Question

find the other two sides of a right angled triangle whose shorter side is 6cm long. the sides of it are in A.P
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Answer 1

Answer:

$\large\boxed{\sf{8\;cm\;\;and\;\;10\;cm}}$

Step-by-step explanation:

It's being given that, in a right angled triangle the shorter side is 6 cm.

And, the other two sides are in AP.

Let, the other sides are (6+d) and (6+2d).

Where, d is common difference.

Therefore, the longer side will be the hypotenus.

Therefore, hypotenus = (6+2d) cm

Now, by pythagoras theorem, we have,

$= > {(6 + 2d)}^{2} = {(6 + d)}^{2} + {6}^{2} \\ \\ = > 36 + 4 {d}^{2} + 24d = 36 +12d + {d}^{2} + 36 \\ \\ = > 4 {d}^{2} - {d}^{2} + 24d - 12d - 36 = 0 \\ \\ = > 3 {d}^{2} + 12d - 36 = 0 \\ \\ = > 3( {d}^{2} + 4d - 12) = 0 \\ \\ = > {d}^{2} + 4d - 12 = 0 \\ \\ = > {d}^{2} + 6d - 2d - 12 = 0 \\ \\ = > d(d + 6) - 2(d + 6) = 0 \\ \\ = > (d + 6)(d - 2) = 0 \\ \\ = > d = - 6 \: \: \: \: and \: \: \: \: 2$

But, if d = -6

=> 6 + d = 6 - 6 = 0

Which can't be possible.

Therefore, d = 2.

Therefore, other sides = (6+2) and (6+4)

Hence, the required sides are 8 cm and 10 cm.