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

1. 9 inches, 12 inches and 15 inches

2. 54 square inches

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 ±
`√(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 =
`√(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 =
`(a +b+c)/(2)`

=
`((9+12+15)/(2)`

= 18

A =
`√(18(18-9)(18-12)(18-15))`

=
`√(18*9*6*3)`

=
`√(2916)`

A = 54

Area of the triangle is 54 square inches.