FINDING AREA OF QUADRILATERAL PARALLELOGRAM AND RHOMBUS WORD PROBLEMS

Problem 1 :

Find the area of the unshaded region.

Solution :

<BDA  =  90

AB =  DA2 + DB2

AB =  122 + 162

AB =  144 + 256

AB =  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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