 3m wide for a gate on one side.
Answer:
Given:
A triangular park ABC with sides:
a = BC = 120 m
b = AC = 80 m
c = AB = 50 m
Rate of fencing = $\textsf{₹}$ 20 per metre.
Width of the gate to be left unfenced = 3 m.
To Find:
1. The area of the park where grass needs to be planted.
2. The cost of fencing the park.
Solution:
1. Area of the park for planting grass:
To find the area of the triangular park, we will use Heron's formula.
First, we need to calculate the semi-perimeter (s) of the triangle.
$s = \frac{a+b+c}{2}$
$s = \frac{120 + 80 + 50}{2} = \frac{250}{2} = 125$ m
Now, we can apply Heron's formula for the area of a triangle:
Area = $\sqrt{s(s-a)(s-b)(s-c)}$
Let's calculate the terms $(s-a)$, $(s-b)$, and $(s-c)$:
$s-a = 125 - 120 = 5$ m
$s-b = 125 - 80 = 45$ m
$s-c = 125 - 50 = 75$ m
Now substitute these values into the formula:
Area = $\sqrt{125 \times 5 \times 45 \times 75}$
To simplify the calculation, let's break down the numbers into their prime factors:
$125 = 5 \times 5 \times 5$
$5 = 5$
$45 = 3 \times 3 \times 5$
$75 = 3 \times 5 \times 5$
Area = $\sqrt{(5 \times 5 \times 5) \times (5) \times (3 \times 3 \times 5) \times (3 \times 5 \times 5)}$
Group the factors in pairs:
Area = $\sqrt{(5 \times 5) \times (5 \times 5) \times (5 \times 5) \times (3 \times 3) \times 3 \times 5}$
Take the pairs out of the square root:
Area = $5 \times 5 \times 5 \times 3 \sqrt{3 \times 5}$
Area = $375 \sqrt{15} \text{ m}^2$
So, the area Dhania needs to plant is $375 \sqrt{15} \text{ m}^2$.
2. Cost of fencing the park:
First, we need to find the perimeter of the park.
Perimeter = $a + b + c = 120 + 80 + 50 = 250$ m.
A space of 3 m is to be left for a gate. So, the length of the boundary to be fenced is:
Length of fence = Perimeter - Width of the gate
Length of fence = $250 - 3 = 247$ m.
The cost of fencing is given at a rate of $\textsf{₹}$ 20 per metre.
Total cost of fencing = Length of fence $\times$ Rate of fencing
Total cost = $247 \times 20$
Total cost = $\textsf{₹ } 4940$
Final Answer:
1. The area Dhania needs to plant is $375 \sqrt{15} \text{ m}^2$.
2. The cost of fencing the park is $\textsf{₹ } 4940$.
Example 3. The sides of a triangular plot are in the ratio of 3 : 5 : 7 and its perimeter is 300 m. Find its area.
Answer:
Given:
The sides of a triangular plot are in the ratio 3 : 5 : 7.
The perimeter of the plot is 300 m.
To Find:
The area of the triangular plot.
Solution:
Let the sides of the triangular plot be $3x$, $5x$, and $7x$ metres, where $x$ is a positive constant.
The perimeter of the triangle is the sum of its sides.
$\text{Perimeter} = 3x + 5x + 7x$
We are given that the perimeter is 300 m.
$300 = (3+5+7)x$
$300 = 15x$
Now, solve for $x$:
$x = \frac{300}{15}$
$x = 20$
The actual lengths of the sides are:
Side 1 = $3x = 3 \times 20 = 60$ m
Side 2 = $5x = 5 \times 20 = 100$ m
Side 3 = $7x = 7 \times 20 = 140$ m
Let $a=60$ m, $b=100$ m, and $c=140$ m.
The semi-perimeter, $s$, is half of the perimeter.
$s = \frac{\text{Perimeter}}{2}$
$s = \frac{300}{2} = 150$ m
Now, calculate the differences $(s-a)$, $(s-b)$, and $(s-c)$.
$(s-a) = 150 - 60 = 90$ m
$(s-b) = 150 - 100 = 50$ m
$(s-c) = 150 - 140 = 10$ m
Using Heron's formula for the area of the triangle:
$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$
Substitute the values:
$\text{Area} = \sqrt{150 \times 90 \times 50 \times 10}$
Factorise the numbers under the square root:
$\text{Area} = \sqrt{(15 \times 10) \times (9 \times 10) \times (5 \times 10) \times 10}$
$\text{Area} = \sqrt{15 \times 9 \times 5 \times 10^4}$
$\text{Area} = \sqrt{(3 \times 5) \times 3^2 \times 5 \times 10^4}$
$\text{Area} = \sqrt{3^{1+2} \times 5^{1+1} \times 10^4}$
$\text{Area} = \sqrt{3^3 \times 5^2 \times 10^4}$
Group perfect squares:
$\text{Area} = \sqrt{(3^2 \times 5^2 \times 10^4) \times 3}$
$\text{Area} = \sqrt{3^2} \times \sqrt{5^2} \times \sqrt{10^4} \times \sqrt{3}$
$\text{Area} = 3 \times 5 \times 10^2 \times \sqrt{3}$
$\text{Area} = 15 \times 100 \times \sqrt{3}$
$\text{Area} = 1500\sqrt{3}$ m$^2$
The area of the triangular plot is $1500\sqrt{3}$ m$^2$.
Exercise 12.1
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:
Given:
A traffic signal board in the shape of an equilateral triangle with side 'a'.
The perimeter of the board is 180 cm.
To Find:
1. The area of the signal board in terms of 'a' using Heron’s formula.
2. The area of the signal board when its perimeter is 180 cm.
Solution:
Part 1: Area of the signal board in terms of 'a'
The signal board is an equilateral triangle with each side of length 'a'.
To use Heron's formula, we first need to find the semi-perimeter (s).
Perimeter = $a + a + a = 3a$.
$s = \frac{\text{Perimeter}}{2} = \frac{3a}{2}$
According to Heron's formula, the area of a triangle is given by:
Area = $\sqrt{s(s-a)(s-b)(s-c)}$
Since the triangle is equilateral, $a=b=c$.
