Question

what is the area of a quadrilateral ABCD in which AB=3cm,BC=4cm,CD=6cm,DA=5cm and diagonal AC =5cm​

Answer 1

Area of quadrilateral ABCD = area of triangle ABC + area of triangle ADC

                                             

Area of triangle ABC = 3 + 4+ 5     = 6 cm^2

                                             2

Area of triangle ADC = 5 + 5 + 6   = 8 cm ^2

                                            2

Area of quadrilateral = 6 + 8 = 14 cm^2.

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Answer 2

Answer:

Dear................................. !

Here is your answer................... √

Hope you wil find it

helpful.................. !

_____________________

[ The above given picture is the

image of the quadrilateral. ]

‡ ABCD is a quadrilateral in which :-

AB = 3 cm BC = 4 cm

CD = 6 cm DA = 5 cm,

‡ Area of quadrilateral :-

= Area of ∆ABC + Area of ∆ADC

Area of ∆ABC •``•

By Heron's Formula :-

= √ s( s - a ) ( s - b ) ( s - c )

{ where S is semi perimeter }

S = a + b + c

2

S = 5 + 3 + 4

2

S = 12

2

S = 6

‡ Now, we have to find the area of ∆ABC

= √ s( s - a ) ( s - b ) ( s - c )

= √ 6( 6 - 5 ) ( 6 - 3 ) ( 6 - 4 )

= √ 6 × 1 × 3 × 2

= √ 36

= 6 cm^2

Area of ∆ADC • `` •

= √ s( s - a ) ( s - b ) ( s - c )

S = a + + c

2

S = 5 + 4 + 5

2

S = 14

2

S = 7

‡ Now, we have to find the area of ∆ADC

= √ s( s - a ) ( s - b ) ( s - c )

= √ 7( 7 - 5 ) ( 7 - 4 ) (7 - 5 )

= √ 7 × 2 × 3 × 2

= 2 × √ 21

= 2 × 4.58

= 9.16 cm^2

Area of quadrilateral ABCD ♦♦♦♦♦

= Area of ∆ABC + Area of ∆ADC

= 6 cm^2 + 9.16 cm^2

= 15.16 cm^2 {answer}