Four equal circles, each of radius a, touch each other. Show that the area between them is 6/7a 2
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Answered by
171
Consider the attached diagram,
the square formed as such is if side 2a
hence area
= area of square - 4× area of a circle quadrant
= (2a)² - 4×πa²/4 = 4a² -
= ![[4 -\frac{22}{7} ]a^{2} = \frac{28-22}{7} a^{2} = \frac{6}{7} a^{2}](https://tex.z-dn.net/?f=%5B4+++-%5Cfrac%7B22%7D%7B7%7D+%5Da%5E%7B2%7D+%3D+%5Cfrac%7B28-22%7D%7B7%7D+a%5E%7B2%7D+%3D+%5Cfrac%7B6%7D%7B7%7D+a%5E%7B2%7D "[4 -\frac{22}{7} ]a^{2} = \frac{28-22}{7} a^{2} = \frac{6}{7} a^{2}")
hence Proved!!
the square formed as such is if side 2a
hence area
= area of square - 4× area of a circle quadrant
= (2a)² - 4×πa²/4 = 4a² -
hence Proved!!
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61
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Step-by-step explanation:
Go Through The Attached Image....
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