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3 years ago

Show work thanks please help me

Mathematics

1 answer:

astra-53 [7]3 years ago

4 0

Answer:

∠ JKL = 32°

Step-by-step explanation:

Since KN bisects ∠ LKM , then

∠ LKN = ∠ NKM = 35°

Thus

∠ JKL = ∠ JKM - ∠ LKN - ∠ NKM

= 102° - 35° - 35°

= 102° - 70°

= 32°

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3 years ago

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 tw

Tresset [83]

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.

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4 years ago

Read 2 more answers

Find x. Leave answers in simplified radical form. (Step by step)

just olya [345]

Answer:

$\sf x = 4\sqrt{2}$