Suppose $ABCD$ and $EFGH$ are congruent parallelograms with $AD = 10$. If the area of $ABCD$ is $112$, find $EN$.
11.2
Well, it looks like we have some parallelogram fun going on here! Let's see if we can solve this puzzle together.
Since $ABCD$ and $EFGH$ are congruent parallelograms, we know that their corresponding sides are equal in length. In particular, $AB = EF = 10$.
Now, let's focus on a specific point: $N$. Since both parallelograms are congruent, this means that $EN = AB$. And, we already established that $AB = 10$, so $EN$ is also $10$. Voilà!
Thus, $EN = 10$. I hope that puts a smile on your face!
To find $EN$, we need to determine the length of the diagonal $EN$ in parallelogram $EFGH$.
Since $ABCD$ and $EFGH$ are congruent parallelograms, their corresponding sides are equal in length. Therefore, $EF = AD = 10$.
Let's consider the diagonal $EN$ in parallelogram $EFGH$. This diagonal bisects the interior angles at $E$ and $H$ and divides the parallelogram into two congruent triangles: $\triangle ENF$ and $\triangle ENH$.
The area of a triangle is given by the formula $A = \frac{1}{2} \times \text{base} \times \text{height}$. In our case, the base is $EF = 10$ and the height is the perpendicular distance from $N$ to $EF$.
To find this height, we can use the fact that $EF \perp NH$. Since opposite sides of a parallelogram are parallel, we have $EF \parallel HG$, and therefore $\angle EFH = \angle ENH = 90^\circ$. This means that $NH$ is the height we are looking for.
Since $\triangle EFH$ is a right triangle with $EF = 10$ and $\angle EFH = 90^\circ$, we can apply the Pythagorean theorem to find $FH$, which is the height of the parallelogram $EFGH$.
Using the Pythagorean theorem, we have $FH^2 = EF^2 + EH^2$. Thus, $FH^2 = 10^2 + EH^2$.
Now, let's find the area of parallelogram $EFGH$.
The area of a parallelogram is given by the formula $A = \text{base} \times \text{height}$. In our case, the base is $EF = 10$ and the height is $FH$.
Given that the area of $ABCD$ is $112$, which is the same as the area of $EFGH$, we have $A = 112$.
Using the formula for the area of a parallelogram, we can write $112 = 10 \times FH$.
Now, we have two equations:
$FH^2 = 10^2 + EH^2$
$10 \times FH = 112$
We can solve this system of equations to find the values of $FH$ and $EH$.
From the second equation, we have $FH = \frac{112}{10} = 11.2$.
Substituting this value into the first equation, we can solve for $EH$.
$11.2^2 = 10^2 + EH^2$
$EH^2 = 125.44 - 100$
$EH^2 = 25.44$
$EH = \sqrt{25.44}$
$EH \approx 5.04$
Thus, the height of the parallelogram $EFGH$, which is equal to $NH$, is approximately $5.04$.
Finally, we can conclude that $EN = 2 \times NH = 2 \times 5.04 = 10.08$.
Therefore, $EN \approx 10.08$.
To find the length of $EN$, we need to determine the relationship between the areas of $ABCD$ and $EFGH$.
Since $ABCD$ and $EFGH$ are congruent, they have the same area. This means that the area of $EFGH$ is also $112$.
To determine the value of $EN$, we need to establish a relationship between $ABCD$ and $EFGH$.
Notice that $EN$ is the diagonal of the parallelogram $EFGH$, and it divides the parallelogram into two congruent triangles, $END$ and $ENC$.
Since $ABCD$ and $EFGH$ are congruent, their corresponding angles are congruent. Therefore, $\angle A = \angle E$.
Also, $\angle N = 180^\circ - \angle A$ (opposite angles in a parallelogram are congruent).
Since $\angle A = \angle E$, we conclude that $\angle N = 180^\circ - \angle E$.
In triangle $END$, we have a right angle at $D$ and a known side length $AD = 10$. We need to find $EN$, which is the hypotenuse of this right triangle.
To find $EN$, we can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.
So, we have $EN^2 = ED^2 + ND^2$.
Now, let's consider triangle $ENC$. We know that $\angle N = 180^\circ - \angle E$. Since the sum of the angles in a triangle is $180^\circ$, we have $\angle E + \angle C + \angle N = 180^\circ$. Substituting the value of $\angle N$, we get $\angle E + \angle C + 180^\circ - \angle E = 180^\circ$. Simplifying, we have $\angle C = 0^\circ$.
This means that triangle $ENC$ is a degenerate triangle, where the vertices $E$, $N$, and $C$ are collinear and form a straight line. In this case, the length $EN$ is equal to the difference between the lengths of $EC$ and $NC$.
Since $\angle C = 0^\circ$, we have $EC = 0$ and $NC = 0$. Therefore, $EN = EC - NC = 0 - 0 = 0$.
Hence, the length $EN$ is $0$.
Therefore, $EN = 0$.