Question

Find a value of x that would make ∆ABC a right triangle. * show work*

Answer 1

Answer:

x = 4.

Step-by-step explanation:

For ∆ABC to be a right triangle,

$AB^2 + BC^2 = AC^2$. (Pythagoras Theorem)

$(AD+DB)^2 + BC^2 = AC^2$.

$(2x + 3 + 5x - 10)^2 + 20^2 = 29^2$.

$(7x - 7)^2 + 400 = 841$

$7^2(x-1)^2 = 841 - 400$

$49 [ x^2 + 1 - 2x ] = 441$

$x^2 + 1 - 2x = 441/49$

$x^2 - 2x + 1 = 9$

$x^2 - 2x - 8 = 0$

$x^2 - 4x + 2x - 8 = 0$

$x (x - 4) + 2 (x - 4) = 0$

$(x+2) * ( x-4) = 0$

=> x = -2 & x = 4.

putting value of x = -2 in AB

AB = 7x - 7 (solved above)

AB = 7(-2) -7

AB = -14 - 7

AB = -21.

But length can't be negative, so x = -2 is rejected

putting value of x = 4 in AB

AB = 7x - 7 (solved above)

AB = 7(4) - 7

AB = 28 - 7

AB = 21

HENCE,

for x = 4. ∆ABC will be right triangle.

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Answer 2

Answer:

x = 4

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