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# 10 CAT Circle [Geometry] Questions with Solutions

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 $\angle A D C = 130 ^ { \circ } ,$ find $\angle$ CAB.

[1] $40 ^ { \circ }$

[2] $50 ^ { \circ }$

[3] $30 ^ { \circ }$

[4] $130 ^ { \circ }$

Option # 1

Since A B C D is a cyclic quadrilateral

Therefore, $\angle A D C + \angle A B C = 180 ^ { \circ }$

$\Rightarrow 130 ^ { \circ } + \angle A B C = 180 ^ { \circ }$

$\Rightarrow \angle A B C = 50 ^ { \circ }$

Also, $\angle A C B = 90 ^ { \circ }$

Therefore,  in $\Delta A B C ,$

$\angle A C B + \angle A B C + \angle C A B = 180 ^ { \circ } ( A S P )$

$\Rightarrow 90 ^ { \circ } + 50 ^ { \circ } + \angle C A B = 180 ^ { \circ } \Rightarrow \angle C A B = 40 ^ { \circ }$

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 $\angle D B C$

[1] $30 ^ { \circ }$

[2] $45 ^ { \circ }$

[3] $50 ^ { \circ }$

[4] $40 ^ { \circ }$

Option # 1

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

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

$\Rightarrow \angle A B D = 90 ^ { \circ }$

Or, $\angle D B C = \angle A B D - \angle A B C = 90 ^ { \circ } - 60 ^ { \circ } = 30 ^ { \circ }$

Question 3:
In the given figure, o is the centre of a circle. If $\angle A O D = 140 ^ { \circ }$ and $\angle C A B = 50 ^ { \circ } ,$ what is $\angle E D B ?$

[1] $70 ^ { \circ }$

[2] $40 ^ { \circ }$

[3] $60 ^ { \circ }$

[4] $50 ^ { \circ }$

Option # 4

$\angle B O D = 180 - \angle A O D = 180 - 140 = 40 ^ { \circ }$

$O B = O D \Rightarrow \angle O B D = \angle O D B = 70 ^ { \circ }$

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

$\Rightarrow 50 ^ { \circ } + 70 ^ { \circ } + \angle O D C = 180 \Rightarrow \angle O D C = 60 ^ { \circ }$

$\angle E D B = 180 ^ { \circ } - \left( 60 ^ { \circ } + 70 ^ { \circ } \right) = 50 ^ { \circ }$

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] $\frac { ( 2 \sqrt { 2 } - 1 ) } { 3 }$

Option # 2

Radius 3$/ 2 \mathrm { cm }$

$\mathrm { AB } = 3 \mathrm { cm }$

$\mathrm { AE } : \mathrm { EB } = 1 : 2$

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

$\mathrm { HL } = 1 / 2$

Similarly OL $= 1 / 2$

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

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

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$r + \sqrt { 3 }$

Option # 1

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

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

Or, $\angle \mathrm { OQS } = \angle \mathrm { OSQ } = 60$

Or, $\mathrm { QS } = \mathrm { r } , \angle \mathrm { POQ } = 120$

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

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

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 $\angle A T C = 30 ^ { \circ }$ and $\angle A C T = 50 ^ { \circ } ,$ then the angle $\angle B O A$ is

[1] $100 ^ { \circ }$

[2] $150 ^ { \circ }$

[3] $80 ^ { \circ }$

[4] Cannot be determined

Option # 1

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

Applying tangent secant theorem

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

$\angle B C A = 50 ^ { \circ } \ldots \angle B O A = 100 ^ { \circ }$.

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

Option # 3

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

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

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

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

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

$x = 50 , x = 10$

Since x cannot be 10 . Therefore x=50.

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

[1] 108$\sqrt { 3 }$

[2] 54$\sqrt { 3 }$

[3] 81$\sqrt { 3 }$

[4] 96$\sqrt { 3 }$

Option # 4

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

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

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

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

$\frac { 1 } { 2 } S P \times P Q = \frac { 1 } { 2 } \times 24 \times 8 \sqrt { 3 } = 96 \sqrt { 3 }$ sq.cm

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 $( \Delta A O C )$ approximately?

[1] 27$\mathrm { cm }$

[2] 30$\mathrm { cm }$

[3] 25$\mathrm { cm }$

[4] 31$\mathrm { cm }$

Option # 1

$\angle O C T = 90 ^ { \circ } , \angle D C T = 45 ^ { \circ }$

OR, $\angle O C B = 45 ^ { \circ }$

OR, $\angle C O B = 45 ^ { \circ } ( \Delta B O C$ is a right angled triangle )

OR, $\angle A O C = 180 ^ { \circ } - 45 ^ { \circ } = 135 ^ { \circ }$

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

$\Rightarrow \mathrm { OC } = 5 \sqrt { 2 } \mathrm { cm } = \mathrm { O } \mathrm { A }$

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

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

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

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

$A C ^ { 2 } \approx 170.70$

$\Rightarrow A C \approx 13 \mathrm { cm }$

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

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

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

Option # 3

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

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

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

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

$= \frac { 1 } { 2 } \times O E \times B C + \frac { 1 } { 2 } \times O D \times A B$

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

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

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

OR, Shortest side = 6 + 7 = 13 cm

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