Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

Mathematics (Maths) Class 9

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Class 9 : Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

The document Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9

APPLICATION OF HERON’S FORMULA IN FINDING AREAS OF QUADRILATERALS

Many fields are in the shape of quadrilaterals. Similarly various surfaces are polygons. To find their areas, we divide the surface in triangular parts and then use the Heron’s formula for finding areas of the triangles.

 

Question 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: 
Let us join B and D, such that ΔBCD is a right-angled triangle.
We have area of ΔBCD = (1/2)x base x altitude
=(1/2)x 12 x 5 m2
= 30 m2
Now, to find the area of ΔABD, we need the length of BD

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

∴ In right ΔBCD, BD2 = BC+ CD2 [Pythagoras theorem]
⇒ BD2 = 12+ 52
⇒ BD2 = 144 + 25 = 169 = 13
⇒ BD = Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev = 13 m
Now, for ΔABD, we have a = 9 m, b = 8 m, c = 13 m
∵        Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

∴ Area of ΔABD

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev
∴ Area of quadrilateral ABCD = area of ΔBCD + area of ΔABD
= 30 m+ 35.5 m2 = 65.5 m2 (appx)


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: 
For ΔABC,

     a= 3 cm, b= 4cm and c = 5 cm

   Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

∴ Area of  ΔABC, 

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

 

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

For ΔACD,

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

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRevEx 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

For ΔACD,

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

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

∴ Area of ΔACD,

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

Now, area of quadrilateral ABCD = area of ΔABC + area of ΔACD
= 6 cm2 + 9.2 cm2 (approx.)
= 15.2 cm(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.

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

Solution: Area of surface I It is an isosceles triangle whose sides are a = 5 cm, b = 5 cm, c = 1 cm

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

= 0.75 x 3.3 cm2 (approx.) = 2.475 cm(approx.)

Area of surface II It is a rectangle with length 6.5 cm and breadth 1 cm.
∴ Area of rectangle II = Length x Breadth
= 6.5 x 1 = 6.5 cm
Area of surface III It is a trapezium whose parallel sides are 1 cm and 2 cm as shown in the adjoining figure. Its height is given by

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

Note: The perpendicular distance between the parallel sides is called the height of the trapezium.

∵ Area of a trapezium = 1/2(sum of the parallel sides) x Height

∴ Area of trapezium III = 

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

Area of surface IV It is a right triangle with base as 6 cm and height as 1.5 cm.
∴ Area of right triangle IV =(1/2)x base x height
=(1/2)x 6 x 1.5 cm2 = 4.5 cm2
Area of surface V
∵ Right triangle V ≌  Right triangle IV
∴ Area of right triangle V = Area of right triangle IV = 4.5 cm
Thus, the total area of the paper used = (area I) + (area II) + (area III) + (area IV) + (area V)
= [2.475 cm2 (approx.)] + [6.5 cm2] + [1.3 cm2 (approx.)] + [4.5 cm2] + [4.5 cm2]
= 19.275 cm= 19.3 cm2 (approx.)


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.

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

Solution: For the given triangle, we have a = 28 cm, b = 30 cm, c = 26 cm

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev
= 6 x 7 x 2 x 4 cm= 336 cm2
∵ [Area of the given parallelogram] = [Area of the given triangle]
∴ [Area of the parallelogram] = 336 cm
∴ [Base of the parallelogram] x [Height of the parallelogram] = 336 cm
⇒ [28 cm] x [h cm] = 336 cm2

⇒ h= (336/28)cm = 12
Thus, the required 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: 
Here, each side of the rhombus = 30 m
One of the diagonal = 48 m
Since a diagonal divides the rhombus into two congruent triangles.

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

Sides of first triangle are a = 30 m, b = 30 m, c = 48 m

∵ 

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

∴ Area of triangle I = 

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev
= 3 x 6 x 4 x 6 m2 = 432 m2
∴ Area of second triangle = 432 m2
⇒ Total area of both triangles = 432 cm2 + 432 m2  = 864 m2
⇒ The area of the rhombus = 864 m2
Thus, area of grass for 18 cows = 864 m
⇒ Area of grass for 1 cow = (864/18)m2 = 48 m2

 

Question 6. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?
 Solution: 
Sides of each triangular piece are

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

a = 20 cm, b = 50 cm, c = 50 cm

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

∴ Area of each triangular piece 

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev  

⇒ Area of 5 triangular pieces of one colour = 5 x 200/√6cm= 1000/√6 cm2 Area of 5 triangular pieces of other colour = 1000/√6 cm2
Note: Total triangular pieces are 10, i.e. five triangular pieces for each colour.

 

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 the figure. How much paper of each shade has been used in it?
 Solution: 
Area of triangle I

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

Diagonal = 32 cm
∴ Height of triangle =(1/2)(32) cm= 16 cm
∵ Area of triangle = (1/2)x base x height
∴ Area of triangle I= (1/2)x 32 x 16 cm2 = 256 cm2
Area of triangle II ∵ Diagonal of a square divides it into two congruent triangles.
∴ Area of triangle II = Area of triangle I
⇒ Area of triangle II = 256 cm2

Area of triangle III The triangle at the base is having sides as a = 8 cm, b = 6 cm, c = 6 cm
∵    Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

∴ Area of triangle III 

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

Thus, the area of different shades are: Area of shade I = 256 cm2,
Area of shade II = 256 cm2
Area of shade III = 17.92 cm2

 

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 figure). Find the cost of polishing the tiles at the rate of 50p per cm2.
 Solution: 
There are 16 equal triangular tiles.
Area of a triangle Sides of the triangle are a = 9 cm, b = 28 cm, c = 35 cm

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

         ∵          Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

=(72/2)cm = 36 cm

∴  Area of the triangle 

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

 Area of 16 triangles: Total area of all the triangles = 16 x 88.2 cm2 (approx.) = 1411.2 cm2 (approx.) Cost of polishing the tiles Rate of polishing = ₹ 0.5 per cm
∴ Cost of polishing all the tiles = ₹ 0.5 x 1411.2
= ₹(5/10) x (14112/10)= ₹705.60 (approx.)

 

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: 
The given field is in the form of a trapezium ABCD such that parallel sides AB = 10 m and DC = 25 m

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

Non-parallel sides are 13 cm and 14 cm.
We draw BE || AD, such that BE = 13 cm.
The given field is divided into two shapes:

(i) parallelogram ABED (ii) DBCE.
Area of ΔBCE: Sides of the triangle are a = 13 m b = 14 m c = 15 m

Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev

Let the height of the ΔBCE corresponding to the side 15 m be ‘h’ metres
∵ Area of a triangle = (1/2)x base x height
∴  (1/2)x 15 x h= 84
Ex 12.2 NCERT Solutions- Heron’s Formula Class 9 Notes | EduRev
x m = (56/5)m
Area of parallelogram ABED
∵ Area of a parallelogram = base x height
∴ Area of parallelogram ABED = 10 x (1/2)m= 2 x 56 m2 = 112 m2
Thus, area of the field = Area of ΔBCE + Area of parallelogram ABED
= 84 m2 + 112 m2 = 196 m2

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