For the triangle marked I : a = 5 cm, b = 5 cm, c = 1 cm
∴ s = (a +b + c)/ 2 = (5 +5+ 1) /2 cm = 11 / 2 cm =5.5 cm
For the rectangle marked II :
Length = 6.5 cm, Breadth = 1 cm
Area of the rectangle = 6.5 × 1 cm2 = 6.5 cm2
For the trapezium marked III :
Draw AF || DC and AE ⊥ BC.
AD = FC = 1 cm, DC = AF = 1 cm
∴ BF = BC – FC = (2 – 1) cm = 1 cm
Hence, ∆ABF is equilateral.
Also, E is the mid-point of BF.
∴ BE = 1/ 2 cm = 0.5 cm
Also, AB2 = AE2 + BE2 [Pythagoras theorem]
⇒ AE2 = 12 – (0.5)2 = 0.75
⇒ AE = 0.9 cm (approx.)
Area of the trapezium = 1/ 2 (sum of the parallel sides) × distance between them.
= 1/ 2 × (BC + AD) × AE = 1/ 2 × (2 + 1) × 0.9 cm2 = 1.4 cm2.
For the triangle marked IV :
It is a right-triangle
∴ Area of the triangle = 1/ 2 × base × height
= 1/ 2 × 6 × 1.5 cm cm2 = 4.5 cm2.
For the triangle marked V :
This triangle is congruent to the triangle marked IV.
Hence, area of the triangle = 4.5 cm2
Total area of the paper used = (2.5 + 6.5 + 1.4 + 4.5 + 4.5) cm2
