[2 Mark Question Answer]

Question 12 Marks

In following fig., PT is a tangent to the circle at $T$ and $PAB$ is a secant to the same circle. If $PB = 9\ cm$ and $AB = S\ cm$, find PT.

Answer

Let $PT = x\ cm$
Since, $PAB$ is a secant and $PT$ is a tangent to the given circle, we have,
$PA · PB = PT^2$
$\Rightarrow 4 · 9 = PT^2$
$\Rightarrow PT^2 = 36$
$\Rightarrow PT = 6\ cm$

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Question 22 Marks

In following fig., $PT$ is a tangent to the circle at $T$ and $PAB$ is a secant to the same circle. If $PA =4 cm$ and $AB = Scm$, find $PT$ .

Answer

Let $PT = x cm$
Since, PAB is a secant and $PT$ is a tangent to the given circle, we have,
$PA · PB = PT^2$
$\Rightarrow 4.9 = PT^2$
$\Rightarrow PT^2 = 36$
$\Rightarrow PT = 6 cm$

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Question 32 Marks

In following fig., $PT$ is a tangent to the circle at $T$ and $PAB$ is a secant to the same circle. If $PA = 4cm$ and $AB = Scm,$ find $PT.$

Answer

Let $PT = x cm$
Since, PAB is a secant and $PT$ is a tangent to the given circle, we have,
$PA · PB = PT^2$
$\Rightarrow 4 · 9 = PT^2$
$\Rightarrow PT^2 = 36$
$\Rightarrow PT = 6cm$

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Question 42 Marks

In a cyclic quadrilateral ABCD , AB || CD and ∠ B = 65 ° , find the remaining angles

Answer

PQ is a diameter of the circle
∴ CPRQ = 90° (angle is a semi cirde)
∠ RPQ = 40° (given)
In Δ PQR,
∠ PRQ + ∠ RQP + ∠ QPR = 180 (Angle sum property)
90 + ∠ RQP + 40 = 180
∠ RQP = 50 °

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Question 52 Marks

In fig., O is the centre of the circle and $\angle AOC = 1500.$ Find $\angle ABC.$

Answer

$\angle AOC =150^{\circ}$
$\text { Reflex } \angle AOC =360^{\circ}-150^{\circ}=210^{\circ}$
$\angle ABC =\frac{1}{2} \text { reflex } \angle AOC =\frac{1}{2}\left(210^{\circ}\right\}$
$\angle ABC =105^{\circ}$

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Question 62 Marks

In following fig., ABCD is a square. A cirde is drawn with centre A so that it cuts AB and AD at Mand N respectively. Prove that Δ DAM ≅ Δ .BAN.

Answer

In Δ DAM and Δ BAN
AN = AM (radii of same circle)
AD = AB (sides of square ABCD)
∠ DAM = ∠ BAN (common)
∴ Δ DAM ≅ Δ BAN.

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Mathematics [2 Mark Question Answer]

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