Heron's Formula – Exercise 12.2 – Class IX

For previous exercise – Heron’s Formula – Exercise 12.1 – Class IX


  1. A park, in the shape of a quadrilateral ABCD, has ∠ C = 90, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

Solution:

Heron's Formula - exercise 12.2. - CLass IX

Given: a park, in the shape of a quadrilateral ABCD, has ∠ C = 90, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m

Construction: Join BD

In ∆DBC, by  Pythagoras theorem, we have,

DB2 = BC2 + DC2

DB2 = (12)2 + (5)2

DB2 = 144 + 25

DB2 = 169

DB = √169 = 13 m

Area  of ∆DBC = 1/2 x base x height

= 1/2 x 12 x 5

= 6 x 5

= 30 m2

In ∆ADB, a = 9m,b  = 8m , c = 13m

Therefore, s = a+b+c/2 = 9+8+13/2 = 15 m

area of ∆ABD = √[s(s-a)(s-b)(s-c)

= √[15(15-9)(15-8)(15-13) m2

= √[15x6x7x2] m2

=√1260 m2

= 35.5 m2

Area of the park = area of ∆DBC + area of ∆ABD

=(30 + 35.5) m2

= 65.5 m2


  1. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

Solution:

Heron's Formula - exercise 12.2. - CLass IXIn ∆ABC, we have,

a = 3 cm, b = 4 cm , c = 5 cm

s = a+b+c/2 = 3+4+5/2 = 12/2 = 6 cm

Area of the triangle = √[s(s-a)(s-b)(s-c)

=√[6(6-3)(6-4)(6-5)

= √[6x3x2x1]

=√[6×6]

= 6 cm2

In ∆ADC, we have

a = 5 cm , b = 5 cm, c = 4 cm

s = a+b+c/2 = 5+5+4/2 = 14/2 = 7 cm

Area of the triangle = √[s(s-a)(s-b)(s-c)

= √[7(7-5)(7-5)(7-4)]

= √[7x2x2x3]

= √84

= 9.2 cm2

Area of the quadrilateral = area of ∆ABC + area of ∆ADC

= (6 + 9.2) cm2

= 15.2 cm2


  1. Herons formula - exercise 12.2 - class IXRadha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used.

 

Solution:

For the triangle marked as I:

a = b = 5 cm, c = 1 cm

s = a+b+c/2 = 5+5+1/2 = 11/2 = 5.5 cm

Area of triangle1 = √[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 m2

≅ 2.5 m2

For the rectangle marked as II:

Area of the rectangle = Length x breadth [Length = 6.5 cm and breadth = 1 cm]

= 6.5 x 1

= 6.5 m2

For the trapezium marked as III:

Draw AF||DC and AE ⊥ BC

Therefore, AD = FC = DC = AF = 1 cm

⸫BF = BC – FC = (2 – 1) cm = 1 cm

Hence, ∆ABF is equilateral

Also, E is the midpoint of BF

⸫ BE = 1/2 cm = 0.5 cm

Also, AB2 = AE2 + BE2

AE2 = 12 – (0.5)2 = 0.75

AE ≅ 0.9 cm

Area of trapezium = 1/2 (sum of the parallel sides) x (distance between them)

= 1/2 x(BC + AD) X AE

= 1/2 X(2 + 1) X 0.9 cm2

= 1.4 cm2

For triangle marked as IV and V:

[Triangle IV and V are congruent to one another as these are the wings of the aeroplane which needs to same for proper figure]

These are right angled triangles, therefore, by Pythagoras theorem, we have,

Area of triangle = 1/2 x base x height

= 1/2 x 6 x 1.5

= 3 x 1.5

= 4.5 cm2

Area of the triangle IV and V are 4.5cm2 and 4.5cm2.

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

= (2.5 + 6.5 + 1.4 + 4.5 + 4.5) cm2

= 19.4 cm2


  1. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.

Solution:

HErons formula exercise 12.2 class IXLet ABCD is a parallelogram and CDE is a triangle with same base CD = 28 cm.

Given area of the triangle = area of the parallelogram.

In triangle ABC,

a = 26 cm , b = 28 cm, c = 30 cm

s = 26+28+30/2 = 84/2 = 42

Area of the triangle = √[s(s-a)(s-b)(s-c)]

= √[42(42-26)(42-28)(42 – 30)

= √[42x16x14x12]

= √112896

=336m2

Since, area of the triangle = area of the parallelogram

Area of parallelogram = base x height

336 = 28 x height

height = 336/28 = 12 cm

Therefore, height of the parallelogram = 12 cm


  1. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?

