CIRCLE MCQ LIST

 Question 1:

In the given figure, ABCD is a cyclic quadrilateral whose side AB is a diameter of the circle through A , B and C . If $\angle ADC={130}^{\circ },$ find $\angle$ CAB.

[1] ${40}^{\circ }$

[2] ${50}^{\circ }$

[3] ${30}^{\circ }$

[4] ${130}^{\circ }$

Answer & Solution

Option # 1

Since A B C D is a cyclic quadrilateral

Therefore, $\angle ADC+\angle ABC={180}^{\circ }$

$\Rightarrow {130}^{\circ }+\angle ABC={180}^{\circ }$

$\Rightarrow \angle ABC={50}^{\circ }$

Also, $\angle ACB={90}^{\circ }$

Therefore,  in $\Delta ABC,$

$\angle ACB+\angle ABC+\angle CAB={180}^{\circ }\left(ASP\right)$

$\Rightarrow {90}^{\circ }+{50}^{\circ }+\angle CAB={180}^{\circ }\Rightarrow \angle CAB={40}^{\circ }$

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Question 2:
PBA and PDC are two secants. AD is the diameter of the circle with centre at $0.\angle A={40}^{\circ },$ $\angle P={20}^{\circ }.$ Find the measure of $\angle DBC$

[1] ${30}^{\circ }$

[2] ${45}^{\circ }$

[3] ${50}^{\circ }$

[4] ${40}^{\circ }$

Answer & Solution

Option # 1

In $\Delta ADP,\angle ADC={40}^{\circ }+{20}^{\circ }={60}^{\circ }$

Therefore, $\angle ABC=\angle ADC={60}^{\circ }$

Since AD is the diameter

$\Rightarrow \angle ABD={90}^{\circ }$

Or, $\angle DBC=\angle ABD-\angle ABC={90}^{\circ }-{60}^{\circ }={30}^{\circ }$

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Question 3:
In the given figure, o is the centre of a circle. If $\angle AOD={140}^{\circ }$ and $\angle CAB={50}^{\circ },$ what is $\angle EDB?$

[1] ${70}^{\circ }$

[2] ${40}^{\circ }$

[3] ${60}^{\circ }$

[4] ${50}^{\circ }$

Answer & Solution

Option # 4

$\angle BOD=180-\angle AOD=180-140={40}^{\circ }$

$OB=OD\Rightarrow \angle OBD=\angle ODB={70}^{\circ }$

Also $\angle CAB+\angle BDC=180$ [ because  A B C D is cyclic ]

$\Rightarrow {50}^{\circ }+{70}^{\circ }+\angle ODC=180\Rightarrow \angle ODC={60}^{\circ }$

$\angle EDB={180}^{\circ }-({60}^{\circ }+{70}^{\circ })={50}^{\circ }$

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Question 4:
In the following figure, the diameter of the circle is 3 cm. AB and MN are two diameters such that MN is perpendicular to AB. In addition, CG is perpendicular to AB such that AE:EB = 1:2, and DF is perpendicular to MN such that NL:LM = 1:2. The length of DH in cm is

[1] $2\sqrt{2}-1$

[2] $\frac{(2\sqrt{2}-1)}{2}$

[3] $\frac{(3\sqrt{2}-1)}{2}$

[4] $\frac{(2\sqrt{2}-1)}{3}$

Answer & Solution

Option # 2

Radius 3$/2cm$

$AB=3cm$

$AE:EB=1:2$

$AE=1$ and $\text{OE}=3/2-1=1/2\text{cm}$

$HL=1/2$

Similarly OL $=1/2$

Let $OH=x$ and $OD=3/2$ radius in $\Delta ODL$ by Pythagoras theorem ${OD}^2={OL}^2+{DL}^2$

${\left(\frac{3}{2}\right)}^2={\left(\frac{1}{2}\right)}^2+(x+\frac{1}{2})\Rightarrow x=\frac{2\sqrt{2}-1}{2}$

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Question 5:
P, Q, S, and R are points on the circumference of a circle of radius r, such that PQR is an equilateral triangle and PS is a diameter of the circle. What is the perimeter of the quadrilateral PQSR?

[1] 2$r(1+\sqrt{3})$

[2] 2$r(2+\sqrt{3})$

[3] $r(1+\sqrt{5})$

[4] 2$r+\sqrt{3}$

Answer & Solution

Option # 1

$\angle QPO={30}^{\circ }$

Or, $\angle QOS=60$ (angle at the center)

Or, $\angle OQS=\angle OSQ=60$

Or, $QS=r,\angle POQ=120$

 

By sine rule $\frac{\sin 30}{r}=\frac{\sin 120}{PQ}$ therefore, $\frac{1}{2r}=\frac{\sqrt{3}}{2\times PQ}\Rightarrow PQ=r\sqrt{3}$

Perimeter $=r\sqrt{3}+r\sqrt{3}+r+r=2r(\sqrt{3}+1)$

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Question 6:
In the figure given below (not drawn to scale), A, B and C are three points on a circle with centre O. The chord BA is extended to a point T such that CT becomes a tangent to the circle at point C. If $\angle ATC={30}^{\circ }$ and $\angle ACT={50}^{\circ },$ then the angle $\angle BOA$ is

[1] ${100}^{\circ }$

[2] ${150}^{\circ }$

[3] ${80}^{\circ }$

[4] Cannot be determined

Answer & Solution

Option # 1

In triangle $ACT,\angle C={50}^{\circ },\angle T={30}^{\circ }$, therefore,  $\angle A={100}^{\circ }.$

Applying tangent secant theorem

$\angle B={50}^{\circ }$ and since $\angle CAT$ is the external angle of the triangle ABC

$\angle BCA={50}^{\circ }\dots \angle BOA={100}^{\circ }$.

