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Standard VI

Mathematics

Area of Square

In the given ...

Question

In the given figure of square $A B C D, P, Q, R$ and $S$ are the midpoints of $¯ ¯¯¯¯¯¯ ¯ A B, ¯ ¯¯¯¯¯¯ ¯ B C, ¯ ¯¯¯¯¯¯¯ ¯ C D$ and $¯ ¯¯¯¯¯¯¯ ¯ D A$ respectively. Find the area of unruled portion.

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Solution

R.E.F image
Given figure
Are of uncovered portion (Area of square)
-4 $(\frac{1}{4}$ area of circle)
$= (14)^{2} - \frac{22}{7} \times 7 = 425$ sq cm

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In the given figure of square ABCD,P,Q,R and S are the midpoints of ¯¯¯¯¯¯¯¯AB,¯¯¯¯¯¯¯¯BC,¯¯¯¯¯¯¯¯¯CD and ¯¯¯¯¯¯¯¯¯DA respectively. Find the area of unruled portion.

R.E.F image Given figure Are of uncovered portion (Area of square)-4(14area of circle)=(14)2−227×7=425 sq cm

In the given figure of square $A B C D, P, Q, R$ and $S$ are the midpoints of $¯ ¯¯¯¯¯¯ ¯ A B, ¯ ¯¯¯¯¯¯ ¯ B C, ¯ ¯¯¯¯¯¯¯ ¯ C D$ and $¯ ¯¯¯¯¯¯¯ ¯ D A$ respectively. Find the area of unruled portion.

R.E.F image
Given figure
Are of uncovered portion (Area of square)
-4 $(\frac{1}{4}$ area of circle)
$= (14)^{2} - \frac{22}{7} \times 7 = 425$ sq cm