Latur School Exam Papers 2026
Raja Narayanlal Lahoti English School
Class 9 | Term Test-I
36 Triangular Park Analysis (Internal Choice)
In my colony a park is situated in front of my house. This park is built in the shape of a triangle (ABC) with the following sides: 120m, 80m and 50m. Nowadays, some animals entered the park to destroy and eat plants. So, our ward member has decided to put a railing around the park for protecting plants and grass. He ordered a gardener to place a railing all around this park and maintain grass inside. Costing is decided at Rs. 10 per meter for railing around the park.[cite: 49]
Based on this information, answer the following:
(i) What is the perimeter of the park? (1 Mark)
(ii) Calculate the semi-perimeter of the triangular park. (1 Mark)
(iii) Calculate the area in which planting is needed? (2 Marks)
(i) What is the perimeter of the park? (1 Mark)
(ii) Calculate the semi-perimeter of the triangular park. (1 Mark)
(iii) Calculate the area in which planting is needed? (2 Marks)
OR
(iii) Find the cost of fencing it with barbed wire at the rate of Rs 20 per metre leaving a space 3m wide for a gate on one side. (2 Marks)
Concept:
1
Perimeter & Semi-perimeter: The perimeter (\(P\)) of a triangle is the sum of its three sides (\(a+b+c\)). The semi-perimeter (\(s\)) is half of the perimeter (\(P/2\)).
2
Heron's Formula: When only the lengths of the three sides of a triangle are known, the area can be calculated using Heron's formula: \( \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \).
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Hint:
1
For the main part (iii), apply Heron's formula. Write down the values for \(s\), \( (s-a) \), \( (s-b) \), and \( (s-c) \) first. Then factorize the numbers inside the square root into prime numbers to easily pull out pairs!
2
For the OR part (iii), remember that the "space for a gate" means you do not put a fence there. Subtract the gate width from the total perimeter before multiplying by the cost rate.
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Mistakes to Avoid:
!
Giant Multiplication in Heron's: Do NOT multiply \(125 \times 5 \times 45 \times 75\) to get \(2,109,375\) and then try to find the square root! Always factorize the smaller numbers instead (e.g., write 75 as \(25 \times 3\)) to group perfect squares.
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Let the sides of the triangular park be:
\( a = 120 \text{ m} \), \( b = 80 \text{ m} \), \( c = 50 \text{ m} \)
PART (i) SOLUTION:
Perimeter (\(P\)) = Sum of all sides
\( P = 120 + 80 + 50 \)
\( P = 250 \text{ m} \) [1 M]
PART (ii) SOLUTION:
Semi-perimeter (\(s\)) = \( \frac{P}{2} \)
\( s = \frac{250}{2} = 125 \text{ m} \) [1 M]
MAIN PART (iii) SOLUTION:
Area to be planted = Area of the triangle.
Using Heron's Formula: Area = \( \sqrt{s(s-a)(s-b)(s-c)} \) [½ M]
First, calculate the differences:
\( (s-a) = 125 - 120 = 5 \text{ m} \)
\( (s-b) = 125 - 80 = 45 \text{ m} \)
\( (s-c) = 125 - 50 = 75 \text{ m} \)
Substitute into the formula:
Area = \( \sqrt{125 \times 5 \times 45 \times 75} \) [½ M]
Factorize to find perfect squares:
Area = \( \sqrt{(25 \times 5) \times 5 \times (9 \times 5) \times (25 \times 3)} \)
Area = \( \sqrt{(25 \times 25) \times (5 \times 5) \times 9 \times (5 \times 3)} \)
Take the square roots of pairs:
Area = \( 25 \times 5 \times 3 \times \sqrt{15} \) [½ M]
Area = \( 375\sqrt{15} \text{ m}^2 \) [½ M]
Area = \(375\sqrt{15} \text{ m}^2\)
OR
OR PART (iii) SOLUTION:
Total perimeter of the park = \( 250 \text{ m} \)
Space left for the gate = \( 3 \text{ m} \)
Total length of wire required for fencing = Perimeter - Gate space
Length = \( 250 - 3 = 247 \text{ m} \) [1 M]
Rate of fencing = Rs. 20 per metre
Total Cost = \( \text{Length} \times \text{Rate} \)
Total Cost = \( 247 \times 20 \)
Total Cost = \( 4940 \) [1 M]
Total Cost = Rs. 4940
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