Asifa cut an aeroplane from a coloured chart paper (as shown in the adjoining figure). Find the total area of the chart paper used, correct to 1 decimal place.

The points are labelled as shown in the figure below:

In △ABK,

a = BK = 1 cm, b = AK = 5 cm and c = AB = 5 cm.

s = $\dfrac{a + b + c}{2} = \dfrac{1 + 5 + 5}{2} = \dfrac{11}{2}$ = 5.5 cm.

Area of △ABK = $\sqrt{s(s - a)(s - b)(s - c)}$

Substituting values we get,

$= \sqrt{5.5 \times (5.5 - 1)(5.5 - 5)(5.5 - 5)} \\[1em] = \sqrt{5.5 \times 4.5 \times 0.5 \times 0.5} \\[1em] = \sqrt{6.1875} \\[1em] = 2.5 \text{ cm}.$

Area of △KIJ = $\dfrac{1}{2} \times$ base × height

= $\dfrac{1}{2}$ × KI × KJ

= $\dfrac{1}{2}$ × 1.5 × 6

= 4.5 cm².

Area of △CBD = $\dfrac{1}{2} \times$ base × height

= $\dfrac{1}{2}$ × BD × CB

= $\dfrac{1}{2}$ × 1.5 × 6

= 4.5 cm².

Area of rectangle BKHE = length × breadth

= BE × EH

= 6.5 × 1 = 6.5 cm².

Let EL and HM be perpendicular to LM.

Since, EFGH is an isosceles trapezium so,

FL = MG = $\dfrac{2 - 1}{2} = \dfrac{1}{2}$ = 0.5 cm.

In right angle △EFL,

⇒ EF² = EL² + FL²

⇒ 1² = EL² + (0.5)²

⇒ EL² = 1² - (0.5)²

⇒ EL² = 1 - 0.25

⇒ EL² = 0.75

⇒ EL = $\sqrt{0.75}$ = 0.87 cm.

Area of trapezium EFGH = $\dfrac{1}{2}$ × (sum of parallel sides) × distance between them

= $\dfrac{1}{2}$ × (EH + FG) × EL

= $\dfrac{1}{2}$ × (1 + 2) × 0.87

= 1.3 cm².

Area of chart paper used = Area of aeroplane = Area of △ABK + Area of △KIJ + Area of △CBD + Area of rectangle BKHE + Area of trapezium EFGH

= 2.5 + 4.5 + 4.5 + 6.5 + 1.3

= 19.3 cm².

Hence, area of chart paper used = 19.3 cm².