triangle pqr is an equilateral triangle point s is on seg qr such that qs=1/3 qr prove that 9ps^2=7pq^2
Answers
Answered by
161
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
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
Attachments:
Answered by
47
Answer: answer is in attachment..
Step-by-step explanation:
Attachments:
Similar questions