(we can use similar solution for proving that midpts of rectangle forms square)
adds this in answer :make sure that since 4 triangle formed ARQ QBR RCS SDP are equilateral right angled triangle hence if we take a condition of angle pqr we can easily find that it's a right angle too as angle pqa and rqb will be 45 and by using total angle in straight line we can reach the soln
Answers & Comments
proved by using SAS congruent method
Step-by-step explanation:
(we can use similar solution for proving that midpts of rectangle forms square)
adds this in answer :make sure that since 4 triangle formed ARQ QBR RCS SDP are equilateral right angled triangle hence if we take a condition of angle pqr we can easily find that it's a right angle too as angle pqa and rqb will be 45 and by using total angle in straight line we can reach the soln
Answer:
The required perimeter =(60+20
2
)
Step-by-step explanation:
Let the side of the outer square ABCD be a cm
⇒ Side of square PQRS =
2
BD
=
2
2a
=
2
a
=PS
And the side of square EFGH
=
2
GE
=
2
PS
=
2
(
2
a
)
=
2
a
cm
∴ Area of square EFGH =
2
a
×
2
a
=
4
a
2
cm
2
Given
4
a
2
=25⇒a
2
=100⇒a=10
∴ Perimeter of ABCD = 4a = 40 cm
Perimeter of PQRS = 4
2
a
=20
2
cm and perimeter of EFGH = 20 cm
∴ The required perimeter =(60+20
2
)