Question

Hi

1-5 already done!!!!!!!!!!

Answer 1

According to tangents secant segments theorem,

11) x(16+x) = (x + 6)^2

x (16+x) = (x + 6)^2

16x + x^2 = x^2 + 2(6)(x) + 6^2

16x + x^2 = x^2 + 12x + 36

16x - 12x + x^2 - x^2 = 36

4x = 36

x = 36/4

x = 4

13) x(x + 5) = (x + 2)^2

x^2 + 5x = x^2 + 2x 2(2)(x) + 2^2

x^2 + 5x = x^2 + 4x + 4

x^2 -x^2 + 5x - 4x = 4

x = 4

14) x + 8( x + 8 + 32) = (3x) ^2

x + 8(x + 40) = 9x^2

x(x + 8) + 40(x + 8) = 9x^2

x^2 + 8x + 40x + 320 = 9x^2

x^2 + 48x + 320 =9x^2

48x + 320 = 9x^2 - x^2

48x + 320 = 8x^2

Dividing the whole eq. by 8,

6x = 40 = x^2

0 = x^2 - 6x - 40

0 = x^2 - 10x + 4x - 40

0 = x(x - 10) + 4(x - 10)

x - 10 = 0 OR x + 4 = 0

x = 10 OR x = - 4

length cannot be negative, so,

x = 10

15) (x + 3) ( x + 3 + 15) = (2x) ^2

(x + 3) (x + 18) = 4x^2

x(x + 18) + 3(x + 18) = 4x^2

x^2 + 18x + 3x + 54 = 4x^2

x^2 + 21x + 54 = 4x^2

0 = 4x^2 - x^2 - 21x - 54

0 = 3x^2 - 21x - 54

Dividing the whole eq. by 3,

x^2 - 7x - 18 = 0

x^2 - 7x - 18 = 0

x^2 - 9x + 2x - 18 = 0

x(x - 9) + 2 (x - 9)

x + 2 = 0 Or x - 9 = 0

x = -2 or x = 9

length cannot be negative,so,

x = 9