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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