Junior Cycle Mathematics Geometry: A triangle has a base length of

A triangle has a base length of $2x$ cm and a perpendicular height of $(x + 3)$ cm - Junior Cycle Mathematics - Question Question 1 - 2012

Question Question 1

A triangle has a base length of $2x$ cm and a perpendicular height of $(x + 3)$ cm. The area of the triangle is 10 cm². Find the distance $x$.

Worked Solution & Example Answer:A triangle has a base length of $2x$ cm and a perpendicular height of $(x + 3)$ cm - Junior Cycle Mathematics - Question Question 1 - 2012

Step 1

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Answer

The area of a triangle is given by the formula:

ext{Area} = rac{1}{2} imes ext{base} imes ext{height}

Substituting the given values (base = $2x$ and height = $(x + 3)$ ), we have:

rac{1}{2} imes (2x) imes (x + 3) = 10

This simplifies to:

x(x + 3) = 10

Step 2

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Answer

Expanding the equation,

x^2 + 3x - 10 = 0

This is a quadratic equation.

Step 3

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Answer

Using the quadratic formula, where $a = 1$ , $b = 3$ , and $c = -10$ :

x = rac{-b imes ext{±} imes ext{√}(b^2 - 4ac)}{2a} = rac{-3 imes ext{±} imes ext{√}(3^2 - 4(1)(-10))}{2(1)}

Calculating the discriminant:

3^2 - 4(1)(-10) = 9 + 40 = 49

Therefore:

x = rac{-3 imes ext{\pm} imes 7}{2} = rac{-3 imes 7}{2} ext{ or } rac{-3 imes -7}{2}

This gives two potential solutions:

x = rac{-21}{2} ext{ (not possible)}, ext{ and } x = 2

Step 4

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Answer

Thus, the distance $x$ is:

x = 2 ext{ cm}