4 Mark Question - Maths STD 9 Questions - Vidyadip

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4 Mark Question

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47 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks

Prove that the line joining the mid-point of a chord to the centre of the circle passes through the mid-point of the corresponding minor arc.

Answer

Given:

C is the mid-point of chord AB. To prove: D is the mid-point of arc AB. Proof: In $\triangle\text{OAC}$ and $\triangle\text{OBC}$ OA = OB [Radius of circle] OC = OC [Common] AC = BC [C is the mid-point of AB] Then $\triangle\text{OAC}\cong\triangle\text{OBC}$ [By SSS condition]$\therefore\angle\text{AOC}=\angle\text{BOC}$
$\Rightarrow\text{m}\overline{\text{A}}\text{D}\cong\text{m}\overline{\text{B}}\text{D}$
$\Rightarrow\overline{\text{A}}\text{D}\cong\overline{\text{B}}\text{D}$
Hence, D is the mid-point of arc AB.

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Question 24 Marks

Suppose you are given a circle. Give a construction to find its centre.

Answer

Steps of Construction:

Take three points A, B and C on the given circle.
Join AB and BC.
Draw the perpendicular bisectors of chord AB and BC which intersect each other at O.
Point O will be the required centre of the circle because we know that the perpendicular bisector of the chord always passes through the centre.

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Question 34 Marks

Two chords AB, CD of lengths 5cm, 11cm respectively of a circle are parallel. If the distance between AB and CD is 3cm, find the radius of the circle.

Answer

Let AB and CD be two parallel chord of the circle with center o such that AB = 5cm and CD = 11cm.let the radius of the circle be r cm.

Draw $\text{OP}\perp\text{AB}$ and $\text{OQ}\perp\text{CD}$ as well as point O, Q and P are collinear. Clearly, PQ = 3cm Let OQ = x then OP = x + 3 In $\triangle\text{OAP}$ and $\triangle\text{OCQ}$ we have$\text{OA}^2=\text{OP}^2+\text{AP}^2$
$\Rightarrow\text{r}^2=(\text{x}+3)^2+\Big(\frac{5}{2}\Big)^2\dots(1)$
And$\text{OC}^2=\text{OQ}^2+\text{CQ}^2$
$\Rightarrow\text{r}^2=\text{x}^2+\Big(\frac{11}{2}\Big)^2\dots(2)$
From (1) and (2) we get$(\text{x}+3)^2+\Big(\frac{5}{2}\Big)^2=\text{x}^2+\Big(\frac{11}{2}\Big)^2$
$\Rightarrow\text{x}^2+\text{6x}+9+\frac{25}{4}=\text{x}^2+\frac{121}{4}$
$\Rightarrow\text{6x}+\frac{61}{4}=\frac{121}{4}$
$\Rightarrow\text{6x}=\frac{121-61}{4}$
$\Rightarrow\text{6x}=\frac{60}{4}$
$\Rightarrow\text{x}=\frac{5}{2}$
Putting the value of x in (2) we get,$\text{r}^2=\Big(\frac{5}{2}\Big)^2+\Big(\frac{11}{2}\Big)^2$
$=\frac{25}{4}+\frac{121}{4}$
$=\frac{146}{4}$
$\Rightarrow\text{r}=\sqrt{\frac{146}{4}}$
$\text{r}=\sqrt{\frac{146}{4}}\text{cm}$

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Question 44 Marks

In a cyclic quadrilateral ABCD, if $\angle\text{A}-\angle\text{C}=60^\circ,$ prove that the smaller of two is $60^{\circ}$

Answer

We have$\angle\text{A}-\angle\text{C}=60^\circ\dots(1)$
Since, ABCD is a cyclic quadrilateral Then $\angle\text{A}+\angle\text{C}=180^\circ\dots(2)$ Add equations (1) and (2)$\angle\text{A}-\angle\text{C}+\angle\text{A}+\angle\text{C}=60^\circ+180^\circ$
$\Rightarrow2\angle\text{A}=240^\circ$
$\Rightarrow\angle\text{A}=\frac{240^\circ}{2}=120^\circ$
Put value of $\angle\text{A}$ in equation (2)$120^\circ+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=180^\circ-120^\circ=60^\circ$

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Question 54 Marks

In the given figure, ABCD is a quadrilateral inscribed in a circle with centre O. CD is produced to E such that $\angle\text{AED} = 95^\circ$ and $\angle\text{OBA} = 30^\circ$ Find $\angle\text{OAC.}$

Answer

We are given ABCD is a quadrilateral with center O, $\angle\text{ADE}=95^\circ$ and $\angle\text{OBA}=30^\circ$ We need to find $\angle\text{OAC}$ We are given the following figure:

Since $\angle\text{ADE}=95^\circ$$\Rightarrow\angle\text{ADC}=180^\circ-95^\circ=85^\circ$
Since squo; ABCD is cyclic quadrilateral This means$\angle\text{ABC}+\angle\text{ADC}=180^\circ$
$\Rightarrow\angle\text{ABO}+\angle\text{OBC}+\angle\text{ADC}=180^\circ$
$\Rightarrow30^\circ+\angle\text{OBC}+85^\circ=180^\circ$
$\Rightarrow\angle\text{OBC}=180^\circ-115^\circ=65^\circ$
Since OB = OC (radius)$\Rightarrow\angle\text{OBC}+\angle\text{OCB}=65^\circ$
In $\triangle\text{OBC}$$\angle\text{BOC}+\angle\text{OBC}+\angle\text{OBC}=180^\circ$
$\angle\text{BOC}+2\angle\text{OBC}=180^\circ$
$\angle\text{BOC}+2\times65^\circ=180^\circ$
$\angle\text{BOC}=180^\circ-130^\circ$
$\angle\text{BOC}=50^\circ$
Since DBAC and DBOC are formed on the same base which is chord. So,$\angle\text{BAC}=\frac{\angle\text{BOC}}{2}$
$=\frac{50^\circ}{2}$
$\angle\text{BAC}=25^\circ$
Consider $\triangle\text{BOA}$ which is isosceles triangle.$\angle\text{OAB}=30^\circ$
$\Rightarrow\angle\text{OAC}+\angle\text{BAC}=30^\circ$
$\Rightarrow\angle\text{OAC}+25^\circ=30^\circ$
$\Rightarrow\angle\text{OAC}=5^\circ$

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Question 64 Marks

A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Answer

We have,

Radius OA = Chord AB
$\Rightarrow OA = OB = AB$
Then triangle OAB is an equilateral triangle.
$\therefore\angle\text{AOB}=60^\circ$ [one angle of equilateral triangle]
By degree measure theorem$\angle\text{AOB}=2\angle\text{APB}$
$\Rightarrow60^\circ=2\angle\text{APB}$
$\Rightarrow\angle\text{APB}=\frac{60^\circ}{2}=30^\circ$
Now, $\angle\text{APB}+\angle\text{AQB}=180^\circ$ [opposite angles of cyclic quadrilateral]
$\Rightarrow300+\angle\text{AQB}=180^\circ$
$\Rightarrow\angle\text{AQB}=180^\circ-30^\circ=150^\circ$
Therefore, Angle by chord AB at minor arc $= 150^\circ$
Angle by chord AB at major arc $= 30^\circ$

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Question 74 Marks

Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius 20m drawn in a park. Ishita throws a ball to Isha, Isha to Nisha and Nisha to Ishita. If the distance between Ishita and Isha and between Isha and Nisha is 24m each, what is the distance between Ishita and Nisha.

Answer

Let R, S and M be the position of Ishita, Isha and Nisha respectively.

$AR=AS=\frac{24}{2}=12 cm$
$OR=OS=OM=20 cm \text { [Radii of circle }] ln$
$\triangle OAR, OA^2+AR^2$
$=OR^2 OA^2+12^2$
$=20^2 OA^2=400-144$
$=256 m^2 OA$
$=16 m$
We know that, in an isosceles triangle altitude divides the base.
$\text { So in } \triangle RSM, \angle RCS=90^{\circ}$
$\text { and } RC=CM \text { Area of } \triangle ORS=\frac{1}{2} \times OA \times RS \Rightarrow \frac{1}{2} \times RC \times OS$
$=\frac{1}{2} \times 16 \times 24$
$\Rightarrow RC \times 20=16 \times 24$
$\Rightarrow RC=19.2$
$\Rightarrow RM=2(19.2)=38.4 m$
So, the distance between Ishita and Nisha is 38.4 m .

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Question 84 Marks

In the given figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If $\angle\text{DBC}=55^\circ$ and $\angle\text{BAC}=45^\circ,$ find $\angle\text{BCD}.$

Answer

Since angles in the same segment of a circle are equal.$\therefore\angle\text{CAD}=\angle\text{DBC}=55^\circ$
$\therefore\angle\text{DAB}=\angle\text{CAD} +\angle\text{BAC}=55^\circ +45^\circ=100^\circ$
But, $\angle\text{DAB}+\angle\text{BCD}=180^\circ$ [Opposite angles of a cyclic quadrilateral]$\therefore\angle\text{BCD}=180^\circ-100^\circ=80^\circ$

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Question 94 Marks

Prove that the line segment joining the mid-point of the hypotenuse of a right triangle to its opposite vertex is half of the hypotenuse.

Answer

Let $\triangle\text{ABC}$ be a right triangle right angled at B. Let P be the mid-point of hypotenuse AC. Draw a circle with centre at P and AC as a diameter. Since, $\angle\text{ABC}=90^\circ.$ Therefore, the circle passes through B.$\therefore$ BP = Radius
Also, AP = CP = Radius$\therefore$ AP = BP = CP
Hence, $\text{BP}=\frac{1}2{}\text{AC}$

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Question 104 Marks

ABCD is a cyclic qudrilateral in which:$\text{BC}\parallel\text{AD},\ \angle\text{ADC}=110^\circ$ and $\angle\text{BAC}=50^\circ.$ Find $\angle\text{DAC}.$

Answer

Since, ABCD is a cyclic quadrilateral.
Then, $\angle\text{ABC}+\angle\text{ADC}=180^\circ$
$\Rightarrow\angle\text{ABC}+110^\circ=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-110^\circ=70^\circ$
Since, AD || BC
Then, $\angle\text{DAB}+\angle\text{ABC}=180^\circ$ [Co-interior angles]
$\Rightarrow\angle\text{DAC}+50^\circ+70^\circ=180^\circ$
$\Rightarrow\angle\text{DAC}=180^\circ-50^\circ-70^\circ=60^\circ$

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Question 114 Marks

In the given figure, $\angle\text{BAD}=78^\circ,\angle\text{DCF}=\text{x}^\circ$ and $\angle\text{DEF}=\text{y}^\circ.$ Find the values of x and y.