$s-a = \frac{3a}{2} - a = \frac{3a-2a}{2} = \frac{a}{2}$
Substituting the values into Heron's formula:
Area = $\sqrt{\frac{3a}{2} \left(\frac{a}{2}\right) \left(\frac{a}{2}\right) \left(\frac{a}{2}\right)}$
Area = $\sqrt{\frac{3a^4}{16}}$
Area = $\frac{\sqrt{3} \sqrt{a^4}}{\sqrt{16}}$
Area = $\frac{\sqrt{3}}{4} a^2$
So, the area of the equilateral signal board with side 'a' is $\frac{\sqrt{3}}{4} a^2$.
Part 2: Area when the perimeter is 180 cm
The perimeter of the equilateral triangle is given as 180 cm.
Perimeter = $3a = 180$ cm
$a = \frac{180}{3} = 60$ cm.
So, each side of the triangle is 60 cm.
Now, we find the semi-perimeter (s):
$s = \frac{\text{Perimeter}}{2} = \frac{180}{2} = 90$ cm
Using Heron's formula:
Area = $\sqrt{s(s-a)(s-b)(s-c)}$
Here, $a=b=c=60$ cm.
$s-a = 90 - 60 = 30$ cm
$s-b = 90 - 60 = 30$ cm
$s-c = 90 - 60 = 30$ cm
Substituting these values into the formula:
Area = $\sqrt{90 \times 30 \times 30 \times 30}$
Area = $\sqrt{(3 \times 30) \times 30 \times 30 \times 30}$
Area = $\sqrt{3 \times 30^4}$
Area = $30^2 \sqrt{3}$
Area = $900\sqrt{3} \text{ cm}^2$
Final Answer:
The area of the signal board, if its perimeter is 180 cm, will be $900\sqrt{3} \text{ cm}^2$.
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 (see Fig. 12.9). 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?
Answer:
Given:
The sides of the triangular side wall of a flyover are:
$a = 122$ m
$b = 22$ m
$c = 120$ m
Rate of earning from advertisements = $\textsf{₹ } 5000$ per m2 per year.
A company hired one wall for a period of 3 months.
To Find:
The amount of rent the company paid.
Solution:
First, we need to find the area of the triangular wall using Heron's formula.
To use Heron's formula, we first calculate the semi-perimeter (s) of the triangle.
$s = \frac{a+b+c}{2}$
$s = \frac{122 + 22 + 120}{2} = \frac{264}{2} = 132$ m
Now, we can apply Heron's formula for the area of a triangle:
Area = $\sqrt{s(s-a)(s-b)(s-c)}$
Let's calculate the terms $(s-a)$, $(s-b)$, and $(s-c)$:
$s-a = 132 - 122 = 10$ m
$s-b = 132 - 22 = 110$ m
$s-c = 132 - 120 = 12$ m
Now substitute these values into the formula:
Area = $\sqrt{132 \times 10 \times 110 \times 12}$
Area = $\sqrt{(12 \times 11) \times 10 \times (11 \times 10) \times 12}$
Area = $\sqrt{12 \times 12 \times 11 \times 11 \times 10 \times 10}$
Area = $\sqrt{12^2 \times 11^2 \times 10^2}$
Area = $12 \times 11 \times 10$
Area = $1320 \text{ m}^2$
The area of the triangular wall is 1320 m2.
Now, we calculate the rent for this area.
The rent is $\textsf{₹ } 5000$ per m2 for 1 year (12 months).
Rent for 1 m2 for 1 year = $\textsf{₹ } 5000$.
Rent for 1 m2 for 1 month = $\frac{\textsf{₹ } 5000}{12}$.
Rent for 1320 m2 for 3 months = $1320 \times \left(\frac{5000}{12}\right) \times 3$
Rent = $1320 \times 5000 \times \frac{3}{12}$
Rent = $1320 \times 5000 \times \frac{1}{4}$
Rent = $330 \times 5000$
Rent = $\textsf{₹ } 16,50,000$
Final Answer:
The company paid $\textsf{₹ } 16,50,000$ as rent.
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” (see Fig. 12.10 ). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.
Answer:
Given:
The sides of the triangular side wall are $a = 15$ m, $b = 11$ m, and $c = 6$ m.
To Find:
The area painted in colour.
Solution:
We will use Heron's formula to find the area of the triangle.
First, calculate the semi-perimeter of the triangle, $s = \frac{a+b+c}{2}$.
$\text{Perimeter} = 15 + 11 + 6 = 32$ m
$s = \frac{32}{2} = 16$ m
Now, calculate the differences $(s-a)$, $(s-b)$, and $(s-c)$.
$(s-a) = 16 - 15 = 1$ m
$(s-b) = 16 - 11 = 5$ m
$(s-c) = 16 - 6 = 10$ m
Using Heron's formula for the area of the triangle:
$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$
Substitute the values:
$\text{Area} = \sqrt{16 \times 1 \times 5 \times 10}$
Simplify the expression under the square root:
$\text{Area} = \sqrt{16 \times 50}$
$\text{Area} = \sqrt{16 \times 25 \times 2}$
$\text{Area} = \sqrt{16} \times \sqrt{25} \times \sqrt{2}$
$\text{Area} = 4 \times 5 \times \sqrt{2}$
$\text{Area} = 20\sqrt{2}$ m$^2$
The area painted in colour is $20\sqrt{2}$ m$^2$.
Question 4. Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.
Answer:
Given:
Let the sides of the triangle be $a$, $b$, and $c$.
Given two sides: $a = 18$ cm and $b = 10$ cm.
Given perimeter: $P = 42$ cm.
To Find:
The area of the triangle.
Solution:
The perimeter of a triangle is the sum of its three sides.
$P = a + b + c$
Substitute the given values:
$42 = 18 + 10 + c$
$42 = 28 + c$
Now, find the third side $c$:
$c = 42 - 28$
$c = 14$ cm
The third side of the triangle is 14 cm.
Now, calculate the semi-perimeter, $s$.
