True False[1 Marks ]

Question 11 Mark

Write True or False and justify your answer in the following:
ABCD is a cyclic quadrilateral such that $\angle\text{A}=90^\circ,\angle\text{B}=70^\circ,\angle\text{C}=95^\circ$ and $\angle\text{D}=105^\circ.$

Answer

False.
Solution:
In a cyclic quadrilateral, the sum of opposite angles is 180°.
Now, $\angle\text{A}+\angle\text{C}=90^\circ+95^\circ=185^\circ\neq180^\circ$
and $\angle\text{B}+\angle\text{D}=70^\circ+105^\circ=175^\circ\neq180^\circ$
Here, we see that, the sum of opposite angles is not equal to 180°. So, it is not a cyclic quadrilateral.

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Question 21 Mark

Write True or False and justify your answer in the following: If $\text{AOB}$ is a diameter of a circle and $C$ is a point on the circle, then $AC^2 + BC^2 = AB^2.$

Answer

Since, any diameter of the circle subtends a right angle to any point on the circle. If $\text{AOB}$ is a diameter of a circle and $C$ is a point on the circle, then $\triangle\text{ACB}$ is right angled at $C.$ In right angled $\triangle\text{ACB}, [$use Pythagoras theorem$] AC^2 + BC^2= AB^2$

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Question 31 Mark

Write True or False and justify your answer in the following:
Two congruent circles with centres O and O′ intersect at two points A and B. Then $\angle\text{AOB}=\angle\text{AO'B}.$

Answer

True.
Solution:

Join AB and OB, O'A and BO'.
In $\triangle\text{AOB}$ and $\triangle\text{AO'B,}$
OA = AO' [both circles have same radius]
OB = BO' [both circles have same radius]
and AB = AB [common chord]
$\triangle\text{AOB}=\triangle\text{AO'B}$ [by SSC congruence rule]
$\Rightarrow\angle\text{AOB}=\angle\text{AO'B}$ [by CPCT]

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Question 41 Mark

Write True or False and justify your answer in the following:
A circle of radius 3cm can be drawn through two points A, B such that AB = 6cm.

Answer

True.
Solution:
Suppose, we consider diameter of a circle is AB = 66m.
Then, radius of a circle $=\frac{\text{AB}}{2}=\frac{6}{2}=3\text{cm},$ which is true.

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Question 51 Mark

Write True or False and justify your answer in the following:
If A, B, C and D are four points such that $\angle\text{BAC}=45^\circ$ and $\angle\text{BDC}=45^\circ,$ then A, B, C, D are concyclic.

Answer

True. Solution: Since, $\angle\text{BAC}=45^\circ$ and $\angle\text{BDC}=45^\circ$

As we know, angles in the same segment of a circle are equal. Hence, A, B, C and D are concyclic.

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Question 61 Mark

Write True or False and justify your answer in the following:
Through three collinear points a circle can be drawn.

Answer

False.
Solution:
Because, circle can pass through only two collinear points but not through three collinear points.

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Question 71 Mark

Write True or False and justify your answer in the following:
Two chords AB and CD of a circle are each at distances 4cm from the centre. Then AB = CD.

Answer

True.
Solution:
Because, the chords equidistant from the centre of circle are equal in length.

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Question 81 Mark

Write True or False and justify your answer in the following:
In Fig. if AOB is a diameter and $\angle\text{ADC}=120^\circ,$ then $\angle\text{CAB}=30^\circ.$

Answer

True.
Solution:
Join CA and CB.

Since, ADCB is a cyclic quadrilateral.
$\angle\text{ADC}+\angle\text{CBA}=180^\circ$ [sum of opposite angles of cyclic quadrilateral is 180°]
$\Rightarrow\angle\text{CBA}=180^\circ-120^\circ=60^\circ\ \ [\therefore\angle\text{ADC}=120^\circ]$
In $\triangle\text{ACB,}\ \angle\text{CAB} + \angle\text{CBA} + \angle\text{ACB} = 180^\circ$ [by angle sum property of a triangle]
$\angle\text{CAB}+60^\circ+90^\circ=180^\circ$ $\big[$triangle formed from diameter to the circle is 90° i.e., $\angle\text{ACB}=90^\circ\big]$
$\Rightarrow\angle\text{CAB}=180^\circ-150^\circ=30^\circ.$

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Question 91 Mark

Write True or False and justify your answer in the following:
If A, B, C, D are four points such that $\angle\text{BAC}=30^\circ$ and $\angle\text{BDC}=60^\circ$ then D is the centre of the circle through A, B and C.

Answer

False.
Solution:
Because, there can be many points D, such that $\angle\text{BDC}=60^\circ$ and each such point cannot be the centre of the circle through A, B and C.

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Question 101 Mark

Write True or False and justify your answer in the following:
Two chords AB and AC of a circle with centre O are on the opposite sides of OA. Then $\angle\text{OAB}=\angle\text{OAC}.$

Answer

False.
Solution:
In figure, AB and AC are two chords of a circle. Join OB and OC.

In $\triangle\text{OAB}$ and $\triangle\text{OAC},$
OA = OA [common side]
OB = OC [both are the radius of circle]
Here, we are not able to show that either the any angle or third side is equal and $\triangle\text{OAB}$ is,
$\therefore\angle\text{OAB}\neq\angle\text{OAC}.$

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