Sec 4 E Maths: Circle Properties, practice questions & worked solutions

$$AB^2=26^2=676$$

$$AC^2+BC^2=24^2+10^2=676=AB^2$$

By the converse of Pythagoras' Theorem, $\triangle ABC$ is a right-angled triangle.
Hence, by the converse of right-angle in semi-circle, $AB$ is the diameter of the circle.

Answer: $AC^2+BC^2=24^2+10^2=676=AB^2$, so $\angle ACB=90^\circ$ (converse of Pythagoras' theorem). A right angle at $C$ on the circle means $AB$ is a diameter (angle in a semicircle).

If a circle can pass through all points $A$, $B$, $C$ and $D$, then Angle DAB + Angle DCB $\boldsymbol{=180^\circ}$ (angles in opp. segment) CHECK:

$\begin{array}{l}\angle ADB=\angle ABD=36^\circ\ \text{(isos. }\Delta)\\\angle DAB=180^\circ-36^\circ-36^\circ=108^\circ\ \text{(sum of angles in }\Delta)\\\angle BDC=180^\circ-121^\circ=59^\circ\ \text{(adj. angles on str. line)}\\\angle DCB=180^\circ-59^\circ-49^\circ=72^\circ\ \text{(sum of angles in }\Delta)\\\quad \therefore\ \angle DAB+\angle DCB=108^\circ+72^\circ=180^\circ\\\quad \textbf{(angles in opp. segment)}\end{array}$

Since the above two angles add up to $180^\circ$, it means that it will be possible to draw a circle that passes through all points $A$, $B$, $C$ and $D$.

Answer: $\angle DAB=108^\circ$ and $\angle DCB=72^\circ$, so $\angle DAB+\angle DCB=180^\circ$. Opposite angles are supplementary, so $ABCD$ is a cyclic quadrilateral and a circle passes through $A$, $B$, $C$ and $D$.

Reflex Angle $AOC=100\times2=200^\circ$
(Angle at centre is twice the angle at circumference)

Obtuse angle $AOC=360^\circ-200^\circ=160^\circ$
(Angles at a point)

Angle $OAC=\dfrac{180^\circ-160^\circ}2=10^\circ$
(Angles of an isosceles triangle $OAC$)

Answer: Angle $OAC=10^\circ$

$$\angle CDA=180^\circ-119^\circ\quad(\angle\text{s in opposite segments})$$

$$=61^\circ$$

$$\angle OAT=90^\circ\quad(\text{tangent}\perp\text{radius})$$

$$\angle OAD=29^\circ\quad(\text{base angles of isosceles triangle})$$

So, $\angle DAT=90^\circ-29^\circ$

$$=61^\circ$$

Since $\angle CDA=\angle DAT$, they are alternate angles and lines $CD$ and $ST$ are parallel.

Answer: $CD\parallel ST$

(a)

$$\begin{aligned}\angle POR&=(180-x)^\circ&&\text{(angles in opp segments)}\\\text{reflex }\angle POR&=360^\circ-(180-x)^\circ&&\text{(angles at a pt)}\\&=(180+x)^\circ\\\angle PSR&=\frac{(180+x)^\circ}{2}&&\text{(angle at centre = $2\times$ angle at circumference)}\end{aligned}$$

(b)

$$\angle PUR=\frac{(180+x)^\circ}{2}\quad\text{(angles in the same segment)}$$

$$105=\frac{180+x}{2}$$

$$x=30$$

Answer: $\text{(a) }\dfrac{(180+x)^\circ}{2}\quad\text{(b) }x=30$

(a)(i)

$\angle CBD=28^\circ$ (angles in same segment)

(a)(ii)

$\angle ABC$

$$\begin{aligned}&=180^\circ-86^\circ\text{( adj }\angle\text{ on a straight line)}\\&=94^\circ\end{aligned}$$

$\angle BDC$

$$\begin{aligned}&=180^\circ-94^\circ-24^\circ(\angle\text{ in opp segments)}\\&=62^\circ\end{aligned}$$

(a)(iii)

