A quadrilateral’s smallest side measures 29 mm and its largest side measures 38 mm. A similar quadrilateral’s largest side measures 95 mm. What is the measure of its smallest side?
Let $r$ be the ratio of side lengths between the two quadrilaterals. Dropping an altitude from a vertex of the larger quadrilateral to its corresponding side, we have the triangle: [asy]
pair A, B, C, D;
D=(0,0);
C=(11.55,0);
B=(8.12,5.36);
A=(2,4.5);
draw(A--B--C--D--A);
draw(A--foot(A,B,C));
label("$R$",(2,5));
label("$P$",foot(A,B,C));
[/asy]The right triangle with sides $P$ and $D$ has hypotenuse $\dfrac{95}{r}$, and the right triangle with sides $\dfrac{P}{r}$ and $D$ has hypotenuse $38$. Setting the expressions equal, we have $\dfrac{P}{r}=\dfrac{\dfrac{95}{r}}{38} \quad \to \quad 95r=38P$, so $P=\dfrac{95r}{38}$. Also, $\sqrt{\dfrac{P^2}{r^2}+D^2}=\sqrt{D^2+29^2}$. Substituting and simplifying, $\dfrac{P^2}{r^2}+D^2=D^2+29^2$, so $\dfrac{P^2}{r^2}=900$. Thus, if the largest length of the similar quadrilateral is $95$, the largest length of the original quadrilateral is $\dfrac{900}{38} \approx \boxed{23.6842}$.
can you dumb that down to an answer?
Yes, of course! The measure of the smallest side of the similar quadrilateral is $\boxed{23.6842}$ mm.
thats incorrect
Apologies for the mistake. Let's solve the problem correctly:
Let $r$ be the ratio of side lengths between the two quadrilaterals. We can set up the following proportion between the corresponding sides:
$\frac{38 \text{ mm}}{29 \text{ mm}} = \frac{95 \text{ mm}}{x}$.
Cross-multiplying gives $x \cdot 38 \text{ mm} = 29 \text{ mm} \cdot 95 \text{ mm}$.
Dividing both sides by 38 mm gives $x = \frac{29 \text{ mm} \cdot 95 \text{ mm}}{38 \text{ mm}} \approx \boxed{72.3684}$ mm.
Therefore, the measure of the smallest side of the similar quadrilateral is approximately 72.3684 mm.