Question

please anyone can put the answer

Answer 1

Area of region 1:

Region 1 is enclosed by triangle of sides a=5cm=b and c=1cm.

Let 2s be the perimeter...

2s=5+5+1

S=11/2cm.

Area of region 1 :

√s(s-a)(s-b)(s-c)

√11/2*(11/2-5)(11/2-5)(11/2-1). When u will solve this you'll be getting 2.49sq.cm

For region 2...

Region 2 is a rectangle of length = 6.5cm and breadth=1cm

Area : l*b= 6.5*1=6.5sq.cm

For region 3 which is an isosceles trapezium..

Ab^2 = Ae^2 + Be^2

1=0.25 + Be^2

Be = √0.75 = √3/4

Area o region 3...

1/2(AD+BC)*BE = 1/2(2+1)*√3/4 sq.cm

When u will solve this u will be getting 3√3/4 sq.cm which is equal to 1.3 sq cm

Region 4

This region forms a right triangle whose sides are 6cm and 1.5cm

So,it's area will be..

1/2*6*1.5 =4.5. (Using area = 1/2*base *height)

For region 5....

Region 4 and 5are congruent.

So area of region 5 will be 4.5 sq.cm

Hence total area of paper used will be equal to sum of areas of all the regions i.e.

2.49+6.5+1.3+4.5+4.5 = 19.29 sq.cm...

Hope this helped.....:)

Answer 2

there are 3 pages of answers ok