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If the lengths of diagonals DF, AG and CE of the cube shown in the adjoining figure are equal to the three sides of a triangle, then the radius of the circle circumscribing that triangle will be

  1. equal to the sides of the cube
  2. $\sqrt{3}$ times the sides of the cube
  3. $\frac{1} {\sqrt{3} }$ times the sides of the cube
  4. impossible to find from the given information
asked in Quantitative Aptitude by (8.1k points) 122 255 439 | 322 views

2 Answers

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Let a be the side of the cube. 

Diagonal = √3*a

since DF= AG = CE 

the triangle will be an equilateral triangle with side=√3 *a

According to the property of equilateral triangle, altitude is √3/2 times hypotenuse of the triangle (hypotenuse is side of equilateral triangle in this case)

Median itself is the altitude in equilateral trinagle. Centroid divides the median in the ratio 2:1.

now, the radius of the circle is the distance between a vertex to centroid of the equilateral triangle

so radius is 2/3 time altitude or median of the triangle

Radius = 2/3 * √3/2 *√3 * a = a  which is side of triangle. option A is the answer

answered by (108 points) 5
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diagonal of cube = √3*a

now since side of triangle is equal to diagonal of cube ; therefore side of triangle =√3*a;

we have to now find radius of circumcircle

circumcircle is the circle in which triangle is inscribed and its radius is a/√3;

incircle is the circle which is inscribed in a triangle  and its radius is a/(√3*2);

so, here rad of circumcircle is √3*a/√3;

hence it is equal to a; ehich is the side of triangle;

hence option A.

answered by (26 points)

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