Answer

We have, $\angle\text{BAD}=78^\circ,\angle\text{DCF}=\text{x}^\circ$ and $\angle\text{DEF}=\text{y}^\circ.$
Since, ABCD is a cyclic quadrilateral.
Then, $\angle\text{BAD}+\angle\text{BCD}=180^\circ$
$\Rightarrow78^\circ+\angle\text{BCD}=180^\circ$
$\Rightarrow\angle\text{BCD}=180^\circ-78^\circ=102^\circ$
Now, $\angle\text{BCD}+\angle\text{DCF}=180^\circ$ [Linear pair of angles]
$\Rightarrow102^\circ=\text{x}^\circ=180^\circ$
$\Rightarrow\text{x}=180^\circ-102^\circ=78^\circ$
Since, DCEF is a cyclic quadrilateral Then, $x+y=180^{\circ} \Rightarrow 78^{\circ}+y=180^{\circ} \Rightarrow y=180^{\circ}-78^{\circ}=102^{\circ}$

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Question 124 Marks

In the given figure, if O is the circumcentre of $\angle\text{ABC},$ then find the value of $\angle\text{OBC} + \angle\text{BAC.}$

Answer

Since, O is the circumcentre of $\triangle\text{ABC,}$ So, O would be centre of the circle passing through points A, B and C.

$\angle\text{ABC}=90^\circ$ (Angle in the semicircle is 90°)

$\Rightarrow\angle\text{OAB}+\angle\text{OBC}=90^\circ\dots(1)$

As OA = OB (Radii of the same circle)

$\therefore\angle\text{OAB}=\angle\text{OBA}$ (Angle opposite to equal sides are equal)or, $\angle\text{BAC}=\angle\text{OBA}$
From (1)
$\angle\text{BAC}+\angle\text{OBC}=90^\circ$

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Question 134 Marks

ABCD is a cyclic qudrilateral in which:$\angle\text{BCD}=100^\circ$ and $\angle\text{ABD}=70^\circ$ find $\angle\text{ADB}.$

Answer

Since, ABCD is a cyclic quadrilateral.
Then, $\angle\text{BAD}+\angle\text{BCD}=180^\circ$
$\Rightarrow\angle\text{BAD}+100^\circ=180^\circ$
$\Rightarrow\angle\text{BAD}=180^\circ-100^\circ=80^\circ$
In by angle sum property
$\angle\text{ABD}+\angle\text{ADB}+\angle\text{BAD}=180^\circ$
$\Rightarrow70^\circ+\angle\text{ADB}+80^\circ=180^\circ$
$\Rightarrow\angle\text{ADB}=180^\circ-70^\circ-80^\circ=30^\circ$

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Question 144 Marks

A line segment AB is of length 5cm. Draw a circle of radius 4cm passing through A and B. Can you draw a circle of radius 2cm passing through A and B? Give reason in support of your answer.

Answer

Draw a line segment AB of 5cm.
Draw the perpendicular bisectors of AB.
With centre A and radius of 4cm, draw an arc which intersects the perpendicular bisector at point O. The point O will be the required centre.
Join OA.
With centre O and radius OA, draw a circle.

No, we cannot draw a circle of radius 2cm passing through A and B because when we draw an arc of radius 2cm with centre A, the arc will not intersect the perpendicular bisector and we will not find the centre.

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Question 154 Marks

ABCD ia a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that:

AD || BC.
EB = EC.

Answer

Given ABCD is a cyclic quadrilateral in which EA = ED
To prove:

AD || BC.
EB = EC.

Proof:

Since EA = ED

Then, $\angle\text{EAD}=\angle\text{EDA}\dots(1)$ [Oppo. angles to equal sides]

Since, ABCD is a cyclic quadrilateral

Then, $\angle\text{ABC}+\angle\text{ADC}=180^\circ$

But $\angle\text{ABC}+\angle\text{EBC}=180^\circ$ [Linear pair of angles]

Then, $\angle\text{ADC}=\angle\text{EBC}\dots(2)$

Compare equations (1) and (2)

$\angle\text{EAD}=\angle\text{EBC}\dots(3)$

Since, corresponding angles are equal

Then, BC || AD

From equation (3)

$\angle\text{EAD}=\angle\text{EBC}\dots(3)$

Similarly $\angle\text{EDA}=\angle\text{ECB}\dots(4)$

Compare equations (1)(3) and (4)

$\angle\text{EBC}=\angle\text{ECB}$

$\Rightarrow \text{EB}=\text{EC}$ [Opposite angles to equal sides]

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Question 164 Marks

Two chords AB and CD of lengths 5cm and 11cm respectively of a circle are parallel to each other and are opposite side of its centre. If the distance between AB and CD is 6cm, find the radius of the circle.

