Circles Test - 20

$$ Given-\\ A,\quad B,\quad C\quad \& \quad D\quad are\quad four\quad points\quad such\quad that\quad \angle BAC={ 30 }^{ o }\quad \& \quad \angle BDC={ 60 }^{ o }.\\ To\quad find\quad out-\\ if\quad D\quad is\quad the\quad centre\quad of\quad the\quad circle\quad which\quad passes\quad through\quad A,\quad B\quad \& \quad C.\\ Solution-\\ \angle BAC={ 30 }^{ o }\quad \& \quad \angle BDC={ 60 }^{ o }.\\ \therefore \quad \angle BDC=2\angle BAC.\\ We\quad know\quad that\quad the\quad angle\quad subtended\quad by\quad a\quad chord\quad at\quad the\quad \\ centre\quad of\quad a\quad circle\quad is\quad double\quad the\quad angle\quad at\quad the\quad circumference\quad \\ subtended\quad by\quad the\quad same\quad chord.\\ \therefore \quad \angle BDC\quad is\quad the\quad angle\quad at\quad the\quad centre\quad and\quad \angle BAC\quad is\quad the\\ angle\quad at\quad the\quad circumference\quad subtended\quad by\quad the\quad chord\quad BC.\\ So\quad D\quad is\quad the\quad centre\quad of\quad the\quad circle\quad which\quad passes\quad through\quad \\ A,\quad B\quad \& \quad C.\\ \therefore \quad The\quad statement\quad is\quad correct\quad by\quad fig\quad I\\ But\quad by\quad fig\quad II\quad this\quad is\quad not\quad true.\\ So\quad the\quad statement\quad is\quad not\quad always\quad true.\quad \quad \\ \\ Ans-\quad Option\quad A. $$

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

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

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 $$O$$. 

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.

$$\angle A+\angle C =180^o$$ .......(The sum of the opposite angles of a cyclic quadrilateral $$={ 180 }^{ o }$$)

$$ Given-\\ AOBEDC\quad is\quad a\quad polygon\quad inscribed\quad in\quad a\quad circle\quad with\quad centre\\ O\quad and\quad diameter\quad AOB.\\ To\quad find\quad out-\\ \angle ACD+\angle BED=?\\ Solution-\\ We\quad join\quad AE\quad \& \quad BC.\\ ACDE\quad is\quad a\quad cyclic\quad quadrilateral.\\ \therefore \quad \angle ACD+\angle AED={ 180 }^{ o }........(i)\\ Similarly\quad BCDE\quad is\quad a\quad cyclic\quad quadrilateral.\\ \therefore \quad \angle BED+\angle BCD={ 180 }^{ o }........(ii).\\ (\because \quad The\quad sum\quad of\quad the\quad opposite\quad pair\quad of\quad angles\quad of\quad \\ a\quad cyclic\quad quadrilateral={ 180 }^{ o }\quad )\\ Again\quad \angle ACB={ 90 }^{ o }=\angle AEB.........(iii\quad \& \quad iv)\quad \\ (\because \quad The\quad angle\quad subtended\quad by\quad the\quad diameter\quad of\quad a\quad circle\\ at\quad the\quad cicumference={ 90 }^{ o }).\\ Adding\quad (i),\quad (ii),\quad (iii)\quad \& \quad (iv)\quad we\quad get\\ \angle ACD+\angle AED+\angle BED+\angle BCD+\angle ACB+\angle AEB={ 180 }^{ o }+{ 180 }^{ o }+{ 90 }^{ o }+{ 90 }^{ o }={ 540 }^{ o }\\ \Longrightarrow (\angle ACD+\angle BED)+(\angle BCD+\angle BCD)+(\angle AED+\angle AEB)={ 540 }^{ o }\\ \Longrightarrow (\angle ACD+\angle BED)+\angle ACD+\angle BED={ 540 }^{ o }\\ \Longrightarrow 2(\angle ACD+\angle BED)={ 540 }^{ o }\\ \Longrightarrow \angle ACD+\angle BED={ 270 }^{ O }.\\ Ans-\quad Option\quad B.\\ \\ \\ \\  $$ 

The  circumference of circle with radius $${ R }_{ 1 }=2\pi { R }_{ 1 }$$.

The area of a triangle is equal to the base times the height.
In a semi circle, the diameter is the base of the semi-circle.
This is equal to $$2\times r$$ (r = the radius)
If the triangle is an isosceles triangle with an angle of $$45^\circ$$ at each end, then the height of the triangle is also a radius of the circle.
A = $$\frac{1}{2} \times b \times h$$ formula for the area of a triangle becomes
A = $$\frac{1}{2}\times 2 \times r \times r$$ because:
The base of the triangle is equal to $$2\times r$$
The height of the triangle is equal to r
A = $$\frac{1}{2} \times 2 \times r \times r$$ becomes:
A = $$r^2$$

Given, $$ABCD$$ is a quadrilateral inscribed in a circle.