Question

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Answer 1

Answer:

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

Figure is in the attachment:

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

=1/2×(AB+DC)×MD

=1/2×(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 × 1/2 × 6 × 1.5cm² = 9cm²

Total area of the paper used = (2.49+ 6.5 + 1.30 + 9)

= 19.3 cm² (approx).

Step-by-step explanation:

COC OP