$s = \frac{\text{Perimeter}}{2}$
$s = \frac{42}{2}$
$s = 21$ cm
We can find the area of the triangle using Heron's formula:
$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$
Substitute the values of $s$, $a$, $b$, and $c$:
$\text{Area} = \sqrt{21(21-18)(21-10)(21-14)}$
$\text{Area} = \sqrt{21(3)(11)(7)}$
$\text{Area} = \sqrt{(3 \times 7) \times 3 \times 11 \times 7}$
$\text{Area} = \sqrt{3^2 \times 7^2 \times 11}$
Take the square root of the perfect squares:
$\text{Area} = \sqrt{3^2} \times \sqrt{7^2} \times \sqrt{11}$
$\text{Area} = 3 \times 7 \times \sqrt{11}$
$\text{Area} = 21\sqrt{11}$
The area of the triangle is $21\sqrt{11}$ cm$^2$.
The area of the triangle is $21\sqrt{11}$ cm$^2$.
Question 5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.
Answer:
Given:
The sides of a triangle are in the ratio 12 : 17 : 25.
The perimeter of the triangle is 540 cm.
To Find:
The area of the triangle.
Solution:
Let the common ratio for the sides of the triangle be $x$.
Then the lengths of the three sides are:
$a = 12x$
$b = 17x$
$c = 25x$
The perimeter of the triangle is the sum of its sides.
Perimeter = $a + b + c$
$540 = 12x + 17x + 25x$
$540 = 54x$
$x = \frac{540}{54} = 10$
Now we can find the actual lengths of the sides:
$a = 12x = 12 \times 10 = 120$ cm
$b = 17x = 17 \times 10 = 170$ cm
$c = 25x = 25 \times 10 = 250$ cm
To find the area, we will use Heron's formula. First, let's find the semi-perimeter (s).
$s = \frac{\text{Perimeter}}{2} = \frac{540}{2} = 270 \text{ cm}$
Now, we apply Heron's formula:
Area = $\sqrt{s(s-a)(s-b)(s-c)}$
Let's calculate the terms inside the square root:
$s-a = 270 - 120 = 150$ cm
$s-b = 270 - 170 = 100$ cm
$s-c = 270 - 250 = 20$ cm
Substitute these values into the formula:
Area = $\sqrt{270 \times 150 \times 100 \times 20}$
Area = $\sqrt{(27 \times 10) \times (15 \times 10) \times 100 \times (2 \times 10)}$
Area = $\sqrt{(3 \times 9 \times 10) \times (3 \times 5 \times 10) \times (10 \times 10) \times (2 \times 10)}$
Area = $\sqrt{(3^4 \times 2^2 \times 5^2 \times 10^2)}$ is not quite right. Let's group prime factors.
Area = $\sqrt{(3 \times 3 \times 3 \times 2 \times 5) \times (3 \times 5 \times 2 \times 5) \times (2 \times 5 \times 2 \times 5) \times (2 \times 2 \times 5)}$
Area = $\sqrt{3^4 \times 2^6 \times 5^6}$
Area = $3^{(4/2)} \times 2^{(6/2)} \times 5^{(6/2)}$
Area = $3^2 \times 2^3 \times 5^3$
Area = $9 \times 8 \times 125$
Area = $72 \times 125$
Area = $9000 \text{ cm}^2$
Final Answer:
The area of the triangle is 9000 cm2.
Question 6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Answer:
Given:
An isosceles triangle with perimeter = 30 cm.
Each of the equal sides = 12 cm.
To Find:
The area of the triangle.
Solution:
Let the sides of the isosceles triangle be $a$, $b$, and $c$. Since it is an isosceles triangle, two sides are equal. Let $a = 12$ cm and $b = 12$ cm.
The perimeter of a triangle is the sum of its three sides.
$\text{Perimeter} = a + b + c$
We are given that the perimeter is 30 cm.
30 cm = 12 cm + 12 cm + c
30 = 24 + c
Now, find the length of the third side, $c$:
c = 30 - 24
c = 6 cm
The sides of the triangle are 12 cm, 12 cm, and 6 cm.
Now, calculate the semi-perimeter, $s$.
s = $\frac{\text{Perimeter}}{2}$
s = $\frac{30}{2}$
s = 15 cm
Now, calculate the differences $(s-a)$, $(s-b)$, and $(s-c)$.
$(s-a) = 15 - 12 = 3$ cm
$(s-b) = 15 - 12 = 3$ cm
$(s-c) = 15 - 6 = 9$ cm
Using Heron's formula for the area of the triangle:
$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$
Substitute the values:
$\text{Area} = \sqrt{15 \times 3 \times 3 \times 9}$
Simplify the expression under the square root:
$\text{Area} = \sqrt{(3 \times 5) \times 3^2 \times 3^2}$
$\text{Area} = \sqrt{3^1 \times 5^1 \times 3^2 \times 3^2}$
$\text{Area} = \sqrt{3^{1+2+2} \times 5}$
$\text{Area} = \sqrt{3^5 \times 5}$
Group the perfect squares:
$\text{Area} = \sqrt{(3^4 \times 3) \times 5}$
$\text{Area} = \sqrt{3^4} \times \sqrt{3 \times 5}$
$\text{Area} = 3^2 \times \sqrt{15}$
$\text{Area} = 9\sqrt{15}$ cm$^2$
The area of the isosceles triangle is $9\sqrt{15}$ cm$^2$.
Example 4 to 6 (Before Exercise 12.2)
Example 4. Kamla has a triangular field with sides 240 m, 200 m, 360 m, where she grew wheat. In another triangular field with sides 240 m, 320 m, 400 m adjacent to the previous field, she wanted to grow potatoes and onions (see Fig. 12.11). She divided the field in two parts by joining the mid-point of the longest side to the opposite vertex and grew patatoes in one part and onions in the other part. How much area (in hectares) has been used for wheat, potatoes and onions? (1 hectare = 10000 m2 )
Answer:
Given:
A triangular field ABC with sides 240 m, 200 m, and 360 m, used for growing wheat.
An adjacent triangular field ACD with sides 240 m, 320 m, and 400 m, used for growing potatoes and onions.
