# Class 9 NCERT Solutions- Chapter 12 Heron’s Formula – Exercise 12.2

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

**Solution:**

Given, a quadrilateral ABCD where âˆ C = 90Âº.

Construction: Join diagonal BD.

As we can see that, â–³DCB is right-angled at C

Hence, BC is the base and CD is height of â–³DCB, so

ar(â–³DCB) =Ã— Base Ã— Height= Ã— 12 Ã— 5

= 30 m^{2}……………………………………(1)As â–³DCB is right angle triangle we can calculate third side by

Pythagoras theoremBD

^{2}= CB^{2}+ CD^{2}BD

^{2}= 12^{2}+ 5^{2}BD = âˆš(144+25)

BD = âˆš169

BD =13 m

Now, Area of â–³DAB can be calculated by Heron’s Formula, where

AB = a = 9 m

AD = b = 8 m

BD = c = 13 m

Semi Perimeter (s) =s =

s = 15 m

ar(â–³DAB) = âˆšs(s-a)(s-b)(s-c)

=âˆš15(15-9)(15-8)(15-13)= âˆš15Ã—(6)Ã—(7)Ã—(2)

= 6âˆš35 m^{2}……………………………………..(2)From (1) and (2), we can conclude that,

ar(ABCD) = ar(â–³DCB)+ar(â–³DAB)

= (30 + 6âˆš35)

= (30 + 35.5)

â‰ˆ 65.5 m^{2}(approx.)

### Question 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:**

Here, we can notice that in â–³ABC

AC

^{2}= AB^{2}+ BC^{2}5

^{2}= 3^{2}+ 4^{2 }25 = 25

LHS = RHS

As this triangle is satisfying the Pythagoras theorem, Therefore, â–³ABC is a

right angle triangle, 90Â° at B.Hence, BC is the base and AB is height of â–³ABC. so

So,

ar(â–³ABC) =Ã— Base Ã— Height

=Ã— 4 Ã— 3

= 6 cm^{2}……………………………….(1)Now, Area of â–³DAC can be calculated by Heron’s Formula, where

AD = a = 5 cm

DC = b = 4 cm

AC = c = 5 cm

Semi Perimeter (s) =s =

s = 7 cm

ar(â–³DAC) = âˆšs(s-a)(s-b)(s-c)= âˆš7(7-5)(7-4)(7-5)

= âˆš7Ã—(2)Ã—(3)Ã—(2)

= 2âˆš21 cm^{2}……………………………………..(2)From (1) and (2), we can conclude that,

ar(ABCD) = ar(â–³ABC)+ar(â–³DAC)= (6 + 2âˆš21)

= (6 + 9.2(approx.))

â‰ˆ 15.2 cm^{2}(approx.)

### Question 3. Radha made a picture of an aeroplane with coloured paper as shown in (Fig). Find the total area of the paper used.

**Solution:**

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

Area I: TriangleNow, Area of triangle can be calculated by Heron’s Formula, where

a = 5 cm

b = 5 cm

c = 1 cm

Semi Perimeter (s) =s =

s = 5.5 cm

ar(â–³) = âˆšs(s-a)(s-b)(s-c)= âˆš5.5(5.5-5)(5.5-5)(5.5-1)

= âˆš5.5Ã—(0.5)Ã—(0.5)Ã—(4.5)

= 0.5Ã—0.5Ã—3âˆš11

= 0.75âˆš11

â‰ˆ 2.5 cm^{2 }……………………………………..(1)

Area II: Rectangle

Area of Rectangle = lengthÃ—Breadth= 1 Ã— 6.5 =

6.5 cm^{2}……………………………….(2)

Area III: Trapezium= Area of parallelogram EFAO + â–³ AFDOD = 2cm

AD = OD-OA = 2-1 = 1 cm

Hence, â–³ AFD is equilateral.

PD = AD = cm

â–³ PFD is right angled at P, Pythagoras theorem is applicable

1

^{2}=h^{2}+()^{2}h = âˆš(1-)

h = âˆš cm

Area III:

=(Base Ã— Height) + ( Ã— Base Ã— Height)= (1 Ã— ) + ( Ã— 1 Ã— )

=

== 1.29 cm^{2}…………………………………(3)

Area IV and V: 2 times Triangle

ar(â–³) =Ã— Base Ã— Height= Ã— 6 Ã— 1.5

= 4.5 cm^{2}Area IV + Area V = 2Ã—ar(â–³)

= 2Ã—4.5

= 9 cm^{2 }…………………………………(4)Hence, Total area of the paper used = (Area I + Area II + Area III + Area IV + Area V)

= (1) + (2) + (3) + (4)

