Math Problem Statement

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

Solution

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:

  • aa and bb be the two shorter sides (legs) of the triangle.
  • The total length of the wire is a+b=20a + b = 20 cm.
  • The area of the triangle is given as 30 cm², so 12ab=30\frac{1}{2}ab = 30 cm².

Step 1: Express one side in terms of the other.

From a+b=20a + b = 20, we can solve for one variable: b=20a.b = 20 - a.

Step 2: Substitute into the area formula.

Using 12ab=30\frac{1}{2}ab = 30, we substitute b=20ab = 20 - a into the equation: 12a(20a)=30.\frac{1}{2}a(20 - a) = 30. Multiply both sides by 2: a(20a)=60.a(20 - a) = 60. Expand the equation: 20aa2=60.20a - a^2 = 60. Rearrange into a standard quadratic form: a220a+60=0.a^2 - 20a + 60 = 0.

Step 3: Solve the quadratic equation.

We will solve a220a+60=0a^2 - 20a + 60 = 0 using the quadratic formula: a=(20)±(20)24(1)(60)2(1)=20±4002402=20±1602.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=20±4102=10±210.a = \frac{20 \pm 4\sqrt{10}}{2} = 10 \pm 2\sqrt{10}. Thus, we have two possible solutions for aa: a1=10+210ora2=10210.a_1 = 10 + 2\sqrt{10} \quad \text{or} \quad a_2 = 10 - 2\sqrt{10}.

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

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Math Problem Analysis

Mathematical Concepts

Geometry
Algebra
Quadratic Equations

Formulas

Area of a right-angled triangle: (1/2)ab
Perimeter of a triangle: a + b + c
Quadratic equation formula: ax^2 + bx + c = 0
Pythagorean theorem: a^2 + b^2 = c^2

Theorems

Quadratic Formula: x = (-b ± √(b^2 - 4ac)) / 2a
Pythagorean Theorem

Suitable Grade Level

Grades 9-10