Exercise (9.1) - Sloutions (Grade 10, New Syllabus)

ပန်းတိုင်း မွှေးပါစေ




Exercise (9.1) ၏ မေးခွန်း ပုစ္ဆာများကို ဖြေရှင်းရန် ပြဌာန်းစာအုပ်ပါ Theorem (1) မှ Theorem (4) အထိ သီအိုရမ် မှန်ကန်ချက်များကို ဦးစွာလေ့လာထားရမည် ဖြစ်ပါသည်။ Theorem (1) မှ Theorem (4) အထိ ရှင်းလင်းချက်များကို ဒီ post မှာ လေ့လာနိုင်ပါသည်။

Ex 9.1-No (1)


The point $O$ is the centre of the given circle. Find the values of $x$, $y$ and $z$.

(a)
$\begin{array}{l} x=\angle A D C(\angle \mathrm{s} \text { on arc } A C) \\\\ \therefore x=70^{\circ}\\\\ A B C D \text { is cyclic quadrilateral. }\\\\ 70^{\circ}+\beta=180^{\circ}\\\\ \therefore \beta=180^{\circ}-70^{\circ}=110^{\circ} \\\\ \text { Since } A D / / \mathrm{BC}, \\\\ \theta=\angle C A D \\\\ \therefore \quad \theta=30^{\circ} \\\\ \therefore \quad y\ =180^{\circ}-(\beta+\theta) \\\\ \quad \quad\quad =180^{\circ}-\left(110^{\circ}+30^{\circ}\right) \\\\ \quad\quad\quad =40^{\circ} \end{array}$
(b)
$\begin{array}{l} \text { Since } A C=A D, \angle A C D=x \\\\ \text{In}\ \Delta A C D, x+x=150^{\circ} \\\\ \therefore\ 2 x=150^{\circ} \Rightarrow x=75^{\circ}\\\\ A B C D \text { is a cyclic quadrilateral. }\\\\ \therefore x+\phi=180^{\circ}\\\\ \therefore\ \ \phi\ =180^{\circ}-x \\\\ \quad\quad =180^{\circ}-75^{\circ} \\\\ \quad\quad =105^{\circ} \\\\ \text { Since } A B=B C, \angle B A C=y . \\\\ \text { In } \Delta A B C, 2 y+\phi=180^{\circ} \\\\ \therefore\ 2 y=180^{\circ}-\phi \\\\ \quad\quad\ =180^{\circ}-105^{\circ} \\\\ \quad\ y\ \ =37.5^{\circ} \end{array}$
(c)


$\begin{aligned} \angle B O D &=360^{\circ}-240^{\circ} \\\\ & =120^{\circ} \\\\ x &=\displaystyle\frac{1}{2} \angle B O D \\\\ &=\displaystyle\frac{1}{2}\left(120^{\circ}\right) \\\\ &=60^{\circ} \\\\ \theta &=\displaystyle\frac{1}{2}\left(240^{\circ}\right) \\\\ \therefore\quad\theta &=120^{\circ} \\\\ \therefore\quad y &=360^{\circ}-\left(120^{\circ}+120^{\circ}+55^{\circ}\right) \\\\ &=65^{\circ} \end{aligned}$
(d)
In $\Delta A D E$,

$\theta=180^{\circ}-\left(30^{\circ}+40^{\circ}\right)=110^{\circ}$

$A C D E$ is a cyclic quadrilateral.

$\begin{aligned} \therefore y+ \theta &=180^{\circ} \\\\ \therefore y \quad &=180^{\circ}-\theta \\\\ &=180^{\circ}-110^{\circ} \\\\ &=70^{\circ} \end{aligned}$

In $\Delta A C D, A C =A D$.

$\therefore\quad \angle A D C=y$

$A C D E$ is a cyclic quadrilateral.

$\begin{aligned} \therefore\quad x+y & =180^{\circ} \\\\ \therefore\quad x \quad &=180^{\circ}-y \\\\ &=180^{\circ}-70^{\circ} \\\\ &=110^{\circ} \end{aligned}$
(e)
Since $O A=O B\ \text{( radii )}$

$\therefore\quad \theta=50^{\circ}$

$A B D E$ is a cyclic quadrilateral.

$\therefore\quad x+60^{\circ}+\theta=180^{\circ}$

$\begin{aligned} \therefore \quad x & =180^{\circ}-\left(60^{\circ}+\theta\right) \\\\ & =180^{\circ}-\left(60^{\circ}+50^{\circ}\right) \\\\ & =70^{\circ} \end{aligned}$

$C E$ is a diameter.

$\therefore \angle C D E=90^{\circ}$

$\begin{aligned} \therefore \quad x+y & =90^{\circ}\\\\ \therefore \quad y & =90^{\circ}-x\\\\ & =90^{\circ}-70^{\circ}\\\\ & =20^{\circ} \end{aligned}$
(f)
$A B D E$ is a cyclic quadrilateral.

