Question

Р
S
100-
X
PQRS is an isosceles trapezium and
QR is extended to X.
If <SRX=100°, find all angles of
PQRS.​

Answer 1

Given :- PQRS is an isosceles trapezium and QR is extended to X. If <SRX=100°, find all angles of PQRS. ?

Solution :-

from image we have :-

• PQRS is an isosceles trapezium with PQ and SR as parallel sides and PS is equal to QR.
• side QR is extended to X.

we have now,

→ Q - R - X is a straight line.

→ ∠SRX = 100° (given) .

So,

→ ∠SRQ = 180° - 100° = 80° (Ans.) { Linear Pair. }

Now,

→ PQ || SR .

So,

→ ∠SRQ + ∠PQR = 180° { Interior angles on the same side of the transversal are supplementary.}

Similarly,

→ ∠RSP + ∠SPQ = 180° { Interior angles between parallel lines.}

So,

→ ∠PQR = 180° - ∠SRQ

→ ∠PQR = 180° - 80° = 100° (Ans.)

Now, in an isosceles ∆ :-

• The two top angles are equal to each other. and, the two bottom angles are equal to each other as well.

Therefore,

→ ∠PQR = ∠SPQ = 100° (Ans.)

→ ∠QRS = ∠PSR = 80° (Ans.)

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