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Asked by Chiranjeevi Last Modified
Answer 
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)
read lessRelated Questions
A park, in the shape of a quadrilateral ABCD, has ∠ C = AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
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