Question

In an equilateral triangle PQR, the side QR is trisected at S. Prove that 9PS square

= 7PQ square​

Answer 1

Answer:

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Answer 2

Given :- In an equilateral triangle PQR, the side QR is trisected at S.

To Prove :- 9•PS² = 7•PQ² .

Formula used :-

• Altitude of an equilateral triangle = (√3/2) × side .
• Altitude of an equilateral triangle bisect the base .
• Pythagoras theorem :- (Perpendicular)² + (Base)² = (Hypotenuse)²

Solution :-

Let us assume that, each side of the given equilateral triangle PQR is equal to a unit and let PT is the altitude .

So,

→ PT = (√3a/2) unit { Altitude = (√3/2)•side } ------- Equation (1)

→ QT = TR = QR/2 = (a/2) unit { since altitude bisect the base }

Now, we have given that, QR is trisected as S .

So,

→ QS = (1/3)•QR

→ QS = (a/3) unit

then,

→ ST = QT - QS

putting values from above in RHS,

→ ST = (a/2) - (a/3)

→ ST = (3a - 2a)/6

→ ST = (a/6) unit -------- Equation (2)

Now, in right angled ∆PTS we have,

→ PS² = PT² + ST²

putting values from Equation (1) and Equation (2) in RHS,

→ PS² = (√3a/2)² + (a/6)²

→ PS² =(3a²/4) + (a²/36)

→ PS² = (27a² + a²)/36

→ PS² = (28a²/36)

→ PS² = (7a²/9)

→ 9•PS² = 7a²

since each side of given equilateral triangle is equal to a unit, finally putting a = PQ in RHS,

→ 9•PS² = 7•PQ² (Proved)

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