question_answer

If \[A\,(-\,5,7),\]\[B\,(-\,4,\,\,-5),\]\[C\,(-1,-\,6)\]and \[D\,(4,5)\] are the vertices of a quadrilateral, then find the area of the quadrilateral ABCD.

A) 72 sq units        

B) 80 sq units

C) 90 sq units                    

D) 92 sq units

Correct Answer: A

Solution :

By joining B and D, we get two triangles \[\Delta ABD\] and \[\Delta BCD.\]

Given, \[{{x}_{1}}=-\,\,5,\]\[={{x}_{2}}=-\,\,4,\]\[{{x}_{3}}=4,\]\[{{y}_{1}}=7\]

\[{{y}_{2}}=-\,\,5\]and \[{{y}_{3}}=5\]

Area of

\[\Delta ABD=\left| \frac{1}{2}[{{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}})]\, \right|\]\[=\frac{1}{2}[-\,\,5\,\,(-\,5-5)+(-\,\,4)(5-7)+4\,\,(7+5)]\]

\[=\frac{1}{2}[50+8+48]=\frac{106}{2}=53\,\,\text{sq}\,\,\text{units}\]

Now, in \[\Delta BCD\]

\[{{x}_{1}}=-\,\,4,\]\[{{x}_{2}}=-\,\,1,\]\[{{x}_{3}}=4,\]

\[{{y}_{1}}=\,\,-5,\]\[{{y}_{2}}=-\,\,6\]and \[{{y}_{3}}=5\]

Area of

\[\Delta BCD=\left| \frac{1}{2}[-\,\,4(-\,\,6-5)-1\,\,(5+5)+4\,\,(-5+6)]| \right.\]

\[=\frac{1}{2}(44-10+4)\]

\[=\frac{1}{2}\times 38=19\]sq units

\[\therefore \]Area of quadrilateral \[\Delta BCD\]

= Area of \[\Delta ABD\] + Area of \[\Delta BCD\]

\[=53+19=72\] sq units