CBSE Questions for Class 9 Maths Circles Quiz 3 - MCQExams.com

Thus, $$\angle AOB=2\times 20=40^°$$

Since, 

it is a circle..

Center of a circle is a point inside the circle and is at an equal distance from all of the points on its circumference.

Angle subtended by an arc(here, $$AQ$$) at the centre is twice the angle subtended by it on the circumference on same side of the chord(here, $$AQ$$)

Center of a circle is a point inside the circle and is at an equal distance from all of the points on its circumference.

Here, the given quadrilateral is cyclic.

Two concentric circles do not intersect. They only share a common centre as shown in the figure.

In the given figure, $$\angle BDC=\angle BAC$$

In the given figure,

In $$\Delta BED$$,

Consider the diagram shown in the question.

 

Given:

$$AB=AC$$

$$\angle BOC=100{}^\circ $$

$$\Rightarrow \angle BAC=50{}^\circ $$ (angle at the circle is half of angle at the centre by same chord $$\left( BC \right)$$)

 

Consider the quadrilateral $$OBAC$$.

$$\angle BOC$$ (of the quadrilateral) $$=360{}^\circ -100{}^\circ =260{}^\circ $$

 

Consider triangle $$OBC$$.

$$\angle OBC=\angle OCB$$ ($$OB$$ and $$OC$$ are radius)

 

Similarly, in $$\Delta ABC$$,

$$\angle ABC=\angle ACB$$ (Since, $$AB=AC$$)

$$ \angle ABO+\angle OBC=\angle ACO+\angle OCB $$

$$ \Rightarrow \angle ABO=\angle ACO $$

 

We know that interior angles of a quadrilateral are $$360{}^\circ $$. Therefore,

$$ 50{}^\circ +260{}^\circ +\angle ABO+\angle ACO=360{}^\circ  $$

$$ 2\angle ACO=50{}^\circ  $$

$$ \angle ACO=25{}^\circ  $$

 

Hence, this is the required result.

$$\text{Theorem: Angles in the same segments are equal}$$
$$\therefore x^{\circ}=30^{\circ}$$

We have,
$$\displaystyle \angle AOB=2\angle ACB$$
[ Angle  of  the  centre  is  twice  at  th  angle  of circumference \right ]
$$\displaystyle 80^{\circ}=2x$$
$$\displaystyle \therefore x=\frac{80}{2}=40^{\circ}.$$

Chords of same length subtend equal angles on the same circle.

A circle is defined by any three non-collinear points. This means that, given any three points that are not on the same line, you can draw a circle that passes through them. It is possible to construct this circle using only a compass and straightedge.

Given- $$A, B$$ &  $$C$$ are points such that $$AB=12cm$$ and $$BC=16cm$$  and $$BC\bot AB$$.

To find out- the radius of the circle passing through $$A, B$$ &  $$C=?$$

Solution- $$BC\bot AB$$ i.e $$\angle ABC={ 90 }^{ 0 }$$ and $$\Delta ABC$$ is a right one with $$AC$$ as hypotenuse. $$\therefore  AC=\sqrt { { AB }^{ 2 }+{ BC }^{ 2 } } =\sqrt { { 12 }^{ 2 }+{ 16 }^{ 2 } } cm=20cm$$. So the circle passing through $$A, B$$ &  $$C$$ will have its  diameter as $$AC$$.

$$\therefore$$  Its radius$$=\dfrac { 1 }{ 2 } \times 20cm=10cm.$$

Ans- Option C.

