3 Marks Question

Question 13 Marks

If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc $\text{PXA}\cong\text{Arc PYB}.$

Answer

Let AB be a chord of a circle having centre at OPQ be the perpendicular bisector of the chord AB, which intersects at M and it always passes through O.

To prove: arc $\text{PXA}\cong\text{Arc PYB}$
Construction: Join AP and BP.
Proof: In $\triangle\text{APM}$ and $\triangle\text{BPM},$
$\text{AM} = \text{MB}$
$\angle\text{PMA}=\angle\text{PMB}$
$\text{PM} = \text{PM}$
$\therefore\triangle\text{APM s}\ \triangle\text{BPM}$
$\therefore\text{PA}=\text{PB}$
$\Rightarrow \text{arc PXA}\cong\text{arc PYB}$

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Question 23 Marks

Two chords AB and AC of a circle subtends angles equal to 90° and 150°, respectively at the centre. Find $\angle\text{BAC},$ if AB and AC lie on the opposite sides of the centre.

Answer

We have,
Reflex $\angle\text{BOC}=90^\circ+150^\circ=240^\circ$
$\therefore\angle\text{BOC}=360^\circ-240^\circ=120^\circ$
Now, $\angle\text{BOC}=2\angle\text{BAC}$
[Since, angle subtended by an arc at the centre is double of the angle subtended by the same arc on the remaining part of the cirlce]

Hence, $\angle\text{BAC}=\frac{1}{2}\angle\text{BOC}=\frac{1}{2}\times120^\circ=60^\circ$

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Question 33 Marks

If BM and CN are the perpendiculars drawn on the sides AC and AB of the triangle ABC, prove that the points B, C, M and N are concyclic.

Answer

As BM and CN are the perpendiculars drawn on the sides AC and AB of the triangle ABC.

$\therefore\angle\text{MBC}=\angle\text{BNC}=90^\circ$
Since, if line segment (here BC) joining two points (here B and C) subtends equal angles $\Big(\text{here }\angle\text{BMC}\ \text{and}\ \angle\text{BNC}\Big)$ at M and N on the same side of the line (here BC) containing the segment, the four points (here B, C M, and N) are concyclic.
Hence, B, M and N are concyclic.

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Question 43 Marks

In Fig. AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of $\angle\text{ACD} + \angle\text{BED.}$

Answer

Join BC.
Since angle in a semicircle is 90°, we have
$\angle\text{ACB}=90^\circ$
As ABCD is a cyclic quadrilateral and opposite angles of a cyclic quadrilateral are supplementary
$\therefore\angle\text{BCD}+\angle\text{BED}=180^\circ$
Now, adding $\angle\text{ACB}$ to both sides, we get
$(\angle\text{BCD}+\angle\text{ACB})+\angle\text{BED}=180^\circ+\angle\text{ACB}$
Hence, $\angle\text{ACD}+\angle\text{BED}=180^\circ+90^\circ=270^\circ$

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