A) Parallelogram
B) Rhombus
C) Rectangle
D) Square
Correct Answer: A
Solution :
| [a] Let the points are A, B, C and D respectively |
| Mid-point of AC is |
| \[\left( \frac{4-1}{2},\frac{7-2}{2},\frac{8+1}{2} \right)\] or \[\left( \frac{3}{2},\frac{5}{2},\frac{9}{2} \right)\]. |
| Mid point of BD is |
| \[\left( \frac{2+1}{2},\frac{3+2}{2},\frac{4+5}{2} \right)\] or \[\left( \frac{3}{2},\frac{5}{2},\frac{9}{2} \right)\]. |
| Thus AC and BD bisect each other. Further, |
| \[AC=\sqrt{{{(4+1)}^{2}}+{{(7+2)}^{2}}+{{(8-1)}^{2}}}\] |
| \[=\sqrt{25+81+49}=\sqrt{155}\] |
| \[BD=\sqrt{{{(2-1)}^{2}}+{{(3-2)}^{2}}+{{(4-5)}^{2}}}\] |
| \[=\sqrt{1+1+1}=\sqrt{3}\] |
| \[\therefore AC\ne BD\]. |
| Hence, ABCD represents a parallelogram. |
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