Circles Test - 22

$$ Given-\\ PQ=30cm\quad \& \quad RS=16cm\quad are\quad the\quad chords\quad of\quad a\quad circle\quad of\quad \\ radius=17cm\quad with\quad centre\quad O.\\ Also\quad they\quad are\quad on\quad the\quad opposite\quad sides\quad of\quad O.\quad \quad \quad \\ PQ\parallel RS.\\ To\quad find\quad out\quad -\quad The\quad distance\quad between\quad PQ\quad \& \quad RS=?\\ the\quad Solution-\\ We\quad join\quad OP\quad \& \quad OR\quad and\quad drop\quad a\quad perpendicular\quad ON\quad from\quad O\\ to\quad RS\quad at\quad N.\\ Since\quad PQ\parallel RS,\quad ON\quad will\quad be\quad perpendicular\quad to\quad PQ\quad at\quad M.\\ \left( \angle ONR={ 90 }^{ o }=\angle OMP\quad as\quad they\quad are\quad corresponding\quad angles. \right) \\ So\quad the\quad distance\quad between\quad PQ\quad \& \quad RS=ON+OM=MN.\\ Here\quad PM=\frac { 1 }{ 2 } PQ=\frac { 1 }{ 2 } \times 30cm=15cm\quad since\quad \\ the\quad perpendicular\quad from\quad the\quad centre\quad of\quad a\quad circle\quad to\quad a\quad chord\quad \\ bisects\quad the\quad latter.\\ \quad Now\quad \Delta OPM\quad is\quad a\quad right\quad one\quad with\quad OP\quad as\quad hypotenuse.\\ \therefore \quad By\quad Pythagoras\quad theorem,\quad we\quad have\quad OM=\sqrt { { OP }^{ 2 }-{ PM }^{ 2 } } \\ =\sqrt { { 17 }^{ 2 }-{ 15 }^{ 2 } } cm=8cm.\\ Again\quad \quad RN=\frac { 1 }{ 2 } RS=\frac { 1 }{ 2 } \times 16cm=8cm\quad since\quad the\quad perpendicular\quad from\quad the\quad \\ centre\quad of\quad a\quad circle\quad to\quad a\quad chord\quad bisects\quad the\quad latter.\\ We\quad have,\quad \Delta ORN\quad is\quad a\quad right\quad one\quad with\quad OR\quad as\quad hypotenuse.\\ \therefore \quad By\quad Pythagoras\quad theorem,\quad we\quad have\quad ON=\sqrt { { OR }^{ 2 }-{ RN }^{ 2 } } \\ =\sqrt { { 17 }^{ 2 }-{ 8 }^{ 2 } } cm=15cm.\\ Now\quad the\quad distance\quad beteen\quad PQ\quad \& \quad RS=MN=ON-OM\\ =(15+8)cm=23cm.\\ Ans-\quad Option\quad C.\\ \\  $$ 

Given: $$O$$ is the center of the circle. $$K$$ and $$L$$ are mid points of chords $$AB$$ and $$CD$$, respectively.
$$\therefore OK \perp AB$$ and $$OL \perp CD$$.             {The line joining the centre of a circle to the mid-point of a chord is perpendicular to the chord}
As $$AB = CD$$.
$$\therefore$$ $$OK = OL$$          ...(Equal chords are equidistant from centre)

Given, $$ABCD$$ is a cyclic quadrilateral.

Given, $$ \angle A =\left ( 2x-1 \right )^{\circ}$$, $$ \angle B =\left ( y+5 \right )^{\circ}$$, $$ \angle C=\left ( 2y+15 \right )^{\circ}$$ and $$\angle D =\left ( 4x-7 \right )^{\circ}$$.

Given- $$\overline { POQ }$$  is a diameter of a given circle. $$PQRS$$ is a cyclic quadrilateral. $$SQ$$ is joined. Also, $$\angle SRQ={ 138 }^{ \circ }$$.

Since, $$ \overline { POQ }$$  is the diameter of the given circle, it subtends $$\angle PSQ$$ to the circumference at $$S$$, i.e. $$  \angle PSQ={ 90 }^{ \circ }$$ ...[ since it is an angle in a semicircle].

Again,

$$\angle QPS+\angle QRS=180^\circ$$ ...[ sum of opposite angles  of a cyclic quadrilateral is $${ 180 }^{ \circ }$$]

$$\Rightarrow \angle QPS={ 180 }^{ \circ }-\angle QRS$$

$$\Rightarrow \angle QPS={ 180 }^{ \circ }-138^{ \circ }$$

$$\Rightarrow\angle QPS={ 42 }^{ \circ}$$.

 In $$\triangle PSQ,$$

$$  \angle PQS +\angle PSQ+\angle QPS=180^\circ$$

                                 $$\angle PQS={ 180 }^{ \circ }-(\angle PSQ+\angle QPS)$$

                                               $$={ 180 }^{ \circ }-({ 90 }^{ \circ }+{ 42 }^{ \circ })$$

                                               $$={ 48 }^{ \circ }$$

Hence, option $$C$$ is correct.

Given, $$ABCD$$ is a cyclic quadrilateral

Given, $$\angle AOC=110^o$$.
By property of circles, we have $$\angle AOC=2\angle ADC$$
$$\Rightarrow \angle ADC=110^o\div 2=55^o$$

$$ABCD $$ is a cyclic quadrilateral. 

Given,
$$\angle BAD=70^{\circ}$$, $$\angle ABD=56^{\circ}$$ and $$\angle ADC=72^{\circ}$$.