Question

H. The altitude of a right triangle is 7 cm less than its base. If the
hypotenuse is 13 cm, find the other two sides.
​

Answer 1

Answer:

do by your own

Step-by-step explanation:

let base be x

.: altitude = x+ 7

given,

         hypotenuse = 13 cm

.: h = √b^2 + h^2

→ 13 = √ x^2 + (x+7)^2

now you can solve it by yourself

Answer 2

Given,

1. Altitude of right triangle is 7 cm less than its base.
2. 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