Q1. Find the area of the quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
Solution:
For triangle ABC
AC^{2} = BC^{2}+AB^{2}
25 = 9 + 16
So, triangle ABC is a right angle triangle right angled at point R
Area of triangle ABC =
= 6 cm^{2}
From triangle CAD
Perimeter = 2s = AC + CD + DA
2s = 5 cm+ 4 cm+ 5 cm
2s = 14 cm
s = 7 cm
By using Heron’s Formula
Area of the triangle CAD
= 9.16cm2
Area of ABCD = Area of ABC + Area of CAD
= (6+9.16) cm^{2}
= 15.16cm^{2}
Q2. The sides of a quadrilateral field, taken in order are 26 m, 27 m, 7 m, 24 m respectively. The angle contained by the last two sides is a right angle. Find its area.
Solution:
Here the length of the sides of the quadrilateral is given as
AB = 26 m, BC = 27 m, CD = 7 m, DA = 24 m
Diagonal AC is joined.
Now, in triangle ADC
By applying Pythagoras theorem
AC^{2} = AD^{2}+CD^{2}
AC^{2} = 142+72
AC = 25 m
Now area of triangle ABC
Perimeter = 2s = AB + BC + CA
2s = 26 m + 27 m + 25 m
s = 39 m
By using Heron’s Formula
The area of a triangle
= 291.84m^{2}
Thus, the area of a triangle is 291.84m^{2}
Now for area of triangle ADC
Perimeter = 2s = AD + CD + AC
= 25 m + 24 m + 7 m
s = 28 m
By using Heron’s Formula
The area of a triangle
= 84m^{2}
Thus, the area of a triangle is 84m^{2}
Therefore, Area of rectangular field ABCD
= Area of triangle ABC + Area of triangle ADC
= 291.84 + 84
= 375.8 m^{2}
Q3. The sides of a quadrilateral, taken in order as 5 m, 12 m, 14 m, 15 m respectively, and the angle contained by first two sides is a right angle. Find its area.
Solution:
Given that the sides of the quadrilateral are
AB = 5 m, BC = 12 m, CD =14 m and DA = 15 m
Join AC
Now area of triangle ABC
= 30 m^{2}
In triangle ABC, By applying Pythagoras theorem
AC = 13 m
Now in triangle ADC,
Perimeter = 2s = AD + DC + AC
2s = 15 m +14 m +13 m
s = 21 m
By using Heron’s Formula,
Area of the triangle PSR =
= 84m^{2}
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ADC
= (30 + 84) m^{2}
= 114m^{2}
Q4. A park in the shape of a quadrilateral ABCD, has angle C = 90^{0}, AB = 9 m, BC = 12 m, CD = 5 m, AD = 8 m. How much area does it occupy?
Solution:
Given sides of a quadrilateral are AB = 9 m, BC = 12 m, CD = 5 m, DA = 8 m.
Let us join BD
In triangle BCD , apply Pythagoras theorem
BD^{2} = BC^{2}+CD^{2}
BD^{2} = 12^{2}+5^{2}
BD = 13 m
Area of triangle BCD =
= 30 m^{2}
Now, in triangle ABD
Perimeter = 2s = 9 m + 8m + 13m
s = 15 m
By using Heron’s Formula,
Area of the triangle ABD =
= 35.49m^{2}
Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD
= (35.496 + 30) m^{2}
= 65.5m^{2}.
Q5. Two parallel sides of a trapezium are 60 m and 77 m and the other sides are 25 m and 26 m. Find the area of the trapezium?
Solution:
Given,
Two parallel sides of trapezium are AB = 77 m and CD = 60 m
The other two parallel sides of trapezium are BC = 26 m, AD = 25m
Join AE and CF
DE is perpendicular to AB and also, CF is perpendicular to AB
Therefore, DC = EF = 60 m
Let AE = x
So, BF = 77 – 60 – x
BF = 17 – x
In triangle ADE,
By using Pythagoras theorem,
DE^{2} = AD2−AE^{2}
DE2 = 25^{2}−x^{2}
In triangle BCF,
By using Pythagoras theorem,
CF^{2} = BC^{2}−BF^{2}
CF^{2} = 26^{2}−(17−x)^{2}
Here, DE = CF
So, DE^{2} = CF^{2}
25^{2}−x^{2} = 26^{2}−(17−x)^{2}
252−x^{2} = 26^{2}−(17^{2}−34x+x^{2})
25^{2}−x^{2} = 26^{2}−17^{2}+34x+x^{2}
25^{2} = 26^{2}−17^{2}+34x
x = 7
DE = 24 m
Area of trapezium
Area of trapezium = 1644m^{2}
Q6. Find the area of a rhombus whose perimeter is 80 m and one of whose diagonal is 24 m.
Solution:
Given,
Perimeter of a rhombus = 80 m
As we know,
Perimeter of a rhombus = 4×side = 4×a
4×a = 80 m
a = 20 m
Let AC = 24 m
Therefore OA
OA = 12 m
In triangle AOB
OB^{2} = AB^{2}−OA^{2}
OB^{2} = 20^{2}−12^{2}
OB = 16 m
Also, OB = OD because diagonal of rhombus bisect each other at 90^{∘}
Therefore, BD = 2 OB = 2 x 16 = 32 m
Area of rhombus
Area of rhombus
Area of rhombus = 384 m^{2}
Q7. A rhombus sheet, whose perimeter is 32 m and whose diagonal is 10 m long, is painted on both the sides at the rate of Rs 5 per meter square. Find the cost of painting.
Solution:
Given that,
Perimeter of a rhombus = 32 m
We know that,
Perimeter of a rhombus = 4×side
4×side = 32 m
4×a = 32 m
a = 8 m
Let AC = 10 m
By using Pythagoras theorem
OB2 = AB^{2}–OA^{2}
OB^{2} = 8^{2}–5^{2}
Area of the sheet
Area of the sheet
Therefore, cost of printing on both sides at the rate of Rs. 5 per m^{2}
= Rs. 625
1. What is Heron's formula? 
2. How is Heron's formula derived? 
3. Can Heron's formula be used for all types of triangles? 
4. Are there any limitations to using Heron's formula? 
5. How is Heron's formula useful in reallife applications? 

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