If \[A\,(-2,1,)\text{ }B\,(a,0,\text{ }C\text{ (}4,b)\] and \[D\text{ (}1,2)\] are the vertices of a parallelogram ABCD, find the values of a and b. Hence find the lengths of its sides. |
OR |
If \[A\,(-5,7),\text{ }B\,(-4,-5),\text{ }C\text{ (}-1,-6)\] and \[D\text{ (}4,5)\]are the vertices of a quadrilateral, find the area of the quadrilateral ABCD. |
Answer:
Given, ABCD is a parallelogram. Midpoint of \[AC=\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{1}}+{{y}_{2}}}{2} \right)\] \[=\left( \frac{-2+4}{2},\frac{1+b}{2} \right)\] \[=\left( \frac{2}{2},\frac{1+b}{2} \right)\] \[=\left( 1,\frac{1+b}{2} \right)\] Midpoint of \[BD=\left( \frac{x{{'}_{1}}+x{{'}_{2}}}{2},\frac{y{{'}_{1}}+y{{'}_{2}}}{2} \right)\] \[=\left( \frac{a+1}{2},\frac{0+2}{2} \right)\] \[=\left( \frac{a+1}{2},\frac{2}{2} \right)\] \[=\left( \frac{a+1}{2},1 \right)\] Since, diagonals of a parallelogram bisect each other, \[\therefore \] \[\left( 1,\frac{1+b}{2} \right)=\left( \frac{a+1}{2},1 \right)\] On comparing, we get \[\therefore \] \[\frac{a+1}{2}=1\] \[\frac{1+b}{2}=1\] \[\Rightarrow \] \[a+1=2\] \[\Rightarrow \] \[1+b=2\] \[\Rightarrow \] \[a=1\] \[\Rightarrow \] \[b=1\] Therefore, the coordinates of vertices of parallelogram ABCD are \[A\,(-2,1),\text{ }B\,(1,0),\,\,C\,(4,1)\]and \[D\,(1,2)\]Length of side \[AB=DC=\sqrt{{{(1+2)}^{2}}+{{(0-1)}^{2}}}\] \[=\sqrt{9+1}=\sqrt{10}\] units And, \[AD=BC=\sqrt{{{(1+2)}^{2}}+{{(2-1)}^{2}}}\] \[=\sqrt{9+1}=\sqrt{10}\] units OR Given ABCD is quadrilateral. By joining points A and C, the quadrilateral is divided into two triangles. Now, Area of quad. ABCD = Area of \[\Delta ABC\]+ Area of \[\Delta \,ACD\] Area of \[\Delta \text{ }ABC\] \[=\frac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]\] \[=\frac{1}{2}\left[ -5\left( -5+6 \right)-4\left( -6-7 \right)-1\left( 7+5 \right) \right]\] \[=\frac{1}{2}[-5(1)-4(-13)-1(12)]\] \[=\frac{1}{2}(-5+52-12)\] \[=\frac{1}{2}(35)=\frac{35}{2}\] sq. units. Area of \[\Delta \text{ }ADC\] \[=\frac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]\] \[=\frac{1}{2}[-5(5+6)+4(-6-7)+(-1)(7-5)]\] \[=\frac{1}{2}[-5(11)+4(-13)-1(2)]\] \[=\frac{1}{2}(-55+52-12)\] \[=\frac{1}{2}|-109|=\frac{109}{2}\] sq. units. Area of quadrilateral ABCD \[=\frac{35}{2}+\frac{109}{2}\] \[=\frac{144}{2}=72\] sq. units.
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