The field ACD is divided into two equal parts by a median.
1 hectare = 10000 m2.
To Find:
The area (in hectares) used for wheat, potatoes, and onions.
Solution:
1. Area for Wheat (Area of $\triangle$ABC)
The sides of the triangular field ABC are:
$a = 240$ m, $b = 200$ m, $c = 360$ m.
We will use Heron's formula to find the area. First, we calculate the semi-perimeter (s):
$s = \frac{a+b+c}{2} = \frac{240+200+360}{2} = \frac{800}{2} = 400 \text{ m}$
Now, we apply Heron's formula:
Area($\triangle$ABC) = $\sqrt{s(s-a)(s-b)(s-c)}$
Area = $\sqrt{400(400-240)(400-200)(400-360)}$
Area = $\sqrt{400 \times 160 \times 200 \times 40}$
Area = $\sqrt{(20 \times 20) \times (16 \times 10) \times (2 \times 100) \times (4 \times 10)}$
Area = $\sqrt{20^2 \times 4^2 \times 2 \times 10^2 \times 2^2 \times 10^2}$
Area = $\sqrt{20^2 \times 4^2 \times 2^2 \times 10^2 \times 10^2 \times 2}$
Area = $20 \times 4 \times 2 \times 10 \times 10 \times \sqrt{2}$
Area = $16000\sqrt{2} \text{ m}^2$
To convert this area to hectares, we divide by 10000:
Area for wheat = $\frac{16000\sqrt{2}}{10000} = 1.6\sqrt{2}$ hectares.
Using $\sqrt{2} \approx 1.414$, the area is approximately $1.6 \times 1.414 = 2.26$ hectares (approx.).
2. Area for Potatoes and Onions (Area of $\triangle$ACD)
The sides of the triangular field ACD are:
$a = 240$ m, $b = 320$ m, $c = 400$ m.
First, we calculate the semi-perimeter (s) for this triangle:
$s = \frac{240+320+400}{2} = \frac{960}{2} = 480 \text{ m}$
Using Heron's formula:
Area($\triangle$ACD) = $\sqrt{s(s-a)(s-b)(s-c)}$
Area = $\sqrt{480(480-240)(480-320)(480-400)}$
Area = $\sqrt{480 \times 240 \times 160 \times 80}$
Area = $\sqrt{(2 \times 240) \times 240 \times (2 \times 80) \times 80}$
Area = $\sqrt{2 \times 240^2 \times 2 \times 80^2} = \sqrt{4 \times 240^2 \times 80^2}$
Area = $2 \times 240 \times 80 = 38400 \text{ m}^2$
Total area for potatoes and onions = $38400 \text{ m}^2$.
3. Area for Potatoes and Onions Separately
The field ACD is divided into two parts by joining the mid-point of the longest side (AD = 400 m) to the opposite vertex C. This line segment is a median of the triangle.
A median divides a triangle into two triangles of equal area.
Area for potatoes = Area for onions = $\frac{\text{Total Area of } \triangle ACD}{2}$
Area for each crop = $\frac{38400}{2} = 19200 \text{ m}^2$
Now, we convert these areas to hectares:
Area for potatoes = $\frac{19200}{10000} = 1.92$ hectares.
Area for onions = $\frac{19200}{10000} = 1.92$ hectares.
Final Answer:
The areas used for growing the crops are:
- Wheat: $1.6\sqrt{2}$ hectares (approximately 2.26 hectares)
- Potatoes: 1.92 hectares
- Onions: 1.92 hectares
Example 5. Students of a school staged a rally for cleanliness campaign. They walked through the lanes in two groups. One group walked through the lanes AB, BC and CA; while the other through AC, CD and DA (see Fig. 12.12). Then they cleaned the area enclosed within their lanes. If AB = 9 m, BC = 40 m, CD = 15 m, DA = 28 m and ∠B = 90º, which group cleaned more area and by how much? Find the total area cleaned by the students (neglecting the width of the lanes).
Answer:
Given:
Triangular lanes AB, BC, CA forming $\triangle$ABC with AB = 9 m, BC = 40 m, $\angle$B = $90^\circ$.
Triangular lanes AC, CD, DA forming $\triangle$ADC with CD = 15 m, DA = 28 m.
To Find:
1. Which group cleaned more area and by how much.
2. The total area cleaned by the students.
Solution:
The park is divided into two triangles, $\triangle$ABC and $\triangle$ADC, by the common side AC.
Area cleaned by the first group (Area of $\triangle$ABC):
In $\triangle$ABC, $\angle$B = $90^\circ$, AB = 9 m, and BC = 40 m.
Since it is a right-angled triangle, the area can be calculated as:
$\text{Area of } \triangle\text{ABC} = \frac{1}{2} \times \text{base} \times \text{height}$
$\text{Area of } \triangle\text{ABC} = \frac{1}{2} \times \text{BC} \times \text{AB}$
$\text{Area of } \triangle\text{ABC} = \frac{1}{2} \times 40 \text{ m} \times 9 \text{ m}$
$\text{Area of } \triangle\text{ABC} = 20 \times 9 = 180 \text{ m}^2$
Area cleaned by the first group is 180 m$^2$.
Area cleaned by the second group (Area of $\triangle$ADC):
To find the area of $\triangle$ADC, we first need the length of side AC. In $\triangle$ABC, AC is the hypotenuse.
Using the Pythagorean theorem in $\triangle$ABC:
$\text{AC}^2 = \text{AB}^2 + \text{BC}^2$
$\text{AC}^2 = (9 \text{ m})^2 + (40 \text{ m})^2$
$\text{AC}^2 = 81 \text{ m}^2 + 1600 \text{ m}^2$
$\text{AC}^2 = 1681 \text{ m}^2$
$\text{AC} = \sqrt{1681} \text{ m} = 41 \text{ m}$
So, the sides of $\triangle$ADC are AD = 28 m, CD = 15 m, and AC = 41 m.
We use Heron's formula to find the area of $\triangle$ADC.
Let the sides be $a=41$ m, $b=15$ m, $c=28$ m.
