Circles Test - 24

$$ Given-\\ O\quad is\quad the\quad centre\quad of\quad a\quad circle\quad with\quad diameter\quad AB.\\ AD\quad \& \quad BC\quad are\quad chords\quad which\quad have\quad been\quad extended\quad  to\quad intersect\quad at\quad E.\\ AB\quad \& \quad CD\quad are\quad joined.\\ \angle BAD={ 35 }^{ o }\quad \& \quad \angle BED={ 25 }^{ o }.\\ \\ To\quad find\quad out-\\ \angle DCB,\quad \angle DBC\quad \& \quad \angle BDC\\ \\ Solution-\\ \angle BAD=\angle DCB={ 35 }^{ o }......(i)\quad [since\quad both\quad of\quad them\quad have\quad been\quad \\ subtended\quad by\quad the\quad chord\quad BD\quad to\quad the\quad circumference\quad at\quad A\quad \& \quad C].\\ Now\quad \angle ACB={ 90 }^{ o }\quad since\quad \angle ACB\quad is\quad angle\quad in\quad a\quad semicircle,\quad AB\\ being\quad the\quad diameter.\\ \therefore \quad \angle ACD=\angle ACB-\angle DCB={ 90 }^{ o }-{ 35 }^{ o }={ 55 }^{ o }.\\ Again\quad \angle ABD=\angle ACD={ 55 }^{ o }\quad [since\quad both\quad of\quad them\quad have\quad been\quad \\ subtended\quad by\quad the\quad chord\quad AD\quad to\quad the\quad circumference\quad at\quad B\quad \& \quad C].\\ Now\quad in\quad \Delta ACE,\quad we\quad have\quad \angle CAD={ 180 }^{ o }-(\angle ACB+\angle BEA)={ 180 }^{ o }-({ 90 }^{ o }+{ 25 }^{ o })=65^{ o }.\\ \therefore \quad \angle BAC=\angle CAD-\angle BAD=65^{ o }-35^{ o }=30^{ o }.\\ \therefore \quad \angle BDC=\angle BAC=30^{ o }.......(ii)\quad [since \quad both\quad of\quad them\quad have\quad been\quad \\ subtended\quad by\quad the\quad chord\quad BC\quad to\quad the\quad circumference\quad at\quad A\quad \& \quad D].\\ Again\quad ACBD\quad is\quad a\quad cyclic\quad quadrilateral.\quad \\ \therefore \quad \angle CAD+\angle DBC={ 180 }^{ o }\quad (The\quad sum\quad of\quad the\quad opposite\quad angles\quad of\quad a\quad cyclic\quad \\ quadrilateral\quad is\quad { 180 }^{ o }).\\ So\quad \angle DBC={ 180 }^{ o }-\angle CAD={ 180 }^{ o }-65^{ o }=115^{ o }.......(iii).\\ \therefore \quad \quad \angle DCB,\quad \angle DBC\quad \& \quad \angle BDC\quad are \quad 35^{ o },\quad 115^{ o }\quad \& \quad 30^{ o }\quad respectively.\\ Hence, \quad option\quad C \quad is \quad correct. $$

Given, $$O$$ is the centre of a circle in which a quadrilateral $$ ABCD$$ has been inscribed.

