What is the largest possible value x could take given that it must be an integer? x < 5
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The wording is unclear without a diagram. As such, there are two possible cases and two possible answers.
Case 1: Diagonal
i<span><span>s formed by connecting the vertices formed by the meeting points of a 25-inch side and a 29-inch side. </span>
</span>Call the intersection point of and 
E. bisects , so 
. Since the diagonals of a kite are perpendicular to each other, 
and 
are both right triangles. One has a hypotenuse of 
, and the other has a hypotenuse of 
, but both share a leg of 
. Using the Pythagorean Theorem, we can get that the length of the other leg in the triangle with a hypotenuse of is 
. Similarly, for the triangle with a hypotenuse of , the other leg has a length of 
. Together, these legs make up , meaning 
, our final answer.
Case 2: Diagonal i<span>s formed by connecting the vertices formed by the meeting points of the sides with equal lengths.
</span>Call the intersection point of and E. We will focus on two triangles, namely 
and 
. Since diagonals intersect perpendicularly, these triangles are right triangles. One of them has a hypotenuse of , and the other has a hypotenuse of . They both share a leg that is half of because bisects . Let 
and the non-shared leg of the right triangle with a hypotenuse of equal 
. Since 
, the non-shared leg of the other right triangle (the one with a hypotenuse of ) has a length of 
. Using the Pythagorean Theorem, we can get the equations 
and 
. These can simplify to 
and 
. Isolating the term 
, we can get 
and 
. The latter can simplify to 
. Using substitution, we can combine the two equations into one and get 
. We can simplify that to 
, meaning 
. However, we are looking for 
(
is only half of ). We can solve for using the Pythagorean Theorem and the triangle with a hypotenuse of and a leg of 
. We get 
, meaning 
, our final answer.
Case 1: Diagonal
</span>Call the intersection point of
Case 2: Diagonal
</span>Call the intersection point of