Chapter 8: Q44. (page 289)

Given: $P Q R S$ is a rectangle.
$∠ A$ is a right angle.
Prove: $B S \cdot R C = P S \cdot Q R = {(P S)}^{2}$

Short Answer

Expert verified

The result is,

$B S \cdot R C = P S \cdot Q R = P S^{2}$ .

Step by step solution

Step 1. Given information.

The triangle is shown below:

$P Q R S$ is rectangle.

Step 2. Consider triangles BSP and PAQ.

$\begin{array}{l} ∠ B S P = ∠ P A Q = 90 ° \\ ∠ P B S = ∠ Q P A (Correspondingangle) \\ ∴ Δ B S P ~ Δ P A Q ...... (1) \end{array}$

Step 3. Consider triangles PAQ and QRC.

$\begin{array}{l} ∠ Q R C = ∠ P A Q = 90 ° \\ ∠ P Q A = ∠ Q C R (Correspondingangle) \\ ∴ Δ Q R C ~ Δ P A Q ...... (2) \end{array}$

From equation $(1)$ and $(2)$ ,

$Δ Q R C ~ Δ B S P$

Hence,

$\begin{array}{l} \frac{B S}{Q R} = \frac{P S}{R C} \\ B S \cdot R C = P S \cdot Q R ...... (3) \end{array}$

Also, $P Q R S$ is rectangle.

Hence, side $P S$ is equal to $Q R$ .

$\begin{matrix} B S \cdot R C = P S \cdot P S \\ B S \cdot R C = P S^{2} ...... (4) \end{matrix}$

Thus, from equations $(3)$ and $(4)$ ,

$B S \cdot R C = P S \cdot Q R = P S^{2}$

Therefore, it is proved that, $B S \cdot R C = P S \cdot Q R = P S^{2}$ .

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Given: $P Q R S$ is a rectangle.
$∠ A$ is a right angle.
Prove: $B S \cdot R C = P S \cdot Q R = {(P S)}^{2}$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Consider triangles BSP and PAQ.

Step 3. Consider triangles PAQ and QRC.

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Given: PQRSis a rectangle. ∠A is a right angle.Prove: BS·RC=PS·QR=PS2

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Consider triangles BSP and PAQ.

Step 3. Consider triangles PAQ and QRC.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

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Theoretical and Mathematical Physics

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Calculus

Study anywhere. Anytime. Across all devices.

Given: $P Q R S$ is a rectangle.
$∠ A$ is a right angle.
Prove: $B S \cdot R C = P S \cdot Q R = {(P S)}^{2}$