Question

Kara is building a sandbox shaped like a kite for her nephew. The top two sides of the sandbox are 29 inches long. The bottom two sides are 25 inches long. The diagonal DB has a length of 40 inches. What is the length of the diagonal AC?

Answer 1

Answer:the length of the diagonal AC is 36. I just did the assignment! hope this helps

Step-by-step explanation:

Answer 2

The wording is unclear without a diagram. As such, there are two possible cases and two possible answers.

Case 1: Diagonal $\overline{DB}$ is formed by connecting the vertices formed by the meeting points of a 25-inch side and a 29-inch side. 
Call the intersection point of $\overline{DB}$ and $\overline{AC}$ E. $\overline{AC}$ bisects $\overline{DB}$, so $DE=BE=20\text{ inches}$. Since the diagonals of a kite are perpendicular to each other, $\triangle AED$ and $\triangle CED$ are both right triangles. One has a hypotenuse of $29$, and the other has a hypotenuse of $25$, but both share a leg of $20$. Using the Pythagorean Theorem, we can get that the length of the other leg in the triangle with a hypotenuse of $29$ is $21$. Similarly, for the triangle with a hypotenuse of $25$, the other leg has a length of $15$. Together, these legs make up $\overline{AC}$, meaning $AC=21+15=36 \text{ inches}$, our final answer.

Case 2: Diagonal $\overline{DB}$ is formed by connecting the vertices formed by the meeting points of the sides with equal lengths. 
Call the intersection point of $\overline{DB}$ and $\overline{AC}$ E. We will focus on two triangles, namely $\triangle ADE$ and $\triangle ABE$. Since diagonals intersect perpendicularly, these triangles are right triangles. One of them has a hypotenuse of $29$, and the other has a hypotenuse of $25$. They both share a leg that is half of $\overline{AC}$ because $\overline{DB}$ bisects $\overline{AC}$. Let $AE=y$ and the non-shared leg of the right triangle with a hypotenuse of $29$ equal $x$. Since $DB=40$, the non-shared leg of the other right triangle (the one with a hypotenuse of $25$) has a length of $40-x$. Using the Pythagorean Theorem, we can get the equations $x^2+y^2=29^2$ and $(40-x)^2+y^2=25^2$. These can simplify to $x^2+y^2=841$ and $1600-80x+x^2 + y^2=625$. Isolating the term $y^2$, we can get $y^2=841-x^2$ and $y^2=625-x^2+80x-1600$. The latter can simplify to $y^2=-975-x^2+80x$. Using substitution, we can combine the two equations into one and get $841-x^2=-975-x^2+80x$. We can simplify that to $80x=1816$, meaning $x=22.7$. However, we are looking for $2y$ ($y$ is only half of $\overline{AC}$). We can solve for $y$ using the Pythagorean Theorem and the triangle with a hypotenuse of $29$ and a leg of $22.7$. We get $y \approx 18.05$, meaning $\overline{AC}=2y \approx 36.1 \text{ inches}$, our final answer.