$\therefore y=110^{\circ}$

Since $\text{arc}\ A D=\text{arc}\ B C,$

$A D=B C$

$\therefore\quad A B D E$ is isosceles trapezium.

$\therefore\quad x=y=110^{\circ}$

$x+z=180^{\circ}$

$\therefore\quad z=180^{\circ}-x$

$\quad\quad =180^{\circ}-110^{\circ}$

$\quad\quad =70^{\circ}$

Ex 9.1-No (2)


$ABC$ is an acute triangle inscribed in $\odot O$, and $OD$ is the perpendicular drawn $O$ to $BC$. Prove that $\angle BOD = \angle BAC$.

$\begin{array}{l} \text { Join }\ O C.\\\\ \text { In }\ \triangle BOD\ \text { and }\ \triangle COD,\\\\ \angle BDO = \angle CDO \ \text{(right angles)}\\\\ OB=OC \ \text{(radii)}\\\\ OD=OD \ \text{(common side)}\\\\ \therefore\quad \triangle B O D \cong \triangle C O D\\\\ \therefore\quad\angle B O D=\angle C O D=\displaystyle\frac{1}{2} \angle B O C\\\\ \text { But }\ \angle B A C=\displaystyle\frac{1}{2} \angle B O C\\\\ \therefore\quad\angle B O D=\angle BAC \end{array}$

Ex 9.1-No (3)


In $\odot O$, two chords $A B$ and $C D$ intersect in the circle at $P$. Show that $\angle A P D=\displaystyle\frac{1}{2}(\angle A O D+\angle B O C)$.

Join $A C$

In $\triangle A P C$,

$\begin{aligned} \angle A P D & =\angle P C A+\angle P A C \\\\ & =\angle D C A+\angle B A C \\\\ & =\displaystyle\frac{1}{2} \angle A O D+\displaystyle\frac{1}{2} \angle B O C \\\\ & =\displaystyle\frac{1}{2}(\angle A O D+\angle B O C) \end{aligned}$

Ex 9.1-No (4)


Two circles intersect at $M, N$. From $M$, diameters $M A, M B$ are drawn in each circle. If $A, B$ are joined to $N$, prove that $A N B$ is a straight line.

$\begin{array}{lll} \text{Given} &: & \odot O\ \text{and}\ \odot P\ \text{intersect at}\ M \text{and}\ N.\\\\ & & M A\ \text{and}\ M B\ \text{are diameters of}\ \odot O\ \text{and}\\\\ & & \odot P\ \text{and respectively.}\\\\ \text{To Prove} &: & A N B\ \text{is a straight line.}\\\\ \text{Proof} &: & \text{Join}\ M N.\\\\ & & \text{In}\ \odot O, \angle A N M=90^{\circ} \quad(AM\ \text{is a diameter} )\\\\ & & \text{In}\ \odot P, \angle B N M=90^{\circ} \quad(BM\ \text{is a diameter} )\\\\ & & \therefore\quad \angle A N M+\angle B N M=180^{\circ}\\\\ & & \therefore\quad A N B\ \text{ is a straight line.} \end{array}$

Ex 9.1-No (5)


$OA$ and $O B$ are two radii of a circle meeting at right angles. From $A, B$ two parallel chords $A X, B Y$ are drawn. Prove that $A Y \perp B X$.

$\begin{array}{lll} \text{Given} &: & O A \perp O B, A X \parallel B Y\\\\ \text{To Prove} &: & A Y \perp B X .\\\\ \text{Proof} &: & \text{Let}\ A Y\ \text{cut}\ B X\ \text{at}\ D \\\\ & & \alpha=\gamma \quad(\angle\mathrm{s}\ \text{stand on arc}\ AB )\\\\ & & \text{But}\ \alpha=\gamma=\displaystyle\frac{1}{2} \angle A O B \\\\ & & \alpha=\gamma=\displaystyle\frac{1}{2}\left(90^{\circ}\right)=45^{\circ}\\\\ & & \text{Since} A X \parallel B Y,\\\\ & & \alpha=\beta \quad (\text{alternating} \angle \mathrm{s})\\\\ & & \therefore\quad \beta=45^{\circ}\\\\ & & \text{Since} \angle A D B=\beta+\gamma,\\\\ & & \angle A D B=90^{\circ}\\\\ & & \therefore\quad A Y \perp B X.\\\\ \end{array}$

Ex 9.1-No (6)


Two circles intersect at $R$ and $S$. Two straight lines $A R B$ and $C S D$ are drawn meeting one circle at $A, C$ and the other at $B, D$. Prove that $A C \parallel B D .$ If $A B \parallel C D$, show that $A B=C D$.

$ACSR$ is a cyclic quadrilateral.

$\therefore\quad \alpha=\theta$

$BDSR$ is also a cyclic quadrilateral.

$\therefore\quad \beta+\theta = 180^{\circ}$

$\therefore\quad \alpha+\theta = 180^{\circ}$

$\therefore\quad A C \parallel B D$

If $A B \parallel C D$, $ABDC$ is a parallelogram.

then we can say $A B=C D$