Question

The area of a square ABCD is 64 sq.cm. Find the area of square obtained by joining the mid-points of the sides of the square ABCD.​

Answer 1

${\huge{\red{\sf{Given}}}}\begin{cases}\leadsto \bf{Area\:of\:a\:square\:is\:64cm^{2}} \\\leadsto \bf{Midpoints\:of\:the\:square\:are\:joined} \end{cases}$

${\huge{\red{\sf{To\:Find}}}}\begin{cases}\leadsto\bf{Area\:made\:by\:the\:new\:enclosed\:figure} \end{cases}$

$\huge\red{\underline{\bf{\green{Answer}}}}$

$\sf{\red{Refer\:to\: attachment\: for\:figure}}$

$\sf{\purple{Let\:the\:square\:be\:ABCD}}$

And,

$\sf{\orange{\mapsto Q\:is\: midpoint\:of\:AB}}$

$\sf{\orange{\mapsto R\:is\: midpoint\:of\:BC}}$

$\sf{\orange{\mapsto S\:is\: midpoint\:of\:CD}}$

$\sf{\orange{\mapsto P\:is\: midpoint\:of\:AD}}$

______________________________________

${\underline{\bf{\red{Now,}}}}$

$\sf{On\:joining\:the\: midpoints\:4\:\triangle s\:are\:formed}$

which are ,

$\sf{\pink{\leadsto \triangle PAQ}}$

$\sf{\pink{\leadsto \triangle BQR}}$

$\sf{\pink{\leadsto \triangle SCR}}$

$\sf{\pink{\leadsto \triangle PDS}}$

______________________________________

And,

$\sf{\underline{\green{\triangle PAQ\cong \triangle BQR\cong \triangle SCR\cong \triangle PAQ\cong \triangle PDS}}}$

$\bf{\blue{(By\:SAS\:congruence\:condition)}}$

______________________________________

${\underline{\bf{\red{So,}}}}$

$\sf{\underline{\green{ar(\triangle PAQ)= ar(\triangle BQR)=ar( \triangle SCR)=ar(\triangle PAQ)=ar(\triangle PDS)}}}$

______________________________________

$\sf{\blue{Area\:of\:triangle\:PAQ}}$

$\sf{\implies ar(\triangle PAQ)=\dfrac{1}{2}\times PA\times AQ}$

$\sf{\implies ar(\triangle PAQ)=\dfrac{1}{2}\times 4cm\times 4cm}$

$\sf{\implies ar(\triangle PAQ)=\dfrac{1}{\cancel{2}}\times \cancel{4cm}^{2}\times 4cm}$

${\underline{\pink{\bf{\longmapsto ar(\triangle PAQ)=8cm^{2}}}}}$

______________________________________

$\sf\purple{So\: total \:area \:of\: 4 \:triangles} = \\\pink{(4\times 8)cm^{2}=32cm^{2}}$

Therefore ,

$\green{\boxed{\orange{\bf{Required\:area=ar(ABCD)-4ar(\triangle PAQ)}}}}$

$\sf{Required\: Area=(64-32)cm^{2}}$

${\underline{\boxed{\red{\bf{\leadsto Required\:Area= 32cm^{2}}}}}}$