3 Mark Question

Question 13 Marks

ABCD is a rectangle. Prove that the centre of the circle through A, B, C, D is the point of intersection of its diagonals.

Answer

ABCD is a rectangle. Let O be the point of intersection of the diagonals AC and BD of rectangle ABCD.

Since the diagonals of a rectangle are equal and bisect each other.$\therefore$ OA = OB = OC = OD
Thus, O is the centre of the circle through A, B, C, D.

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Question 23 Marks

In the given figure, O is the canter of the circle and $\angle\text{AOB}=70^\circ.$ Calculate the values of

$\angle\text{OCA}$
$\angle\text{OAC}$

Answer

The angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.

Thus, $\angle\text{AOB}=2\angle\text{OCA}$

$\Rightarrow\ \angle\text{OCA}=\Big(\frac{\angle\text{AOB}}{2}\Big)=\Big(\frac{70^\circ}{2}\Big)$

$=35^\circ$

OA = OC [Radii of a circle]

$\angle\text{OAC}=\angle\text{OCA}$ [Base angles of an isosceles triangle are equal]

$=35^\circ$

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Question 33 Marks

In the given figure, O is the centre of the circle and arc ABC subtends an angle of 130° at the centre. If AB is extended to P, find $\angle\text{PBC}.$

Answer

Take a points D on the major arc CA and join AD and DC.$\therefore\ \angle{2}=2\angle{1}$ [Angle subtended by arc is twice the angle subtended by it on the circumference in the alternate segment]
$\therefore\ 130^\circ=2\angle{1}$
$\Rightarrow\ \angle{1}=65^\circ\dots(\text{i})$
$\angle\text{PBC}=\angle{1}$ $[\because$ exterior angle of a cyclic quadrilateral interior opposite angle$]$
$\therefore\ \angle\text{PBC}=65^\circ$

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Question 43 Marks

In the given figure, AB is a chord of a circle with centre O and AB is produced to C such that BC = OB. Also, CO is joined and produced to meet the circle in D. If $\angle\text{ACD}=\text{y}^\circ$ and $\angle\text{AOD}=\text{x}^\circ,$ prove that x = 3y.

Answer

We have:$\text{OB}=\text{OC},$ $\angle\text{BOC}=\angle\text{BCO}=\text{y}$
External $\angle\text{OBA}=\angle\text{BOC}+\angle\text{BCO}=(2\text{y})$ Again, $\text{OA}=\text{OB},$ $\angle\text{OAB}=\angle\text{OBA}=(2\text{y})$ External $\angle\text{AOD}=\angle\text{OAC}+\angle\text{ACO}$ Or $\text{x}=\angle\text{OAB}+\angle\text{BCO}$ Or $\text{x}=(2\text{y})+\text{y}=32\text{y}$ Hence, $\text{x}=3\text{y}$

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Question 53 Marks

In the given figure, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that $\angle\text{DBC}=60^\circ$and $\angle\text{BAC}=40^\circ.$ Find

$\angle\text{BCD}$
$\angle\text{CAD}$

Answer

$\angle\text{BDC}=\angle\text{BAC}=40^\circ$ [Angles in the same segment]

In $\triangle\text{BCD},$ we have:

$\angle\text{BCD}+\angle\text{DBC}+\angle\text{BDC}=180^\circ$ [Angle sum property of a triangle]

$\Rightarrow\ \angle\text{BCD}+60^\circ+40^\circ=180^\circ$

$\Rightarrow\ \angle\text{BCD}=(180^\circ-100^\circ)=80^\circ$

$\angle\text{CAD}=\angle\text{CBD}$ [Angles in the same segment]

$=60^\circ$

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Question 63 Marks

In the given figure $\triangle\text{ABC}$ is an isosceles triangle in which AB = AC and a circle passing through B andC intersects AB and AC at D and E respectively. Prove that DE || BC.

Answer

$\triangle\text{ABC}$ is an isosceles triangle in which AB = AC and circle passing through B and C intersects AB and AC at D and E.Since AB = AC
$\therefore\ \angle\text{ACB}=\angle\text{ABC}$
So, exterior
$\ \angle\text{ADE}=\angle\text{ACB}=\angle\text{ABC}$
$\therefore\ \angle\text{ADE}=\angle\text{ABC}$
$\Rightarrow\ \text{DE }||\text{ BC}$

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Question 73 Marks

On a common hypotenuse AB, two right triangles ACB and ADB are situated on opposite sides. Prove that $\angle\text{BAC}=\angle\text{BDC}.$

Answer

AB is the common hypotenuse of $\triangle\text{ACB}$ and $\triangle\text{ADB}.$$\Rightarrow\ \angle\text{ACB}=90^\circ$ and $\angle\text{BDC}=90^\circ$
$\Rightarrow\ \angle\text{ACB}+\angle\text{BDC}=180^\circ$
⇒ The opposite angles of quadrilateral ACBD are supplementary. Thus, ACBD is a cyclic quadrilateral. This means that a circle passes through the points A, C, B and D.$\Rightarrow\ \angle\text{BAC}=\angle\text{BDC}$ [angles in the same segment]

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