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
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
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 triangle are given below. Determine it is a right triangle or not? In case of a right triangle, write the length of its hypotenuse. 7 cm, 24 cm, 25 cm
Tick the correct answer and justify: In ΔABC, AB = `6sqrt3` cm, AC = 12 cm and BC = 6 cm.
The angle B is:
In the given figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC2 = AB2 + BC2 − 2BC.BD.
In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:
`"AC"^2 = "AD"^2 + "BC"."DM" + (("BC")/2)^2`
Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, ho much string does she have out (see Figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?
Identify, with reason, if the following is a Pythagorean triplet.
(4, 9, 12)
In ∆ABC, ∠BAC = 90°, seg BL and seg CM are medians of ∆ABC. Then prove that:
4(BL2 + CM2) = 5 BC2
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.
ABC is a triangle, right-angled at B. M is a point on BC.
Prove that: AM2 + BC2 = AC2 + BM2
Prove that in a right angle triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.
In the right-angled ∆LMN, ∠M = 90°. If l(LM) = 12 cm and l(LN) = 20 cm, find the length of seg MN.
From a point O in the interior of aΔABC, perpendicular OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that: AF2 + BD2 + CE2 = OA2 + OB2 + OC2 - OD2 - OE2 - OF2
In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AB2 = AD2 - BC x CE + `(1)/(4)"BC"^2`
AD is perpendicular to the side BC of an equilateral ΔABC. Prove that 4AD2 = 3AB2.
The perimeters of two similar triangles ABC and PQR are 60 cm and 36 cm respectively. If PQ = 9 cm, then AB equals ______.
The top of a broken tree touches the ground at a distance of 12 m from its base. If the tree is broken at a height of 5 m from the ground then the actual height of the tree is ______.
Two angles are said to be ______, if they have equal measures.
