Advertisements
Advertisements
प्रश्न
M andN are the mid-points of the sides QR and PQ respectively of a PQR, right-angled at Q.
Prove that:
(i) PM2 + RN2 = 5 MN2
(ii) 4 PM2 = 4 PQ2 + QR2
(iii) 4 RN2 = PQ2 + 4 QR2(iv) 4 (PM2 + RN2) = 5 PR2
Advertisements
उत्तर
We draw, PM, MN, NR
Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
Since M and N are the mid-points of the sides QR and PQ respectively, therefore, PN = NQ, QM = RM
(i) First, we consider the ΔPQM, and applying Pythagoras theorem we get,
PM2 = PQ2 + MQ2
= ( PN + NQ )2 + MQ2
= PN + NQ2 + 2PN . NQ + MQ2
= MN2+ PN2 + 2PN.NQ ...[From, ΔMNQ, MN2 = NQ2 + MQ2] ......(i)
Now, we consider the ΔRNQ, and applying Pythagoras theorem we get,
RN2 = NQ2 + RQ2
= NQ2 + ( QM + RM )2
= NQ2 + QM2 + RM2 + 2QM .RM
= MN2 + RM2 + 2QM . RM .......(ii)
Adding (i) and (ii) we get,
PM2 + RN2 = MN2 + PN2 + 2PN.NQ + MN2 + RM2 + 2QM. RM
PM2 + RN2 = 2MN2 + PN2 + RM2 + 2PN.NQ + 2QM.RM
PM2 + RN2 = 2MN2 + NQ2 + QM2 + 2(QN2 ) + 2(QM2 )
PM2 + RN2 = 2MN2 + MN2 + 2MN2
PM2 + RN2 = 5MN2
Hence Proved.
(ii) We consider the ΔPQM, and applying Pythagoras theorem we get,
PM2 = PQ2 + MQ2
4PM2 = 4PQ2 + 4MQ2 ...[ Multiply both sides by 4]
4PM2 = 4PQ2 + 4.`(1/2 "QR")^2` ...[ MQ = `1/2` QR ]
4PM2 = 4PQ2 + 4PQ + 4 . `1/4` QR2
4PM2 = 4PQ2 + QR2
Hence Proved.
(iii) We consider the ΔRQN, and applying Pythagoras theorem we get,
RN2 = NQ2 + RQ2
4RN2 = 4NQ2 + 4QR2 ...[ Multiplying both sides by 4]
4RN2 = 4QR2 + 4 .(1/2 PQ)2 ...[ NQ = `1/2` PQ ]
4RN2 = 4QR2 + 4 .`1/4` PQ2
4RN2 = PQ2 + 4QR2
Hence Proved.
(iv) First, we consider the ΔPQM, and applying Pythagoras theorem we get,
PM2 = PQ2 + MQ2
= ( PN + NQ )2 + MQ2
= PN2 + NQ2 + 2PN.NQ + MQ2
= MN2 + PN2 + 2PN.NQ ...[ From, ΔMNQ, = MN2 = NQ2 + MQ2 ] ......(i)
Now, we consider the ΔRNQ, and applying Pythagoras theorem we get,
RN2 + NQ2 + RQ2
= NQ2 + ( QM + RM )2
= NQ2 + QM2 + RM2 + 2QM .RM
= MN2 + RM2 + 2QM . RM .......(ii)
Adding (i) and (ii) we get,
PM2 + RN2 = MN2 + PN2 + 2PN . NQ + MN2 + RM2 + 2QM. RM
PM2 + RN2 = 2MN2 + PN2 + RM2 + 2PN . NQ + 2QM . RM
PM2 + RN2 = 2MN2 + NQ2 + QM2 + 2(QN2 ) + 2(QM2 )
PM2 + RN2 = 2MN2 + MN2 + 2MN2
PM 2 + RN2 = 5MN2
4( PM2 + RN2 ) = 4.5. (NQ2 + MQ2)
4( PM2 + RN2 ) = 4.5. `[ ( 1/2 "PQ" )^2 + ( 1/2 "RQ" )^2 ] ....[ ∵ "NQ" = 1/2 "PQ" , "MQ" = 1/2 "QR" ]`
4 ( PM2 + RN2 ) = 5PR2
Hence Proved.
APPEARS IN
संबंधित प्रश्न
ABC is a right-angled triangle, right-angled at A. A circle is inscribed in it. The lengths of the two sides containing the right angle are 5 cm and 12 cm. Find the radius of the circle
Sides of triangles are given below. Determine it is a right triangles? In case of a right triangle, write the length of its hypotenuse. 3 cm, 8 cm, 6 cm
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM . MR
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
In an equilateral triangle ABC, D is a point on side BC such that BD = `1/3BC` . Prove that 9 AD2 = 7 AB2
Which of the following can be the sides of a right triangle?
2.5 cm, 6.5 cm, 6 cm
In the case of right-angled triangles, identify the right angles.
A man goes 40 m due north and then 50 m due west. Find his distance from the starting point.
In equilateral Δ ABC, AD ⊥ BC and BC = x cm. Find, in terms of x, the length of AD.
Prove that `(sin θ + cosec θ)^2 + (cos θ + sec θ)^2 = 7 + tan^2 θ + cot^2 θ`.
The sides of the triangle are given below. Find out which one is the right-angled triangle?
11, 60, 61
From the given figure, find the length of hypotenuse AC and the perimeter of ∆ABC.
A ladder 25m long reaches a window of a building 20m above the ground. Determine the distance of the foot of the ladder from the building.
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.
Find the unknown side in the following triangles
Prove that the area of the semicircle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the semicircles drawn on the other two sides of the triangle.
In an equilateral triangle PQR, prove that PS2 = 3(QS)2.
Lengths of sides of a triangle are 3 cm, 4 cm and 5 cm. The triangle is ______.
The foot of a ladder is 6 m away from its wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its top reach?