Answer

Draw $\text{OM}\perp\text{AB}$ and $\text{ON}\perp\text{CD}.$

Join OB and OD.$\text{BM}=\frac{\text{AB}}{2}=\frac{5}{2}$ [Perpendicular from the centre bisects the chord]
$\text{ND}=\frac{\text{CD}}{2}=\frac{11}{2}$
Let ON be x, so OM will be 6 - x.$\triangle\text{MOB}$
$\text{OM}^2+\text{MB}^2=\text{OB}^2$
$(6-\text{x})^2+\Big(\frac{5}{2}\Big)^2+\text{OB}^2$
$36+\text{x}^2-12\text{x}+\frac{25}{4}=\text{OB}^2\dots(\text{i})$
In $\triangle\text{NOD}$
$\text{ON}^2+\text{ND}^2=\text{OD}^2$
$\text{x}^2+\Big(\frac{11}{2}\Big)^2=\text{OD}^2$
$\text{x}^2+\frac{121}{4}=\text{OD}^2\dots(\text{ii})$
We have OB = OD. [Radii of same circle] So, from equation (i) and (ii).$36+\text{x}^2-\text{12x}+\frac{25}{4}=\text{x}^2+\frac{121}{2}$
$\Rightarrow\text{12x}=36+\frac{25}{4}-\frac{121}{4}$
$=\frac{144+25-121}{4}$
$=\frac{48}{4}=12$
$\text{x}=1$
From equation (ii)$(1)^2+\Big(\frac{121}{4}\Big)=\text{OD}^2$
$\text{OD}^2=1+\frac{121}{4}=\frac{125}{4}$
$\text{OD}=\frac{5\sqrt{5}}2{}$
So, the radius of the circle is found to be $55\sqrt{2}\text{cm}.$

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Question 174 Marks

In the given figure, O is the center of the circle. Find $\angle\text{CBD}.$

Answer

We have, $\angle\text{AOC}=100^\circ$ By degree measure theorem$\angle\text{AOC}=2\angle\text{APC}$
$\Rightarrow100^\circ=2\angle\text{APC}$
$\Rightarrow\angle\text{APC}=\frac{100^\circ}{2}=50^\circ$
$\therefore\angle\text{APC}+\angle\text{ABC}=180^\circ$ [Opposite angles of cyclic quad.]
$\Rightarrow50^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-50^\circ=130^\circ$
$\therefore\angle\text{ABC}+\angle\text{CBD}=180^\circ$ [Linear pair of angles]
$\Rightarrow130^\circ+\angle\text{CBD}=180^\circ$
$\Rightarrow\angle\text{CBD}=180^\circ-130^\circ=50^\circ$

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Question 184 Marks

ABCD is a cyclic trapezium with AD || BC. If $\angle\text{B}=70^\circ,$determine other three angles of the trapezium.

Answer

We have
ABCD is a cyclic trapezium with AD || BC and $\angle\text{B}=70^\circ.$
Since, ABCD is a cyclic quadrilateral
Then, $\angle\text{B}+\angle\text{D}=180^\circ$
$\Rightarrow70^\circ+\angle\text{D}=180^\circ$
$\Rightarrow\angle\text{D}=180^\circ-70^\circ=110^\circ$
Since, AD || BC
Then, $\angle\text{A}+\angle\text{B}=180^\circ$ [Co-interior angles]
$\Rightarrow\angle\text{A}+70^\circ=180^\circ$
$\Rightarrow\angle\text{A}=180^\circ-70^\circ=110^\circ$
Since, ABCD is a cyclic quadrilateral
Then, $\angle\text{A}+\angle\text{C}=180^\circ$
$\Rightarrow110^\circ+\angle\text{C}=180^\circ$

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Question 194 Marks

Give a method to find the centre of a given circle.

Answer

Steps of Construction:

Take three points A, B and C on the given circle.
Join AB and BC.
Draw the perpendicular bisectors of the chord AB and BC which intersect each other at O.
Point O will give the required circle because we know that, the Perpendicular bisectors of chord always pass through the centre.

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Question 204 Marks

If the two sides of a pair of opposite sides of a cyclic quadrilateral ae equal, prove that its diagonals are equal.

Answer

Given ABCD is a cyclic quadrilateral in which AB = DC To prove AC = BD Proof In $\triangle\text{PAB}$ and $\triangle\text{PDC}$$\text{AB}=\text{DC}$ [Given]
$\angle\text{BAP}=\angle\text{CDP}$ [Angles in the same segment]
$\angle\text{PBA}=\angle\text{PCD}$ [Angles in same segment]
Then, $\triangle\text{PAB}\cong\triangle\text{PDC}$ [By ASA condition]$\therefore\text{PA}=\text{PD}\dots(1)$ [C.P.C.T.]
and $\text{PC}=\text{PB}\dots(2)$ [C.P.C.T.] Add equation (1) and (2) PA + PC = PD + PB ⇒ AC = BD

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Question 214 Marks

In a cyclic quadrilateral ABCD if $\text{m}\angle\text{A}=\big(\text{m}\angle\text{C}\big).$ Find $\text{m}\angle\text{A}.$

Answer

We have, $\angle\text{A}=3\angle\text{C}$
Let $\angle\text{C}=\text{x}$
Then, $\angle\text{A}=\text{3x}$
$\therefore\angle\text{A}+\angle\text{C}=180^\circ$ [Opposite angles of cyclic quad.]
$\Rightarrow\text{3x}+\text{x}=180^\circ$
$\Rightarrow\text{4x}=180^\circ$
$\Rightarrow\text{x}=\frac{180^\circ}{4}=45^\circ$
$\therefore\angle\text{A}=\text{3x}$
$=3\times45^\circ$
$=135^\circ$

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Question 224 Marks

Prove that the angle in a segment shorter than a semicircle is greater than a right angle.

