# Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used.

**Solution:**

The picture of the aeroplane with coloured paper is a combination of multiple diagrams: triangle, rectangle, trapezium.

By using Heron’s formula, we can calculate the area of a triangle.

Heron's formula for the area of a triangle is: Area = √s(s - a)(s - b)(s - c)

Where a, b and c are the sides of the triangle, and s = Semi-perimeter = Half the Perimeter of the triangle

i) For the triangle marked as I:

It is an isosceles triangle, therefore a = 5 cm, b = 5 cm, c = 1 cm

Semi Perimeter: (s) = (Perimeter/2)

s = (a + b + c)/2

= (5 + 5 + 1)/2

= 11/2

= 5.5 cm

By using Heron’s formula,

Area of triangle marked I = √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 cm

= 2.5 cm^{2} (approx.)

Area of triangle (I) = 2.5 cm^{2}

(ii) For the rectangle marked as II:

The measures of the sides are 6.5 cm, 1 cm, 6.5 cm, and 1 cm.

Area of rectangle = length × breadth

= 6.5 cm × 1 cm

= 6.5 cm^{2}

Area of a rectangle (II) = 6.5 cm^{2}

(iii) For the trapezium marked as III

Name it as ABCD.

Draw AE ⊥ BC from A on BC and AF parallel to DC

AD = FC = 1 cm (opposite sides of a parallelogram)

AB = DC = 1 cm

BC = 2 cm

AF = DC = 1 cm (opposite sides of parallelogram)

BF = BC - FC = 2 cm - 1 cm = 1 cm

Here ∆ABF is an equilateral triangle, Hence E will be the mid-point of BF.

So, BE = EF = BF/2

EF = 1/2 = 0.5 cm

In ∆AEF,

AF^{2} = AE^{2} + EF^{2} [Pythagoras theorem]

1^{2} = AE^{2 }+ 0.5^{2}

AE^{2} = √1^{2} - 0.5^{2}

AE^{2} = √0.75

AE = 0.9 cm (approx.)

Area of trapezium = 1/2 × sum of parallel sides × distance between them

= 1/2 × (BC + AD) × AE

= 1/2 × (2 + 1) × 0.9

= 1/2 × 3 × 0.9

= 1.4 cm^{2} (approx.)

Areaof trapezium (III) = 1.4 cm^{2}

(iv) For the triangle marked as IV and V

Triangles IV and V are congruent right-angled triangles with base 6 cm and height 1.5 cm.

Area of the triangle = 1/2 × base × height

= 1/2 × 6 × 1.5

= 4.5 cm^{2}

Area of two triangles (IV and V) = 4.5 cm^{2} + 4.5 cm^{2} = 9 cm^{2}

Total area of the paper used = Area I + Area II + Area III + Area IV + Area V

= 2.5 cm^{2} + 6.5 cm^{2} + 1.4 cm^{2} + 9 cm^{2} = 19.4 cm^{2}

Thus, the total area of the paper used is 19.4 cm^{2} (approx.).

**Video Solution:**

## Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used.

### Class 9 Maths NCERT Solutions - Chapter 12 Exercise 12.2 Question 3:

**Summary:**

It is given that Radha has made a picture of an aeroplane with coloured paper. We have found that the total area of paper used is 19.4 cm^{2}.