Try this beautiful problem from Geometry: Ratio of the area of cube’s cross section . You may use sequential hints to solve the problem.

## Area of cube’s cross section – AMC-8, 2018 – Problem 24

In the cube ABCDEFGH with opposite vertices C and E ,J and I are the mid points of segments FB and HD respectively .Let R be the ratio of the area of the cross section EJCI to the area of one of the faces of the cube .what is $R^2$ ?

- $\frac{5}{4}$
- $\frac{3}{2}$
- $\frac{4}{3}$

**Key Concepts**

Geometry

Area

Pythagorean theorem

## Check the Answer

But try the problem first…

Answer:$\frac{3}{2}$

AMC-8(2018) Problem 24

Pre College Mathematics

## Try with Hints

First hint

EJCI is a rhombus by symmetry

Can you now finish the problem ……….

Second Hint

Area of rhombus is half product of its diagonals….

can you finish the problem……..

Final Step

Let Side length of a cube be x.

then by the pythagorean theorem$ EC=X \sqrt {3}$

$JI =X \sqrt {2}$

Now the area of the rhombus is half product of its diagonals

therefore the area of the cross section is $\frac {1}{2} \times (EC \times JI)=\frac{1}{2}(x\sqrt3 \times x\sqrt2)=\frac {x^2\sqrt6}{2}$

This shows that $R= \frac{\sqrt6}{2}$

i.e$ R^2=\frac{3}{2}$