Naresh Sawlani

For finding the area of the paper used, determine the area of each part separately and then find the sum of the areas to get the area of used paper.

For region I (Triangle)

Length of the sides of the triangle section I = a=5cm, b=1cm and c=5cm

Semi Perimeter of the triangle,
s =( a+b+c)/2
s=(5 + 5 + 1)/2= 11/2cm

Semi perimeter = 11/2 cm = 5.5cm

Using heron’s formula,
Area of section I = √s (s-a) (s-b) (s-c)

= √5.5(5.5 – 5) (5.5 – 5) (5.5 – 1) cm²

= √5.5 × 0.5 × 0.5 × 4.5 cm²

= √5.5 × 0.5 × 0.5 × 4.5 cm²

= 0.75√11 cm²= 0.75 ×3.32 cm²
= 2.49 cm² (approx)

Section II( rectangle)

Length of the sides of the rectangle of section II = 6.5cm and 1cm

Area of section II = l ×b= 6.5 × 1
= 6.5cm²

Section III is an isosceles trapezium

In ? AMD
AD = 1cm (given)
AM + NB = AB – MN = 1cm
Therefore, AM = 0.5cm
Now,AD² =AM² +MD²
MD²= 1² – 0.5²
MD²= 1- 0.25= 0.75
MD = √0.75= √75/100=√3/4cm

Now, area of trapezium = ½(sum of parallel sides)×height

=½×(AB+DC)×MD

=½×(2+1)×√3/4

= ½(3)×√(3/4)= ½×3×√3×2=(3/4)√3

= (3/4)×1.73= 1.30cm²(approx)
[√3=1.73....]

Hence, area of trapezium = 1.30cm²

Section IV and V are 2 congruent right angled triangles with base 6cm and height 1.5cm

Area of region IV and V = 2 (½ × 6 × 1.5)cm² = 9cm²

Total area of the paper used = Area I + Area II + Area III + (Area IV + Area V) = (2.49+ 6.5 + 1.30 + 9)

= 19.3 cm² (approx)

Heron's Formula for the area of a triangle(Hero's Formula)

A method for calculating the area of a triangle when you know the lengths of all three sides.

Let a=15m,b=11m,c=6m be the lengths of the sides of a triangle. The area is given by:

Area = √p(p−a)(p−b)(p−c)

where p is half the perimeter, or (a+b+c)/2 = (15+11+6)/2 =16m

Therefore, Area = √16(16-15)(16-11)(16-6) = √16(1)(5)(10) =20√2 = 20 × 1.414 = 28.28 m²