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

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

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

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

AMC-8(2018) Problem 24

Pre College Mathematics

## Try with Hints

EJCI is a rhombus by symmetry

Can you now finish the problem ……….

Area of rhombus is half product of its diagonals….

can you finish the problem……..

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$R^2=\frac{3}{2}$