# Math Help - Quadratics Problems

Question No. 1

In a right angled triangle, the second to longest side is 5 cm longer than the shortest side, and the hypotenuse is three times longer than the shortest side. Find exact length of the hypotenuse.

(The following image is what I've visualized it to be - is it correct?)

This is how I tried solving it and the answer was incorrect:

a^2+b^2=c^2
x^2+(x+5)^2=3x^2
x^2+x^2+10x+25=3x^2

Question No. 2

A straight length of wire is 20 cm long. It is bent at the right angles to form the two shorter sides of a right angled triangle. If the triangle's area is 30cm^2, find:

1. the length of the hypotenuse
2. The triangles perimeter

^ I have no idea how to do either of those questions and am not sure how to visualize it either.

Thank you so much.

2. For number 1, you forgot to square 3.

3. $x^2+ (x+5)^2 = (3x)^2$

$x^2 +x^2 +10x +25 = 9x^2$

$0 =7x^2-10x-25$

Use quad formula to solve for x

4. I'll list the given for number 2 and give you some tips:
let a and b denote the "legs" of the right-angled triangle, c denotes the hypotenuse.

$a + b = 20~~~ [1]$

$\frac{1}{2}ab = 30 \Rightarrow ab = 60$

Square equation number one, then remember the Pythagorean Theorem:
$a^2 + b^2 = c^2$

Once you have found the hypotenuse, the perimeter of the triangle is:
$P = a+b+c$

5. Originally Posted by juliak
Question No. 1

In a right angled triangle, the second to longest side is 5 cm longer than the shortest side, and the hypotenuse is three times longer than the shortest side. Find exact length of the hypotenuse.

(The following image is what I've visualized it to be - is it correct?)

This is how I tried solving it and the answer was incorrect:

a^2+b^2=c^2
x^2+(x+5)^2=3x^2
x^2+x^2+10x+25=3x^2

Question No. 2

A straight length of wire is 20 cm long. It is bent at the right angles to form the two shorter sides of a right angled triangle. If the triangle's area is 30cm^2, find:

1. the length of the hypotenuse
2. The triangles perimeter

^ I have no idea how to do either of those questions and am not sure how to visualize it either.

Thank you so much.
For question 2, you're only using the wire to create the two shorter sides.

So call x the length of one of those sides. The other is therefore 20 - x.

The area of a triangle is given by

$A=\frac{1}{2}bh$.

So we find $30=\frac{1}{2}x(20-x)$.

$30=10x-\frac{1}{2}x^2$.

Solve for x, then you have the two shorter lengths. Use Pythagoras to find the hypotenuse, and then add the 3 side lengths together to find the triangle's perimeter.