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 |
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