Answer

Given: $\angle\text{ACB}$ is an angle in minor segment. To prove: $\angle\text{ACB}>90^\circ$ Proof: By degree measure theorem Reflex $\angle\text{AOB}=2\angle\text{ACB}$ And reflex $\angle\text{AOB}>180^\circ$ Then, $2\angle\text{ACB}>180^\circ$$\Rightarrow\angle\text{ACB}>\frac{180^\circ}{2}$
$\Rightarrow\angle\text{ACB}>90^\circ$

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Question 234 Marks

In figure, O is the centre of the circle, then prove that $\angle\text{x}=\angle\text{y}+\angle\text{z}.$

Answer

We have,$\angle3=\angle4$ [Angles in same segment]
$\therefore\angle\text{x}=2\angle3$ [By degree measure theorem]
$\Rightarrow\angle\text{X}=\angle3+\angle3$
$\Rightarrow\angle\text{X}=\angle3+\angle4\dots(\text{i})$ $[\angle3=\angle4]$
But $\angle\text{y}=\angle3+\angle1$ [By exterior angle property]$\Rightarrow\angle3=\angle\text{y}-\angle1\dots(\text{ii})$
From (i) and (ii)$\angle\text{x}=\angle\text{y}-\angle1+\angle4$
$\Rightarrow\angle\text{x}=\angle\text{y}+\angle4-\angle1$
$\Rightarrow\angle\text{x}=\angle\text{y}+\angle\text{z}+\angle1-\angle1$ [By exterior angle property]
$\Rightarrow\angle\text{x}=\angle\text{y}+\angle\text{z}$

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Question 244 Marks

In the given figure, O is the center of the circle. If $\angle\text{CEA}=30^\circ,$ find the value of x, y and z.

Answer

We have, $\angle\text{CEA}=30^\circ$ Since, quad. ABCE is a cyclic quadrilateral. Then, $\angle\text{ABC}+\angle\text{CEA}=180^\circ$$\Rightarrow\text{x}+30^\circ=180^\circ$
$\Rightarrow\text{x}=180^\circ-30^\circ=150^\circ$
By degree measure theorem$\angle\text{AOC}=2\angle\text{CEA}$
$\Rightarrow\text{y}=2\times30^\circ=60^\circ$
$\therefore\angle\text{ADC}=\angle\text{CEA}$ [Angle in same segment]
$\Rightarrow\text{z}=30^\circ$

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Question 254 Marks

In the given figure, $\triangle\text{PQR}$ is an isosceles triangle with PQ = PR and $\text{m}\angle\text{PQR}=35^\circ.$ Find $\text{m}\angle\text{QSR}$ and $\text{m}\angle\text{QTR}.$

Answer

We have, $\angle\text{PQR}=35^\circ$ Since, $\triangle\text{PQR}$ is an isosceles triangle with PQ = PR. Then, $\angle\text{PQR}=\angle\text{PRQ}=35^\circ$ In $\triangle\text{PQR},$ by angle sum property$\angle\text{p}+\angle\text{PQR}+\angle\text{PRQ}=180^\circ$
$\Rightarrow\angle\text{p}+35^\circ+35^\circ=180^\circ$
$\Rightarrow\angle\text{p}=180^\circ-35^\circ-35^\circ=110^\circ$
$\therefore\angle\text{QSR}=\angle\text{p}=110^\circ$ [Angles in same segment]
Now, $\angle\text{QSR}+\angle\text{QTR}=180^\circ$ [Opposite angles of cyclic quad.]$\Rightarrow110^\circ+\angle\text{QTR}=180^\circ$
$\Rightarrow\angle\text{QTR}=180^\circ-110^\circ=70^\circ$

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Question 264 Marks

Find the length of a chord which is at a distance of 4cm from the centre of a circle of radius 6cm.

Answer

Given that,

Distance $(O C)=4 cm$ Radius of the circle $(O A)=6 cm$ In $\triangle OCA$, by Pythagoras theorem $AC ^2+ OC ^2= OA ^2$
$\Rightarrow A C^2+4^2=6^2$
$\Rightarrow A C^2=36-16$
$\Rightarrow A C^2=20$
$\Rightarrow A C=\sqrt{20}$
$\Rightarrow A C=4.47 cm$
We know that the perpendicular distance from centre to chord bisects the chord. $AC = BC =4.47 cm$ Then $AB = 4.47 + 4.47 = 8.94cm$

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Question 274 Marks

The lengths of two parallel chords of a circle are 6cm and 8cm. If the smaller chord is at a distance of 4cm from the centre, what is the distance of the other chord from the centre?

Answer

Distance of smaller chord AB from centre of circle = 4cm, OM = 4cm$\text{MB}=\frac{\text{AB}}2{}=\frac{6}{2}=3\text{cm}$

In $\triangle\text{OMB }$ $OM^2 + MB^2
= OB^2 4^2 + 9^2
= OB^2 16 + 9
= OB^2$ $\text{OB}=\sqrt{25}$
OB = 5cm In $\triangle\text{OND }$ OD= OB = 5cm [Radii of same circle]$\text{ND}=\frac{\text{CD}}{2}=\frac{8}{2}=4\text{cm}$
$ON^2 + ND^2 = OD^2 ON^2 + 4^2 = 5^2 ON^{2} $
$= 25 - 16$
$\text{ON}=\sqrt{9}$
$ON = 3cm$
So, the distance of bigger chord from the circle is 3cm.

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Question 284 Marks

Prove that a diameter of a circle which bisects a chord of the circle also bisects the angle subtended by the chord at the centre of the circle.

