Question

In triangle ABC, angle B = 90°. Find the sides of the triangle, if AB = x cm, BC = (4x+4) cm, and AC = (4x+5) cm​

Answer 1

Answer:

Step-by-step explanation:

As per Pythagoras  theorem,

$AC^{2} = AB^{2} + BC^{2}$

(4x+5)² = x² + (4x+4)²

=> 16x² + 25 + 40 = x² + 16x² + 16 + 32x

=> 65 = x² + 16 + 32x

=> x² + 32x - 49 = 0

on solving you get x = 1.46 and x = -33.46

we don't take the negative value because the length of a side cannot be negative,

so AB = 1.46cm

BC = 9.84

AC = 10.84

Answer 2

Answer:

We are given that in △ABC,∠=90 degree

tyhe sides of △ABC is given by

AB=x cm,BC=(4x+4)cm &

AC=(4x+5) cm

From the pythagoras theorem,

AC ^2

=AB ^2

+BC ^2

So,

(4x+5) ^2

−(x) ^2

+(4x+4) ^2

16x ^2

+40x+25=x ^2

+16x ^2

+22z−16

∴ 16x ^2

+40x+25=17x ^2

+32x+16

∴ x ^2

−8x−9=0

∴ (x−9)(x+1)=0

∴ x=9 or x=−1

but distance cannot be negative

So, x=9

Now, sides are AB=x=9 cm

BC=4(9)+4=40 cm

AC=4(9)+5=41 cm

Step-by-step explanation: