Question

In the given figure, CD and AB are diameters of circle and AB and CD are perpendicular to each other. LQ and SR are perpendiculars to AB abd CD respectively. Radius of circle is 5 cm, PB : PA = 2 : 3 and CN: ND = 2 : 3. What is the length (in cm) of SM ?

A) [(5√3) – 3] B) [(4√3) – 2] C) [(2√5) – 1] D) [(2√6) – 1]

Answer 1

Given:

CD and AB are diameters of a circle

AB and CD are perpendicular to each other

LQ ⊥ AB

SR ⊥ CD

The radius of the circle = 5 cm

PB : PA = 2 : 3

CN: ND = 2 : 3

To find:

The length of SM (in cm)

Solution:

We have (from the figure attached below),

O = centre of the circle

Radius = AO = OB = OC = OD = 5 cm

LQ cuts AB at P

SR cuts CD at N

LQ & SR intersects at M

Finding the length of diameter AB:

∵ PB : PA = 2 : 3 ..... (given)

So, we can assume PB = 2x and PA = 3x

∴ Diameter, AB = PA + PB = 3x + 2x = 5x

Also, AB = OA + OB = 5 + 5 = 10 cm

∴ 5x = 10

⇒ x = $\frac{10}{5}$

⇒ x = 2 cm ..... (i)

∴ PB = 2 × 2 = 4 cm and PA = 3 × 2 = 6 cm

Finding the length of diameter CD:

∵ CN : ND = 2 : 3 ..... (given)

So, we can assume CN = 2x and ND = 3x

∴ Diameter, CD = CN + CD = 3x + 2x = 5x

Also, CD = OC + OD = 5 + 5 = 10 cm

Similarly, as shown in eq. (i), we get

∴ CN = 2 × 2 = 4 cm and ND = 3 × 2 = 6 cm

Finding the length of OP, ON & NM:

OP = PA - OA = 6 - 5 = 1 cm

ON = OC - CN = 5 - 4 = 1 cm

From the figure,

In quad MNOP,

LQ ⊥ AB & SR ⊥ CD ....... (given)

⇒ ∠ MNO = ∠MPO = 90°

⇒ MNOP is square  

⇒ OP = PM = NM = ON = 1 cm

Finding the length of SM:

Construction: Let's join O and S

Now, in Δ SON, using the Pythagoras Theorem, we get

$OS^2 = ON^2 + SN^2$

$\implies SN = \sqrt{OS^2 - ON^2}$

by substituting OS = radius = 5 cm & On = 2 cm

$\implies SN = \sqrt{5^2 - 1^2}$

$\implies SN = \sqrt{24}$

$\implies SN = 2\sqrt{6}\: cm$

∴ SM = SN - NM = $\bold{[2\sqrt{6} - 1]\: cm}$ ← option (D)

Thus, the length (in cm) of SM is $\underline{[2\sqrt{6} - 1]\: cm }$.

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