**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:

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,

DB^{2} = BC^{2} + DC^{2}

DB^{2} = (12)^{2} + (5)^{2}

DB^{2} = 144 + 25

DB^{2} = 169

DB = √169 = 13 m

Area of ∆DBC = ^{1}/_{2} x base x height

= ^{1}/_{2} x 12 x 5

= 6 x 5

= 30 m^{2}

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) m^{2}

= √[15x6x7x2] m^{2}

=√1260 m^{2}

= 35.5 m^{2}

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

=(30 + 35.5) m^{2}

= 65.5 m^{2}

**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:

In ∆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 cm^{2}

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 cm^{2}

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

= (6 + 9.2) cm^{2}

= 15.2 cm^{2}

**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:

**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 m^{2}

≅ 2.5 m^{2}

**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 m^{2}

**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, AB^{2} = AE^{2} + BE^{2}

AE^{2} = 1^{2} – (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 cm^{2}

= 1.4 cm^{2}

**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 cm^{2}

Area of the triangle IV and V are 4.5cm^{2} and 4.5cm^{2}.

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) cm^{2}

= 19.4 cm^{2}

**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:

Let 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

=336m^{2}

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

- 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:

Given: . 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 m^{2}

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

= 2 x 432

= 864 m^{2}

Area of the grass field each cow gets = ^{area of rhombus}/_{no. of cows}

= ^{864}/_{18}

= 48 m^{2}

**An 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 m^{2}

Therefore, area of 10 triangular pieces = 200 √6 x 10 = 2000 √6 m^{2}

Hence, cloth required for each color = ^{2000 √6}/_{2} = 1000 √6 m^{2}

**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 cm^{2}

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 cm^{2}

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 cm^{2}

Therefore, paper needed for shade I = 256cm^{2}, paper needed for shade II = 256cm^{2}, paper needed for shade III = 8√5 cm^{2}

**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 cm**^{2}.

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 cm^{2}

Area of 16 such tiles = 88.18 x 16 = 1410.88 cm^{2}

Given: the cost of polishing the tiles at the rate of 50p per cm^{2} i.e., the cost of polishing the tiles for 1 cm^{2} is 50p

Then, the cost of polishing the tiles for 1410.88 cm^{2} will be = 1410.88 x .50 = Rs. 705.44

**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:

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 m^{2}

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 m^{2}

Hence area if the field = 196 m^{2}

## 1 thought on “Heron’s Formula – Exercise 12.2 – Class IX”

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