(a) Using a ruler and a pair of compasses only, (i) construct \(\Delta\) XYZ such that |XY| = 8 cm and < YXZ = < ZYX = 45°. (ii) locate a point P inside the...

(a) Construction of the triangle and its incircle

1. Draw the line segment \(XY=8\text{ cm}\).
2. At \(X\), construct a perpendicular to \(XY\), then bisect the \(90^\circ\) angle to make a ray at \(45^\circ\) to \(XY\).
3. At \(Y\), construct a perpendicular to \(YX\), then bisect the \(90^\circ\) angle to make a ray at \(45^\circ\) to \(YX\). The two \(45^\circ\) rays meet at \(Z\).
4. Bisect \(\angle YXZ\) and \(\angle XYZ\). Their intersection is \(P\).

A point on an angle bisector is equidistant from the two sides of that angle. Therefore, \(P\) is equidistant from \(XY\), \(XZ\), and \(YZ\): it is the incentre of the triangle.

To construct the circle, draw a perpendicular from \(P\) to any side, for example \(XY\). Use the perpendicular distance as the radius and draw a circle with centre \(P\). Because \(P\) is equidistant from all three sides, this circle touches all three sides.

The triangle is right-angled at \(Z\), with hypotenuse \(XY=8\text{ cm}\). Its two equal shorter sides are

\[ 8\cos45^\circ=4\sqrt{2}\text{ cm}. \]

For a right-angled triangle, the inradius is

\[ r=\frac{a+b-c}{2}. \]

Hence,

\[ r=\frac{4\sqrt2+4\sqrt2-8}{2} =4\sqrt2-4 =4(\sqrt2-1) \approx1.66\text{ cm}. \]

Therefore, the measured radius is approximately \(1.7\text{ cm}\).

(b) Add all six side lengths to make the perimeter:

\[ (x-5)+2x+2x+(2x+7)+2x+(2x-1)=144. \]

Collecting like terms gives

\[ 11x+1=144 \]

\[ 11x=143 \]

\[ x=\frac{143}{11}=13. \]

Therefore, \(x=13\).

Examination reminder: When finding a perimeter, include every side exactly once and keep brackets around expressions such as \(x-5\) and \(2x+7\) before simplifying.