In the given figure, and ∠a = ∠b; show that △ PQS ~ △ TQR.

In Δ PQR, ⇒ ∠a = ∠b (Given) ⇒ PR = PQ (In a triangle sides opposite to equal angles are equal) Given, In Δ PQS and Δ TQR ⇒ ∠PQS = ∠TQR = ∠b (Same angle) ⇒ ...... [From (1)] ⇒ Δ PQS ~ Δ TQR (By S.A.S. axiom) Hence, proved that Δ PQS ~ Δ TQR.

Mathematics

In the given figure, $\dfrac{QR}{QS} = \dfrac{QT}{PR}$ and ∠a = ∠b; show that △ PQS ~ △ TQR.

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Answer

In Δ PQR,

⇒ ∠a = ∠b (Given)

⇒ PR = PQ (In a triangle sides opposite to equal angles are equal)

Given,

$\phantom{\therefore} \dfrac{QR}{QS} = \dfrac{QT}{PR} \\[1em] \therefore \dfrac{QR}{QS} = \dfrac{QT}{PQ} \quad ……[1]$

In Δ PQS and Δ TQR

⇒ ∠PQS = ∠TQR = ∠b (Same angle)

⇒ $\dfrac{QR}{QS} = \dfrac{QT}{PQ}$ …… [From (1)]

⇒ Δ PQS ~ Δ TQR (By S.A.S. axiom)

Hence, proved that Δ PQS ~ Δ TQR.

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Class-interval	Frequency
0-20	17
20-40	m
40-60	32
60-80	24
80-100	19

Items	Marked price	Discount	No. of items	GST
A	₹ 1000	10%	2	0%
B	₹ 2000	20%	1	5%

Mathematics

In the given figure, $\dfrac{QR}{QS} = \dfrac{QT}{PR}$ and ∠a = ∠b; show that △ PQS ~ △ TQR.

Similarity

Answer

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In the given figure, QRQS=QTPR\dfrac{QR}{QS} = \dfrac{QT}{PR}QSQR​=PRQT​ and ∠a = ∠b; show that △ PQS ~ △ TQR.

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