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# Problem on Cube | AMC 10A, 2008 | Problem 21

Try this beautiful problem from Geometry:Squarefrom AMC-10A (2008) You may use sequential hints to solve the problem.

Try this beautiful problem from Geometry: Problem on Cube.

## Problem on Cube – AMC-10A, 2008- Problem 21

A cube with side length 1 is sliced by a plane that passes through two diagonally opposite vertices $A$ and $C$ and the midpoints $B$ and $D$ of two opposite edges not containing $A$ or $C$, as shown. What is the area of quadrilateral $A B C D ?$

,

i

• $\frac{\sqrt{6}}{2}$
• $\frac{5}{4}$
• $\sqrt{2}$
• $\frac{5}{8}$
• $\frac{3}{4}$

Geometry

Square

Pythagoras

## Check the Answer

But try the problem first…

Answer: $\frac{\sqrt{6}}{2}$

Source

AMC-10A (2008) Problem 21

Pre College Mathematics

## Try with Hints

First hint

The above diagram is a cube and given that side length $1$ and $$B$$ and $$D$$ are the mid points .we have to find out area of the $$ABCD$$.Now since $A B=A D=C B=C D=\sqrt{\frac{1}{2}^{2}}+1^{2},$ it follows that $A B C D$ is a rhombus. can you find out area of the rhombus?

Can you now finish the problem ……….

Second Hint

The area of the rhombus can be computed by the formula $A = \frac 12 d_1d_2$, where $d_1,\,d_2$ are the diagonals of the rhombus (or of a kite in general). $BD$ has the same length as a face diagonal, or $\sqrt{1^{2}+1^{2}}=\sqrt{2} \cdot A C$ is a space diagonal, with length $\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}$

can you finish the problem……..

Final Step

Therefore area $A=\frac{1}{2} \times \sqrt{2} \times \sqrt{3}=\frac{\sqrt{6}}{2}$

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