Problem 1 :
Find the area of the unshaded region.
Solution :
<BDA = 90
AB2 = DA2 + DB2
AB2 = 122 + 162
AB2 = 144 + 256
AB2 = 400
AB = 20
Area of unshaded region = Area of ACB - Area of ADB
Area of ACB :
AC = 34 cm, BC = 42 cm and AB = 20
s = (a + b + c)/2
s = (34 + 42 + 20)/2
s = 48
s - a = 48 - 34 = 14, s - b = 48 - 42 = 6
and s - c = 48 - 20 = 28
= √s(s - a) (s - b) (s - c)
= √48 ⋅ 14 ⋅ 6 ⋅ 28
= 336 cm2
Area of ADB :
DB = 16 cm, DA = 12 cm and AB = 20
s = (16 + 12 + 20)/2
s = 48/2 = 24
s - a = 24 - 16 = 8, s - b = 24 - 12 = 12
and s - c = 24 - 20 = 4
= √24 ⋅ 8 ⋅ 12 ⋅ 4
= 96 cm2
Area of unshaded region = 336 - 96
= 240 cm2
Problem 2 :
Find the area of a quadrilateral ABCD whose sides are AB = 13cm, BC = 12cm, CD = 9cm, AD = 14cm and diagonal BD = 15cm.
Solution :
Area of quadrilateral ABCD = Area of ADB+Area of DBC
Area of ADB :
a = 13, b = 15, c = 14
s = (13 + 15 + 14)/2 = 21
= √21(21 - 13) (21 - 15) (21 - 14)
= √21 ⋅ 8 ⋅ 6 ⋅ 7
= 84 cm2
Area of BDC :
a = 12, b = 9, c = 15
s = (12 + 9 + 15)/2 = 18
= √18(18 - 12) (18 - 9) (18 - 15)
= √18 ⋅ 6 ⋅ 9 ⋅ 3
= 54 cm2
Area of quadrilateral = 84 + 54
= 138 cm2
Problem 3 :
A park is in the shape of a quadrilateral. The sides of the park are 15 m, 20 m, 26 m and 17 m and the angle between the first two sides is a right angle. Find the area of the park.
Solution :
In triangle ABC,
AC2 = AB2 + BC2
AC2 = 152 + 202
AC2 = 225 + 400
AC = √625 = 25
Area of quadrilateral ABCD = Area of ABC+Area of ADC
Area of ABC :
a = 15, b = 20, c = 25
s = (15 + 20 + 25)/2 = 30
= √30(30 - 15) (30 - 20) (30 - 25)
= √30 ⋅ 15 ⋅ 10 ⋅ 5
= 150 cm2
Area of ADC :
a = 17, b = 26, c = 25
s = (17 + 26 + 25)/2 = 34
= √34(34 - 17) (34 - 26) (34 - 25)
= √34 ⋅ 17 ⋅ 8 ⋅ 9
= 204 cm2
Area of quadrilateral = 150 + 204
= 354 cm2
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