Question

In the following rightABC. Angle B= 90°. AB = (x+8)cm, BC = (x+1)cm and AC = (x+15)cm.
Find the lengths of the sides of the triangle.

Answer 1

Answer:

Using Pythagoras Theorem:

AC²=AB²+BC²

(x+15)²=(x+8)²+(x+1)²

x²+30x+225=x²+16x+64+x²+2x+1

x²+30x+225=2x²+18x+65

x²-2x²+30x-18x+225-65=0

-x²+12x+160=0

x²-12x-160=0

x²-(20-8)x-160=0

x²-20x+8x-160=0

x(x-20)+8(x-20)=0

(x-20)(x+8)=0

Therefore, either x-20=0 or x+8=0

i.e., x=20 or x=-8

Now, length of AB=x+8

either 20 +8. or -8+8

= 28cm or 0cm

Since length of AB cannot be 0cm, x=-8 is discarded.

Length of BC = x+1

= 20+1

= 21cm

Length of AC = x+15

= 20+15

= 35cm