Question

In triangle ABC, AB= (x-3)cm BC=(x+4)cm and AC=(x+6)cm. If angle B=90 degree. Find the sides of the triangle.​

Answer 1

Answer:

Step-by-step explanation:for right angle triangle

by Pythagoras theorem

ab^2 + bc^2 = ac^2

(x-3)^2 + (x+4)^2 = (x+6)^2

x^2 -6x +9 + x^2 + 8x + 16 = x^2 + 12x + 36

2x^2 +2x + 25 = x^2 + 12x + 36

x^2 -10x -11 = 0

x = (-b+-√(b^2 -4ac))/2a

x = {-(-10) +- √(100+44)}/2

=(10 +- √144)/2

= (10 +- 12)/2

x=(10+12)/2 or x = (10 -12)/2

x= 22/2 or x = -2/2

x =11 or x = -1

since side cannot be negative

so x = 11

there fore sides of triangle are

x-3 = 11-3=8

x+4=11+4=15

x+6=11+6=17

Answer 2

Answer:

Not able to solve BRO

Very Difficult Question