A right-angled triangle ABC has legs AB=6 cm and BC=8 cm. An altitude BD is drawn from the vertex B to the hypotenuse AC. What is the length of the altitude BD?

Correct — A. Find the triangle's area with each pair of base and height, and set the results equal.

The legs meet at B, so AC is the hypotenuse
Pythagoras: AC = √(6² + 8²) = √100 = 10 cm
Area from the legs: ½ × 6 × 8 = 24 cm²

Area with AC as base: ½ × 10 × BD = 24
Solve: BD = 48 ÷ 10 = 4.8 cm → option (a)

Why the others are wrong

• (b)
  5.6 cm — 5.6 cm fails the area check: ½ × 10 × 5.6 = 28 cm², but the legs give 24 cm². The two areas must match.
• (c)
  6.4 cm — 6.4 cm is longer than the leg AB = 6 cm, impossible for the perpendicular from B, the shortest path to AC. 6.4 cm is DC = 8² ÷ 10, a piece of the hypotenuse.
• (d)
  7.2 cm — 7.2 cm is longer than AB = 6 cm. The perpendicular from B to AC must be shorter than both legs, which also run from B to that line.

The altitude to the hypotenuse falls out of the area. With the legs as base and height the area is ½ × AB × BC; with the hypotenuse as base it is ½ × AC × BD.

Equate the two and BD = (AB × BC) ÷ AC, the product of the legs divided by the hypotenuse: 48 ÷ 10 = 4.8 cm.

Key facts

• Altitude to the hypotenuse = (leg × leg) ÷ hypotenuse.
• 6-8-10 is the 3-4-5 triple doubled.
• BD² = AD × DC, where D is the foot of the altitude on the hypotenuse.

Common traps

• Answering 6.4 cm, the segment DC of the hypotenuse, instead of the altitude BD.
• Adding the legs, 6 + 8 = 14, instead of using √(6² + 8²) = 10 for the hypotenuse.

No directly related past PYQ was found.