RBSE Solutions Class 9 Maths Chapter 11 Area of Plane Figures Exercise 11.1

Question 1. The base and height of a triangle are 20 cm and 6 cm. Find the area of the triangle.
Answer: The area of a triangle is found by multiplying half of its base by its height. We use the formula: Area of triangle \( = \frac{1}{2} \times \text{base} \times \text{altitude} \).
So, for a base of 20 cm and height of 6 cm:
Area \( = \frac{1}{2} \times 20 \times 6 = 60 \text{ cm}^2 \).
Knowing the formula helps quickly solve these problems.
In simple words: To find the area of a triangle, multiply half of the base by the height.

🎯 Exam Tip: Remember the formula for the area of a triangle, \( \frac{1}{2} \times \text{base} \times \text{height} \), and include the correct units (\( \text{cm}^2 \), \( \text{m}^2 \), etc.) in your final answer.

Question 2. Find the area of the triangle whose sides are 15 cm, 25 cm and 30 cm respectively.
Answer: To find the area of a triangle when all three side lengths are known, we use Heron's formula. Let the sides be \( a = 15 \text{ cm} \), \( b = 25 \text{ cm} \), and \( c = 30 \text{ cm} \).
First, calculate the semi-perimeter 's':
\( s = \frac{a+b+c}{2} = \frac{15+25+30}{2} = \frac{70}{2} = 35 \text{ cm} \).
Next, substitute 's' and the side lengths into Heron's formula for the area:
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
\( = \sqrt{35(35-15)(35-25)(35-30)} \)
\( = \sqrt{35 \times 20 \times 10 \times 5} \)
\( = \sqrt{(7 \times 5) \times (4 \times 5) \times (2 \times 5) \times 5} \)
\( = \sqrt{7 \times 5 \times 2^2 \times 5 \times 2 \times 5 \times 5} \)
\( = \sqrt{2^3 \times 5^4 \times 7} \) (The source calculation simplifies differently; following the source steps exactly)
\( = \sqrt{7 \times 5 \times 4 \times 5 \times 2 \times 5 \times 5} \)
\( = \sqrt{5 \times 5 \times 4 \times 7 \times 2 \times 5 \times 5} \)
\( = 5 \times 5 \times 2 \sqrt{7 \times 2} \)
\( = 50\sqrt{14} \text{ cm}^2 \).
This method is especially useful when the height is not given.
In simple words: First, find half the perimeter of the triangle. Then, use Heron's formula with the side lengths to get the area.

🎯 Exam Tip: Make sure to correctly calculate the semi-perimeter 's' and carefully simplify the square root for the final area. Always check for common factors to simplify the radical expression.

Question 3. Find the area of an isosceles triangle, the measure of one of its equal sides being 8 cm and the third side is 4 cm.
Answer: For an isosceles triangle with two equal sides of 8 cm and a third side (base) of 4 cm, we can find its area using the formula for an isosceles triangle. If \( a \) is the length of the equal sides and \( b \) is the length of the base, the area is given by \( \frac{b}{4}\sqrt{4a^2 - b^2} \).
Here, \( a = 8 \text{ cm} \) and \( b = 4 \text{ cm} \).
Area \( = \frac{4}{4}\sqrt{4(8)^2 - (4)^2} \)
\( = 1 \times \sqrt{4 \times 64 - 16} \)
\( = \sqrt{256 - 16} \)
\( = \sqrt{240} \)
\( = \sqrt{16 \times 15} \)
\( = 4\sqrt{15} \text{ cm}^2 \).
This calculation gives us an exact value for the area in square centimeters.
In simple words: Find the area of an isosceles triangle where two sides are 8 cm and the third side is 4 cm by using its specific area formula.

🎯 Exam Tip: You can either use Heron's formula or the specific isosceles triangle area formula \( \frac{b}{4}\sqrt{4a^2 - b^2} \) (where 'b' is the unequal side and 'a' is an equal side) to solve this type of problem. Choose the one you find easier to remember and apply correctly.

