A right-angled isosceles triangle has an area of 50 square units. Its hypotenuse is (in units):

Correct — D. In a right-angled isosceles triangle the two equal sides are the legs enclosing the right angle, so each one is the height for the other.

Area = a²⁄2 = 50
a² = 100 → a = 10 units
h² = 10² + 10² = 200
h = √200 = √(100 × 2) = 10√2 units

The four choices are printed as images; the one reading 10√2 is option (d).

Why the others are wrong

• (a)
  Option (a) reads 5√5, i.e. √125. A 45-45-90 triangle with that hypotenuse has legs of 5√5⁄√2 and an area of 31.25 square units, not 50.
• (b)
  Option (b) reads 10√3 = √300, the hypotenuse pattern of a 30-60-90 triangle. Here the sides run 1 : 1 : √2, and 10√3 would mean an area of 75 square units.
• (c)
  Option (c) reads 5√2 — what you get by dropping the ½ and solving a² = 50, then reporting that leg as the hypotenuse. A hypotenuse of 5√2 means legs of 5 and an area of 12.5.

A right-angled isosceles triangle is the 45-45-90 triangle, sides in the ratio 1 : 1 : √2.

Because the two equal sides are perpendicular to each other, either serves as the height for the other — so area = a²⁄2, with no separate altitude to find.

That gives a one-line route from area straight to hypotenuse: a = √(2 × area) and h = a√2, so h = 2√(area). With area 50 that is 2√50 = 10√2. The last step is pure surd work — pull the perfect square out of √200.

Key facts

• A right-angled isosceles triangle has angles 45°, 45°, 90°.
• Its sides are in the ratio 1 : 1 : √2.
• With legs a: area = a²⁄2 and hypotenuse = a√2, so hypotenuse = 2√(area).
• √200 = √(100 × 2) = 10√2 ≈ 14.14.

Common traps

• Treating the hypotenuse as the base in ½ × base × height
• Dropping the ½, solving a² = 50, and reporting the leg 5√2 as the hypotenuse
• Stopping at √200 or 2√50 and then matching no option on screen

No directly related past PYQ was found.