Many fields are in the shape of quadrilaterals. Similarly, various surfaces are polygons. To find their areas, we divide the surface into triangular parts and then use Heron’s formula for finding areas of the triangles.
Q1. 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?
Ans: Let us join B and D, such that ΔBCD is a rightangled triangle.
We have area of ΔBCD = (1/2)x base x altitude
=(1/2)x 12 x 5 m^{2}
= 30 m^{2}
Now, to find the area of ΔABD, we need the length of BD
∴ In right ΔBCD, BD^{2} = BC^{2 }+ CD^{2} [Pythagoras theorem]
⇒ BD^{2} = 12^{2 }+ 5^{2}
⇒ BD^{2} = 144 + 25 = 169 = 13^{2 }
⇒ BD = = 13 m
Now, for ΔABD, we have a = 9 m, b = 8 m, c = 13 m
∵
∴ Area of ΔABD
∴ Area of quadrilateral ABCD = area of ΔBCD + area of ΔABD
= 30 m^{2 }+ 35.5 m^{2} = 65.5 m^{2} (appx)
Q2. 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.
Ans: For ΔABC,
a= 3 cm, b= 4cm and c = 5 cm
∴ Area of ΔABC,
For ΔACD,
a=5 cm, b = 4cm and c= 5cm
For ΔACD,
a=5 cm, b = 4cm and c= 5cm
∴ Area of ΔACD,
Now, area of quadrilateral ABCD = area of ΔABC + area of ΔACD
= 6 cm^{2} + 9.2 cm^{2} (approx.)
= 15.2 cm^{2 }(approx.)
Q3. Radha made a picture of an aeroplane with coloured paper as shown in Fig. Find the total area of the paper used.
Ans: Area of surface I It is an isosceles triangle whose sides are a = 5 cm, b = 5 cm, c = 1 cm
= 0.75 x 3.3 cm^{2} (approx.) = 2.475 cm^{2 }(approx.)
Area of surface II It is a rectangle with length of 6.5 cm and breadth of 1 cm.
∴ Area of rectangle II = Length x Breadth
= 6.5 x 1 = 6.5 cm^{2 }
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
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 =
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 cm^{2} = 4.5 cm^{2}
Area of surface V
∵ Right triangle V ≌ Right triangle IV
∴ Area of right triangle V = Area of right triangle IV = 4.5 cm^{2 }
Thus, the total area of the paper used = (area I) + (area II) + (area III) + (area IV) + (area V)
= [2.475 cm^{2} (approx.)] + [6.5 cm^{2}] + [1.3 cm^{2} (approx.)] + [4.5 cm^{2}] + [4.5 cm^{2}]
= 19.275 cm^{2 }= 19.3 cm^{2} (approx.)
Q4. 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.
Ans: For the given triangle, we have a = 28 cm, b = 30 cm, c = 26 cm
= 6 x 7 x 2 x 4 cm^{2 }= 336 cm^{2}
∵ [Area of the given parallelogram] = [Area of the given triangle]
∴ [Area of the parallelogram] = 336 cm^{2 }
∴ [Base of the parallelogram] x [Height of the parallelogram] = 336 cm^{2 }
⇒ [28 cm] x [h cm] = 336 cm^{2}
⇒ h= (336/28)cm = 12
Thus, the required height of the parallelogram = 12 cm
Q5. 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?
Ans: Here, each side of the rhombus = 30 m
One of the diagonal = 48 m
Since a diagonal divides the rhombus into two congruent triangles.
Sides of first triangle are a = 30 m, b = 30 m, c = 48 m
∵
∴ Area of triangle I =
= 3 x 6 x 4 x 6 m^{2} = 432 m^{2}
∴ Area of second triangle = 432 m^{2}
⇒ Total area of both triangles = 432 cm^{2} + 432 m^{2} = 864 m^{2}
⇒ The area of the rhombus = 864 m^{2}
Thus, area of grass for 18 cows = 864 m^{2 }
⇒ Area of grass for 1 cow = (864/18)m^{2} = 48 m^{2}
Q6. 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?
Ans: Sides of each triangular piece are
a = 20 cm, b = 50 cm, c = 50 cm
∴ Area of each triangular piece
⇒ Area of 5 triangular pieces of one colour = 5 x 200/√6cm^{2 }= 1000/√6 cm^{2} Area of 5 triangular pieces of other colour = 1000/√6 cm^{2}
Note: Total triangular pieces are 10, i.e. five triangular pieces for each colour.
Q7. 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?
Ans: Area of triangle I
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 cm^{2} = 256 cm^{2}
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 cm^{2}
Area of triangle III The triangle at the base is having sides as a = 8 cm, b = 6 cm, c = 6 cm
∵
∴ Area of triangle III
Thus, the area of different shades are: Area of shade I = 256 cm^{2},
Area of shade II = 256 cm^{2}
Area of shade III = 17.92 cm^{2}
Q8. 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 cm^{2}.
Ans: There are 16 equal triangular tiles.
Area of a triangle Sides of the triangle are a = 9 cm, b = 28 cm, c = 35 cm
∵
=(72/2)cm = 36 cm
∴ Area of the triangle
Area of 16 triangles: Total area of all the triangles = 16 x 88.2 cm^{2} (approx.) = 1411.2 cm^{2} (approx.) Cost of polishing the tiles Rate of polishing = ₹ 0.5 per cm^{2 }
∴ Cost of polishing all the tiles = ₹ 0.5 x 1411.2
= ₹(5/10) x (14112/10)= ₹705.60 (approx.)
Q9. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The nonparallel sides are 14 m and 13 m. Find the area of the field.
Ans: The given field is in the form of a trapezium ABCD such that parallel sides AB = 10 m and DC = 25 m
Nonparallel 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
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
⇒
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 }= 2 x 56 m^{2} = 112 m^{2}
Thus, area of the field = Area of ΔBCE + Area of parallelogram ABED
= 84 m^{2} + 112 m^{2} = 196 m^{2}
1402 docs679 tests

1. Can Heron's Formula be used to find the area of any quadrilateral? 
2. How do you determine if a quadrilateral is cyclic? 
3. Can Heron's Formula be used to find the area of a rectangle or square? 
4. Are there any other formulas that can be used to find the area of noncyclic quadrilaterals? 
5. How is Heron's Formula derived for finding the area of a triangle? 
1402 docs679 tests


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