Question:medium

The ratios of sides in a triangle ABC are $5 : 12 : 13$ and its area is 270 . Then sides of the triangle are

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Always scale ratio using area condition: Area $\propto k^2$.
Updated On: May 14, 2026
  • $5, 12, 13$
  • $10, 24, 26$
  • $15, 36, 39$
  • $20, 48, 52$
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
We are provided the ratio of the side lengths of a triangle. The numbers 5, 12, and 13 are a very common Pythagorean triple. This instantly tells us the triangle is a right-angled triangle. This property allows us to use the simple area formula for right triangles instead of the more complex Heron's formula. Step 2: Key Formula or Approach:
1. Pythagorean Theorem test: If $a^2 + b^2 = c^2$, the triangle is right-angled. 2. Area of a right-angled triangle $= \frac{1}{2} \times \text{base} \times \text{height}$. 3. Introduce a scaling factor $x$ for the side ratios. Step 3: Detailed Explanation:
Let the actual lengths of the sides of the triangle be $5x$, $12x$, and $13x$, where $x$ is a positive constant scale factor. First, verify it's a right-angled triangle: Check if $(5x)^2 + (12x)^2 = (13x)^2$: LHS $= 25x^2 + 144x^2 = 169x^2$ RHS $= (13x)^2 = 169x^2$ Since LHS = RHS, it is a right-angled triangle. The legs forming the right angle are $5x$ and $12x$, and the hypotenuse is $13x$. The area of this right-angled triangle is: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] \[ \text{Area} = \frac{1}{2} \cdot (5x) \cdot (12x) = 30x^2 \] We are given that the area is exactly 270. Set up the equation: \[ 30x^2 = 270 \] Solve for $x^2$: \[ x^2 = \frac{270}{30} = 9 \] Taking the principal square root (as side lengths are positive): \[ x = 3 \] Now, substitute the value of $x$ back to find the actual lengths of the sides: Side 1 $= 5x = 5(3) = 15$ Side 2 $= 12x = 12(3) = 36$ Side 3 $= 13x = 13(3) = 39$ The sides of the triangle are 15, 36, and 39. Step 4: Final Answer:
The sides are 15, 36, 39.
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