SSC CHSL 19 Mar 2020 Shift 2 Paper

Question : 68 of 100

Marks: +1, -0

Two circles of radii

20 {\cm}

and

5 {\cm}

, respectively, touch each other externally at the point P. AB is the direct common tangent of those two circles of centres R and S, respectively. The length of AB is equal to

$15 {\cm}$
$5 {\cm}$
$10 {\cm}$
$20 {\cm}$

Validate

Solution:

Given,

AB is the tangent to both the circles.
⇒

\AB \⊥\AR

Let SC be the line parallel to AB.
⇒

\SC \⊥\AR

and

\AC=\BS=5 {\cm}

and

\AB=\SC

Radius of the bigger circle,

\AR=20 {\cm}

⇒

\AC+\CR=20 \⇒5+\CR=20

⇒

\CR=20-5=15 {\cm}

and

\RS=25 {\cm}

\ln \Δ\RCS

\SC^{2}=\RS^{2}-\CR^{2}=25^{2}-15^{2}

\SC=\√{625-225}

=\√{400}=20 {\cm}

∴

\AB=\SC=20 {\cm}

Go to Question:

Prev Question

Next Question