Question

16*. In the figure 2.35, A PQR is an
equilatral triangle. Point S is on
seg QR such that
QS = QR.
Prove that : 9 PS? = 7 PQ?​

Answer 1

Answer:

Let the side length of an equilateral triangle ∆PQR is a and PT be the altitude of ∆PQR as shown in figure.

Now, PQ = QR = PR = a

We know, in case of equilateral triangle, altitude divide the base in two equal parts. e.g., QT = TR = QR/2 = a/2

Given, QS = QR/3 = a/3

∴ ST = QT - QS = a/2 - a/3 = a/6

Also PT = √(PQ² - QT²) = √(a² -a²/4) = √3a/2

Now, use Pythagoras theorem for ∆PST

PS² = ST² + PT²

⇒PS² = (a/6)² + (√3a/2)² = a²/36 + 3a²/4

= (a² + 27a²)/36 = 28a²/36 = 7a²/9

⇒9PS²  = 7PQ² 

Hence proved