Question

A PQR is an equilatral triangle. Point S is on seg QR such that QS = 1/3 QR.
Prove that : 9 PS² = 7 PQ²

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Answer 1

Step-by-step explanation:

Let the side of equilateral triangle ∆PQR be x.

PT be the altitude of the ∆PQR.

We know that, in equilateral triangle, altitude divides the base in two equal parts.

∴ QT = TR =

12QR=x2

Given: QS =

13

QR =

x3

∴ST=QT−QS=x2−x3=x6

According to Pythagoras theorem,

In ∆PQT

PQ2=QT2+PT2

⇒(x)2=(x2)2+PT2

⇒x2=x24+PT2

⇒PT2=x2−x24

⇒PT2=3x24

⇒PT=3x2

In ∆PST

PS2=ST2+PT2

⇒PS2=(x6)2+(3x2)2

⇒PS2=x236+3x24

⇒PS2=x2+27x236

⇒PS2=28x236

⇒PS2=7x29

⇒9PS2=7PQ2

Hence, 9 PS2 = 7 PQ2.