$$\angle{AOC}=\angle{AOB}+\angle{BOC}$$ $$=60^{\circ}+30^{\circ}=90^°$$

$$ Given-\\ Two\quad circles\quad intersect\quad at\quad P\quad \& \quad Q.\\ The\quad lines\quad APB\quad \& \quad \quad DQC\quad intersect\quad the\quad circles\quad at\\ A,\quad P,\quad B\quad and\quad D,\quad Q,\quad C\quad respectively.\\ AD\quad \& \quad BC\quad have\quad been\quad joined.\\ \angle ADQ={ 80 }^{ o }\quad and\quad \angle DAP={ 84 }^{ o }.\\ To\quad find\quad out-\\ \angle QBC=?\\ \angle BCP=?\\ Solution-\\ We\quad join\quad PQ.\\ ADPQ\quad is\quad a\quad cyclic\quad quadrilateral.\\ \therefore \quad \angle PQA={ 180 }^{ o }-\angle ADP={ 180 }^{ o }-{ 84 }^{ o }=96^{ o }\quad [since \quad \\ the\quad sum\quad of\quad the\quad opposite\quad angles\quad of\quad a\quad cyclic\quad \\ quadrilateral={ 180 }^{ o }].\\ Also\quad \angle PQB={ 180 }^{ o }-\angle PQA={ 180 }^{ o }-{ 96 }^{ o }={ 84 }^{ o }(linear\quad pair).\\ Again\quad QBCP \quad is\quad a\quad cyclic\quad quadrilateral.\\ \therefore \quad \angle BCP={ 180 }^{ o }-\angle BQP={ 180 }^{ o }-{ 84 }^{ o }=96^{ o }.....(i)\quad [since\\ the\quad sum\quad of\quad the\quad opposite\quad angles\quad of\quad a\quad cyclic\quad \\ quadrilateral={ 180 }^{ o }].\\ Similarly, \quad ADPQ\quad is\quad a\quad cyclic\quad quadrilateral.\\ \therefore \quad \angle QPD={ 180 }^{ o }-\angle QAD={ 180 }^{ o }-{ 80 }^{ o }=100^{ o }\quad [since\\ the\quad sum\quad of\quad the\quad opposite\quad angles\quad of\quad a\quad cyclic\quad \\ quadrilateral={ 180 }^{ o }].\\ Also,\quad \angle QPC={ 180 }^{ o }-\angle QPD={ 180 }^{ o }-{ 100 }^{ o }={ 80 }^{ o }(linear\quad pair).\\ Again\quad QBCP\quad \quad is\quad a\quad cyclic\quad quadrilateral.\\ \therefore \quad \angle QBC={ 180 }^{ o }-\angle QPC={ 180 }^{ o }-{ 80 }^{ o }=100^{ o }......(ii)\quad [since\\ the\quad sum\quad of\quad the\quad opposite\quad angles\quad of\quad a\quad cyclic\quad \\ quadrilateral={ 180 }^{ o }].\\ \therefore \quad \quad \angle QBC=100^{ o }\\ and\quad \angle BCP=96^{ o }.\\ Hence, \quad option\quad A \quad is \quad correct.\\ \\ \\  $$

Since $$AB$$ is diameter, $$\angle AEB = 90^\circ$$.

In $$\triangle AEB$$,

$$\angle ABE + \angle EAB +\angle AEB=180^o$$ ...[Angle sum property]

$$\implies$$ $$\angle ABE = 180^o - \angle EAB - \angle AEB$$

$$\implies$$ $$\angle ABE = 180^o – 65^o -90^o = 25^\circ$$.

Given, $$ED \parallel AB$$,

$$\implies$$ $$ \angle DEB = \angle EBA = 25^\circ$$ ....[Alternate interior angles].

Since $$EDCB$$ is a cyclic quadrilateral,

$$\angle EAB + \angle EDB  = 180^\circ$$ ...[Opposite angles of cyclic quadrilateral are supplementary]

$$\implies$$ $$\angle EDB = 180^o – 65^o = 115^\circ $$.

In $$\triangle EDB$$,

$$\angle EBD + \angle DEB+ \angle BDE=180^o$$

$$\implies$$ $$\angle EBD = 180 - \angle DEB - \angle BDE$$

$$\implies$$ $$\angle ABE = 180 – 25 -115 = 40^\circ$$.

Hence, option $$D$$ is correct.

$$\textbf{Step - 1: Verify, If we join any points on a circle we get a diameter of the circle}$$

                   $$\text{A Chord is a line segment that joins any two points of the circle.}$$

                   $$\text{The endpoints of this line segments lie on the circumference of the circle.}$$

                   $$\therefore \text{Option (A) is false}$$

$$\textbf{Step - 2: Verify, A diameter of a circle contains the center of the circle}$$

                   $$\text{Any interval joining two points on the circle and passing through the center is called a}$$

                   $$\text{diameter of the circle.}$$

                   $$\therefore \text{Option (B) is True}$$

$$\textbf{Step - 3: Verify, A semicircle is an arc}$$

                   $$\text{The arc of a circle consists of two points on the circle and all of the points on the circle that lie}$$

                   $$\text{between those two points.}$$

                   $$\text{It's like a segment that was wrapped partway around a circle.}$$

                   $$\text{An arc whose measure equals 180 degrees is called a semicircle since it divides the circle in two}$$

                   $$\therefore \text{Option (C) is True}$$

$$\textbf{Step - 4: Verify, the length of a circle is called its circumference}$$

                   $$\text{A Circle is a round closed figure where all its boundary points are equidistant from a fixed point}$$

                   $$\text{called the center.}$$ 

                   $$\text{The two important metrics of a circle is the area of a circle and the circumference of a circle.}$$

                   $$\therefore \text{Option (D) is True}$$

$$\textbf{Hence, option A is correct as it is false}$$        

We know, opposite angles of a cyclic quadrailateral are supplementary.
$$\therefore 3x+x=180^{\circ}$$ and $$2y+y=180^{\circ}$$
$$\Rightarrow 4x=180^{\circ}$$ and $$3y=180^{\circ}$$
$$\Rightarrow x=45^{\circ}$$ and $$y=60^{\circ}$$.

The perpendicular drawn from the center of the circle to the chord bisects the chord. So, it divides the chord in $$1:1$$ 

Construction: Join A and C. This forms a right triangle ACB   [ angle in a semicircle is $$90^{\circ}$$]