Question

If PQRS is a square and point T can lie anywhere on the line PQ, then how many equilateral and right-angle triangles can be formed with SR as base? ​

Answer 1

Answer:

Because PQRS is a square

∠PSR=∠QRS=90

∘

Now In △SRT

∠TSR=∠TRS=60

∘

∠PSR+∠TSR=∠QRS+∠TRS

⟹∠TSP=∠TRQ

Now in △TSP and △TRQ

TS=TR

∠TSP=∠TRQ

PS=QR

Therefore , △TSP≡△TRQ

So PT=QT.

Answer 2

According the question, PQRS is a square

∠PSR=∠QRS= ∠RQP= ∠ QPS=90°

Now we take a point T in the middle of line PQ .

∴ T will be the midpoint of PQ.

Let us join TR  & TS

Now In △SRT

From Euclid's axiom, things which are equal to the same thing are equal.

Hence TS =SR =TR

Therefore  △SRT is a equilateral Δ

and  sum of three angles will be 60°.

so ∠TSR=∠TRS=∠ STR= 60∘

Now in △TSP and △TRQ

• ∠SPT = ∠ RQT = Right angle= 90°
• PT = TQ                  ( T is the mid point on line PQ)
• PS = QR                   ( four sides of a square will be equal )

∴  △TSP and △TRQ   both  are  right-angle triangles.

Hence in square PQRS there are one equilateral triangle named  △SRT, and two right angle Δs named △TSP and △TRQ formed.

Ans :- In square PQRS there are one equilateral triangle named  △SRT , and two right angle Δs named △TSP and △TRQ formed.

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