$\angle EBD$

$$\begin{aligned}&=48^\circ\ (\angle\text{ in the same segment)}\end{aligned}$$

$\angle DBC$

$$\begin{aligned}&=28^\circ(\angle\text{ in the same segment)}\end{aligned}$$

$\angle ADE$

$$\begin{aligned}&=\angle ABE\ (\angle\text{ in the same segment)}\\&=94^\circ-28^\circ-48^\circ\\&=18^\circ\end{aligned}$$

(b)

$$\begin{aligned}\angle BCD&=180^\circ-\angle CBD-\angle CDB\ (\angle\text{ sum of triangle)}\\&=180^\circ-62^\circ-28^\circ\\&=90^\circ\end{aligned}$$

Since $\angle BCD=90^\circ$, $BD$ is the diameter by the property rt. $\angle$ in semicircle.

Answer: (a)(i) $28^\circ$   (ii) $62^\circ$   (iii) $18^\circ$   (b) $\angle BCD=90^\circ$, so $BD$ subtends a right angle at the circumference and is a diameter

(a(i))

$\angle FOG$

$$=180^\circ-64^\circ-64^\circ\text{ (base angles of isosceles triangle)}$$

$$=52^\circ$$

(a(ii))

$\angle FEG$

$$=\frac{52^\circ}{2}\text{ (angle at centre}=2\times\text{angle at circumference)}$$

$$=26^\circ$$

$$\angle FGE=90^\circ\text{ (angle in a semicircle)}$$

$$\angle EGI=180^\circ-90^\circ-26^\circ-25^\circ\text{ (angles in opposite segments)}$$

$$=39^\circ$$

$$\angle IGH=180^\circ-90^\circ-39^\circ\text{ (adjacent angles on a straight line)}$$

$$=51^\circ$$

(b)

$$\angle EIG=180^\circ-64^\circ=116^\circ$$

$$\angle GIH=180^\circ-116^\circ=64^\circ$$

$$\frac{IH}{\sin51^\circ}=\frac{2.93}{\sin64^\circ}$$

$$IH=2.53343$$

$$\angle IOG=25^\circ\times2=50^\circ$$

Length of major arc $GFEI$

$$=\frac{360^\circ-50^\circ}{360^\circ}\times2\pi(3.5)$$

$$=18.93682$$

Perimeter

$$=18.93682+2.53343+2.93$$

$$=24.40025$$

$$=24.4\text{ cm}$$

Alternative method/solution:

$$IG^2=3.5^2+3.5^2-2(3.5)(3.5)\cos50^\circ$$

$$IG=2.958327$$

$$\frac{IH}{\sin51^\circ}=\frac{2.958327}{\sin65^\circ}$$

$$IH=2.5367$$

Perimeter

$$=18.93682+2.5367+2.93$$

$$=24.40352$$

$$=24.4\text{ cm}$$

Answer: $\text{(a)(i) }52^\circ\quad\text{(ii) }51^\circ\quad\text{(b) }24.4\text{ cm}$

(a)

$$\begin{aligned}\angle OPV=\angle ORV&=90^\circ\ \ (\text{tan}\perp\text{rad})\\\angle POR&=360-90-90-40\ \ (\text{sum of quad})\\&=140\\\textit{reflex}\ \angle POR&=360-140\ \ (\angle\text{ at a pt.})\\&=220\\\angle PQR&=\frac{220}{2}\ \ (\angle\text{ at ctr}=2\times\angle\text{ at circum})\\&=110\end{aligned}$$

(b)

$$\begin{aligned}\angle OPR&=\frac{180-140}{2}\ \ (\text{sum of isos }\Delta)\\&=20\\\angle SPR&=\angle STR\ \ (\angle\text{ in same seg})\\&=43\\\angle OPU&=43-20\\&=23\\\angle POU&=180-140\ \ (\angle\text{ on str. line})\\&=40\\\angle TUP&=\angle OPU+\angle POU\ \ (\text{ext. }\angle\text{ of }\Delta)\\&=23+40\\&=63^\circ\end{aligned}$$

