Question

the altitide of a right triangle is 7 less than its base and its hypothenuse is 13cm then find the other two sides.​

Answer 1

Let base be x cm , then altitude will be (x-7) cm.

Now ,

By using pythagoras theorem-

132 = x2 + (x-7)2

169 = x2 + x2 + 49 - 14x

2x2 - 14x - 120 = 0

x2 - 7x - 60 = 0

x2 - 12x + 5x - 60 = 0

x ( x - 12 ) + 5 ( x - 12 ) = 0

Hence x = 12

sO , Altitude is 5 cm

AND. ,

base = 12cm

Answer 2

Correct Question:-

the altitide of a right triangle is 7 less than its base and its hypothenuse is 13cm then find the other two sides.

Solution :-

Let the base (BC) of the right triangle = x cm .

So, altitude (A B) = (x-7) cm .

And hypotenuse (AC) = 13cm .

So , we will going to use pythagoras theorem,

$a \: c \: {}^{2} = a \: b {}^{2} + b \: c \: {}^{2}$

$= > (13) {}^{2} = (x - 7) {}^{2} + (x) {}^{2}$

$= > 169 =x {}^{2} + 49 - 14x + x {}^{2}$

$= 169 = 2x {}^{2} - 14x + 49$

$= > 2x {}^{2} - 14x + 49 - 120 = 0$

$= > x {}^{2} - 7x - 60 = 0$

$= > x {}^{2} - 12x + 5x - 60 = 0$

$= > x(x - 12) + 5(x - 12) = 0$

$= > (x + 5)(x - 12) = 0$

$= > x + 5 = 0$

Since , side of triangle is never negative

So, x =12

Now Required sides of triangle are

Ab = x - 7

= 12 - 7 ==5 cm and bc = x = 12cm .

Thanks !