In △PQR, ∠PQR=90°, seg QD ⊥ seg PR then find x, y and z.

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Solution

According to geometric mean,
$Q S^{2} = P R \times S R \Rightarrow x^{2} = 10 c m \times 8 c m = 80 c m^{2} \Rightarrow x = 4 \sqrt{5} c m$

In $△ Q S R, 8^{2} + x^{2} = y^{2} \Rightarrow 64 + 80 = y^{2} \Rightarrow y^{2} = 144 \Rightarrow y = 12 c m$

In $△ P S Q, 10^{2} + x^{2} = z^{2} \Rightarrow 100 + 80 = z^{2} \Rightarrow z^{2} = 180 \Rightarrow z = 6 \sqrt{5} c m$

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△ P Q R, ∠ P Q R = 90^{\circ}

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