Answer

Given:

PQ is a diameter of circle which bisects the chord AB at C.
To Prove: PQ bisects $\angle\text{AOB}$
Proof:
In $\angle\text{AOC}$ and $\angle\text{BOC}$
OA = OB [Radius of circle]
OC = OC [Common]
AC = BC [Given]
Then $\triangle\text{AOC}\cong\triangle\text{BOC}$ [By SSS condition]
$\angle\text{AOC}=\angle\text{BOC}$ [C.P.C.T]
Hence PQ bisects $\angle\text{AOB}.$

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Question 294 Marks

On a semi-circle with AB as diameter, a point C is taken, so that $\text{m}\big(\angle\text{CAB}\big)=30^\circ.$ Find $\text{m}\big(\angle\text{ACB}\big)$ and $\text{m}\big(\angle\text{ABC}\big).$

Answer

We have, $\angle\text{CAB}=30^\circ$
$\therefore\angle\text{ACB}=90^\circ$ [Angle in semicircle]
In $\triangle\text{ABC},$ by angle sum property
$\angle\text{CAB}+\angle\text{ACB}+\angle\text{ABC}=180^\circ$
$\Rightarrow30^\circ+90^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-90^\circ-30^\circ=60^\circ$

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Question 304 Marks

Prove that the angle in a segment greater than a semi-circle is less than a right angle.

Answer

Given: $\angle\text{ACB}$ is an angle in minor segment.
To prove: $\angle\text{ACB}<90^\circ$
Proof: By degree measure theorem
$\angle\text{AOB}=2\angle\text{ACB}$
And $\angle\text{AOB}<180^\circ$
Then, $2\angle\text{ACB}<180^\circ$
$\Rightarrow\angle\text{ACB}<\frac{180^\circ}{2}$
$\Rightarrow\angle\text{ACB}<90^\circ$

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Question 314 Marks

A circular park of radius 40m is situated in a colony. Three boys Ankur, Amit and Anand are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.

Answer

Given that AB = BC = CA

So, ABC is an equilateral triangle OA (radius) = 40m Medians of equilateral triangle pass through the circum centre (O) of the equilateral triangle ABC. We also know that median intersect each other at the ratio 2 : 1 As AD is the median of equilateral triangle ABC, we can write:$\frac{\text{OA}}{\text{OD}}=\frac{2}{1}$
$\Rightarrow\frac{\text{4OM}}{\text{OD}}=\frac{2}{1}$
$\Rightarrow\text{OD}=20\text{m}$
Therefore, AD = OA + OD = (40 + 20)m = 60m In $\triangle\text{ADC}$ By using Pythagoras theorem$\text{AC}^2 = \text{AD}^2 + \text{DC}^2$
$\text{AC}^2=60^2+\big(\text{AC}^2\big)^2$
$\text{AC}^2=3600+\frac{\text{AC}^2}{4}$
$\Rightarrow\frac{3}{4}\text{AC}^2=3600$
$\Rightarrow\text{AC}^2=4800$
$\Rightarrow\text{AC}=40\sqrt{3}\text{m}$
So, length of string of each phone will be $40\sqrt{3}\text{m}.$

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Question 324 Marks

In the given figure, two congruent circles with centres O and O' intersect at A and B. If $\angle\text{AOB} = 50^\circ,$ then find $\angle\text{APB.}$

Answer

Since both the circles are congruent, they will have equal radii. Let their radii be ‘r’ So, from the given figure we have, OA = OB = O'A = O'B = r

Now, since all the sides of the quadrilateral OBO’A are equal it has to be a rhombus. One of the properties of a rhombus is that the opposite angles are equal to each other. So, since it is given that $\angle\text{AO}'\text{B}=50^\circ$ we can say that the angle opposite it, that is to say that $\angle\text{AO}\text{B}$ should also have the same value. Hence we get $\angle\text{AO}\text{B}=50^\circ$ Now, consider the first circle with the centre ‘O’ alone. ‘AB’ forms a chord and it subtends an angle of 50° with its centre, that is $\angle\text{AO}\text{B}=50^\circ.$ A property of a circle is that the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle. This means that,$\angle\text{APB}=\frac{\angle\text{AOB}}{2}$
$=\frac{50^\circ}{2}$
$=25^\circ$
Hence the measure of $\angle\text{APB}$ is 25°.

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Question 334 Marks

In the given figure, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If $\angle\text{APB} = 70^\circ,$ find $\angle\text{ACB.}$

Answer

Consider the smaller circle whose centre is given as ‘O’. The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

So, here we have$\angle\text{AOB}=2\angle\text{APB}$
$=2(70^\circ)$
$\angle\text{AOB}=140^\circ$
Now consider the larger circle and the points ‘A’, ‘C’, ‘B’ and ‘O’ along its circumference. ‘ACBO’ form a cyclic quadrilateral. In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to 180°.$\angle\text{AOB}+\angle\text{ACB}=180^\circ$
$\angle\text{ACB}=180^\circ-\angle\text{AOB}$
$=180^\circ-140^\circ$
$\angle\text{ACB}=40^\circ$
Hence ,the measure of $\angle\text{ACB}$ is $40^\circ.$

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Question 344 Marks

Find the length of a chord which is at a distance of $5cm$ from the centre of a circle of radius $10cm.$

Answer

Given that,

Distance $(O C)=5 cm$ Radius of the circle $(O A)=10 cm In$
$\triangle OCA$, by Pythagoras theorem $OC ^2+ AC ^2= OA ^2$
$\Rightarrow 5^2+A C^2=10^2$
$\Rightarrow 25+A C^2=100$
$\Rightarrow A C^2=100-25$
$\Rightarrow A C^2=75$
$\Rightarrow A C=\sqrt{75}$
$\Rightarrow A C=8.66 cm$
We know that, the perpendicular from the centre to chord bisects the chord Therefore, $AC = BC =8.66 cm$ Then the chord $AB =8.66+8.66=17.32 cm$

