Question

(60 Points)Hey Mate!!Answer this to be the brainliest
No.42 and 42...
I want step by step process

Answer 1

Hope it helps............please mark me brainliest ....please

Answer 2

Q.42                                PART 1 AREA

In isosceles right triangle ABE we have

AB = 10 cm

it is a isosceles triangle so,

EA = EB = x cm (say)

by Pythagoras theorem we have

x^2 + x^2 = 10^2

2x^2 = 100

x^2 = 50

x = 5√2 cm

hence

EA = EB = 5√2 cm

similarly GC = GD = 5√2 cm

We know that

area of triangle  = 1/2 * base * height

area of triangle ABE  = 1/2 * 5√2 * 5√2

= 25 cm^2

similarly area of triangle DCG = 25 cm^2

again we know that ,

area of equilateral triangle is √3/4 a^2

area of equilateral triangle AHD = √3/4 8^2

= 16√3 cm^2

similarly area of equilateral triangle BFC =  16√3 cm^2

again we know that area of rectangle is length * breadth

area of rectangle ABCD =  8 * 10

= 80 cm^2

area of whole figure = {area of rectangle ABCD + area of equilateral triangle BFC + area of equilateral triangle AHD + area of triangle DCG +area of triangle ABE} cm^2

= {80 + 16√3 + 16√3 + 25 + 25} cm^2

= {130 + 32√3} cm^2

= {130 + 55.42562584220407} cm^2

= 185.42562584220407 cm^2

                  PART 2 PERIMETER

Perimeter of figure = {AE + EB + BF + CF + CG + GD + DH + HA} cm

={5√2 +5√2 + 8 + 8 + 5√2 + 5√2 + 8 + 8} cm

= {32 + 20√2} cm

= {32 + 28.2842712474619} cm

= 60.2842712474619 cm

I  have tried my best for that and Q.no 43 has no figure so i could not solve that Please mark as brainliest