Question

△ABC is a right triangle with right angle C. Side AC¯¯¯¯¯¯¯¯ is 6 units longer than side BC¯¯¯¯¯¯¯¯. If the hypotenuse has length 217−−√ units, find the length of AC¯¯¯¯¯¯¯¯.

Answer 1

Given:

In △ABC is a right angle triangle.

AC is 6 units longer than side BC.

$Hypotenuse=2\sqrt{17}$

To find:

The length of AC.

Solution:

Let the length of BC be x.

So, Length of AC = x+6

According to the Pythagoras theorem, in a right angle triangle

$Hypotenuse^2=Base^2+Perpendicualar^2$

△ABC is a right angle triangle and AC is hypotenuse, so

$(2\sqrt{17})^2=(x)^2+(x+6)^2$

$68=x^2+x^2+12x+36$            $[\because (a+b)^2=a^2+2ab+b^2]$

Subtract 68 from both sides.

$0=2x^2+12x+36-68$

$0=2x^2+12x-32$

$0=2(x^2+6x-16)$

Divide both sides by 2.

$x^2+6x-16=0$

Splitting the middle term, we get

$x^2+8x-2x-16=0$

$x(x+8)-2(x+8)=0$

$(x+8)(x-2)=0$

$x=-8,2$

Side cannot be negative, so x=2 only.

Now,

$AC=x+6$

$AC=2+6$

$AC=8$

Therefore, the length of AC is 8 units.