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Question 354 Marks

O is the circumference of the triangle ABC and OD is perpendicular on BC. Prove that $\angle\text{BOD}=\angle\text{A}.$

Answer

Given O is the circum centre of triangle ABC and $\text{OD}\perp\text{BC}$

To prove $\angle\text{BOD}=2\angle\text{A}$ Proof: In $\triangle\text{OBD}$ and $\triangle\text{OCD}$$\angle\text{ODB}=\angle\text{ODC}$ [Each 90°]
OB = OC [Radius of circle] OD = OD [Common] Then $\triangle\text{OBD}\cong\triangle\text{OCD}$ [By RHS Condition].$\therefore\angle\text{BOD}=\angle\text{COD}\dots(\text{i})$ [PCT].
By degree measure theorem$\angle\text{BOC}=2\angle\text{BAC}$
$\Rightarrow2\angle\text{BOD}=2\angle\text{BAC}$ [By using (i)]
$\Rightarrow\angle\text{BOD}=\angle\text{BAC}$

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Question 364 Marks

In figure, O is the centre of the circle, BO is the bisector of $\angle\text{ABC}.$ Show that AB = AC.

Answer

Given, BO is the bisector of $\angle\text{ABC}$ To prove AB = BC Proof: Since, BO is the bisector of $\angle\text{ABC}$ Then, $\angle\text{ABO}=\angle\text{CBO}\dots(\text{i})$ Since, OB = OA [Radius of circle] Then, $\angle\text{ABO}=\angle\text{DAB}\dots(\text{ii})$ [opposite angles to equal sides] Since OB = OC [Radius of circle] Then, $\angle\text{OAB}=\angle\text{OCB}\dots(\text{iii})$ [opposite angles to equal sides] Compare equations (i), (ii) and (iii)$\angle\text{OAB}=\angle\text{OCB}\dots(\text{iii})$
In $\triangle\text{OAB}$ and $\triangle\text{OCB}$$\angle\text{OAB}=\angle\text{OCB}$ [From (iv)]
$\angle\text{OBA}=\angle\text{OBC}$ [Given]
OB = OB [Common] Then, $\triangle\text{OAB}\cong\triangle\text{OCB}$ [By AAS condition]$\therefore\text{AB}=\text{BC}$ [C.P.C.T.]

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Question 374 Marks

In the given figure, ABCD is a cyclic quadrilateral. Find the value of x.

Answer

$\angle\text{EDC}+\angle\text{CDA}=180^\circ$ [Linear pair of angles]$\Rightarrow80^\circ+\angle\text{CDA}=180^\circ$
$\Rightarrow\angle\text{CDA}=180^\circ-80^\circ=100^\circ$
Since, ABCD is a cyclic quadrilateral.
$\angle\text{ADC}+\angle\text{ABC}=180^\circ$
$\Rightarrow100^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-100^\circ=80^\circ$
Now, $\angle\text{ABC}+\angle\text{ABF}=180^\circ$ [Linear pair of angles]
$\Rightarrow80^\circ+\text{x}^\circ=180^\circ$
$\Rightarrow\text{x}=180^\circ-80^\circ=100^\circ$

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Question 384 Marks

In figure, O is the centre of a circle and PQ is a diameter. If $\angle\text{ROS} = 40^\circ, $ find $\angle\text{RTS}.$

Answer

Since PQ is diameter Then,$\angle\text{PRQ} = 90^\circ$ [Angle in semicircle]
$\therefore\angle\text{PRQ}+\angle\text{TRQ}=180^\circ$ [Linear pair of angle]
$900+\angle\text{TRQ}=180^\circ$
$\angle\text{TRQ}=180^\circ-90^\circ=90^\circ$
By degree measure theorem$\angle\text{ROS}=2\angle\text{RQS}$
$\Rightarrow40^\circ=2\angle\text{RQS}$
$\Rightarrow\angle\text{RQS}=\frac{40^\circ}{2}=20^\circ$
In $\triangle\text{RQT},$ by Angle sum property$\angle\text{RQT}+\angle\text{QRT}+\angle\text{RTS}=180^\circ$
$\Rightarrow20^\circ+90^\circ+\angle\text{RTS}=180^\circ$
$\Rightarrow\angle\text{RTS}=180^\circ-20^\circ-90^\circ=70^\circ$

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Question 394 Marks

An equilateral triangle of side 9cm is inscribed in a circle. Find the radius of the circle.

Answer

Let $A B C$ be an equilateral triangle of side 9 cm and let $A D$ is one of its median.

Let $G$ be the centroid of $\triangle ABC$
Then $AG : GD =2: 1 We$ know that in an equilateral triangle, centroid coincides with the circum centre.
Therefore, G is the centre of the circumference with circum radius GA. Also G is the centre and GD is perpendicular to BC .
Therefore, In right triangle $A D B$,
we have $A B^2=A D^2+D B^2 \Rightarrow 9^2=A D^2+D B^2$
$\Rightarrow AD =\sqrt{81-\frac{81}{4}}=\frac{9 \sqrt{3}}{2} cm$
$\therefore$ Radius $= AG =\frac{2}{3} AD$
$=3 \sqrt{3} cm$.

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Question 404 Marks

In a cyclic quadrilateral ABCD if AB || CD and $\angle\text{B}=70^\circ,$ find the remaining angles.

