Question

The altitude of a right triangle is 7 cm less than its base.if the hypoteneuse is 13 cm ,find the other two sides?

Answer 1

Let the base be x
Then altitude be x-7
we know that h²=p²-b²
So (13)²=( x-7)² - (x) ²
169 =x²-2x7 +7² - x²
169=-2x7+49
169 -49 =2x7
120=2x7
120/2 =x7
60=x7
So x=60/7

Answer 2

Given,

• Altitude of right triangle is 7 cm less than its base.
• Hypotenuse is 13 cm.

To find,

• The other two sides.

Solution,

Let x be the base of the triangle

Then altitude will be (x-7)

We know that,

$\sf{Base^2+Altitude^2=Hypotenuse^2}$

So, by pythagoras theorem,

${x}^{2} + ( {x - 7)}^{2} = {13}^{2} \\ \\ ⟹2 {x}^{2} - 14x + 49 = 169 \\ \\ ⟹2 {x}^{2} - 14x + 49 - 169 = 0 \\ \\ ⟹2 {x}^{2} - 14x - 120 = 0 \\ \\ ⟹2( {x}^{2} - 7x - 60) = 0 \\ \\ ⟹ {x}^{2} - 7x - 60 = \frac{0}{2} \\ \\⟹ {x }^{2} - 7x - 60 = 0 \\ \\ ⟹ {x}^{2} - 12x + 5x - 60 = 0 \\ \\ ⟹x(x - 12) + 5(x - 12) = 0 \\ \\ ⟹(x - 12)(x + 5) = 0$

So, x = 12 or x = -5

Since,the side of a triangle cannot be negative,so the base of the triangle is 12 cm.

And the altitude will be (12-7) = 5 cm