= 2.5 + 6.5 + 1.29 +9

= 19.29 cm^{2}

### Question 4. 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:**

Now, Area of â–³AEB can be calculated by Heron’s Formula, where

AE = a = 28 cm

EB = b = 30 cm

AB = c = 26 cm

Semi Perimeter (s) =

s = (28+30+26)/2s = 42 cm

ar(â–³AEB) = âˆšs(s-a)(s-b)(s-c)= âˆš42(42-28)(42-30)(42-26)

= âˆš42Ã—(14)Ã—(12)Ã—(16)

= 336 cm^{2}As it is given, ar(â–³AEB) = ar(parallelogram ABCD)

336 = Base Ã— Height

336 = 28 Ã— h

h =

h = 12 cm

Hence, the height of the parallelogram = 12 cm

### Question 5. 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:**

ABCD is a rhombus having diagonal AC = 48 cm

side AB=BC=CD=AD=30 cm

Diagonal of Rhombus divides the area into two equal parts.

Now, ar(â–³ABC) can be calculated by Heron’s Formula, where

AB = a = 30 m

BC = b = 30 m

AC = c = 48 m

Semi Perimeter (s) =s =

s = 54 m

ar(â–³ABC) = âˆšs(s-a)(s-b)(s-c)= âˆš54(54-30)(54-30)(54-48)

= âˆš54Ã—(24)Ã—(24)Ã—(6)

= 432 m

^{2}Hence, Area of rhombus =

2 Ã— (ar(â–³))= 2 Ã— 432 = 864 m

^{2}Area for 18 cows = 864 m

^{2}

Area for each cow = 864 / 18 = 48 m^{2}

### Question 6. 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:**

Let’s consider for each triangle.

Now, for each ar(â–³)can be calculated by Heron’s Formula, where

a = 50 cm

b = 50 cm

c = 20 cm

Semi Perimeter (s) =s =

s = 60 cm

ar(â–³) = âˆšs(s-a)(s-b)(s-c)= âˆš60(60-50)(60-50)(60-20)

= âˆš60Ã—(10)Ã—(10)Ã—(40)

= 200âˆš6 cm

^{2}Hence, the

Total Area= 5Ã—200âˆš6

= 1000âˆš6 cm^{2}

### Question 7. 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. How much paper of each shade has been used in it?

**Solution:**

As the area of kite is in the square, it area will be

Area of kite =Ã—(diagonal)^{2}= Ã—32Ã—32

= 512 cm

^{2}Diagonal divides area into equal areas.

Area of kite = Area I + Area II512 = 2 Ã— Area I

Area I =Area II = 256 cm^{2}……………………………..(1)

Area III: Area of triangleNow, for each ar(â–³)can be calculated by Heron’s Formula, where

a = 6 cm

b = 6 cm

c = 8 cm

Semi Perimeter (s) =s =

s = 10 cm

ar(â–³) = âˆšs(s-a)(s-b)(s-c)= âˆš10(10-6)(10-6)(10-8)

= âˆš10Ã—(4)Ã—(4)Ã—(2)

= 8âˆš5 cm^{2}………………………………(2)

### Question 8. 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). Find the cost of polishing the tiles at the rate of 50p per cm^{2}.

**Solution:**

Area for

each trianglewill be:Now, for each ar(â–³)can be calculated by Heron’s Formula, where

a = 9 cm

b = 28 cm

c = 35 cm

Semi Perimeter (s) =s =

s = 36 cm

ar(â–³) = âˆšs(s-a)(s-b)(s-c)= âˆš36(36-9)(36-28)(36-35)

= âˆš36Ã—(27)Ã—(8)Ã—(1)

= 36âˆš6 cm

^{2}As there are 16 tiles, so total area = 16 Ã— 36âˆš6

= 1410.906 cm^{2}As 1 cm

^{2 }= 50 p = â‚¹ 0.51410.906 cm

^{2}= â‚¹ 0.5 Ã—1410.906

= â‚¹ 705.45

### Question 9. 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:**

AB = 25 m

EB = AB-AE = 25-10 = 15 m

Now, for ar(â–³ECB) can be calculated by Heron’s Formula, where

a = 13 m

b = 14 m

c = 15 m

Semi Perimeter (s) =s =

s = 21 m

ar(â–³ECB) = âˆšs(s-a)(s-b)(s-c)= âˆš21(21-3)(21-14)(21-15)

= âˆš21Ã—(18)Ã—(7)Ã—(6)

= 84 m^{2 }………………………….(1)

ar(â–³ECB) =Ã— Base Ã— Height84 m

^{2}=Ã— 15 Ã— hh = m

h = 11.2 m

Total Area = Area of parallelogram AECD + ar(â–³ECB)

= (Base Ã— Height)+84m^{2 }= 10 Ã— 11.2 + 84

Total Area = 196 m^{2}

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