(a) (i) In the diagram, semi-circles are drawn on the sides of right-angled $\triangle X Y Z$, as shown. If the area of the semi-circle with diameter $Y Z$ is $50 \pi$ and the area of the semi-circle with diameter $X Z$ is $288 \pi$, determine the area of the semi-circle with diameter $X Y$.

![](https://cdn.mathpix.com/cropped/2024_04_30_9fccce2562493283ec28g-3.jpg?height=369&width=420&top_left_y=1103&top_left_x=1338)

(ii) In the diagram, squares with side lengths $p, q$ and $r$ are constructed on the sides of right-angled $\triangle P Q R$, as shown. Diagonals $Q T, R U$ and $P S$ have lengths $a, b$ and $c$, respectively. Prove that the triangle formed with sides of lengths $a, b$ and $c$ is a rightangled triangle.

![](https://cdn.mathpix.com/cropped/2024_04_30_9fccce2562493283ec28g-3.jpg?height=664&width=553&top_left_y=1506&top_left_x=1228)

(b) In the diagram, $\triangle X Y Z$ is right-angled at $Z$ and $A Z$ is the altitude from $Z$ to $X Y$. Also, segments $A D$ and $A B$ are altitudes in $\triangle A X Z$ and $\triangle A Y Z$, respectively, and segments $D E$ and $B C$ are altitudes in $\triangle A D X$ and $\triangle A B Y$, respectively. Prove that $A E=A C$.

![](https://cdn.mathpix.com/cropped/2024_04_30_9fccce2562493283ec28g-3.jpg?height=287&width=569&top_left_y=2182&top_left_x=1209)
(a) (i) The semi-circles with diameter $Y Z, X Z$ and $X Y$ have diameters of length $x, y$ and $z$, and so have radii of length $\frac{1}{2} x, \frac{1}{2} y$ and $\frac{1}{2} z$, respectively.

![](https://cdn.mathpix.com/cropped/2024_04_30_1271530ef38a72b7fb44g-06.jpg?height=377&width=426&top_left_y=278&top_left_x=988)

In terms of $x$, the area of a semi-circle with radius $\frac{1}{2} x$ is $\frac{1}{2} \pi\left(\frac{1}{2} x\right)^{2}=\frac{1}{8} \pi x^{2}$. (This is one-half of the area of the circle with radius $\frac{1}{2} x$.)

Similarly, in terms of $y$ and $z$, the areas of the semi-circles with diameter $X Z$ and $X Y$ are $\frac{1}{8} \pi y^{2}$ and $\frac{1}{8} \pi z^{2}$, respectively.

Therefore, $\frac{1}{8} \pi x^{2}=50 \pi$ and $\frac{1}{8} \pi y^{2}=288 \pi$ and we want to determine $\frac{1}{8} \pi z^{2}$.

Since $\triangle X Y Z$ is right-angled at $Z$, then $Y Z^{2}+X Z^{2}=X Y^{2}$, by the Pythagorean Theorem, and so $x^{2}+y^{2}=z^{2}$.

Multiplying both sides of this equation by $\frac{1}{8} \pi$, we obtain $\frac{1}{8} \pi x^{2}+\frac{1}{8} \pi y^{2}=\frac{1}{8} \pi z^{2}$.

Therefore, $50 \pi+288 \pi=\frac{1}{8} \pi z^{2}$ and so $\frac{1}{8} \pi z^{2}=338 \pi$.

Thus, the area of the semi-circle with diameter $X Y$ is $\frac{1}{8} \pi z^{2}=338 \pi$.

(Note that we did not need to determine $x, y$ or $z$.)

(ii) A square with side length $s$ has diagonal of length $\sqrt{2} s$. This is because the diagonal is the hypotenuse of an isosceles right-angled triangle with equal sides of length $s$.

![](https://cdn.mathpix.com/cropped/2024_04_30_1271530ef38a72b7fb44g-06.jpg?height=616&width=526&top_left_y=1378&top_left_x=954)

Therefore, $a=\sqrt{2} p, b=\sqrt{2} q$ and $c=\sqrt{2} r$.

Since $\triangle P Q R$ is right-angled at $R$, then $Q R^{2}+P R^{2}=P Q^{2}$, by the Pythagorean Theorem, and so $p^{2}+q^{2}=r^{2}$.

