Question

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

Don't spam pleech ☺​

Answer 1

Answer:

base =x

altitude =(x-7)

hyp =13

13^2=(x-7)^2+x^2

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

solve..

hopes it helps you

Answer 2

$\fbox{\huge\pink{A}\blue{N}\purple{S}\green{W}\red{E}\orange{R}}$

QUESTION :- 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 :-

Let $x$ be the base of the triangle, then the altitude will be $(x-7)$.

By Pythagoras Theorem,

p² + b² = h²

Here,

b = base

p = perpendicular height

h = hypotenuse

${x}^{2} + {(x - 7)}^{2} = {13}^{2}$

${2x}^{2} - 14x + 49 - 169 = 0$

${2x}^{2} - 14x - 120 = 0$

$\frac{ {2x}^{2} - 14x - 120 }{2} = 0$

Canceling the term to lowest term possible.

${x}^{2} - 7x - 60 = 0$

${x}^{2} + 12x - 5x - 60 = 0$

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

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

$\fbox {x = 12}$ and $\fbox {x = - 5}$

Since the side of the triangle cannot be negative, so the base of the triangle is 12 cm and the altitude of the triangle will be 12 - 7 = 5 cm.

$\fbox{Base = 12 cm}$

$\fbox{Altitude = 5 cm}$