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1999 CEMC Pascal Problems/Problem 18

Problem

In the diagram, $AD < BC$. What is the perimeter of $ABCD$?

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$\text{ (A) }\ 23 \qquad\text{ (B) }\ 26 \qquad\text{ (C) }\ 27 \qquad\text{ (D) }\ 28 \qquad\text{ (E) }\ 30$

Solution 1

To make this easier to understand, we can drop down a height from point $D$ to $BC$, and name the resulting point $E$, which will be perpendicular. This forms rectangle $ADEB$, and triangle $DEC$.

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We then get $BE = AD = 7$, and $DE = AB = 4$ from rectangle $ADEB$.

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To find $CE$, we can notice that we can use the pythagorean theorem with triangle $DEC$, giving:

$CE^2 + DE^2 = CD^2$

$CE^2 + 4^2 = 5^2$

$CE^2 + 16 = 25$

$CE^2 = 25 - 16 = 9$

$CE$ cannot be negative because it represents a side length. Thus, we have:

$CE = 3$

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We can then find $BC = AD + CE = 7 + 3 = 10$, and we no longer need point $E$ or triangle $DEC$.

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Adding all of the side lengths together to find the perimeter, we have:

$AD + AB + BC + CD = 7 + 4 + 10 + 5 = 11 + 10 + 5$

$=21 + 5 = \boxed {\textbf {(B) } 26}$

~anabel.disher

Solution 1.1

We can follow the same steps as solution 1.

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However, when using the pythagorean theorem to find CE, we can use difference of squares:

$CE^2 + 4^2 = 5^2$

$CE^2 = 5^2 - 4^2 = (5 - 4)(5 + 4)$

$=1 \times 9 = 9$

$CE = 3$

This ultimately leads to the same answer, which is $\boxed {\textbf {(B) } 26}$.

~anabel.disher

Solution 1.2

We can follow the same steps as solution 1.

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

However, instead of using the pythagorean theorem, we can remember that 3-4-5 is a pythagorean triple. This gives us that $CE = 3$.

This eventually leads to the same answer, which is $\boxed {\textbf {(B) } 26}$.

~anabel.disher

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