Question

Find the area of quadrilateral ABCD whose vertices are A
(-4, -2), B (-3, -5), C (3, -2), D (2, 3).

Answer 1

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Step-by-step explanation:

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Answer 2

Given,

Vertices of quad. ABCD

• $\sf A(-4,-2)$
• $\sf B(-3,-5)$
• $\sf C(3,-2)$
• $\sf D(2,3)$

To find,

• Area of quadrilateral ABCD

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$\:$

Solution,

$\sf \:\:\:\:\:\:\:\:\:\:\:\:Join \:A\: and\: C.\: Then,\:\:\:\:\:\:\:\:\:\:\:\:[fig个]$

$\sf ar\: of \:quad. ABCD=$ $\sf ar(∆ABC)+ar(∆ACD)$

$\:\:$

${\bold { \underline{\normalsize{Area\: of \:∆ABC}}}}$

$\:\:$

$\bf\dfrac{1}{2}$ $\sf •|{(-4)•(-5+2)-3(-2+2)+3(+3)}|$

$\:\:$

$\:\:\:\:\:\:\:$$\leadsto$$\bf\dfrac{1}{2}$ $\sf •|(-4)•(-3)-3×0+3×3|$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$\leadsto$$\bf\dfrac{1}{2}$ $\sf •|12-0+9|$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$\leadsto$$\bf\dfrac{21}{2}$$\sf sq\:units.$

$\:\:$

${\bold { \underline{\normalsize{Area\: of \:∆ACD}}}}$

$\:\:$

$\bf\dfrac{1}{2}$$\sf •|{(-4)•(-2-3)+3(5)+2(-2+2)}|$

$\:\:$

$\:\:\:\:\:\:\:\:$$\leadsto$$\bf\dfrac{1}{2}$$\sf •|{(-4)•(-5)+3×5+2×0}|$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$\leadsto$$\bf\dfrac{1}{2}$$\sf •|20+15+0|$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$\leadsto$$\bf\dfrac{35}{2}$$\sf sq\:units.$

$\:\:$

$\therefore$ $\sf area\:of\:quad.ABCD=$$\bigg($ $\bf\dfrac{21}{2}$+$\bf\dfrac{35}{2}$$\bigg)$$\sf sq\:u$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$\underline{\boxed{\sf{=\frak{\purple{28\:sq\:units.}}}}}$