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.