Learn Exercise 12.2 with Free Lessons & Tips

Let ABCD be a rhombus-shaped field.

For ΔBCD,

Semi-perimeter, = 54 m

By Heron’s formula,

Area of triangle

Therefore, area of ΔBCD

Area of field = 2 × Area of ΔBCD

= (2 × 432) m² = 864 m²

Area for grazing for 1 cow = 48 m²

Each cow will get 48 m² area of grass field.

For triangle

Perimeter of triangle = (26 + 28 + 30) cm = 84 cm

2s = 84 cm

s = 42 cm

By Heron’s formula,

Area of triangle

Area of triangle

= 336 cm²

Let the height of the parallelogram be h.

Area of parallelogram = Area of triangle

h × 28 cm = 336 cm²

h = 12 cm

Therefore, the height of the parallelogram is 12 cm.

Let us join BD.

In ΔBCD, applying Pythagoras theorem,

BD² = BC² + CD²

= (12)² + (5)²

= 144 + 25

BD² = 169

BD = 13 m

Area of ΔBCD

For ΔABD,

By Heron’s formula,

Area of triangle

Area of ΔABD 

Area of the park = Area of ΔABD + Area of ΔBCD

= 35.496 + 30 m² = 65.496 m² = 65.5 m² ( Approximate Value)

For ΔABC,

AC² = AB² + BC²

(5)² = (3)² + (4)²

Therefore, ΔABC is a right-angled triangle, right-angled at point B.

Area of ΔABC

For ΔADC,

Perimeter = 2s = AC + CD + DA = (5 + 4 + 5) cm = 14 cm

s = 7 cm

By Heron’s formula,

Area of triangle

Area of ABCD = Area of ΔABC + Area of ΔACD

= (6 + 9.166) cm² = 15.166 cm² = 15.2 cm^{2 ( Approximate Value)}

For each triangular piece,

Semi-perimeter, 

By Heron’s formula,

Area of triangle

Since there are 5 triangular pieces made of two different coloured cloths,

Area of each cloth required

It can be observed that

Semi-perimeter of each triangular-shaped tile, 

By Heron’s formula,

Area of triangle

Area of each tile 

= (36 × 2.45) cm²

= 88.2 cm²

Area of 16 tiles = (16 × 88.2) cm²= 1411.2 cm²

Cost of polishing per cm² area = 50 p

Cost of polishing 1411.2 cm² area = Rs (1411.2 × 0.50) = Rs 705.60

Therefore, it will cost Rs 705.60 while polishing all the tiles

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 I = 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 × 0.5 × 0.5 × 4.5 cm²

= √5.5 × 0.5 × 0.5 × 4.5 cm²

= 0.75√11 cm²= 0.75 ×3.32 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 = l ×b= 6.5 × 1
= 6.5cm²

Section III is an isosceles trapezium

In ? AMD
AD = 1cm (given)
AM + NB = AB – MN = 1cm
Therefore, AM = 0.5cm
Now,AD² =AM² +MD²
MD²= 1² – 0.5²
MD²= 1- 0.25= 0.75
MD = √0.75= √75/100=√3/4cm

Now, area of trapezium = ½(sum of parallel sides)×height

=½×(AB+DC)×MD

=½×(2+1)×√3/4

= ½(3)×√(3/4)= ½×3×√3×2=(3/4)√3

= (3/4)×1.73= 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

Area of region IV and V = 2 (½ × 6 × 1.5)cm² = 9cm²

Total area of the paper used = Area I + Area II + Area III + (Area IV + Area V) = (2.49+ 6.5 + 1.30 + 9)

= 19.3 cm² (approx)

A kite is in the shape of square thats why ; diagonals are equal and divided in to two equal halfs 

 ∴area of 1st part is in shape of triangle base 32cm and height 16cm 

Area of square (base x height)

=16×16

=256 sq.cm

Therefore, the area of paper required in each shape is 256 cm².

Similarly area of 2nd part also 256 sq.cm

Area of 3rd part using with heron's formula 

Area of triangle

a=8cm

b=6cm

c=6cm

S=a+b+c/2

 =8+6+6/2

= 20/2

 =10cm

Area of III^rd triangle

Area of paper required for III^rd shade = 17.92 cm²

Total area is 256+256+17.92

 ( approximately)