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 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.).
☛ Check: NCERT Solutions Class 9 Maths Chapter 12
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 cm2.
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