Question

The altitude of a right triangle is 7 cm less rhen it's base . If the hypotanios is 13 cm . Then find its other two sides.​

Answer 1

Answer:

let other sides be x and x-7

acc. to question :( x) ^2 +(x-7) ^2= (13) ^2

x^2+x^2+49-14x= 169

2(x^2) +14x = 120

x^2 +7x =60

x^2+7x-60 =0

x^2+10x-3x-60 = 0

x(x+10) -3(x+10) = 0

(x-3) = 0; x=3

(x+10) =0 ; x=-10

Since side cannot be negative x=3

Therefore, the sides are 3 cm and 10 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