Question

A wire bent in the form of equilateral triangle encloses an area of 36√3cm2.Find the area enclosed by the same wire in the form of
(i) a square
(ii)A rectangle whose length is 2cm more than it's width​

Answer 1

Answer:

side of triangle becomes 12 cm, thus total length of wire is 36 cm.

a. side of square-9 cm.area of square-36 cm

B. length of rectangle-10 cm and breadth-8 cm

area is going to be 80 cm

Answer 2

Answer:

(i) 81cm^2

(ii) 80cm^2

Step-by-step explanation:

(i)area of equi. ∆ = √3a^2/4

=> 36√3 = √3a^2/4

=> a= 12 cm

so, perimeter of equi. ∆ would be, 3a

=> 3a = 3x12 = 36cm

Same wire is now enclosed to form a square,

so inorder to get sides of square.

=> perimeter of sq. = 4xside

=> 36=4xside

=> side=9cm

area of sq. =sidexside

=> area of sq = 9x9 = 81cm^2

(ii) let's assume breadth = b, then length =b+2

Similarly, Same wire is now enclosed to form a rectangle .

permiter of sq. = perimeter of rect.

=> perimeter of rect. = 2(l+b)

=> 36=2(b+2+b)

=> b =8 cm.

=> l=b+2= 10cm.

area of rect. =lxb

=> area of rect. = 8x10 = 80cm^2

Thanks, ❤️