In the figure-QR-QS - QT-PR and- 1 - - 2-Show that - PQS - TQR-

In Δ PQR, ∠ PQR = ∠ PRQ (Given) nbsp;∴ PQ = PR ...(i) (sides opposite to equal angles are equal) Given, QR/QS = QT/PR Using (i), we get nbsp;QR/QS = Q ...

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Standard X

Mathematics

Pythagoras Theorem

In the figure...

Question

In the figure,
$\frac{Q R}{Q S} = \frac{Q T}{P R} a n d$
$∠ 1 = ∠ 2$ .
$S h o w t h a t Δ P Q S \sim Δ T Q R$ .

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Solution

$I n Δ P Q R, ∠ P Q R = ∠ P R Q (G i v e n)$

$∴ P Q = P R . . . (i)$ (sides opposite to equal angles are equal)

$G i v e n, \frac{Q R}{Q S} = \frac{Q T}{P R}$

Using ( $i$ ), we get
$\frac{Q R}{Q S} = \frac{Q T}{Q P} . . . (i i)$

In $Δ P Q S$ and $Δ T Q R$ ,
$\frac{Q R}{Q S} = \frac{Q T}{Q P}$ [using ( $i i$ )]

$∠ Q = ∠ Q$ (Common)

$∴ Δ P Q S \sim Δ T Q R$ [SAS similarity criterion]

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In the figure, QRQS=QTPR and ∠1=∠2. Show that ΔPQS∼ΔTQR.

In ΔPQR,∠PQR=∠PRQ (Given) ∴PQ=PR...(i) (sides opposite to equal angles are equal) Given, QRQS=QTPR Using (i), we get QRQS=QTQP...(ii) In ΔPQS and ΔTQR, QRQS=QTQP [using (ii)] ∠Q=∠Q (Common) ∴ΔPQS∼ΔTQR [SAS similarity criterion]

In the figure,
$\frac{Q R}{Q S} = \frac{Q T}{P R} a n d$
$∠ 1 = ∠ 2$ .
$S h o w t h a t Δ P Q S \sim Δ T Q R$ .