Question

find the area of a square pqrs formed by joining the midpoints of sides of another square ABCD of 9 root 2 centimetre
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Answer 1

Answer:

Area of PQRS is 81 cm^2

Step-by-step explanation:

Lengths of side of ABCD = 9√2 cm

Length of side of PQRS( which is formed by joining the midpoints of ABCD ) :

By Using Pythagoras Theorem :

= > Side of PQRS = √( half of side of ABCD^2 + half of side of ABCD^2 )

= > Side of PQRS = √{ ( 9√2 / 2 )^2 + ( 9√2 / 2 )^2 } cm

= > Side of PQRS = √( 2( 81 x 2 / 4 ) cm = √81 cm

= > Side of PQRS = 9 cm

From the properties of quadrilaterals :

• Area of square : side of square^2

= > Area of square PQRS = ( 9 cm )^2

= > Area of square PQRS = 81 cm^2

Hence the area of PQRS is 81 cm^2 .

Answer 2

Answer : 81 cm²

$\rule{100}2$

Given :

• In a square ABCD, an another square PQRS is formed by joining its midponts.
• Side of square ABCD = 9√2 cm

To Find :

• Area of square PQRS

Formula used :

• Area of square = ( side )²
• Pythagoras theorem = ( hypotenuse)² = ( perpendicular )² + ( base )²

SolUtion :

As we know, all the four sides of a square are equal, i.e

In square ABCD, AB = BC = CD = DA = 9 √2 cm

From the given diagram, it is clear that

RC = QC = 9 √2 / 2

→ Use Pythagoras theorem ( stated above ).

$\sf {RC}^2= \dfrac{(9\sqrt 2)^2}{2^2}+\dfrac{(9\sqrt 2)^2}{2^2}$

$\sf {RC}^2= \dfrac{162}{4}+\dfrac{162}{4}$

$\sf {RC}^2 =\dfrac{324}{4}$

$\sf {RC}^2 = 81$

$\sf {RC} =\sqrt{81}$

$\sf\therefore {RC}=9$

Hence,

Each Side of the square PQRS is 9 cm.

Now,

Area of the square PQRS = ( 9 )² = 81 cm²

Answer : Area of the square PQRS is 81 cm².

$\rule{200}2$

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