Circles Test

Result

Circles Test - 17

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

In Fig 10.4 if $$\displaystyle \angle ABC = 20^{\circ}$$ then $$\displaystyle \angle AOC $$ is equal to

A
$$\displaystyle 20^{\circ}$$

B
$$\displaystyle 40^{\circ}$$

C
$$\displaystyle 60^{\circ}$$

D
$$\displaystyle 10^{\circ}$$

Solution

We know that the angle at the centre of the circle is twice the angle at the circumference subtended by the same arc.
Thus,
$$\angle AOC=2 \times \angle ABC$$
$$\implies \angle AOC=2\times 20^0$$
$$\implies \angle AOC=40^0$$
Question 2
1 / -0

In Fig 10.7 if $$\displaystyle \angle DAB=60^{\circ},\angle ABD= 50^{\circ}$$ then $$\displaystyle \angle ACB$$ is equal to

A
$$\displaystyle 60^{\circ}$$

B
$$\displaystyle 50^{\circ}$$

C
$$\displaystyle 70^{\circ}$$

D
$$\displaystyle 80^{\circ}$$

Solution

In $$\triangle ADC,$$ we have
$$\angle DAC=60^0$$ and $$\angle ABD=50^0$$
By angle sum property, in $$\triangle ABD$$
$$\angle DAB+\angle ABD+\angle ADB=180^0$$
$$=>60^°+50^°+\angle ADB=180^°$$
$$=>\angle ADB=180^°-110^°$$
$$=>\angle ADB=70^°$$
We know that the angles subtended at the circumference by the same arc are equal.
$$\therefore \angle ADB=\angle ACB$$
$$=>\angle ACB=70^0$$
Question 3
1 / -0

In the given figure $$O$$ is the centre of the circle and $$\displaystyle \angle AOC=120^{\circ}$$. What is the value of $$\displaystyle \angle APC+\angle ABC$$?

A
$$\displaystyle 120^{\circ}$$

B
$$\displaystyle 140^{\circ}$$

C
$$\displaystyle 160^{\circ}$$

D
$$\displaystyle 180^{\circ}$$

Solution

As $$APCB$$ is a cyclic quadrilateral, sum of its opposite angles are always equal to $$180^o$$.
Here, $$\angle APC$$ and $$\angle ABC$$ are the opposite angles of the cyclic quadrilateral.
Then clearly,
$$\angle APC+\angle ABC=180^o$$.

Therefore, option $$D$$ is correct.
Question 4
1 / -0

Fixed point in the circle is called _____ of the circle.

A
radius

B
centre

C
diameter

D
none

Solution

A circle is the set of all those point in a plane whose distance from a fixed point remains constant. Then, this fixed point is called the centre of the circle.
Hence, option $$B$$ is correct.
Question 5
1 / -0

In the given figure, $$\displaystyle \triangle ABC$$ is an equilateral triangle. What is the value of $$\displaystyle \angle BEC$$?

A
$$\displaystyle 80^{\circ}$$

B
$$\displaystyle 100^{\circ}$$

C
$$\displaystyle 120^{\circ}$$

D
$$\displaystyle 130^{\circ}$$

Solution

Here, $$ABEC$$ is a cyclic quadrilateral.
We know, sum of its opposite angles are equal to $$180^o$$.

$$\Rightarrow \angle BAC+\angle BEC =180^{\circ}$$

$$\Rightarrow 60^o+\angle BEC =180^{\circ}$$

$$\Rightarrow \angle BEC =180^{\circ}-60^{\circ}=120^{\circ}$$.

Hence, option $$C$$ is correct.
Question 6
1 / -0

In the figure, $$O$$ is the center and $$\angle ACB=20^°$$. Then the $$\angle AOB$$ is equal to

A
$$40^°$$

B
$$60^°$$

C
$$20^°$$

D
$$30^°$$

Solution

Theorem: Angle subtended by an arc at the centre is double the angle subtended by the same arc at the circumference of a circle
Thus, $$\angle AOB=2\times 20=40^°$$
Question 7
1 / -0

In the figure, $$O$$ is the center, then $$\angle ACD$$ equals

A
$$20^°$$

B
$$60^°$$

C
$$45^°$$

D
$$30^°$$

Solution

Since,
$$\angle AOD=90^°\\ \angle ACD\quad is\quad substended\quad by\quad the\quad same\quad arc.\quad Therefore\\ \angle ACD=\dfrac{\angle AOD}{2}= \frac { 90 }{ 2 } =45^°$$
Question 8
1 / -0

What is name of following shape ?

A
Cylinder

B
Cone

C
Circle

D
Cube

Solution

it is a circle..
Question 9
1 / -0

In the figure shown above, the radius $$OA$$ is equal to the chord $$AB$$. Then, what is $$\angle APB$$?

A
$$30^{\circ}$$

B
$$60^{\circ}$$

C
$$15^{\circ}$$

D
$$45^{\circ}$$

Solution

$$OAB$$ is an equilateral triangle(all sides are equal). So $$\angle AOB = 60^{\circ}$$
$$\Rightarrow \angle APB = \dfrac {1}{2} \angle AOB = 60^{\circ} \div 2 = 30^{\circ}$$.
Question 10
1 / -0

From the figure, identify the center of the circle.

A
$$A$$

B
$$D$$

C
$$B$$

D
$$C$$

Solution

Center of a circle is a point inside the circle and is at an equal distance from all of the points on its circumference.
In the given figure, $$C$$ is such point.
So, $$C$$ is the center of the given circle.
Hence, option $$D$$ is correct.