Radha made a picture ofan aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used.
Answers
Answer:
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White paper used = 6.5 sq-cm
Blue paper used= 2.5+6.5+1.29
= 10.29 sq-cm
Solution:
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Region I:
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Isosceles Triangle (two equal side a = 5, base b = 1)
Area of Isosceles Triangle =
or h ≈ 5 cm
Area of region 1 =
2) Area of Region V,IV :
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Both have same dimensions ,these are right angle triangle
Base =1.5 cm
Height= 6 cm
Area =
Total area of both region V,IV = 2(4.5) = 9 sq-cm
3) Area of Region II:
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III region is a rectangle: length = 6.5 cm
breadth = 1 cm
Area of rectangle=
Area of third region = 6.5 sq-cm
4) Area of Region III:
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III region is a trapezium
Area of trapezium= 1/2(sum of parallel sides)*h
= 1/2(3)(.86)
= 1.29 sq-cm
Hope it helps you.
Part I: - Isosceles triangle:
Sides are a = b = 5 cm, c = 1 cm.
We know that s = (a + b + c)/2
= (5 + 5 + 1)/2
= 11/2
= 5.5 cm.
We know that by heron's formula:
⇒ √s(s - a)(s - b)(s - c)
⇒ √(5.5)(5.5 - 5)(5.5 - 5)(5.5 - 1)
⇒ √(5.5)(0.5)(0.5)*(4.5)
⇒ √6.1875
⇒ 2.5 m².
Part II: - Rectangle:
Length of the Sides are 6.5 cm and 1 cm.
We know that Area of rectangle = length * breadth
⇒ 6.5 * 1
⇒ 6.5 cm².
Part III: - Trapezium:
Length of non-parallel sides are 1 cm and 2 cm.
⇒ Area of equilateral triangle + Area of trapezium
⇒ √3/4 a^2 + Base * Height
⇒ √3/4 (1)^2 + 1 * √3/2
⇒ 3√3/4
⇒ 1.3 cm².
Part - IV: - triangles:
Sides are 6 cm and height is 1.5 cm.
Area = (1/2) * base * height
= (1/2) * 6 * (1.5)
= 4.5 cm².
Area of IV and V are 4.5 cm² and 4.5 cm².
⇒ Area of the paper used = 2.5 + 6.5 + 1.3 + 4.5 + 4.5
⇒ 19.3 cm².
Therefore, total area of the paper used = 19.3 cm².
Hope this helps!