Question:
In a triangle ABC, the sides are in the ratio 3 : 4 : 5. If the area of the triangle is 96 sq units, what is its perimeter?
In a triangle ABC, the sides are in the ratio 3 : 4 : 5. If the area of the triangle is 96 sq units, what is its perimeter?
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Recognizing Pythagorean triplets like (3, 4, 5), (5, 12, 13), (8, 15, 17), etc., can save a lot of time by immediately identifying a triangle as right-angled, which simplifies area calculations.
Updated On: Jul 7, 2026
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Correct Answer: 48
Approach Solution - 1
Approach: A 3 : 4 : 5 triangle is automatically right-angled (Pythagorean triple), so the two smaller sides are base and height \(-\) no need for Heron's formula at all.
Step 1: Let the sides be \(3x, 4x, 5x\). Since \((3x)^2 + (4x)^2 = 25x^2 = (5x)^2\), the angle between the sides \(3x\) and \(4x\) is \(90^\circ\).
Step 2: So area \(= \tfrac{1}{2}(3x)(4x) = 6x^2\). Set equal to 96:
\[ 6x^2 = 96 \implies x^2 = 16 \implies x = 4. \]
Step 3: Perimeter \(= 3x + 4x + 5x = 12x = 12 \times 4 = 48\) units.
(Sides are \(12, 16, 20\); check area \(= \tfrac12 \times 12 \times 16 = 96\). Correct.)
Final Answer: Perimeter \(=\) 48 units.
Step 1: Let the sides be \(3x, 4x, 5x\). Since \((3x)^2 + (4x)^2 = 25x^2 = (5x)^2\), the angle between the sides \(3x\) and \(4x\) is \(90^\circ\).
Step 2: So area \(= \tfrac{1}{2}(3x)(4x) = 6x^2\). Set equal to 96:
\[ 6x^2 = 96 \implies x^2 = 16 \implies x = 4. \]
Step 3: Perimeter \(= 3x + 4x + 5x = 12x = 12 \times 4 = 48\) units.
(Sides are \(12, 16, 20\); check area \(= \tfrac12 \times 12 \times 16 = 96\). Correct.)
Final Answer: Perimeter \(=\) 48 units.
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Approach Solution -2
Right-triangle recognition method: A \( 3:4:5 \) ratio is the classic Pythagorean triple, so the triangle with sides \( 3k, 4k, 5k \) is right-angled, with the two shorter sides as the legs. The area of a right triangle is half the product of its legs:
\[ \text{Area}=\frac{1}{2}(3k)(4k)=6k^2. \] Given the area is \( 96 \): \( 6k^2=96 \Rightarrow k^2=16 \Rightarrow k=4 \).
So the sides are \( 12, 16, 20 \), and the perimeter is:
\[ \boxed{12+16+20=48} \]
\[ \text{Area}=\frac{1}{2}(3k)(4k)=6k^2. \] Given the area is \( 96 \): \( 6k^2=96 \Rightarrow k^2=16 \Rightarrow k=4 \).
So the sides are \( 12, 16, 20 \), and the perimeter is:
\[ \boxed{12+16+20=48} \]
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