The semi-perimeter, $s$, is:
$s = \frac{a+b+c}{2}$
$s = \frac{41 + 15 + 28}{2} = \frac{84}{2} = 42 \text{ m}$
Now, calculate $(s-a)$, $(s-b)$, and $(s-c)$:
$(s-a) = 42 - 41 = 1 \text{ m}$
$(s-b) = 42 - 15 = 27 \text{ m}$
$(s-c) = 42 - 28 = 14 \text{ m}$
Using Heron's formula for the area of $\triangle$ADC:
$\text{Area of } \triangle\text{ADC} = \sqrt{s(s-a)(s-b)(s-c)}$
$\text{Area of } \triangle\text{ADC} = \sqrt{42 \times 1 \times 27 \times 14}$
Simplify the expression under the square root:
$\text{Area of } \triangle\text{ADC} = \sqrt{(6 \times 7) \times 1 \times (3 \times 9) \times (2 \times 7)}$
$\text{Area of } \triangle\text{ADC} = \sqrt{(2 \times 3 \times 7) \times 3 \times 3^2 \times 2 \times 7}$
$\text{Area of } \triangle\text{ADC} = \sqrt{2^2 \times 3^4 \times 7^2}$
Take the square root of the perfect squares:
$\text{Area of } \triangle\text{ADC} = \sqrt{2^2} \times \sqrt{3^4} \times \sqrt{7^2}$
$\text{Area of } \triangle\text{ADC} = 2 \times 3^2 \times 7$
$\text{Area of } \triangle\text{ADC} = 2 \times 9 \times 7$
$\text{Area of } \triangle\text{ADC} = 18 \times 7 = 126 \text{ m}^2$
Area cleaned by the second group is 126 m$^2$.
Comparison of Areas:
Area cleaned by the first group = 180 m$^2$.
Area cleaned by the second group = 126 m$^2$.
Since $180 > 126$, the first group cleaned more area.
Difference in area = Area of $\triangle$ABC - Area of $\triangle$ADC
Difference = $180 \text{ m}^2 - 126 \text{ m}^2 = 54 \text{ m}^2$
The first group cleaned 54 m$^2$ more area than the second group.
Total Area Cleaned:
The total area cleaned is the sum of the areas of the two triangles.
$\text{Total Area} = \text{Area of } \triangle\text{ABC} + \text{Area of } \triangle\text{ADC}$
$\text{Total Area} = 180 \text{ m}^2 + 126 \text{ m}^2 = 306 \text{ m}^2$
The first group cleaned more area by 54 m$^2$.
The total area cleaned by the students is 306 m$^2$.
Example 6. Sanya has a piece of land which is in the shape of a rhombus (see Fig. 12.13). She wants her one daughter and one son to work on the land and produce different crops. She divided the land in two equal parts. If the perimeter of the land is 400 m and one of the diagonals is 160 m, how much area each of them will get for their crops?
Answer:
Given:
The land is in the shape of a rhombus.
Perimeter of the rhombus = 400 m.
Length of one diagonal = 160 m.
To Find:
The area each child will get for their crops.
Solution:
A rhombus has all four sides equal in length.
Let the side of the rhombus be 'a'.
Perimeter of the rhombus = 4 $\times$ side
400 m = 4a
a = $\frac{400}{4}$ m
a = 100 m
So, each side of the rhombus is 100 m.
Let the rhombus be ABCD, and the given diagonal be AC = 160 m.
The land is divided into two equal parts by the diagonal AC. These parts are two triangles, $\triangle$ABC and $\triangle$ADC.
The diagonal of a rhombus divides it into two congruent triangles. Therefore, the area of $\triangle$ABC is equal to the area of $\triangle$ADC.
Each child gets one of these triangles.
Consider $\triangle$ABC. Its sides are AB, BC, and AC.
AB = 100 m (side of the rhombus)
BC = 100 m (side of the rhombus)
AC = 160 m (given diagonal)
We can find the area of $\triangle$ABC using Heron's formula.
Let the sides be $a' = 100$ m, $b' = 100$ m, and $c' = 160$ m.
The semi-perimeter, $s'$, of $\triangle$ABC is:
s' = $\frac{a' + b' + c'}{2}$
s' = $\frac{100 + 100 + 160}{2}$ m
s' = $\frac{360}{2}$ m = 180 m
Now, calculate the differences $(s'-a')$, $(s'-b')$, and $(s'-c')$.
$(s'-a') = 180 - 100 = 80$ m
$(s'-b') = 180 - 100 = 80$ m
$(s'-c') = 180 - 160 = 20$ m
Using Heron's formula for the area of $\triangle$ABC:
$\text{Area of } \triangle\text{ABC} = \sqrt{s'(s'-a')(s'-b')(s'-c')}$
Substitute the values:
$\text{Area of } \triangle\text{ABC} = \sqrt{180 \times 80 \times 80 \times 20}$
Simplify the expression under the square root:
$\text{Area} = \sqrt{(18 \times 10) \times (8 \times 10) \times (8 \times 10) \times (2 \times 10)}$
$\text{Area} = \sqrt{18 \times 8 \times 8 \times 2 \times 10^4}$
$\text{Area} = \sqrt{(2 \times 9) \times 8^2 \times 2 \times 10^4}$
$\text{Area} = \sqrt{2^2 \times 9 \times 8^2 \times 10^4}$
$\text{Area} = \sqrt{2^2} \times \sqrt{9} \times \sqrt{8^2} \times \sqrt{10^4}$
$\text{Area} = 2 \times 3 \times 8 \times 10^2$
$\text{Area} = 6 \times 8 \times 100$
$\text{Area} = 4800$ m$^2$
The area of $\triangle$ABC is 4800 m$^2$. Since the land is divided into two equal parts by the diagonal, each child gets an area equal to the area of one of these triangles.
$\text{Area each child gets} = 4800 \text{ m}^2$
Each child will get 4800 m$^2$ of area for their crops.
Exercise 12.2
Question 1. A park, in the shape of a quadrilateral ABCD, has ∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
Answer:
Given:
A quadrilateral park ABCD with sides AB = 9 m, BC = 12 m, CD = 5 m, and AD = 8 m.
