Respuesta :
The wording is unclear without a diagram. As such, there are two possible cases and two possible answers.
Case 1: Diagonal [tex]\overline{DB}[/tex] 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 [tex]\overline{DB}[/tex] and [tex]\overline{AC}[/tex] E. [tex]\overline{AC}[/tex] bisects [tex]\overline{DB}[/tex], so [tex]DE=BE=20\text{ inches}[/tex]. Since the diagonals of a kite are perpendicular to each other, [tex]\triangle AED[/tex] and [tex]\triangle CED[/tex] are both right triangles. One has a hypotenuse of [tex]29[/tex], and the other has a hypotenuse of [tex]25[/tex], but both share a leg of [tex]20[/tex]. Using the Pythagorean Theorem, we can get that the length of the other leg in the triangle with a hypotenuse of [tex]29[/tex] is [tex]21[/tex]. Similarly, for the triangle with a hypotenuse of [tex]25[/tex], the other leg has a length of [tex]15[/tex]. Together, these legs make up [tex]\overline{AC}[/tex], meaning [tex]AC=21+15=36 \text{ inches}[/tex], our final answer.
Case 2: Diagonal [tex]\overline{DB}[/tex] is formed by connecting the vertices formed by the meeting points of the sides with equal lengths.
Call the intersection point of [tex]\overline{DB}[/tex] and [tex]\overline{AC}[/tex] E. We will focus on two triangles, namely [tex]\triangle ADE[/tex] and [tex]\triangle ABE[/tex]. Since diagonals intersect perpendicularly, these triangles are right triangles. One of them has a hypotenuse of [tex]29[/tex], and the other has a hypotenuse of [tex]25[/tex]. They both share a leg that is half of [tex]\overline{AC}[/tex] because [tex]\overline{DB}[/tex] bisects [tex]\overline{AC}[/tex]. Let [tex]AE=y[/tex] and the non-shared leg of the right triangle with a hypotenuse of [tex]29[/tex] equal [tex]x[/tex]. Since [tex]DB=40[/tex], the non-shared leg of the other right triangle (the one with a hypotenuse of [tex]25[/tex]) has a length of [tex]40-x[/tex]. Using the Pythagorean Theorem, we can get the equations [tex]x^2+y^2=29^2[/tex] and [tex](40-x)^2+y^2=25^2[/tex]. These can simplify to [tex]x^2+y^2=841[/tex] and [tex]1600-80x+x^2 + y^2=625[/tex]. Isolating the term [tex]y^2[/tex], we can get [tex]y^2=841-x^2[/tex] and [tex]y^2=625-x^2+80x-1600[/tex]. The latter can simplify to [tex]y^2=-975-x^2+80x[/tex]. Using substitution, we can combine the two equations into one and get [tex]841-x^2=-975-x^2+80x[/tex]. We can simplify that to [tex]80x=1816[/tex], meaning [tex]x=22.7[/tex]. However, we are looking for [tex]2y[/tex] ([tex]y[/tex] is only half of [tex]\overline{AC}[/tex]). We can solve for [tex]y[/tex] using the Pythagorean Theorem and the triangle with a hypotenuse of [tex]29[/tex] and a leg of [tex]22.7[/tex]. We get [tex]y \approx 18.05[/tex], meaning [tex]\overline{AC}=2y \approx 36.1 \text{ inches}[/tex], our final answer.
Case 1: Diagonal [tex]\overline{DB}[/tex] 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 [tex]\overline{DB}[/tex] and [tex]\overline{AC}[/tex] E. [tex]\overline{AC}[/tex] bisects [tex]\overline{DB}[/tex], so [tex]DE=BE=20\text{ inches}[/tex]. Since the diagonals of a kite are perpendicular to each other, [tex]\triangle AED[/tex] and [tex]\triangle CED[/tex] are both right triangles. One has a hypotenuse of [tex]29[/tex], and the other has a hypotenuse of [tex]25[/tex], but both share a leg of [tex]20[/tex]. Using the Pythagorean Theorem, we can get that the length of the other leg in the triangle with a hypotenuse of [tex]29[/tex] is [tex]21[/tex]. Similarly, for the triangle with a hypotenuse of [tex]25[/tex], the other leg has a length of [tex]15[/tex]. Together, these legs make up [tex]\overline{AC}[/tex], meaning [tex]AC=21+15=36 \text{ inches}[/tex], our final answer.
Case 2: Diagonal [tex]\overline{DB}[/tex] is formed by connecting the vertices formed by the meeting points of the sides with equal lengths.
Call the intersection point of [tex]\overline{DB}[/tex] and [tex]\overline{AC}[/tex] E. We will focus on two triangles, namely [tex]\triangle ADE[/tex] and [tex]\triangle ABE[/tex]. Since diagonals intersect perpendicularly, these triangles are right triangles. One of them has a hypotenuse of [tex]29[/tex], and the other has a hypotenuse of [tex]25[/tex]. They both share a leg that is half of [tex]\overline{AC}[/tex] because [tex]\overline{DB}[/tex] bisects [tex]\overline{AC}[/tex]. Let [tex]AE=y[/tex] and the non-shared leg of the right triangle with a hypotenuse of [tex]29[/tex] equal [tex]x[/tex]. Since [tex]DB=40[/tex], the non-shared leg of the other right triangle (the one with a hypotenuse of [tex]25[/tex]) has a length of [tex]40-x[/tex]. Using the Pythagorean Theorem, we can get the equations [tex]x^2+y^2=29^2[/tex] and [tex](40-x)^2+y^2=25^2[/tex]. These can simplify to [tex]x^2+y^2=841[/tex] and [tex]1600-80x+x^2 + y^2=625[/tex]. Isolating the term [tex]y^2[/tex], we can get [tex]y^2=841-x^2[/tex] and [tex]y^2=625-x^2+80x-1600[/tex]. The latter can simplify to [tex]y^2=-975-x^2+80x[/tex]. Using substitution, we can combine the two equations into one and get [tex]841-x^2=-975-x^2+80x[/tex]. We can simplify that to [tex]80x=1816[/tex], meaning [tex]x=22.7[/tex]. However, we are looking for [tex]2y[/tex] ([tex]y[/tex] is only half of [tex]\overline{AC}[/tex]). We can solve for [tex]y[/tex] using the Pythagorean Theorem and the triangle with a hypotenuse of [tex]29[/tex] and a leg of [tex]22.7[/tex]. We get [tex]y \approx 18.05[/tex], meaning [tex]\overline{AC}=2y \approx 36.1 \text{ inches}[/tex], our final answer.
Answer:the length of the diagonal AC is 36. I just did the assignment! hope this helps
Step-by-step explanation:
