# PRACTICAL PROBLEMS USING PYTHAGOREAN THEOREM

Practical Problems Using Pythagorean Theorem

Here we are going to see some practice problems on pythagorean theorem.

## Practical Problems Using Pythagorean Theorem

Question 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  =  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

Hence the required distance is 0.8 m.

Question 2 :

The perpendicular PS on the base QR of a tiranlge 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

Question 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 = AB² + (2x)²

AE² = AB² + 4x²

In Δ ABC,

AC² = AB² + BC²

AC² = AB + (3x)²

AC² = AB² + 9x²

Now,

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

=  8AB² + 32x²

=  8(AB² + 4x²)

= 8AE²

Hence proved.

After having gone through the stuff given above, we hope that the students would have understood, "Pythagorean Theorem Word Problems for Grade 10".

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