Question:medium

The perimeter of a right-angled triangle ABC, right angled at A, is (3+√3) cm. What is the area of the triangle?

Statement 1: AC ≠ AB.
Statement 2: ∠ABC = 30°.

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A right triangle's area is fixed once you know the perimeter AND one non-right angle (which fixes the full shape via 30-60-90 side ratios); check what each statement actually fixes.
Updated On: Jul 20, 2026
  • If the data in statement (1) alone is sufficient to answer the question, but the data in statement (2) alone is not sufficient.
  • If the data in statement (2) alone is sufficient to answer the question, but the data in statement (1) alone is not sufficient.
  • If the data in both the statements together are needed to answer the question.
  • If either statement (1) alone or statement (2) alone is sufficient to answer the question.
  • If the data in neither statement (1) nor statement (2) is sufficient to answer the question, and more data is needed.
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The Correct Option is B

Solution and Explanation

A right triangle's shape (not just its size) needs to be pinned down before area is computable from perimeter alone, because the same perimeter can be shared by many differently-shaped right triangles. Shape is fixed once we know one of the acute angles (the right angle is already fixed at A).

Statement 1 (AC ≠ AB) only rules out one specific shape - the 45-45-90 case - but leaves every other possible acute-angle combination open. A 20-70-90 triangle, a 10-80-90 triangle, and so on could all have the same perimeter of (3+√3) cm and all satisfy AC ≠ AB, yet each would enclose a different area. So this statement cannot pin a single numeric area.

Statement 2 (angle B = 30 degrees) fully fixes the shape: with the right angle at A already fixed, angle C is forced to be 60 degrees, making it a 30-60-90 triangle. Using trigonometry directly with hypotenuse BC = h: AB = h cos(30) = h(√3/2), and AC = h sin(30) = h/2. Perimeter = h(√3/2) + h/2 + h = h[(√3+1+2)/2] = h(√3+3)/2. Setting this equal to (3+√3): h(√3+3)/2 = 3+√3, so h = 2. Then AB = 2(√3/2) = √3, AC = 2(1/2) = 1. Area = (1/2)(AB)(AC) = (1/2)(√3)(1) = √3/2 cm² - one exact number.

Because the perimeter is already fixed in the question and statement 2 supplies the missing angle needed to lock down the whole triangle's shape, statement 2 alone is sufficient while statement 1 alone is not; the answer here works out to (b), which differs from the (e) shown in the source key - this is flagged rather than silently overridden, since the trigonometric check (1²+√3²=2²) confirms the triangle is valid and the area unique.
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