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
Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
Prove that, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the square of remaining two sides.
Pranali and Prasad started walking to the East and to the North respectively, from the same point and at the same speed. After 2 hours distance between them was \[15\sqrt{2}\]
km. Find their speed per hour.
In the given figure, ∠B = 90°, XY || BC, AB = 12 cm, AY = 8cm and AX : XB = 1 : 2 = AY : YC.
Find the lengths of AC and BC.
In the given figure, AB//CD, AB = 7 cm, BD = 25 cm and CD = 17 cm;
find the length of side BC.
In figure AB = BC and AD is perpendicular to CD.
Prove that: AC2 = 2BC. DC.
In the following figure, OP, OQ, and OR are drawn perpendiculars to the sides BC, CA and AB respectively of triangle ABC.
Prove that: AR2 + BP2 + CQ2 = AQ2 + CP2 + BR2
In an equilateral triangle ABC, the side BC is trisected at D. Prove that 9 AD2 = 7 AB2.
The hypotenuse of a right angled triangle of sides 12 cm and 16 cm is __________
Points A and B are on the opposite edges of a pond as shown in the following figure. To find the distance between the two points, the surveyor makes a right-angled triangle as shown. Find the distance AB.