The angle at vertex C is $\angle C = 90^\circ$.
To Find:
The total area of the park ABCD.
Solution:
To find the total area, we can split the quadrilateral into two triangles by drawing the diagonal BD.
Area of park ABCD = Area of $\triangle BCD$ + Area of $\triangle ABD$.
1. Area of $\triangle BCD$:
Since $\angle C = 90^\circ$, $\triangle BCD$ is a right-angled triangle. Its area can be calculated as:
Area of $\triangle BCD = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BC \times CD$
Area of $\triangle BCD = \frac{1}{2} \times 12 \text{ m} \times 5 \text{ m} = 30 \text{ m}^2$.
2. Area of $\triangle ABD$:
To find the area of $\triangle ABD$, we need the lengths of all its three sides. We have AB = 9 m and AD = 8 m. We can find the length of the third side, BD, using the Pythagorean theorem in the right-angled $\triangle BCD$.
$BD^2 = BC^2 + CD^2$
$BD^2 = (12)^2 + (5)^2 = 144 + 25 = 169$
$BD = \sqrt{169} = 13$ m.
Now, we can find the area of $\triangle ABD$ with sides 9 m, 8 m, and 13 m using Heron's formula.
First, find the semi-perimeter (s):
$s = \frac{9 + 8 + 13}{2} = \frac{30}{2} = 15$ m.
Area of $\triangle ABD = \sqrt{s(s-a)(s-b)(s-c)}$
Area = $\sqrt{15(15-9)(15-8)(15-13)}$
Area = $\sqrt{15 \times 6 \times 7 \times 2} = \sqrt{(3 \times 5) \times (2 \times 3) \times 7 \times 2}$
Area = $\sqrt{2^2 \times 3^2 \times 35} = 2 \times 3 \sqrt{35} = 6\sqrt{35} \text{ m}^2$.
Using $\sqrt{35} \approx 5.916$, the area is approximately $6 \times 5.916 = 35.496 \text{ m}^2$.
3. Total Area of the Park:
Total Area = Area of $\triangle BCD$ + Area of $\triangle ABD$
Total Area = $30 \text{ m}^2 + 6\sqrt{35} \text{ m}^2 \approx 30 + 35.496 = 65.496 \text{ m}^2$.
The park occupies an area of $(30 + 6\sqrt{35}) \text{ m}^2$, which is approximately 65.5 m2.
Question 2. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
Answer:
Given:
A quadrilateral ABCD with side lengths AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm, and diagonal AC = 5 cm.
To Find:
The area of the quadrilateral ABCD.
Solution:
The diagonal AC divides the quadrilateral into two triangles: $\triangle ABC$ and $\triangle ADC$. The total area will be the sum of the areas of these two triangles.
1. Area of $\triangle ABC$:
The sides of $\triangle ABC$ are 3 cm, 4 cm, and 5 cm. Let's check if this is a right-angled triangle using the converse of the Pythagorean theorem.
$3^2 + 4^2 = 9 + 16 = 25$
$5^2 = 25$
Since $AB^2 + BC^2 = AC^2$, $\triangle ABC$ is a right-angled triangle with the right angle at B.
Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 4 = 6 \text{ cm}^2$.
2. Area of $\triangle ADC$:
The sides of $\triangle ADC$ are 5 cm, 4 cm, and 5 cm. We can find its area using Heron's formula.
First, find the semi-perimeter (s):
$s = \frac{5 + 4 + 5}{2} = \frac{14}{2} = 7$ cm.
Area of $\triangle ADC = \sqrt{s(s-a)(s-b)(s-c)}$
Area = $\sqrt{7(7-5)(7-4)(7-5)}$
Area = $\sqrt{7 \times 2 \times 3 \times 2} = \sqrt{84} = \sqrt{4 \times 21} = 2\sqrt{21} \text{ cm}^2$.
Using $\sqrt{21} \approx 4.58$, the area is approximately $2 \times 4.58 = 9.16 \text{ cm}^2$.
3. Total Area of the Quadrilateral:
Total Area = Area of $\triangle ABC$ + Area of $\triangle ADC$
Total Area = $6 + 2\sqrt{21} \text{ cm}^2 \approx 6 + 9.16 = 15.16 \text{ cm}^2$.
The area of the quadrilateral ABCD is $(6 + 2\sqrt{21}) \text{ cm}^2$, which is approximately 15.2 cm2.
Question 3. Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used.
Answer:
Given:
A picture of an aeroplane made of paper, divided into five regions with given dimensions.
To Find:
The total area of the paper used.
Solution:
We will calculate the area of each of the five regions and then add them together.
Region I: This is an isosceles triangle with sides 5 cm, 5 cm, and 1 cm.
Using Heron's formula: $s = (5+5+1)/2 = 5.5$ cm.
Area$_1 = \sqrt{5.5(5.5-5)(5.5-5)(5.5-1)} = \sqrt{5.5 \times 0.5 \times 0.5 \times 4.5} \ $$ = 0.75\sqrt{11} \approx 2.488 \text{ cm}^2$.
Region II: This is a rectangle with length 6.5 cm and width 1 cm.
Area$_2 = 6.5 \times 1 = 6.5 \text{ cm}^2$.
Region III: This is a trapezium. To find its area, we can draw a line parallel to one of the non-parallel sides to form a parallelogram and a triangle. Let's find its height. The parallel sides are 1 cm and 2 cm. The non-parallel sides are both 1 cm. Let's draw perpendiculars from the vertices of the 1 cm side to the 2 cm side. The height (h) can be found using Pythagoras' theorem on the resulting right triangles with hypotenuse 1 cm and base $(2-1)/2 = 0.5$ cm.
$h^2 + (0.5)^2 = 1^2 \implies h^2 = 1 - 0.25 = 0.75 \implies h = \sqrt{0.75} = \frac{\sqrt{3}}{2}$ cm.
Area$_3 = \frac{1}{2} \times (\text{sum of parallel sides}) \times h = \frac{1}{2} \times (1+2) \times \frac{\sqrt{3}}{2} \ $$ = \frac{3\sqrt{3}}{4} \approx 1.3 \text{ cm}^2$.
