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What is the ratio of the area of the semicircle to the quarter circle . |
| Answer» REQUIRED ratio of semi - circle to quarter circle is 2 / 3.Step-by-step explanation:In the FIGURE, a square is there whose length is equal to the length of radius of the semi - circle. Let the radius of semi circle is a, and, diameter or AC is 2a.Also, let that the length of radius of quarter circle is r, In TRIANGLE ABC, length of AB and BC is equal to the length of radius of quarter circle or r, and length of AC is 2a( from above ).In triangle AOB and triangle COB, AO = OC ( radius of semi circle )AB = BC ( radius of quarter circle )OB = OB ( common )Hence triangle AOB is congruent to triangle COB. Thus, angle AOB or angle COB is 90*.Therefore, by Pythagoras theorem, = > AB^2 = AO^2 + OB^2= > r^2 = a^2 + OB^2= > r^2 - a^2 = OB^2 …( i )From the properties of square we know that the area formed by joining the two tangents ( also , radius ) is a square. Thus, by Pythagoras theorem,= > side^2 + side^2 = OB^2= > a^2 + a^2 = OB^2 ...( ii )Comparing the values of OB^2 from ( i ) and ( ii ),= > √{ r^2 - a^2 } = a√2= > r^2 - a^2 = 2a^2= > r^2 = 2a^2 + a^2= > r^2 = 3a^2= > 1 / 3 = a^2 / r^2= > 2 / 3 = (2 π a^2 ) / ( π r^2 )= > 2 / 3 = ( π a^2 ) / ( π r^2 / 2 )= > 2 / 3 = ( π a^2 / 2 ) / { π r^2 / ( 2 x 2 ) } = > 2 / 3 = ( π a^2 / 2 ) / ( π r^2 / 4 )= > 2 / 3 = Area of semi - circle / area of quarter circleHence the required ratio of semi - circle to quarter circle is 2 / 3. | |