Question

Find the area of quadrilateral whose vertices are A(0,4), B(4,0),C(-4,0),D(0,4)? ​

Answer 1

Heya mate...

Answer:

Area of quadrilateral= ar.(∆ABC) + ar.(∆ADC).

Area of triangle ABC

= 1/2[x1(y2-y3) + x2(y3-y1) + x3(y1-y2).

=1/2[0(0-0) + 4(0-(-4) + (-4)(-4-0)

=1/2[0+ 4(4) + (-4)(-4)

=1/2(16+16)

=16 sq. units.

Area of triangle ACD

=|1/2[x1(y2-y3) + x2(y3-y1) + x3(y1-y2).|

=|1/2 [0(0-4) + (-4)(4-(-4) + 0(-4-0)|

=|1/2[0-4(4+4)+0]|

=|1/2×-32|

=16 sq.units.

Then the area of quadrilateralABCD = 16 +16 = 32 sq.units

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Ur question has some mistake, A(0,-4) is correct.

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i hope it helps you...✌️

.Thanks.

Answer 2

Answer:

Heya mate...

Area of quadrilateral = ar.(∆ABC) + ar.(∆ADC)

Area of triangle ABC,

=1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)

=1/2[0(0-0)+4(0-(-4)+(-4)(-4-0)

=1/2[0 +4(4)+(-4)(-4)

=1/2(16+16)

=16 sq.units

Area of triangle ADC,

=|1/2[0(0-4)+(-4)(4-(-4)+0(-4-0)]|

=|1/2[0-4(4+4)+0]|

=|1/2×-32|

=16 sq.units.

Then the area of quadrilateral ABCD= 16+16 = 32 sq. units.

___________________

Ur question has some silly mistake, A(0,-4) is correct.

I Hope It HelPs you♣️✌️