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62. A poor artist on the street makes funny cartoons for children and earns his living. Once he made a comic face by drawing a circle within a circle, theradius of the bigger circle being 30 cm and that of smaller being 20 cm asshown in the figure. What is the area of the cap given in this figure? |
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Answer» Answer: Let A and B be the centres of the smaller and bigger circles,respectively.We have,BC=BD=BF=BE=30 cm,AP=AF=20 cmar(∆CDE)=?Let ∆CDE be an isosceles TRIANGLE representing the capAs, BF=30 cm⇒AB+AF=30⇒AB+20=30⇒AB=30−20=10 cmSo, BP=AP−AB=20−10=10 cm .....(i)Also, BE=30 cm⇒BP+PE=30⇒10+PE=30⇒PE=30−10⇒PE=20 cm ..... (ii)Since, CD is a tangent to the smaller circleSo, BP⊥CD (Tangent is perpendicular to the radius at the point of contact)Also, CD is a CHORD of the bigger circleSo, CP=DP (Perpendicular drawn from CENTRE to the chord bisects the chord) .....(iii)Now, in ∆BPC, using pythagoras theoremCP^2=BC^2−BP^2⇒CP=302−102−−−−−−−−√ [As, BC=30 cm, given and BP=10 cm, from (i)]⇒CP=900−100−−−−−−−−√=800−−−√⇒CP=20√2cm So, CD=2CD [From (iii)]⇒CD=2(202√)=40√2 cmNow, ar(∆CED)=12×CD×PE=12×40√2*20⇒ar(∆CED)=400√2cm^2 Hence, the area of the cap is 400√2 cm^2. |
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