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
In a right triangle ABC, right-angled at B, BC = 12 cm and AB = 5 cm. The radius of the circle inscribed in the triangle (in cm) is
(A) 4
(B) 3
(C) 2
(D) 1
Sides of triangle are given below. Determine it is a right triangle or not? In case of a right triangle, write the length of its hypotenuse. 13 cm, 12 cm, 5 cm
In the given figure, ABC is a triangle in which ∠ABC> 90° and AD ⊥ CB produced. Prove that AC2 = AB2 + BC2 + 2BC.BD.
Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.
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.
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
Identify, with reason, if the following is a Pythagorean triplet.
(24, 70, 74)
Find the length diagonal of a rectangle whose length is 35 cm and breadth is 12 cm.
Prove that `(sin θ + cosec θ)^2 + (cos θ + sec θ)^2 = 7 + tan^2 θ + cot^2 θ`.
The sides of a certain triangle is given below. Find, which of them is right-triangle
6 m, 9 m, and 13 m
A boy first goes 5 m due north and then 12 m due east. Find the distance between the initial and the final position of the boy.
In the right-angled ∆PQR, ∠ P = 90°. If l(PQ) = 24 cm and l(PR) = 10 cm, find the length of seg QR.
In an equilateral triangle ABC, the side BC is trisected at D. Prove that 9 AD2 = 7 AB2.
A point OI in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. Prove that OB2 + OD2 = OC2 + OA2
∆ABC is right-angled at C. If AC = 5 cm and BC = 12 cm. find the length of AB.
Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels at a speed of `(20 "km")/"hr"` and the second train travels at `(30 "km")/"hr"`. After 2 hours, what is the distance between them?
In a quadrilateral ABCD, ∠A + ∠D = 90°. Prove that AC2 + BD2 = AD2 + BC2
[Hint: Produce AB and DC to meet at E.]
Lengths of sides of a triangle are 3 cm, 4 cm and 5 cm. The triangle is ______.
Two squares are congruent, if they have same ______.
