Advertisements
Advertisements
Question
The perpendicular PS on the base QR of a ∆PQR intersects QR at S, such that QS = 3 SR. Prove that 2PQ2 = 2PR2 + QR2
Advertisements
Solution
Given QS = 3SR
QR = QS + SR
= 3SR + SR = 4SR
SR = `1/4` QR ...(1)
QS = 3SR
SR = `"QS"/3` ...(2)
From (1) and (2) we get
`1/4 "QR" = "QS"/3`
∴ QS = `3/4` QR ...(3)
In the right ∆PQS,
PQ2 = PS2 + QS2 ...(4)
Similarly in ∆PSR
PR2 = PS2 + SR2 ...(5)
Subtract (4) and (5)
PQ2 – PR2 = PS2 + QS2 – PS2 – SR2
= QS2 – SR2
PQ2 – PR2 = `[3/4 "QR"]^2 - ["QR"/4]^2`
From (3) and (1)
= `(9"QR"^2)/16 - "QR"^2/16`
= `(8"QR"^2)/16`
PQ2 – PR2 = `1/2 "QR"^2`
2PQ2 – 2PR2 = QR2
2PQ2 = 2PR2 + QR2
Hence the proved.
APPEARS IN
RELATED QUESTIONS
ABCD is a rectangle whose three vertices are B (4, 0), C(4, 3) and D(0,3). The length of one of its diagonals is
(A) 5
(B) 4
(C) 3
(D) 25
If the sides of the triangle are in the ratio 1: `sqrt2`: 1, show that is a right-angled triangle.
In triangle ABC, given below, AB = 8 cm, BC = 6 cm and AC = 3 cm. Calculate the length of OC.
In the given figure, BL and CM are medians of a ∆ABC right-angled at A. Prove that 4 (BL2 + CM2) = 5 BC2.
In triangle PQR, angle Q = 90°, find: PQ, if PR = 34 cm and QR = 30 cm
The foot of a ladder is 6m away from a wall and its top reaches a window 8m above the ground. If the ladder is shifted in such a way that its foot is 8m away from the wall to what height does its tip reach?
In a right-angled triangle ABC,ABC = 90°, AC = 10 cm, BC = 6 cm and BC produced to D such CD = 9 cm. Find the length of AD.
In a square PQRS of side 5 cm, A, B, C and D are points on sides PQ, QR, RS and SP respectively such as PA = PD = RB = RC = 2 cm. Prove that ABCD is a rectangle. Also, find the area and perimeter of the rectangle.
The hypotenuse of a right angled triangle of sides 12 cm and 16 cm is __________
The perimeter of the rectangle whose length is 60 cm and a diagonal is 61 cm is ______.
