PRACTICAL PROBLEMS USING PYTHAGOREAN THEOREM

Problem 1 :

5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.

Solution :

In triangle ABC,

AC2  =  AB2 + BC2

52  =  AB2 + 42

AB2  =  25 - 16

AB2  =  9

AB  =  3

DB  =  AB - AD

DB  =  3 - 1.6

DB  =  1.4

In triangle DBE,

ED2  =  EB2 + DB

52  =  EB2 + (1.4)

EB2  =  25 - 1.96

EB = √23.04

EB  =  4.8

EC  =  EB - BC

=  4.8 - 4

EC  =  0.8 m

So, the required distance is 0.8 m.

Problem 2 :

The perpendicular PS on the base QR of a triangle PQR intersects QR at S, such that QS = 3 SR. Prove that 2PQ2 = 2PR2 + QR2

Solution :

QR   =   QS + RS

QR   =   3RS + RS

QR  =  4RS

RS  =  QR/4

By applying Pythagorean theorem in a triangle PQS, we get

PQ2  =  PS2 + QS2

PS=  PQ2 - QS -----(1)

By applying Pythagorean theorem in a triangle PRS, we get

PR2  =  PS2 + SR2

PS=  PR- SR2 -----(2)

PQ2 - QS  =  PR- SR2

PQ2 - QS  =  PR- SR2

PQ2   =  PR- SR2   + QS

PQ2   =  PR- SR2   + (QR - RS)

PQ2   =  PR- SR2  + QR2 + RS2 - 2QR x RS

PQ2   =  PR+ QR2 - 2QR x RS

PQ2   =  PR+ QR2 - 2 QR x (QR/4)

PQ2   =  PR+ QR2 - (QR2/2)

PQ2   =  (2PR+ 2QR2 - QR2)/2

2PQ2   =  2PR+ QR2

Problem 3 :

In the adjacent figure, ABC is a right angled triangle with right angle at B and points D, E trisect BC. Prove that 8AE2 = 3AC2 + 5AD2

Solution :

AE2 = AB+ (2x)2

AE= AB+ 4x2

In Δ ABC,

AC= AB2 + BC2  

AC2 = AB + (3x)2 

AC2 = AB2 + 9x2

Now,

3AC2 + 5AD2 = 3(AB2 + 9x2) + 5(AB2 + x2

=  8AB2 + 32x2

=  8(AB2 + 4x2)

= 8AE2

8AE2 = 3AC2 + 5AD2

Hence proved.

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