Question 4. Find the area of an equilateral triangle whose side is 20 cm.
Answer: To calculate the area of an equilateral triangle, where all sides are equal, we use a special formula: Area \( = \frac{\sqrt{3}}{4} (\text{side})^2 \).
For a triangle with sides of 20 cm:
Area \( = \frac{\sqrt{3}}{4} \times (20)^2 \)
\( = \frac{\sqrt{3}}{4} \times 400 \)
\( = 100\sqrt{3} \text{ cm}^2 \).
This formula simplifies finding the area directly.
In simple words: An equilateral triangle has equal sides. Use its special formula (\( \frac{\sqrt{3}}{4} \times \text{side}^2 \)) to find its area.

🎯 Exam Tip: Always remember the specific formula for the area of an equilateral triangle; it is often quicker than using Heron's formula for these specific cases.

Question 5. A triangular plot whose two sides are 8 cm and 15 cm and its perimeter is 40 cm, then find the area of the triangular plot.
Answer: First, find the length of the third side of the triangular plot by subtracting the two given sides from the total perimeter. The perimeter is the sum of all three sides.
Given: Perimeter = 40 cm, Side 1 = 8 cm, Side 2 = 15 cm.
\( 40 = 8 + 15 + \text{third side} \)
\( \implies 40 = 23 + \text{third side} \)
\( \implies \text{Third side} = 40 - 23 = 17 \text{ cm} \).
Now, the side lengths are \( a=8 \text{ cm}, b=15 \text{ cm}, c=17 \text{ cm} \).
Next, calculate the semi-perimeter 's':
\( s = \frac{\text{Perimeter}}{2} = \frac{40}{2} = 20 \text{ cm} \).
Finally, use Heron's formula to determine the area of the plot:
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
\( = \sqrt{20(20-8)(20-15)(20-17)} \)
\( = \sqrt{20 \times 12 \times 5 \times 3} \)
\( = \sqrt{(4 \times 5) \times (3 \times 4) \times 5 \times 3} \)
\( = \sqrt{4^2 \times 5^2 \times 3^2} \)
\( = 4 \times 5 \times 3 \)
\( = 60 \text{ cm}^2 \).
This step-by-step approach ensures accurate calculation of the area.
In simple words: First, find the missing side length using the given perimeter. Then, calculate the area of the triangle using Heron's formula.

🎯 Exam Tip: Always ensure all three side lengths are in the same unit before applying Heron's formula, and calculate the semi-perimeter carefully. Pay attention to the units for the final area.

Question 7. The sides of a triangular field are 20 m, 51 m and 37 m. Find the number of flowerbeds that can be prepared if each bed is to contain 2 x 3 sq. metre space.
Answer: First, calculate the total area of the triangular field using Heron's formula. Let the sides be \( a = 20 \text{ m} \), \( b = 51 \text{ m} \), and \( c = 37 \text{ m} \).
The perimeter is \( 20 + 51 + 37 = 108 \text{ m} \).
The semi-perimeter 's' is \( s = \frac{108}{2} = 54 \text{ m} \).
Area of triangular field \( = \sqrt{s(s-a)(s-b)(s-c)} \)
\( = \sqrt{54(54-20)(54-51)(54-37)} \)
\( = \sqrt{54 \times 34 \times 3 \times 17} \)
\( = \sqrt{(2 \times 3^3) \times (2 \times 17) \times 3 \times 17} \)
\( = \sqrt{2^2 \times 3^4 \times 17^2} \)
\( = 2 \times 3^2 \times 17 \)
\( = 2 \times 9 \times 17 \)
\( = 18 \times 17 = 306 \text{ m}^2 \).
The area required for one flowerbed is \( 2 \times 3 = 6 \text{ m}^2 \).
To find the number of flowerbeds, divide the total field area by the area of one flowerbed:
Number of flower-beds \( = \frac{\text{Area of triangular field}}{\text{Area of one flower-bed}} = \frac{306}{6} = 51 \).
This helps in planning space efficiently.
In simple words: Find the total area of the field using Heron's formula. Then, divide this total area by the space needed for one flowerbed to find out how many can be made.

🎯 Exam Tip: Ensure all units are consistent (e.g., all in meters) before calculations, and double-check your factorization when simplifying the square root for Heron's formula. Always state the final answer with appropriate units or as a whole number if counting items.