Answer

We have, $\angle\text{B}=70^\circ$
Since, ABCD is a cyclic quadrilateral
Then, $\angle\text{B}+\angle\text{D}=180^\circ$
$\Rightarrow70^\circ+\angle\text{D}=180^\circ$
$\Rightarrow\angle\text{D}=180^\circ-70^\circ=110^\circ$
Since, AB || DC
Then, $\angle\text{B}+\angle\text{C}=180^\circ$ [Co-interior angles]
$\Rightarrow70^\circ+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=180^\circ-70^\circ=110^\circ$
Now, $\angle\text{A} +\angle\text{C}=180^\circ$ [Opposite angles of cyclic quad.]
$\Rightarrow\angle\text{A}+110^\circ=180^\circ$
$\Rightarrow\angle\text{A}=180^\circ-110^\circ=70^\circ$

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Question 414 Marks

In figure, O is the centre of the circle. If $\angle\text{APB}=50^\circ,$ find $\angle\text{AOB}$ and $\angle\text{OAB}.$

Answer

$\angle\text{APB}=50^\circ$By degree measure theorem
$\angle\text{AOB}=2\angle\text{APB}$
$\Rightarrow\angle\text{APB}=2\times50^\circ=100^\circ$ since OA = OB [Radius of circle]
Then $\angle\text{OAB}=\angle\text{OBA}$ [Angles opposite toequalsides]
Let $\angle\text{OAB}=\text{x}$
In $\triangle\text{OAB},$ by angle sum property
$\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^\circ$
$\Rightarrow\text{x}+\text{x}+100^\circ=180^\circ$
$\Rightarrow\text{2x}=180^\circ-100^\circ$
$\Rightarrow\text{2x}=80^\circ$
$\Rightarrow\text{x}=40^\circ$
$\angle\text{OAB}=\angle\text{OBA}=40^\circ$

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Question 424 Marks

Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

Answer

Each pair of circles have 0, 1 or 2 points in common.
The maximum number of points in common is ‘2’.

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Question 434 Marks

In figure, O is the centre of the circle. Find $\angle\text{BAC}.$

Answer

We have $\angle\text{AOB}=80^\circ$ And $\angle\text{AOC}=110^\circ$ Therefore, $\angle\text{AOB}+\angle\text{AOC}+\angle\text{BOC}=360^\circ$ [Complete angle]$\Rightarrow80^\circ+100^\circ+\angle\text{BOC}=360^\circ$
$\Rightarrow\angle\text{BOC}=360^\circ-80^\circ-110^\circ$
$\Rightarrow\angle\text{BOC}=170^\circ$
By degree measure theorem$\angle\text{BOC}=2\angle\text{BAC}$
$\Rightarrow170^\circ=2\angle\text{BAC}$
$\Rightarrow\angle\text{BAC}=\frac{170^\circ}{2}=85^\circ$

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Question 444 Marks

In the given figure, O and O' are centers of two circles intersecting at B and C. ACD is a straight line, find x.

Answer

By degree measure theorem$\angle\text{AOB} = 2\angle\text{ACB}$
$\Rightarrow130^\circ=2\angle\text{ACB}$
$\Rightarrow\angle\text{ACB}=\frac{130^\circ}{2}=65^\circ$ [Liner a pair of angles]
$\Rightarrow65^\circ+\angle\text{BCD}=180^\circ$
$\Rightarrow\angle\text{BCD}=180^\circ-65^\circ=115^\circ$ By degree measure theorem reflex
$\angle\text{BOD}=2\angle\text{BCD}$
$\Rightarrow\text{reflex }\angle\text{BOD}=2\times115^\circ=230^\circ$
Now, reflex $\angle\text{BOD}+\angle\text{BO}'\text{D}=360^\circ$ [Complex angle]$\Rightarrow230^\circ+\text{x}=360^\circ$
$\Rightarrow\text{x}=360^\circ-230^\circ$
$\therefore\text{x}=130^\circ$

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Question 454 Marks

Given an arc of a circle, complete the circle.

Answer

Steps of Construction:

Take three points A, B and C on the given arc.
Join AB and BC.
Draw the perpendicular bisectors of chords AB and BC which intersect each other at point O. Then O will be the required centre of the required circle.
Join OA.
With centre O and radius OA, complete the circle.

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Question 464 Marks

Circles are described on the sides of a triangle as diameters. Proved that the circle on any two sides intersect each other on the third side (or third side produced).

Answer

Since, AB is a diameter Then, $\angle\text{ADB}=90^\circ\dots(1)$ [Angles in semicircle] Since, AC is a diameter Then, $\angle\text{ADC}=90^\circ\dots(2)$ [Angles in semicircle] Add equations (1) and (2)$\angle\text{ADB}+\angle\text{ADC}=90^\circ+90^\circ$
$\Rightarrow\angle\text{BDC}=180^\circ$
Then, BDC is a line Hence, the circles on any two sides intersect each other on the third side.

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Question 474 Marks

Prove that the circles described on the four sides of a rhombus as diameters, pass through the point of intersection of its diagonals.

Answer

LetABCD be a rhombus such that its diagonals AC and BD intersect at O. Since, the diagonals of a rhombus intersect at right angle.$\therefore\angle\text{AOB}=\angle\text{BOC}$
$\angle\text{COD}=\angle\text{DOA}=90^\circ$
Now, $\angle\text{AOB}=90^\circ\Rightarrow$ circle described on AB as diameter will pass through O. Similarly, all the circles described on BC, AD and CD as diameters pass through O.

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4 Mark Question - Maths STD 9 Questions - Vidyadip