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Question 7:
In the figure below, the rectangle at the corner measures 10 cm × 20 cm. The corner A of the rectangle is also a point on the circumference of the circle. What is the radius of the circle in cm?

[1] 10 cm

[2] 40 cm

[3] 50 cm

[4] None of these

Answer & Solution

Option # 3

$(x-20{)}^2+(x-10{)}^2=x^2$

$x^2+400-40x+x^2+100-20x=x^2$

$x^2-60x+500=0$

$x^2-50x-10x+500=0$

$x(x-50)-10(x-50)=0$

$x=50,x=10$

Since x cannot be 10 . Therefore x=50.

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Question 8:
Given below is a circle with centre $O$ and four points $-P,Q,R$ and $S-on$ the circle. If the chords SQ and PR intersect each other at 0 and the radius of the circle is $8\sqrt{3}cm,$ find area (in sq.cm) of $\Delta PSQ$

[1] 108$\sqrt{3}$

[2] 54$\sqrt{3}$

[3] 81$\sqrt{3}$

[4] 96$\sqrt{3}$

Answer & Solution

Option # 4

$\Delta SPQ$ is right-angled (angle in a semicircle)

$\angle POQ={180}^{\circ }-{120}^{\circ }={60}^{\circ }$ and $OP=OQ=$ radius (i.e. 8$\sqrt{3}cm⟩$ . Hence $\Delta POQ$ is equilateral and $PQ=8\sqrt{3}cm$

Now in $\Delta SPQ,SP=\sqrt{S{Q}^2-P{Q}^2}=24cm=\sqrt{(2\times 8\sqrt{3}{)}^2-(8\sqrt{3}{)}^2}$

$\Rightarrow$ Area of $\Delta SPQ$, right-angled at P , will be

$\frac{1}{2}SP\times PQ=\frac{1}{2}\times 24\times 8\sqrt{3}=96\sqrt{3}$ sq.cm

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Question 9:
In the given diagram CT is tangent at C, making an angle of $\frac{\pi }{4}$ with CD. O is the centre of the circle. CD = 10 cm. What is the perimeter of the shaded region $\left(\Delta AOC\right)$ approximately?

[1] 27$cm$

[2] 30$cm$

[3] 25$cm$

[4] 31$cm$

Answer & Solution

Option # 1

$\angle OCT={90}^{\circ },\angle DCT={45}^{\circ }$

OR, $\angle OCB={45}^{\circ }$

OR, $\angle COB={45}^{\circ }(\Delta BOC$ is a right angled triangle )

OR, $\angle AOC={180}^{\circ }-{45}^{\circ }={135}^{\circ }$

Now, Because $CD=10\Rightarrow BC=5cm=OB$

$\Rightarrow OC=5\sqrt{2}cm=OA$

Again, ${AC}^2={OA}^2+{OC}^2-2OA\cdot OC\cos {135}^{\circ }$

$=2(OA{)}^2-2(OA{)}^2\cdot \cos {135}^{\circ }$

$=2(5\sqrt{2}{)}^2-2(5\sqrt{2}{)}^2\times (-\frac{1}{\sqrt{2}})$

$=100+\frac{100}{\sqrt{2}}$

$A{C}^2\approx 170.70$

$\Rightarrow AC\approx 13cm$

OR,  Perimeter of $\Delta OAC=OA+OC+AC$

$=5\sqrt{2}+5\sqrt{2}+13=27cm.$

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Question 10:
The radius of the incircle of a $\Delta$ is 4 cm and the segments into which one side is divided by the point of contact are 6 cm and 8 cm, then the length of the shortest side of the $\Delta$ is

[1] 12 cm

[2] 15 cm

[3] 13 cm

[4] 14 cm

Answer & Solution

Option # 3

BD = BE = 6 cm and AB = (x + 6) cm, BC = (16 + 8)cm = 14cm AC = (x + 8)cm

Hence, $S=\frac{a+b+c}{2}=\frac{2x+28}{2}=x+14$

Now ar. ($\Delta$ ABC) = ar.( $\Delta$ OBC) + ar.( $\Delta$ OCA) + ar.( $\Delta$ OAB)

$\Rightarrow \sqrt{S(S-a)(S-b)(S-c)}$

$=\frac{1}{2}\times OE\times BC+\frac{1}{2}\times OD\times AB$

$\Rightarrow 4\sqrt{3x^2+42x}=4(14+x)$

$\Rightarrow 2x^2-14x-196=0$ or $x^2-7x-98=0$

Therefore,  x = 7 , x = - 14 ( not possible )

OR, Shortest side = 6 + 7 = 13 cm