Solution:

HErons formula - exercise 12.2. class IXGiven: . If each side of the rhombus is 30 m and its longer diagonal is 48 m.

In rhombus ABCD, we have 2 equal triangles i.e., ∆ABC and ∆ADC

Therefore, area of ∆ABC = area of ∆ADC ————(1)

area of the rhombus = area of ∆ABC + area of  ∆ADC

= 2 x area of ∆ABC [from (1)]

In ∆ABC and ∆ADC,

a = b = 30 m ; c = 48 m

s = a+b+c/2 = 30+30+48/2 = 108/2 = 54 m

area of ∆ABC = √[s(s-a)(s-b)(s-c)]

=√[54(54-30)(54-30)(54-48)]

=√[54*24*24*6]

=√1,86,624

=432 m2

Therefore, area of rhombus = 2 x area of ∆ABC

= 2 x 432

= 864 m2

Area of the grass field each cow gets = area of rhombus/no. of cows

= 864/18

= 48 m2


  1. Herons Formula Exercise 12.2. class IXAn umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig. 12.16), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

Solution:

Given: An umbrella is made by stitching 10 triangular pieces of cloth of two different colours, each piece measuring 20 cm, 50 cm and 50 cm.

Sides of each triangle are 20cm, 50cm and 50cm i.e., a = 20 cm, b = 50 cm, c = 50 cm

s = 20+50+50/2 = 120/2 = 60

Area of each triangle = √[s(s-a)(s-b)(s-c)]

= √[60(60-20)(60-50)(60-50)]

= √[60*40*10*10]

= 10√[20x3x20x2]

=200 √6 m2

Therefore, area of 10 triangular pieces = 200 √6 x 10 = 2000 √6 m2

Hence, cloth required for each color = 2000 √6/2 = 1000 √6 m2


  1. A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Fig. 12.17. How much paper of each shade has been used in it?A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Fig. 12.17. How much paper of each shade has been used in it?

 

 

Solution:

Area of a kite = diagonal^2/2 = 32^2/2 = 1024/2 = 512 cm2

Area of  shades in a kite are equal . Therefore area of two equal shades of a kite = area of  a kite/2 = 512/2 = 256 cm2

Given an isosceles triangle with base 8 cm and 6cm. Therefore, a = 8 cm, b = 6 cm and c = 6 cm

s = a+b+c/2= 8+6+6/2 = 20/2 = 10 cm

Area of an equilateral triangle = √[s(s-a)(s-b)(s-c)

= √[10(10-8)(10-6)(10-6)

= √[10*2*4*4]

= √[2*5*2*4*4]

= 8√5 cm2

Therefore, paper needed for shade I = 256cm2, paper needed for shade II = 256cm2, paper needed for shade III = 8√5 cm2


  1. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see Fig. 12.18). Find the cost of polishing the tiles at the rate of 50p per cm2.A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see Fig. 12.18). Find the cost of polishing the tiles at the rate of 50p per cm2.

Solution:

Given: A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm.

Therefore, in each triangle a = 9 cm , b = 28 cm and c = 35 cm

s = a+b+c/2 = 9+28+35/2 = 36 cm

Area of each triangle = √[s(s-a)(s-b)(s-c)

= √[36(36-9)(36-28)(36-35)]

= √[36*27*8*1]

= √7776

= 88.18 cm2

Area of 16 such tiles = 88.18 x 16 = 1410.88 cm2

Given: the cost of polishing the tiles at the rate of 50p per cm2 i.e., the cost of polishing the tiles for 1 cm2 is 50p

Then, the cost of polishing the tiles for 1410.88 cm2 will be = 1410.88 x .50 = Rs. 705.44


  1. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

Solution:

A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see Fig. 12.18). Find the cost of polishing the tiles at the rate of 50p per cm2.Let ABCD be trapezium.

Draw a line segment from B parallel to AD which intersects DC at F and BE parallel to DC.

In ∆BFC,

a = 13m , b = 14m , c = 15m

s = 13+14+15/2 = 42/2 = 21 m

area of triangle BFC = √[s(s-a)(s-b)(s-c)

=√[21(21-13)(21-14)(21-15)]

=√[21*8*7*6]

=√7056

= 84 m2

In ∆BFC,

Area of ∆BFC = 1/2 x base x height

84 =  1/2 X 15 x BE

84 x 2/15 = BE

BE = 11.2 m

Area of trapezium = 1/2 (sum of parallel sides)x(distance between them)

= 1/2 x (25 + 10) x (11.2)

= 196 m2

Hence area  if the field = 196 m2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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