Discussion
If ΔABC is an isosceles triangles such that AB = BC = 8 cm and ∠ABC = 90°, then the length of the altitude drawn from B to AC is :
Answer: D
The altitude is \(4\sqrt{2}\) cm.
With AB = BC = 8 cm and the right angle at B, the hypotenuse is:
\[ AC = \sqrt{8^{2} + 8^{2}} = 8\sqrt{2} \]
The altitude from the right angle to the hypotenuse is the product of the legs divided by the hypotenuse:
\[ \begin{aligned} h &= \frac{8 \times 8}{8\sqrt{2}} \\ &= \frac{8}{\sqrt{2}} = 4\sqrt{2} \end{aligned} \]
In an isosceles right triangle this altitude is also the median, half the hypotenuse, which gives the same \(4\sqrt{2}\), about 5.66 cm. 3 cm, \(2\sqrt{2}\) and \(3\sqrt{2}\) are all too short.
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