Question

2. In ABC,B = 90°. Find the sides of the
triangle, if :
(i) AB = (x – 3) cm, BC = (x + 4) cm and
AC = (x + 6) cm
(ii) AB = x cm, BC = (4x + 4) cm and
AC = (4x + 5) cm​

Answer 1

Step-by-step explanation:

HEY MATE YOUR ANSWER IS HERE ^_^

by the Pythagoras theorem

AB^2 + BC^2 = AC^2

(1)

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

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

x^2-10x-11=0

x^2-11x+x-11=0

x(x-11)+1(x-11)=0

(x-11)(x+1)=0

x = 11 or -1

side can't be negative therefore x=11

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Answer 2

Answer:

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