Region IV and V: These are two identical right-angled triangles with base 1.5 cm and height 6 cm.
Area of one triangle (Area$_4$) = $\frac{1}{2} \times \text{base} \times \text{height} \ $$ = \frac{1}{2} \times 1.5 \times 6 \ $$ = 4.5 \text{ cm}^2$.
Since Region V is identical, Area$_5 = 4.5 \text{ cm}^2$.
Total Area:
Total Area = Area$_1$ + Area$_2$ + Area$_3$ + Area$_4$ + Area$_5$
Total Area $\approx 2.488 + 6.5 + 1.3 + 4.5 + 4.5$
Total Area $\approx 19.288 \text{ cm}^2$
The total area of the paper used is approximately 19.3 cm2.
Question 4. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram
Answer:
Given:
A triangle with sides 26 cm, 28 cm, and 30 cm.
A parallelogram with a base of 28 cm.
The triangle and the parallelogram have the same base (28 cm) and the same area.
To Find:
The height of the parallelogram.
Solution:
First, we need to calculate the area of the triangle using Heron's formula.
The sides of the triangle are $a = 26$ cm, $b = 28$ cm, and $c = 30$ cm.
The semi-perimeter (s) is:
$s = \frac{26 + 28 + 30}{2} = \frac{84}{2} = 42$ cm.
Area of Triangle = $\sqrt{s(s-a)(s-b)(s-c)}$
Area = $\sqrt{42(42-26)(42-28)(42-30)}$
Area = $\sqrt{42 \times 16 \times 14 \times 12}$
Area = $\sqrt{(6 \times 7) \times 16 \times (2 \times 7) \times (2 \times 6)}$
Area = $\sqrt{6^2 \times 7^2 \times 2^2 \times 16} = \sqrt{6^2 \times 7^2 \times 2^2 \times 4^2}$
Area = $6 \times 7 \times 2 \times 4 = 336 \text{ cm}^2$.
We are given that the area of the parallelogram is the same as the area of the triangle.
Area of Parallelogram = 336 cm2.
The formula for the area of a parallelogram is: Area = base $\times$ height.
The base of the parallelogram is given as 28 cm.
$336 = 28 \times \text{height}$
height = $\frac{336}{28}$
height = 12 cm.
The height of the parallelogram is 12 cm.
Question 5. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
Answer:
Given:
A rhombus-shaped field with each side measuring 30 m.
The length of the longer diagonal is 48 m.
The field provides grass for 18 cows.
To Find:
The area of the grass field that each cow gets.
Solution:
First, we need to find the total area of the rhombus-shaped field.
A diagonal divides a rhombus into two congruent triangles. Let the rhombus be ABCD and the longer diagonal be AC = 48 m. This divides the rhombus into two triangles, $\triangle ABC$ and $\triangle ADC$.
Let's find the area of one of these triangles, say $\triangle ABC$.
The sides of $\triangle ABC$ are AB = 30 m, BC = 30 m, and AC = 48 m.
We can use Heron's formula to find the area of $\triangle ABC$.
The semi-perimeter (s) is:
$s = \frac{30 + 30 + 48}{2} = \frac{108}{2} = 54$ m.
Area of $\triangle ABC = \sqrt{s(s-a)(s-b)(s-c)}$
Area = $\sqrt{54(54-30)(54-30)(54-48)}$
Area = $\sqrt{54 \times 24 \times 24 \times 6} = \sqrt{(9 \times 6) \times 24^2 \times 6}$
Area = $\sqrt{9 \times 6^2 \times 24^2} = 3 \times 6 \times 24 = 432 \text{ m}^2$.
Since the two triangles are congruent, the total area of the rhombus is twice the area of one triangle.
Total Area of Rhombus = $2 \times$ Area of $\triangle ABC = 2 \times 432 = 864 \text{ m}^2$.
This total area of 864 m2 is available for 18 cows.
Area per cow = $\frac{\text{Total Area}}{\text{Number of Cows}} = \frac{864}{18} = 48 \text{ m}^2$.
Each cow will be getting 48 m2 of the grass field.
Question 6. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig. 12.16), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?
Answer:
Given:
An umbrella made of 10 triangular pieces of cloth.
There are two different colours, so there are 5 pieces of each colour.
The sides of each triangular piece are 20 cm, 50 cm, and 50 cm.
To Find:
The total area of cloth required for each colour.
Solution:
First, we calculate the area of one triangular piece of cloth using Heron's formula.
The sides of the triangle are $a = 20$ cm, $b = 50$ cm, and $c = 50$ cm.
The semi-perimeter (s) is:
$s = \frac{20 + 50 + 50}{2} = \frac{120}{2} = 60$ cm.
Area of one piece = $\sqrt{s(s-a)(s-b)(s-c)}$
Area = $\sqrt{60(60-20)(60-50)(60-50)}$
Area = $\sqrt{60 \times 40 \times 10 \times 10} = \sqrt{2400 \times 100}$
Area = $\sqrt{24 \times 100 \times 100} = \sqrt{4 \times 6 \times 100^2}$
Area = $2 \times 100 \times \sqrt{6} = 200\sqrt{6} \text{ cm}^2$.
The area of one triangular piece is $200\sqrt{6} \text{ cm}^2$.
There are 5 pieces of cloth of the first colour.
Area of cloth of the first colour = $5 \times$ Area of one piece
Area = $5 \times 200\sqrt{6} = 1000\sqrt{6} \text{ cm}^2$.
Similarly, there are 5 pieces of cloth of the second colour.
Area of cloth of the second colour = $5 \times$ Area of one piece
Area = $5 \times 200\sqrt{6} = 1000\sqrt{6} \text{ cm}^2$.
The cloth required for each colour is $1000\sqrt{6} \text{ cm}^2$.
Question 7. A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Fig. 12.17. How much paper of each shade has been used in it?
Answer:
Given:
A kite made of three shades of paper.
Shades I and II are triangles forming a square with a diagonal of 32 cm.
Shade III is an isosceles triangle with a base of 8 cm and equal sides of 6 cm each.
To Find:
The area of paper used for each shade.
