AB is one of the direct common tangent of two circles ofradii 12cm and 4cmrespectively touching each other find the - Maths - Areas Related to Circles

Question

AB is one of the direct common tangent of two circles ofradii
12cm and 4cmrespectively touching each other find the area enclosed
by the circles and the tangent

Ajanta Trivedi · Accepted Answer

AB is the common tangent to the circles with center M and N.

let the circles touch each other externally at O.

the radius of the circles AM = 12 cm and BN = 4 cm

since MA ⊥AB and NB⊥AB therefore MA is parallel to NB.

draw a line NP parallel to AB, NPAB is a rectangle

PM=AM-AP=AM-BN=12-4= 8 cm

MN=MO+ON=12+4=16 cm

in the triangle MPN, using pythagoras thm.

$P M^{2} + P N^{2} = M N^{2} 8^{2} + P N^{2} = 16^{2} P N^{2} = 16^{2} - 8^{2} = 256 - 64 = 192 P N = \sqrt{192} = 8 \sqrt{3} c m$

∠PMN= 60 deg [since PM/MN=8/16 = 1/2]

and ∠PNM = 30 deg

required area of shaded region = area of trapezium ABNM - (area of the sector AMO+area of sector BNO)

area of trapezium ABNM = 1/2*AB*(AM+BN) [since AB=PN ]

$= \frac{1}{2} * 8 \sqrt{3} * (12 + 4) = 4 \sqrt{3} * 16 = 64 \sqrt{3} c m^{2} = 64 * 1.732 = 110.848 c m^{2}$

area of the sector AMO = $π * 12^{2} * \frac{60}{360} = 12 * 12 * \frac{1}{6} π = 24 π c m^{2}$

area of the sector BNO = $π * 4^{2} * \frac{30 + 90}{360} = 4 * 4 * \frac{120}{360} π = \frac{16}{3} π c m^{2}$

area of the sector AMO +area of the sector BNO = $24 π + \frac{16}{3} π = \frac{88}{3} * \frac{22}{7} = 92.19 c m^{2}$

thus the required area = 110.848-92.19=18.658 sq cm

hope this helps you.

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AB is one of the direct common tangent of two circles ofradii 12cm and 4cmrespectively touching each other.find the area enclosed by the circles and the tangent.