Question

Radha made a picture of an aeroplane with paper as shown in figure calculate the total area of the paper used. ​

Answer 1

Step-by-step explanation:

Please see the images attached...

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Answer 2

Answer:

Step-by-step explanation:

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 cm2 (approx.)

Area of triangle (I) = 2.5 cm2

(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 cm2

Area of a rectangle (II) = 6.5 cm2

(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,

AF2 = AE2 + EF2  [Pythagoras theorem]

12 = AE2 + 0.52

AE2 = √12 - 0.52

AE2 = √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 cm2 (approx.)

Areaof trapezium (III) = 1.4 cm2

(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 cm2

Area of two triangles (IV and V) = 4.5 cm2 + 4.5 cm2 = 9 cm2

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

= 2.5 cm2 + 6.5 cm2 + 1.4 cm2 + 9 cm2 = 19.4 cm2

Thus, the total area of the paper used is 19.4 cm2 (approx.).