Question

Find the area of the quadrilateral ABCD.​

Answer 1

$\sf\huge\bold{Answer:}$

Given :

A quadrilateral ABCD in cartesian plane

A(-3,3)

B(-2,-4)

C(4,-1)

D(3,4)

To find :

Area of ABCD

Construction :

Join BD to divide quadrilateral ABCD into two triangles ∆ ABD and ∆CBD

Solution :

From figure,

AB = (4+3) units = 7units $\:\:$[along y - axis]

BC = (2+4) units = 6units $\:\:$[along x - axis]

CD = (1+4) units = 5units $\:\:$[along y - axis]

DA = = (3+3) units = 6units $\:\:$[along x - axis]

To find BD

$\fbox{Distance\:Formula}$

$\boxed{\tt \red{d = \sqrt{(x_2 - x_1) {}^{2} + ( y_2 - y_1) {}^{2} } } }$

where,

$\tt\ BD = \sqrt{(3 - ( - 2) ){}^{2} + (4 - ( - 4)) {}^{2} }$

$\tt\ BD = \sqrt{(3 + 2) {}^{2} + (4 + 4) {}^{2} }$

$\tt\ BD = \sqrt{ {5}^{2} + {8} {}^{2} }$

$\tt\ BD = \sqrt{25 + 64}$

$\tt\ BD = \sqrt{89}$

$\tt\ BD≈9.4$

Now,

ar.(ABCD) = ar.(∆ABD) + ar.(∆CBD)

To find area of triangles using Heron's Formula

$\boxed{ \tt\red{ A = \sqrt{s(s - a)(s - b)(s - c)} }}$

• In ∆ABD

Semiperimeter

= Perimeter/2

= (AB+BD+AD)/2

= (7+9.4+6)/2

= 22.4/2

=11.2

Let AB = a = 7

BD = b = 9.4

AD = c = 6

Inserting values in Heron's Formula :

$\tt\ A = \sqrt{11.2(11.2 - 7)(11.2 - 9.4)(11.2 - 6)}$

$\tt \: A = \sqrt{11.2(4.2)(1.8)(5.2)}$

$\tt\ A = \sqrt{440.2944}$

$\tt\ A≈20.98 \: sq.units$

ar.∆ABD ≈ 20.98 sq.units

• In ∆CBD

Semiperimeter

= Perimeter/2

= (CB+BD+CD)/2

= (6+9.4+5)/2

= 20.4/2

=10.2

Let CB = a = 6

BD = b = 9.4

CD = c = 5

Inserting values in Heron's Formula :

$\tt\ A = \sqrt{10.2(10.2 - 6)(10.2 - 9.4)(10.2 - 5)}$

$\tt\ A = \sqrt{10.2(4.2)(0.8)(5.2)}$

$\tt\ A = \sqrt{178.2144}$

$\tt\ A≈13.35 \: sq. \: units$

ar.∆CBD ≈ 13.35 sq.units

Area of quadrilateral ABCD = ar.(∆ABD) + ar.(∆CBD)

Area of quadrilateral ABCD = 20.98 + 13.35 sq. units

Area of quadrilateral ABCD = 34.33 sq. units.

$\tt\ Area = \huge{\boxed{\purple{34.33}}} \: sq.units$

Answer 2

Answer:

Area =34.33 sq unit

Step-by-step explanation:

the answer is correct

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