Review question

# Can we find the areas within the semicircles? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6699

## Solution

State the theorem of Pythagoras concerning the three sides of a right-angled triangle.

Pythagoras’s theorem states that, if we are given a right-angled triangle with sides $a, b$ and $c$, where $c$ is the side opposite the right angle, then $a^2 + b^2 = c^2.$

In fact the converse to this is true also, so we can say that a triangle with sides $a, b$ and $c$, where $c$ is the longest side, is right-angled if and only if $a^2 + b^2 = c^2.$

The diagram shows an isosceles triangle $ABC$ right-angled at $B$. Semicircles $S_1$ and $S_2$ are drawn on $AB$ and $AC$ as diameter, as shown.

1. Explain why $S_2$ passes through $B$.

We know the circle theorem, ‘The angle in a semicircle is a right angle’. We are given here that $AC$ is a diameter of $S_2$, and that $B$ is a right angle. If $B$ is inside $S_2$, then we can find a point $B'$ on $S_2$ where the angle $AB'C$ is less than a right angle, which is a contradiction.

Similarly, if $B$ is outside $S_2$, then we can find a point $B'$ on $S_2$ where the angle $AB'C$ is more than a right angle, which is also a contradiction. Thus $B$ is on $S_2$.

Or alternatively we can say that $AB = BC, \quad AO = OC$ (where $O$ is the centre of $S_2$), and angle $BAO$ equals angle $BCO$, and so the triangles $ABO$ and $BCO$ are congruent right-angled isosceles triangles.

This means $AO = BO$ = radius of $S_2$, and $B$ must be on the semicircle.

1. If $AB = \quantity{2x}{cm}$, show that the area shaded vertically is $\quantity{\tfrac{1}{2}(\pi - 2)x^2}{cm^2}$.

We’ll begin with the following two diagrams.

We can write $\text{area of the vertically-shaded region} = \text{area of the blue region} - \text{area of the red region}.$

The blue region is a quarter of the area of the circle that includes $S_2$. We know that $\triangle AOB$ is an isosceles triangle, with $AB$ as the hypotenuse. Since $AB$ has length $2x$, and since $OA=OB=r$, where $r$ is the radius of $S_2$, by Pythagoras’ theorem we know that $(2x)^2 = OA^2 + OB^2 = 2r^2,$ which implies that $r^2 = 2x^2$. Thus \begin{align*} \text{area of the blue region} &= \frac{1}{4} \times \text{area of the circle} \\ &= \frac{1}{4} \times \pi r^2\\ &= \frac{\pi x^2}{2}. \end{align*} The red region is an isosceles, right-angled triangle, so \begin{align*} \text{area of the red region} &= \frac{1}{2} \times \text{base} \times \text{height} \\ &= \frac{1}{2} r^2 \\ &= x^2. \end{align*} So we have \begin{align*} \text{area of the vertically-shaded region} &= \frac{\pi x^2}{2} - x^2 \\ &= \quantity{\frac{1}{2} \left( \pi - 2 \right) x^2}{cm^2}. \end{align*}
1. Prove that the area shaded horizontally is half the area of the triangle $ABC$.
We can write $\text{area of the horizontally-shaded region} = \text{area of S_1} - \text{area of vertically-shaded region}.$ We have just calculated the area of the vertically-shaded region. For the area of $S_1$, as it is a semicircle with diameter $AB$, we have that $\dfrac{\pi}{2}x^2$, and thus \begin{align*} \text{area of the horizontally-shaded region} &= \frac{\pi}{2} x^2 - \frac{1}{2} (\pi x^2 - 2 x^2) \\ &= \quantity{x^2}{cm^2}. \end{align*}

It is easy to see the area of $\triangle ACB$ is $\quantity{2x^2}{cm^2}$, which is twice the area of the horizontally-shaded region.