Question

the area of squre is 64 cm2 . find the area of the squre formed by joining the midpoints of the given squre​

Answer 1

Diagram:

$\setlength{\unitlength}{1cm} \begin{picture}(6,6) \put(2, 2){\line(0, 1){2}} \put(2, 4){\line(1, 0){2}} \put(4, 4){\line(0, -1){2}} \put(4, 2){\line(-1, 0){2}}\put(4,3){\line(-1,1){1}} \put(4, 2){\line(-1, 0){2}}\put(4,3){\line( - 1, - 1){1}} \put(4, 2){\line(-1, 0){2}}\put(3,2){\line( - 1,1){1}} \put(4, 2){\line(-1, 0){2}} \put(2,3){\line(1,1){1}} \put(4,4){\sf B} \put(1.7,4){\sf A} \put(1.7,2){\sf C} \put(4.1,2){\sf D}\put(3,1.7){\sf E} \put(3,4.1){\sf G} \put(4.1,3){\sf F} \put(1.7,3){\sf H}\end{picture}$

Given:

• ABCD is a square having area 64cm²

Find:

• Area of square formed by joining the mid-points.

Solution:

F1stly Let a is the length of the side of Square

So, we know that

$\boxed{ \rm Area \: of \: square = {(a)}^{2} }$

where,

• Area of square = 64cm²

So,

$\dashrightarrow \rm Area \: of \: square = {(a)}^{2}$

$\dashrightarrow \rm 64 = {(a)}^{2}$

$\dashrightarrow \rm {(a)}^{2} = 64$

$\dashrightarrow \rm a = \sqrt{64} = 8cm$

$\dashrightarrow \rm a = 8cm$

$\therefore \rm a = 8cm$

________________________

Here, EFGH area the mid points of sides of square

Then, AH = 8/2 = 4cm = AG

In $\triangle$ AGH

By Pythogoras Theorem

$\boxed{ \rm H^2 = P^2 + B^2}$

$\rm\multimap {GH}^2 = {AG}^2 + {AH}^2$

where,

• AG = 4cm
• AH = 4cm

So,

$\rm\multimap {GH}^2 = {4}^2 + {4}^2$

$\rm\multimap {GH}^2 = 16 + 16$

$\rm\multimap {GH}^2 = 32$

$\rm\multimap GH = \sqrt{32} cm$

$\rm\multimap GH = \sqrt{16 \times 2} cm$

$\rm\multimap GH = 4\sqrt{2} cm$

So, GH = 4√2cm

Again using

$\boxed{ \rm Area \: of \: square = {(a)}^{2} }$

where,

• Side, a = 4√2cm

So,

$\to \rm Area \: of \: square = {(a)}^{2}$

$\implies \rm Area \: of \: square = {(4 \sqrt{2} )}^{2}$

$\implies \rm Area \: of \: square = 16 \times 2$

$\implies \rm Area \: of \: square = 32{cm}^2$

Hence, Area of Square EFGH = 32cm²