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

We are given a wire that is 20 cm long and is bent at right angles to form the two shorter sides of a right-angled triangle. The area of the triangle is 30 cm².

### Let:
- \( a \) and \( b \) be the two shorter sides (legs) of the triangle.
- The total length of the wire is \( a + b = 20 \) cm.
- The area of the triangle is given as 30 cm², so \( \frac{1}{2}ab = 30 \) cm².

#### Step 1: Express one side in terms of the other.
From \( a + b = 20 \), we can solve for one variable:
\[
b = 20 - a.
\]

#### Step 2: Substitute into the area formula.
Using \( \frac{1}{2}ab = 30 \), we substitute \( b = 20 - a \) into the equation:
\[
\frac{1}{2}a(20 - a) = 30.
\]
Multiply both sides by 2:
\[
a(20 - a) = 60.
\]
Expand the equation:
\[
20a - a^2 = 60.
\]
Rearrange into a standard quadratic form:
\[
a^2 - 20a + 60 = 0.
\]

#### Step 3: Solve the quadratic equation.
We will solve \( a^2 - 20a + 60 = 0 \) using the quadratic formula:
\[
a = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(1)(60)}}{2(1)} = \frac{20 \pm \sqrt{400 - 240}}{2} = \frac{20 \pm \sqrt{160}}{2}.
\]
Simplifying the square root:
\[
a = \frac{20 \pm 4\sqrt{10}}{2} = 10 \pm 2\sqrt{10}.
\]
Thus, we have two possible solutions for \( a \):
\[
a_1 = 10 + 2\sqrt{10} \quad \text{or} \quad a_2 = 10 - 2\sqrt{10}.
\]

Since both values satisfy the equation \( a + b = 20 \), we can proceed to find the hypotenuse using these approximate values of \( a \) and \( b \).

Let me know how to proceed to find the

A straight length of wire is 20 cm long. It is bent at right angles to form the two shorter sides of a right-angled triangle. If the triangle's area is 30 cm², find: i. the length of the hypotenuse ii. the triangle's perimeter.

Solve a right-angled triangle problem involving a 20 cm wire bent to form two shorter sides and an area of 30 cm². Learn to find the hypotenuse and perimeter using algebra and the Pythagorean theorem. Suitable for Grades 9-10.

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