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ratelena [41]
3 years ago
8

Charlene puts together two isosceles triangles so that they share a base, creating a kite. The legs of the triangles are 10 inch

es and 17 inches, respectively. If the length of the base for both triangles is 16 inches long, what is the length of the kite’s other diagonal?
Mathematics
2 answers:
enot [183]3 years ago
7 0
From the information given you have:

1) Smaller diagonal of the kite: 16 inches


2) Larger diagonal of the kite: height of one triangle (h1) + height of the other triangle (h2)


3) Calculation of the height of the smaller triangle, h1:


10^2 = (16/2)^2 + (h1)^2 => h1 = √ [10^2 - 8^2] = 6


4) Calculation of the height of the larger triangle, h2


17^2 = (16/2)^2 + (h2)^2 => h2 = √[17^2  - 8^2] = 15


5) Larger diagonal = h1 + h2 = 6 + 15 = 21


Answer: 21 inches
kenny6666 [7]3 years ago
4 0

Answer:

21 inches

Step-by-step explanation:

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\textsf{Hey there!}

\textsf{Put these numbers in order from GREATEST to LEAST}

\bullet{\mathsf{ \ 1\dfrac{6}{8}}}

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\bullet{\textsf{ 1}}

\star\textsf{ We know that fractions, decimals, negatives are BELOW 0 (on the left} \textsf{of 0)}

\star\textsf{ \& we also know that positives and real numbers is/are ABOVE 0 (to the right} \textsf{of 0)}

\textsf{Now we can answer your question}

\blacksquare \ \mathsf{1\dfrac{6}{8} = \dfrac{14}{8} = 1.75}

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\blacksquare\ \mathsf{- \dfrac{18}{25} = -0.72}

\blacksquare \ \mathsf{1 = 1}

\textsf{Look carefully at the stuff converted to the decimal or regular number \& } \textsf{you will  have your answer from \underline{greatest} to \underline{least}}

\boxed{\textsf{If you solved correctly you  SHOULD have: \boxed{\mathsf{\dfrac{14}{8}, \ 1, \ - \dfrac{10}{25}\ ,\  -\dfrac{18}{25}}}}}\checkmark

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~\frak{LoveYourselfFirst:)}

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