One leg of a triangle is 28 inches longer than the smaller leg, and the hypotenuse is 32 inches longer than the smaller leg. Find the lengths of the sides of the triangle.

The sides of the triangle are 20 in, 48 in and 52 in

Explanation:

Given:

Let x be the length of smaller leg

Hypotenuse, H = x + 32

Height, h = x + 28

Length of the sides of a triangle = ?

If the triangle is a right angle triangle then we use pythagoras theorm to solve the question.

So,

(Hypotenuse)² = (height)² + (Base)²

(x + 32)² = (x + 28)² + (x)²

x² + 1024 + 64x = x² + 784 + 56x + x²

240 + 8x - x² = 0

x² - 8x - 240 = 0

Solving the quadratic equation:

x² + 12x - 20x - 240 = 0

x(x+12) - 20(x+12) = 0

(x-20) (x+12) = 0

(x-20) = 0

x = 20 in

Hypotenuse, H = x + 32

                    H = 20 + 32 in

                    H = 52 in

Height, h = x + 28

            h = 20 + 28 in

            h = 48 in

Therefore, the sides of the triangle are 20 in, 48 in and 52 in

sides of triangle is 20 inches , ( 20 +28) = 48 inches , ( 20 +32 ) = 52 inches !

Step-by-step explanation:

Here we have , one leg of a triangle is 28 inches longer than the smaller leg, and the hypotenuse is 32 inches longer than the smaller leg. We need to Find the lengths of the sides of the triangle. Let's find out:

Let the smaller side be x  ,So

one leg of a triangle is 28 inches longer than the smaller leg

We have the following length of side as :

⇒

the hypotenuse is 32 inches longer than the smaller leg

We have the following length of hypotenuse as :

⇒

Now , By Pythagoras Theorem :

⇒

⇒

⇒

Simplifying we get :

⇒

⇒

⇒

⇒

⇒

Taking square root both sides we get :

⇒

Since , side can't be negative hence :

⇒

⇒

Therefore , sides of triangle is 20 inches , ( 20 +28) = 48 inches , ( 20 +32 ) = 52 inches !

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this