# CLASS-8PYTHAGORAS THEOREM - PROBLEM & SOLUTION

There are some example are given below for your better understanding and all these example may help you to solve the other type of problem on basics of Pythagoras Theorem.

Example.1) Calculate the length of the unknown side in each of the following triangles

Ans.)  As per the Pythagoras theorem AC² = AB² + BC²

As per the above given picture AB = 5 cm and BC = 12 cm

So, AC² = 5² + 12² = 25 + 144 = 169

Or, AC  =  √169  = √13² =  13   (Ans.)

Ans.) As per the Pythagoras theorem  QR² = PQ² + RP²

As per the above given picture PQ = 9 cm and  QR = 15 cm

So,   15² =  9² + PR²

Or,   PR² =  15²- 9²

Or,   PR² =  225 – 81 = 144

Or,   PR  =  √144  =  √12² =  12  (Ans.)

Example.2) Calculate the length of BC in the adjoining figure. Is ∆ ABD is a right angled triangle ?

Ans.)  In ∆ ABCAB² =  AC²+ BC², by Pythagoras theorem

Here from the above given picture we can find, AB = 15 cm, BC = 9 cm

So, AB² = AC² + BC²

Or,     15² =  AC² +  9²

Or,     225 = AC² + 81

Or,     AC² = 225 – 81 =  144

Or,     AC =  √144 =  √12² =  12 cm

Now, in ∆ ACD, AD² = AC² + CD², by Pythagoras theorem

Here we can find from the above given picture, AC = 12 cm, AD = 20 cm.

So,    AD² =  AC² + CD²

Or,    20² =  12² + CD²

Or,    400 = 144 + CD²

Or,    CD² =  400 – 144 = 256

Or,     CD = √256 =  √16² = 16 cm

Now,   BD = BC + CD

Or,         = 9 + 16 = 25

We can observe that, BD² = AB² + AD²

= 15² + 20² = 225 + 400 = 625

BD =  √625 =  √25²  =  25

In, ∆ ABD is a right angled triangle in which BD is the hypotenuse, that is, √A = 90⁰            (Ans.)

Example.3) Find PQ in the following figure.

Ans.)  In the right angled ∆ PRS, PS² = PR² + RS², from the above given picture we can find  -

PR = 4 cm, and  RS = 3 cm

Or,  PS² =  4² + 3²  =  16 + 9 = 25

Or,   PS  =  √25  =  √5²  =  5  cm

In the right angled ∆ PRS, PQ² = PS² + SQ², from the above given picture we can find  -

PS = 5 cm, and  SQ = 12 cm

Or,   PQ² =  5² + 12² =  25 + 144 =  169

Or,    PQ  =  √169 =  √13² =  13 cm

So,  we can find that, PQ = 13 cm               (Ans.)

Example.4) Calculate ‘x’ in below figure

Ans.) In ∆ ABC, AC² = AB²+ BC², by Pythagoras theorem

From above given picture, AB = 48 cm, and AC = 50 cm

So,  AC² =  AB² + BC²

Or,  50² = 48² + BC²

Or,  BC² = 50² - 48²

= 2500 – 2304 = 196

Or,   BC² = 196

Or,   BC = √196 = √14² =  14

Now, in ∆ ECDED² = EC²+ CD², by Pythagoras theorem

From the above given picture, ED = 75 cm, and EC = 45 cm

So,  ED² =  EC² + CD²

Or,   75² = 45² + CD²

Or,   5625 =  2025 + CD²

Or,   CD² = 5625 – 2025 =  3600

Or,   CD = √3600 = √60² = 60

Now, as we can find from the above picture, x = BC + CD

Here we find, BC = 14 cm, and CD = 60 cm.

Now,  x = BC + CD = 14 + 60 = 74 cm

So, the desired answer is 74 cm      (Ans.)