Question 10
In given figure PQR is a right triangle, right angled at Q and QS $⊥$ PR. If PQ = 6cm and PS = 4cm, then find QS, RS and QR.

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Solution

Given, $Δ P Q R$ in which $∠ Q = 90^{\circ}$ , $Q S ⊥ P R$ and PQ = 6cm, PS = 4cm.

In $Δ S Q P$ and $Δ S R Q$ ,
$∠ P S Q = ∠ R S Q$ [each equal to $90^{\circ}$ ]
$∠ S P Q = ∠ S Q R$ [each equal to $90^{\circ} - ∠ R$ ]
$∴$ $Δ S Q P \sim Δ S R Q$ [ by AAA similarity criterion]
Then, $\frac{S Q}{P S} = \frac{S R}{S Q}$
$\Rightarrow S Q^{2} = P S \times S R$ ..........(i)

In right angled $Δ P S Q$ ,
$P Q^{2} = P S^{2} + Q S^{2}$ [by Pythagoras theorem]
$\Rightarrow (6)^{2} = (4)^{2} + Q S^{2}$
$\Rightarrow 36 = 16 + Q S^{2}$
$\Rightarrow Q S^{2} = 36 - 16 = 20$
$∴$ $Q S = \sqrt{20} = 2 \sqrt{5} c m$
On putting the value of QS in Eq..(i), we get
$(2 \sqrt{5})^{2} = 4 \times S R$
$\Rightarrow S R = \frac{4 \times 5}{4} = 5 c m$

In right angled $Δ Q S R$ ,
$Q R^{2} = Q S^{2} + S R^{2}$
$\Rightarrow Q R^{2} = (2 \sqrt{5})^{2} + (5)^{2}$
$\Rightarrow Q R^{2} = 20 + 25$
$∴$ $Q R = \sqrt{45} = 3 \sqrt{5} c m$
Hence, $Q S = 2 \sqrt{5} c m, R S = 5 c m$ and $Q R = 3 \sqrt{5} c m$

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Similar questions

In a triangle PQR right angled at Q if QS = SR then prove that PR² = 4PS²- PQ²

In quadrilateral PQRS, PQ = 4 cm, QR = 3 cm, RS = 5.9 cm, QS = 4.8 cm and PR = 5.5 cm. Find the value of PS using constructions.

Δ

(A) (P R) (Q R) = R S^{2}

(B) Q S^{2} + R S^{2} = Q R^{2}

(C) P R^{2} + Q R^{2} = P Q^{2}

(D) P S^{2} + R S^{2} = P R^{2}

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