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}$
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