Multiplying both sides of this equation by 2 , we obtain $2 p^{2}+2 q^{2}=2 r^{2}$, and so $(\sqrt{2} p)^{2}+(\sqrt{2} q)^{2}=(\sqrt{2} r)^{2}$ or $a^{2}+b^{2}=c^{2}$.

Thus, the lengths $a, b$ and $c$ satisfy the Pythagorean equation, which tells us that the triangle formed with side lengths $a, b$ and $c$ is a right-angled triangle.

Note that the Pythagorean Theorem has two parts:

Suppose that $\triangle X Y Z$ has side lengths $Y Z=x, X Z=y$ and $X Y=z$. Then:

- If the triangle is right-angled at $Z$, then $x^{2}+y^{2}=z^{2}$, and
- If $x^{2}+y^{2}=z^{2}$, then the triangle is right-angled at $Z$.
(b) Solution 1

Let $A Z=h$ and let $\angle Z X Y=\theta$.

Since $\triangle X D E$ is right-angled at $E$, then

$$
\angle X D E=180^{\circ}-\angle D X E-\angle X E D=180^{\circ}-\theta-90^{\circ}=90^{\circ}-\theta
$$

Since $\angle X D A=90^{\circ}$, then $\angle E D A=90^{\circ}-\angle X D E=90^{\circ}-\left(90^{\circ}-\theta\right)=\theta$.

Similarly, $\angle E A D=90^{\circ}-\theta, \angle D A Z=\theta, \angle D Z A=90^{\circ}-\theta, \angle A Z B=\theta, \angle Z A B=90^{\circ}-\theta$, and $\angle B A C=\theta$.

![](https://cdn.mathpix.com/cropped/2024_04_30_1271530ef38a72b7fb44g-07.jpg?height=265&width=544&top_left_y=602&top_left_x=888)

Since $A Z=h$ and $\triangle A D Z$ is right-angled at $D$, then $A D=A Z \cos (\angle D A Z)=h \cos \theta$. Since $A D=h \cos \theta$ and $\triangle D E A$ is right-angled at $E$, then

$$
A E=A D \sin (\angle E D A)=(h \cos \theta) \sin \theta=h \cos \theta \sin \theta
$$

Since $A Z=h$ and $\triangle Z B A$ is right-angled at $B$, then $A B=A Z \sin (\angle A Z B)=h \sin \theta$. Since $A B=h \sin \theta$ and $\triangle A C B$ is right-angled at $C$, then

$$
A C=A B \cos (\angle B A C)=(h \sin \theta) \cos \theta=h \cos \theta \sin \theta
$$

Therefore, $A E=A C$.

Solution 2

Let $A Z=h$ and let $\angle Z X Y=\theta$.

Since $\triangle X D E$ is right-angled at $E$, then

$$
\angle X D E=180^{\circ}-\angle D X E-\angle X E D=180^{\circ}-\theta-90^{\circ}=90^{\circ}-\theta
$$

Since $\angle X D A=90^{\circ}$, then $\angle E D A=90^{\circ}-\angle X D E=90^{\circ}-\left(90^{\circ}-\theta\right)=\theta$.

Similarly, $\angle E A D=90^{\circ}-\theta, \angle D A Z=\theta, \angle D Z A=90^{\circ}-\theta, \angle A Z B=\theta, \angle Z A B=90^{\circ}-\theta$, and $\angle B A C=\theta$.

Construct rectangle $J P C E$ with side $J P$ passing through $Z, J$ on $E D$ extended and $P$ on $C B$ extended.

![](https://cdn.mathpix.com/cropped/2024_04_30_1271530ef38a72b7fb44g-07.jpg?height=271&width=544&top_left_y=1957&top_left_x=888)

Now $A B Z D, J Z A E$ and $Z P C A$ are all rectangles since they each have three right angles. Therefore, $J Z=E A$. We will show that $J Z=A C$, which will tell us that $A E=A C$. Since $A B Z D$ is a rectangle, then $A B=D Z$.

Since $J Z$ is parallel to $A C$ and $D Z$ is parallel to $A B$, then $\angle J Z D=\angle B A C$.

Since $D J$ is parallel to $C B$ and $D Z$ is parallel to $A B$, then $\angle J D Z=\angle C B A$.

Therefore, $\triangle Z D J$ is congruent to $\triangle A B C$ (angle-side-angle).

Thus, $J Z=C A$ and so $A E=J Z=A C$, as required.