Solution:
Area of Shade I and Shade II:
The top part of the kite is a square. The diagonal of a square divides it into two congruent right-angled isosceles triangles. These triangles represent Shade I and Shade II.
The area of a square can be calculated as $\frac{1}{2} \times (\text{diagonal})^2$. Since both diagonals of a square are equal, the area is $\frac{1}{2} \times d_1 \times d_2$.
Total Area of the square = $\frac{1}{2} \times 32 \times 32 = 512 \text{ cm}^2$.
The area of Shade I is half the area of the square.
Area of Shade I = $\frac{1}{2} \times 512 = 256 \text{ cm}^2$.
The area of Shade II is also half the area of the square.
Area of Shade II = $\frac{1}{2} \times 512 = 256 \text{ cm}^2$.
Area of Shade III:
The bottom part of the kite is an isosceles triangle with sides 8 cm, 6 cm, and 6 cm.
We can find its area using Heron's formula.
The semi-perimeter (s) is:
$s = \frac{8 + 6 + 6}{2} = \frac{20}{2} = 10$ cm.
Area of Shade III = $\sqrt{s(s-a)(s-b)(s-c)}$
Area = $\sqrt{10(10-8)(10-6)(10-6)}$
Area = $\sqrt{10 \times 2 \times 4 \times 4} = \sqrt{16 \times 20} = \sqrt{16 \times 4 \times 5}$
Area = $4 \times 2 \sqrt{5} = 8\sqrt{5} \text{ cm}^2$.
Using $\sqrt{5} \approx 2.24$, the area is approximately $8 \times 2.24 = 17.92 \text{ cm}^2$.
The area of paper used for each shade is:
- Shade I: 256 cm2
- Shade II: 256 cm2
- Shade III: $8\sqrt{5} \text{ cm}^2$ (approx. 17.92 cm2)
Question 8. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see Fig. 12.18). Find the cost of polishing the tiles at the rate of 50p per cm2.
Answer:
Given:
A floral design made of 16 triangular tiles.
The sides of each triangular tile are 9 cm, 28 cm, and 35 cm.
The cost of polishing is 50 paise per cm2.
To Find:
The total cost of polishing all 16 tiles.
Solution:
First, we need to find the area of a single triangular tile using Heron's formula.
The sides of the triangle are $a = 9$ cm, $b = 28$ cm, and $c = 35$ cm.
The semi-perimeter (s) is:
$s = \frac{9 + 28 + 35}{2} = \frac{72}{2} = 36$ cm.
Area of one tile = $\sqrt{s(s-a)(s-b)(s-c)}$
Area = $\sqrt{36(36-9)(36-28)(36-35)}$
Area = $\sqrt{36 \times 27 \times 8 \times 1} = \sqrt{36 \times (9 \times 3) \times (4 \times 2)}$
Area = $\sqrt{36 \times 9 \times 4 \times 6} = \sqrt{36} \times \sqrt{9} \times \sqrt{4} \times \sqrt{6}$
Area = $6 \times 3 \times 2 \times \sqrt{6} = 36\sqrt{6} \text{ cm}^2$.
The floral design is made up of 16 such tiles.
Total Area = $16 \times$ Area of one tile = $16 \times 36\sqrt{6} = 576\sqrt{6} \text{ cm}^2$.
Using $\sqrt{6} \approx 2.45$, the total area is approximately $576 \times 2.45 = 1411.2 \text{ cm}^2$.
The cost of polishing is 50 paise per cm2, which is equal to $\textsf{₹ } 0.50$ per cm2.
Total Cost = Total Area $\times$ Rate of polishing
Total Cost = $1411.2 \text{ cm}^2 \times \textsf{₹ } 0.50/\text{cm}^2$
Total Cost = $\textsf{₹ } 705.60$.
The total cost of polishing the tiles is $\textsf{₹ } 705.60$.
Question 9. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.
Answer:
Given:
A trapezium-shaped field with parallel sides of 25 m and 10 m.
The non-parallel sides are 14 m and 13 m.
To Find:
The area of the field.
Solution:
Let the trapezium be ABCD with AB $\parallel$ DC. Let AB = 25 m, DC = 10 m, AD = 13 m, and BC = 14 m.
To find the area, we need the height of the trapezium. We can find this by dividing the trapezium into a parallelogram and a triangle.
Draw a line CE from C parallel to AD, meeting the side AB at E.
Now, AECD is a parallelogram since AE $\parallel$ DC and AD $\parallel$ CE.
Therefore, AE = DC = 10 m and CE = AD = 13 m.
The remaining part of the base AB is EB = AB - AE = 25 m - 10 m = 15 m.
Now consider the triangle $\triangle BCE$. Its sides are 13 m, 14 m, and 15 m.
We can find the area of $\triangle BCE$ using Heron's formula.
The semi-perimeter (s) is:
$s = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21$ m.
Area of $\triangle BCE = \sqrt{s(s-a)(s-b)(s-c)}$
Area = $\sqrt{21(21-13)(21-14)(21-15)}$
Area = $\sqrt{21 \times 8 \times 7 \times 6} = \sqrt{(3 \times 7) \times (2 \times 4) \times 7 \times (2 \times 3)}$
Area = $\sqrt{3^2 \times 7^2 \times 2^2 \times 4} = 3 \times 7 \times 2 \times 2 = 84 \text{ m}^2$.
The area of $\triangle BCE$ can also be expressed as $\frac{1}{2} \times \text{base} \times \text{height}$. Let's use the base EB = 15 m. The height (h) of this triangle will be the height of the trapezium.
Area = $\frac{1}{2} \times EB \times h$
$84 = \frac{1}{2} \times 15 \times h$
$h = \frac{84 \times 2}{15} = \frac{168}{15} = \frac{56}{5} = 11.2$ m.
Now we can find the area of the trapezium ABCD.
Area of Trapezium = $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$
Area = $\frac{1}{2} \times (10 + 25) \times 11.2$
Area = $\frac{1}{2} \times 35 \times 11.2 = 35 \times 5.6 = 196 \text{ m}^2$.
The area of the field is 196 m2.