Question

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

samaz me nahi aa raha hai​

Answer 1

For finding the area of the paper used determine the area of each part separately and then find the sum of the areas to get the area of used paper.

For region I (Triangle)

Length of the sides of the triangle section 1 = a=5cm, b=1cm and c=5cm

Semi Perimeter of the triangle,

s =(a+b+c)/2

s=(5 + 5 + 1)/2= 11/2cm

Semi perimeter = 11/2 cm = 5.5cm

Using heron's formula, Area of section I = √s (s-a) (s-b) (s-c)

= √5.5(5.5 - 5) (5.5-5) (5.5 - 1) cm2

= √5.5 x 0.5 x 0.5 x 4.5 cm2

= √5.5 x 0.5 x 0.5 x 4.5 cm2

= 0.75-√11 cm²= 0.75 x3.32 cm² = 2.49 cm² (approx)

Using heron's formula,

Area of section I = √s (s-a) (s-b) (s-c)

= √5.5(5.5 - 5) (5.5-5) (5.5 - 1) cm2

= √5.5 x 0.5 x 0.5 x 4.5 cm2

= √5.5 x 0.5 x 0.5 x 4.5 cm2

= 0.75√11 cm²= 0.75 x3.32 cm² = 2.49 cm² (approx)

Section II( rectangle)

Length of the sides of the rectangle of section II = 6.5cm and 1cm

Area of section II = I xb= 6.5 x 1 6.5cm²

Section III is an isosceles trapezium Figure is in the attachment:

In Δ AMD

AD = 1cm (given)

AM + NB = AB - MN = 1cm

Therefore, AM = 0.5cm

Now,AD² =AM² +MD²

In Δ AMD

AD = 1cm (given)

AM + NB = AB - MN = 1cm -

Therefore, AM = 0.5cm MD²= 1-0.25= 0.75

Now,AD² =AM² +MD²

MD²= 1² - 0.5²

MD = √0.75= √75/100=√3/4cm

Now, area of trapezium = 1/(sum of parallel sides)xheight

=1/2×(AB+DC) MD

=1/2×(2+1)x√3/4

= 1/2(3)x-√√(3/4)= 12×3×√√3×2=(3/4)-√3

= (3/4)x1.73=1.30cm²(approx) [√3=1.73....]

Hence, area of trapezium = 1.30cm²

Section IV and V are 2 congruent right angled triangles with base 6cm and height 1.5cm