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Explore the Back-StoryTry this beautiful Problem on Geometry from Area of rectangle from (AMC 10 A, 2012).

Let points $A=(0,0,0), B=(1,0,0), C=(0,2,0),$ and $D=(0,0,3)$. Points $E, F, G,$ and $H$ are midpoints of line segments $\overline{B D}, \overline{A B}, \overline{A C},$ and $\overline{D C}$ respectively. What is the area of rectangle $E F G H ?$

,

- $\sqrt{2}$
- $\frac{2 \sqrt{5}}{3}$
- $\frac{3 \sqrt{5}}{4}$
- $\sqrt{3}$
- $\frac{2 \sqrt{7}}{3}$

Tetrahedron

Area of rectangle

Co -ordinate geometry

Pre College Mathematics

AMC-10A, 2012, Problem 21

$\frac{3 \sqrt{5}}{4}$

We have to find out the area of the rectangle \(EFGH\). so we have to compute the co-ordinate of the points \(E\), \(F\), \(G\), \(H\) . Next we have to find out the length of the sides \(EF\), \(FG\) , \(GH\), \(EH\). Next since rectangle area will be \(EF\) \(\times FG\)

Can you solve the problem?

Now co-ordinates of the points are $E(0.5,0,1.5), F(0.5,0,0), G(0,1,0), H(0,1,1.5)$. The vector $E F$ is (0,0,-1.5) , while the vector $H G$ is also (0,0,-1.5) , meaning the two sides $E F$ and $G H$ are parallel. Similarly, the vector $F G$ is (-0.5,1,0) , while the vector $E H$ is also (-0.5,1,0) . Again, these are equal in both magnitude and direction, so $F G$ and $E H$ are parallel. Thus, figure $E F G H$ is a parallelogram.The area of the Hexagon = $(6)(0.5)(2 \sqrt{3})=6 \sqrt{3}$. But we have to find out the area area enclosed by the curve. Then add the areas of the three sectors outside the hexaqon and subtract the areas of the three sectors inside the hexagon but outside the fiqure to get the area enclosed in the curved figure

Can you solve the problem?

Taking the dot product of vector $E F$ and vector $F G$ gives $0 \cdot-0.5+0 \cdot 1+-1.5 \cdot 0=0,$ which means the two vectors are perpendicular. (Alternately, as above, note that vector $E F$ goes directly down on the z-axis, while vector $F G$ has no z-component and lie completely in the xy plane.) Thus, the figure is a parallelogram with a right angle, which makes it a rectangle.

Using the distance formula we get \(E F=\frac{3}{2} \text { and } F G=\frac{\sqrt{5}}{2}\)

Therefore area of the rectangle \(EFGH\)=\(EF \times GH\)=$\frac{3}{2} \cdot \frac{\sqrt{5}}{2}$=$\frac{3 \sqrt{5}}{4}$

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=k3eXbgwcNRw

Try this beautiful Problem on Geometry from Area of rectangle from (AMC 10 A, 2012).

Let points $A=(0,0,0), B=(1,0,0), C=(0,2,0),$ and $D=(0,0,3)$. Points $E, F, G,$ and $H$ are midpoints of line segments $\overline{B D}, \overline{A B}, \overline{A C},$ and $\overline{D C}$ respectively. What is the area of rectangle $E F G H ?$

,

- $\sqrt{2}$
- $\frac{2 \sqrt{5}}{3}$
- $\frac{3 \sqrt{5}}{4}$
- $\sqrt{3}$
- $\frac{2 \sqrt{7}}{3}$

Tetrahedron

Area of rectangle

Co -ordinate geometry

Pre College Mathematics

AMC-10A, 2012, Problem 21

$\frac{3 \sqrt{5}}{4}$

We have to find out the area of the rectangle \(EFGH\). so we have to compute the co-ordinate of the points \(E\), \(F\), \(G\), \(H\) . Next we have to find out the length of the sides \(EF\), \(FG\) , \(GH\), \(EH\). Next since rectangle area will be \(EF\) \(\times FG\)

Can you solve the problem?

Now co-ordinates of the points are $E(0.5,0,1.5), F(0.5,0,0), G(0,1,0), H(0,1,1.5)$. The vector $E F$ is (0,0,-1.5) , while the vector $H G$ is also (0,0,-1.5) , meaning the two sides $E F$ and $G H$ are parallel. Similarly, the vector $F G$ is (-0.5,1,0) , while the vector $E H$ is also (-0.5,1,0) . Again, these are equal in both magnitude and direction, so $F G$ and $E H$ are parallel. Thus, figure $E F G H$ is a parallelogram.The area of the Hexagon = $(6)(0.5)(2 \sqrt{3})=6 \sqrt{3}$. But we have to find out the area area enclosed by the curve. Then add the areas of the three sectors outside the hexaqon and subtract the areas of the three sectors inside the hexagon but outside the fiqure to get the area enclosed in the curved figure

Can you solve the problem?

Taking the dot product of vector $E F$ and vector $F G$ gives $0 \cdot-0.5+0 \cdot 1+-1.5 \cdot 0=0,$ which means the two vectors are perpendicular. (Alternately, as above, note that vector $E F$ goes directly down on the z-axis, while vector $F G$ has no z-component and lie completely in the xy plane.) Thus, the figure is a parallelogram with a right angle, which makes it a rectangle.

Using the distance formula we get \(E F=\frac{3}{2} \text { and } F G=\frac{\sqrt{5}}{2}\)

Therefore area of the rectangle \(EFGH\)=\(EF \times GH\)=$\frac{3}{2} \cdot \frac{\sqrt{5}}{2}$=$\frac{3 \sqrt{5}}{4}$

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=k3eXbgwcNRw

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