It is given that, Diameter, $$AD = 34$$ cm and chord, $$AB = 30$$ cm

Draw a perpendicular $$ON$$ from $$O$$ to $$AB$$. It meets $$AB$$ at $$N$$.

$$ \therefore$$ Radius, $$AO=\dfrac { 1 }{ 2 } AD=\dfrac { 1 }{ 2 } \times 34$$ cm $$=17$$ cm.

$$\because ON\bot AB$$

$$ \therefore ON$$ bisects $$AB$$ at $$N$$ and $$\angle ANO=90^0$$

So, $$AN=\dfrac { 1 }{ 2 }AB=\dfrac { 1 }{ 2 } \times 30$$ cm $$=15$$ cm

Now, $$AO=17$$ cm, $$AN=15$$ cm and $$\angle ANO=90^0$$ 

$$\therefore \Delta AON$$ is a right one with hypotenuse $$AO.$$ 

So, by pythagoras theorem, we have

$${ AO }^{ 2 }-{ AN }^{ 2 }={ ON }^{ 2 }\Rightarrow { ON }^{ 2 }={ (17 }^{ 2 }{ -15 }^{ 2 })\ { cm }^{ 2 }$$ 

$$ \Rightarrow ON=8$$ cm

So, $$AB$$ is at a distance of $$8$$ cm from the centre $$O$$. 

$$\angle ACB = 90^o$$ as it is inscribed in a semi circle.

Given- $$ AB$$ and $$AC$$ are the chords of a circle with centre $$O$$.

A circle with $$AB=6cm$$ as diameter will have its radius equal to $$3cm$$.

Given:

$$ Given-\quad \\ AXB\quad \& \quad CYD\quad are\quad congruent\quad arcs\quad of\quad a\quad circle.\\ to\quad find\quad out-\\ the\quad ratio\quad AB\quad :\quad CD=?\\ Solution-\\ We\quad join\quad AB\quad \& \quad CD.\quad \\ AXB\quad \& \quad CYD\quad are\quad congruent\quad arcs\quad of\quad the\quad circle.\\ \therefore \quad Their\quad repective\quad chords\quad will\quad be\quad equal.\\ So\quad AB=CD.\\ \therefore AB:CD=1:1.\\ Ans-\quad Option\quad A.\\  $$

$$ AB\quad is\quad the\quad chord\quad of\quad a\quad circle\quad with\quad centre\quad O.\\ AB\quad subtends\quad \angle AOB={ 90 }^{ o }\quad to\quad the\quad centre\quad and\quad \angle ACB\quad to\quad \\ the\quad circumference.\\ \angle ABC={ 30 }^{ o }.\\ To\quad find\quad out\quad -\\ \angle CAO=?\\ Solution-\\ AB\quad subtends\quad \angle AOB={ 90 }^{ o }\quad to\quad the\quad centre\quad and\quad \angle ACB\quad to\quad \\ the\quad circumference.\\ \therefore \quad \angle ACB=\frac { 1 }{ 2 } \angle AOB=\frac { 1 }{ 2 } \times { 90 }^{ o }=45^{ o }.(The\quad angle,\quad subtended\quad \\ by\quad a\quad chord\quad to\quad the\quad circumference\quad of\quad a\quad circle,\quad is\quad half\quad of\quad \\ that\quad subtended\quad to\quad the\quad centre).\\ Now\quad in\quad \Delta ACB\quad \quad we\quad have\\ \angle ACB+\angle ABC\quad =45^{ o }+30^{ o }=75^{ o }.\\ \therefore \quad \angle BAC=\quad 180^{ o }-(\angle ACB+\angle ABC)=\quad 180^{ o }-75^{ o }=105^{ o }.\\ (angle\quad sum\quad property\quad of\quad triangles)\\ Now\quad in\quad \Delta AOB\quad we\quad have\quad AO=BO\quad (radii\quad of\quad the\quad same\quad circle).\\ \therefore \quad \angle OAB=\angle OBA\quad i.e\quad \angle OAB+\angle OBA=2\angle OAB.\\ So\angle AOB+\angle OAB+\angle OBA=\angle AOB+2\angle OAB=180^{ o }.\\ (angle\quad sum\quad property\quad of\quad triangles).\\ i.e\quad { 90 }^{ o }+2\angle OAB=180^{ o }\Longrightarrow \angle OAB={ 45 }^{ o }.\\ \therefore \quad \angle CAO=\angle BAC-\angle OAB=105^{ o }-45^{ o }=60^{ o }.\\ Ans-\quad Option\quad B.\\  $$