Question

One leg of a right triangle is more than 3 more inches than the shorter leg. The hypotenuse is 15 inches. What are the lengths of the legs of the triangle and what is the area?

Answer 1

Answer:

1. 9 inches, 12 inches and 15 inches

2. 54 square inches

Step-by-step explanation:

1. The first part of this question would lead to a quadratic equation. Let the shorter leg be represented by x.

shorter leg = x

other leg = x + 3

hypotenuse = 15 inches

Applying the Pythagoras theorem, we have;

$/15/^{2}$ = $/x/^{2}$ + $/x+3/^{2}$

225 = $x^{2}$ + $(x+3)^{2}$

225 = $x^{2}$ + $x^{2}$ + 6x + 9

       = 2$x^{2}$ + 6x + 9

2$x^{2}$ + 6x + 9 - 225 = 0

2$x^{2}$ + 6x - 216 = 0

divide through by 2 to have

$x^{2}$ + 3x - 108 = 0

From the quadratic formula;

x = (-b ± $\sqrt{b^{2}-4ac }$ ) ÷ 2a

but, a = 1, b = 3, c = -108

x = (-3 ± $\sqrt{(3)^{2}-4(1)(-108)}$) ÷ 2

  = (-3 ± $\sqrt{441}$) ÷ 2

= (-3 ± 21) ÷ 2

Thus,

x = (-3 + 21) ÷ 2 OR x = (-3 - 21) ÷ 2

x = 9 OR x = -12

So that, x = 9 inches

The shorter leg is 9 inches, and the other leg is 12 inches.

2. The area of the triangle can be determined by applying Heron's formula:

A = $\sqrt{s(s-a)(s-b)(s-c)}$

where s is the average value of the sum of the three sides a, b, and c.

Let, a = 9, b = 12 and c = 15

s = $\frac{a +b+c}{2}$

  = $\frac{(9+12+15}{2}$

  = 18

A = $\sqrt{18(18-9)(18-12)(18-15)}$

   = $\sqrt{18*9*6*3}$

   = $\sqrt{2916}$

A = 54

Area of the triangle is 54 square inches.