Question

given triangle
1.) Find the area of the quadrilateral whose vertices, taken in order, are (-4,-2), (-3,-5)
(3,-2) and (2, 3).
:
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Answer 1

Step-by-step explanation:

let A =( -4 , -2) B = (-3 , -5) C= (3,-2) D = (2,3)

now in triangle ABD

area = 1/2 { 2*(-2 + 5) - 4(-5 -3)-3(3 + 2)}

1/2 { 2*3 4* (-8) 3 * 5 )

1/2{ 6 + 32 - 15}

23/2

therefore area of triangle ABD = 23/2 unit -1 square now area of triangle BCD area = 1/2{ -3(-2 -3) + 3(3 + 5) + 2(-5 +2)}

1/2 {-3 * -5 + 3 * 8 + 2 * (- 3)}

1/2 {15 + 24 - 6}

33/2

therefore area of triangle BCD = 33/2 unit square--------- 2

adding 1 & 2

23/2 + 33/2

( 23 + 33 )/2

56 / 2

28

therefore totall area = 28 unit square

Answer 2

$\huge\mathfrak\green{Answer:}$

Given:

• We have been given that the vertices of a quadrilateral ABCD are A(-4,-2), B(-3,-5) C(3,-2) and D(2, 3).

To Find:

• We need to find the area of the quadrilateral.

Solution:

The given vertices of the quadrilateral are A(-4,-2), B(-3,-5), C(3,-2) and D(2, 3).

Now, area of △ABC

$\sf{ = \dfrac{1}{2} [ x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]}$

We have,

x1 = -4

y1 = -2

x2 = -3

y2 = -5

x3 = 3

y3 = -2

Now, substituting the values, we have

$\sf{ \dfrac{1}{2} [- 4( - 5 + 2) + +( - 3)( - 2 + 2) + 3( - 2 + 5)] }$

$\sf \implies{ \dfrac{1}{2} ( - 4 \times 3) + ( - 3 \times 0) + (3 \times 3))}$

$\sf \implies{ \frac{1}{2} [(12 + 0 + 9)]}$

$\sf \implies{ \dfrac{1}{2} \times 21}$

$\sf{ = 10.5 units}$

Now, area of △ADC

$\sf{ = \dfrac{1}{2} [x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]}$

Here, we have

x1 = -4

y1 = -2

x2 = 3

y2 = -2

x3 = 2

y3 = 3

Substututing the values, we have

$\sf \implies{ \dfrac{1}{2} [( - 4)( - 2 - 3) + 3(3 + 2) + 2( - 2 + 2)]}$

$\sf \implies{ \dfrac{1}{2} [- 4 \times ( - 5)) + (3 \times 5) + (2 \times 0)]}$

$\sf \implies{ \dfrac{1}{2} [20 + 15 + 0]}$

$\sf \implies{ \dfrac{1}{2} \times 35}$

$\sf{ = 17.5 \: units}$

Now, area of quadrilateral ABCD is:

Area of △ABC + Area of △ADC

= 10.5 + 17.5

= 28 units

Hence, the area of quadrilateral ABCD is 28 sq. units.