Question

a wire bent in form of square encloses an area of 576 cm^2. find largest area enclosed when bent to form 1. an equilateral triangle 2.a rectangle whose adjacent sides differ by 4 cm

Answer 1

Area of square=(side) ²=576cm²

Side= 24 CM hence total length of wire=4*24=96cm

1. Side of the equilateral triangle =96/3=32cm
Hence, area=(3)½/4*side² =443.405cm²

2. Therefore perimeter of rectangle =2(l+b)=96
Also given l=b+4
2(2b+4)=96
b=(48-4)/2=22cm
l=22+4=26cm

Area of rectangle =l*b=26*22=572cm²

Answer 2

Answer:

1. 256√3

2. 572 cm^2

Step-by-step explanation:

Area of square = 576 cm^2

(side)^2 = 576

side = √576

side = 24 cm

Perimeter of square= 4*side

= 4*24

= 96 cm

Length of wire = Perimeter of square

(i) Perimeter of equilateral triangle = 3*side

96 = 3*side

side= 32 cm

Area of an equilateral triangle = √3/4* (side)^2

= √3/4*32*32

= 256√3 cm^2

(ii) Let length be (x+4) and breadth be x

Then,

Perimeter of rectangle = Length of wire

2(l+b) = 96 cm

2(x+4+x) = 96

2x+4 = 96/2

2x = 48-4

2x = 44

x = 22 cm

Length = x+4= 22+4 = 26 cm

Breadth = x = 22 cm

Now,

Area of rectangle = l*b

= 22*26

= 576 cm^2