(c)

$$\begin{aligned}\angle SOR&=2\times\angle STR\ \ (\angle\text{ at ctr}=2\times\angle\text{ at circum})\\&=2\times43\\&=86\\\text{reflex}\ \angle SOR&=360-86\\&=274\\\text{area of minor sector}&=\frac{86}{274}\times\text{area of major sector}\\&=0.31387\times\text{area of major sector}\end{aligned}$$

No because the area of the minor sector is 0.314 of the area of major sector, which is less than $\dfrac13$ (0.333)

Answer: (a) $110^\circ$ (b) $63^\circ$ (c) No

(a)(i)

$\angle ABC=\dfrac{100}2=50^\circ$ (angle at centre $=2\times$ angle at circumference)

(a)(ii)

$\angle ADC=180^\circ-50^\circ=130^\circ$ (angles in opp seg)

(a)(iii)

$$\begin{aligned}\angle ACO&=\frac{180^\circ-100^\circ}2=40^\circ\quad\text{(base angles of isos triangle)}\\\angle COT&=40^\circ\quad\text{(alt angles, }AC\parallel OT)\\\angle OCT&=90^\circ\quad\text{(tan perpendicular to rad)}\\\angle OTC&=180^\circ-90^\circ-40^\circ=50^\circ\quad\text{(angle sum of triangle)}\end{aligned}$$

(b)

$$\begin{aligned}\angle ADC+\angle AOC&=130^\circ+100^\circ\\&=230^\circ\\&\ne180^\circ\end{aligned}$$

They are not angles in opposite segment.
A circle cannot be drawn passing through the points $A$, $O$, $C$ and $D$.

(c)

$$\begin{aligned}\tan50^\circ&=\frac{OC}{4.8}\\OC&=5.72042\end{aligned}$$

Minor segment = minor sector $-$ triangle

$$\begin{aligned}&=\frac{100}{360}\times\pi(5.72042)^2-\frac12(5.72042)^2\sin(100^\circ)\\&=12.4\text{ cm}^2\end{aligned}$$

Answer: (a)(i) $50^\circ$ (ii) $130^\circ$ (iii) $50^\circ$ (c) $12.4$ cm$^2$

(a)(i)(i)

angle $EBC=64^\circ$ (angles in the same segment)

(a)(i)(ii)

angle $EOC=2\times64^\circ=128^\circ$ (angles at centre $=2\times$ angles at circumference)

reflex angle $EOC=360^\circ-128^\circ=232^\circ$ (angles at a point)

(a)(i)(iii)

angle $EBC=64^\circ$ (angles in the same segment)

angle $EDC=180^\circ-64^\circ=116^\circ$ (angles in opposite segments)

angle $EDG=180^\circ-116^\circ-46^\circ=18^\circ$ (adj angles on a str line)

(a)(ii)

$CH=DH$ (tangent from ext point)

angle $CHD=180^\circ-2(46^\circ)=88^\circ$

Since angle $CHD$ does not form a right angle in a semi circle, therefor a semicircle with $CD$ as diameter will not pass through $H$.

(b)(i)

A: angle $ORS=$ angle $PRQ$ (common angle)

Angle $OSR=x^\circ$ (base angle of isosceles triangle)
Angle $OSP=180^\circ-x^\circ$ (adj angles on a str line)
Angle $PQR=x^\circ$ (angles in opposite segments)

A: Angle $PQR=$ angle $OSR=x^\circ$

By AA, triangles $ORS$ and $PRQ$ are similar.

(b)(ii)

Since $x^\circ=60^\circ$, triangle $ORS$ is equilateral

$OR:QR$
$1:2$

triangle $ORS$ : triangle $PRQ$
$1:4$

area of triangle $ORS$ and quadrilateral $PQOS$.
$1:3$

Answer: (a)(i)(a) $64^\circ$   (a)(i)(b) $232^\circ$   (a)(i)(c) $18^\circ$   (a)(ii) $\angle CHD=88^\circ$, not $90^\circ$   (b)(ii) $1:3$