Each of the two equal sides of an isosceles triangle is 14 cm long and the angle between them is 90°. What is the length of the shortest altitude of the triangle?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(7\sqrt{2}\) cm
In a right triangle the two legs are altitudes of each other, and the third altitude falls on the hypotenuse. Area \(=\frac{1}{2}\times\) base \(\times\) height for any side, so the altitude on the longest side is the shortest.
Hypotenuse \(=\sqrt{14^2+14^2}=14\sqrt{2}\) cm
Area \(=\displaystyle\frac{1}{2}\times14\times14=98\) cm2
Altitude on the hypotenuse: \(\displaystyle\frac{1}{2}\times14\sqrt{2}\times h=98\), so \(h=\displaystyle\frac{98}{7\sqrt{2}}=\frac{14}{\sqrt{2}}=7\sqrt{2}\) cm
The other two altitudes are the legs, 14 cm each; \(7\sqrt{2}\approx9.9\) cm is the shortest
Why the other options are incorrect
Option (1) 14 cm — Incorrect. This is the altitude to each equal side, the longest altitude of this triangle.
Option (2) 7 cm — Incorrect. This is half of a leg; no altitude of the triangle has this length.
Option (3) \(14\sqrt{2}\) cm — Incorrect. This is the length of the hypotenuse itself, not an altitude.