[1 Mark Question Answer]

Question 11 Mark

If O is the centre of the circle, find the value of x in each of the following figures

Answer

$APC =\frac{1}{2} AOC =\frac{1}{2} \times 120^{\circ}=60^{\circ}$
Since ABCD is a cyclic quadrilateral.
CBD = x is the exterior angle
So, x = APC = 60°.

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Question 21 Mark

If O is the centre of the circle, find the value of x in each of the following figures

Answer

We have AOC = 135°
then COB = 180° - 135° = 45°but x =$\frac{ COB }{2}=\frac{45^{\circ}}{2}=22 \frac{1}{2}^{\circ}$

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Question 31 Mark

In the figure given below, $PT$ is a tangent to the circle. Find $PT$ if $AT = 16\ cm$ and $AB = 12\ cm$.

Answer

$PT$ is tangent.
Hence by theorem,
$PT^2 = AT x BT$
$PT^2= 16 x (AT - AB)$
$PT^2= 16 x (16 - 12)$
$PT^2= 16 x 4 = 64$
$PT = 8 cm.$

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Question 41 Mark

In Fig., chords AB and CD of the circle intersect at O. AO = 5 cm, BO = 3 cm and CO = 2.5 cm. Determine the length of DO.

Answer

Clearly, Chords AB and CD intersect at O.
∴ OA x OB = OC x OD
⇒ 5 x 3 = 2.5 x OD
⇒ OD = $\left(\frac{5 \times 3}{2.5}\right)$ = 6 cm.

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Question 51 Mark

In Fig. AT is a tangent to the circle. If m∠ABC = 50°, AC = BC, Find ∠BAT.

Answer

AC = BC
⇒ ∠CBA = ∠CAB
⇒ ∠CAB = 50°
∴ ∠ACB = 180° - (50° + 50°) = 80°
Now, ∠BAT = ∠BCA ....(Angles are in alternate segments)
⇒ ∠BAT = 80°.

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Question 61 Mark

In the Figure, PT is a tangent to a circle. If m(∠BTA) = 45° and m(∠PTB) = 70°. Find m(∠ABT).

Answer

∠ ATP = ∠ PTB - ∠ BTA
= 70° - 45° = 25°
∴ ∠ ABT = ∠ ATP ...(Angles are in alternate segments)
⇒ ∠ ABT = 25°.

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Mathematics [1 Mark Question Answer]

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