Area of a garden

Question

In my maths homework I have this question (Image here) which asks me to find the area of a garden. It tells me that it's made of an isosceles triangle and a semi-circle. Is this even possible to work out, or did my teacher just make a mistake on the question?

Answer 1

In geometry, an isosceles triangle is a triangle that has two sides of equal length. Sometimes it is specified as having two and only two sides of equal length.

So, in your right angled triangle we can use the Pythagorean Theorem, which states that:

$$\text{a}^2+\text{b}^2=\text{c}^2$$

So:

1. When we know that $\text{a}=6$, set $\text{b}=6$: $$6^2+6^2=\text{c}^2\Longleftrightarrow\text{c}=\pm6\sqrt{2}$$
2. When we know that $\text{a}=6$, set $\text{c}=6$: $$6^2+\text{b}^2=6^2\Longleftrightarrow\text{b}=0$$

So, we know that the area of the triangle equals:

$$\mathcal{A}_{\text{triangle}}=\frac{6\cdot6}{2}=18$$

And for the semicircle, the diameter is $6$, so for the area:

$$\mathcal{A}_{\text{semicircle}}=\frac{\pi\cdot\left(\frac{6}{2}\right)^2}{2}=\frac{9\pi}{2}$$

So, for the area of the garden:

$$\mathcal{A}_{\text{garden}}=\mathcal{A}_{\text{triangle}}+\mathcal{A}_{\text{semicircle}}=18+\frac{9\pi}{2}$$

Answer 2

Yes, it is possible to work out, however, it is a poorly drawn diagram for the question. As the triangle is isosceles, you know that the bottom and left-hand sides of the triangle both equal 6, as does the diameter of the circle, as the left-hand side is shared. This means that to work out the area, you need to do $\frac{base \times height}{2}+ \frac{\pi (\frac{diameter}{2})^2}{2}$ in this case that gives $\frac{6\times 6}{2}+ \frac{\pi (\frac{6}{2})^2}{2}=\frac{36}{2}+ \frac{3^2\pi}{2}=18+\frac{9\pi}{2}$ if you need to give an answer to 3s.f. that would be $32.1m$