Question

The altitude of a right triangle is 7 cm less than its base if hypotenuse is 13 cm find out the other two sides

Answer 1

$\bf\huge\underline{Question}$

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

$\bf\huge\underline{Answer}$

Let the base of the given right traingle be x cm.

∴ Its height = (x - 7) cm

∵ Hypotenuse = $\sqrt{(Base)^2 + (Height)^2}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀[By Pythagoras theorem]

∴ 13 = $\sqrt{x^2+(x-7)^2}$

Squaring both sides, we get

⠀ 169 = x² + (x - 7)²

=> 169 = x² + x² - 14x + 49

=> 2x² - 14x + 49 - 169 = 0

=> 2x² - 14x - 120 = 0

=> x² - 7x - 60 = 0

=> x² - 12x + 5x - 60 = 0

=> x(x - 12) + 5(x - 12) = 0

=> (x - 12)(x + 5) = 0

⠀⠀Either x - 12 = 0 or x + 5 = 0

=> x = 12 or x = -5

But the sides of a traingle can never be negetive

=> x = -5 is rejected.

∴ x = 12

∴ Length of base = 12 cm

=> Length of altitude = (12 - 7)cm = 5 cm

Thus, the required base = 12 cm and altitude = 5 cm

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