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horsena [70]
3 years ago
11

What is the inradius of a triangle with side lengths 9, 40, and 41?

Mathematics
1 answer:
Mars2501 [29]3 years ago
6 0

Answer: 20.5

Step-by-step explanation:

From the measure of the length of the sides of the triangle, 9, 40 and 41 we can infer that the triangle is a right angled triangle. 9-40-41 is a Pythagorean triplet.

In a right angled triangle, the circumradius is half the hypotenuse.

In the given triangle, the hypotenuse = 41.

Therefore, the circumradius is  41÷2= 20.5 units.

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Makovka662 [10]
14.86 is the answer to this question
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What is the area, in square units, of trapezoid AEDC shown below?
just olya [345]
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2 years ago
If you can answer this u are the best
spayn [35]

Answer:

47

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5 0
2 years ago
Simplify √49 + [√81 - x(9x = 14)]
eimsori [14]

\longrightarrow{\green{- 9 {x}^{2}   +  14x + 16}} 

\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\orange{:}}}}}

\sqrt{49}  + [ \sqrt{81} - x \: (9x   -  14) ] \\ \\   =  \sqrt{7 \times 7}  + [ \sqrt{9 \times 9}   - 9 {x}^{2}   +  14x] \\  \\  =  \sqrt{( {7})^{2} }  + [ \sqrt{ ({9})^{2} }  - 9 {x}^{2}  +  14x ] \\  \\  (∵ \sqrt{ ({x})^{2} } = x ) \\  \\  = 7 + (9 - 9 {x}^{2}   + 14x) \\  \\  = 7 + 9 - 9 {x}^{2}   + 14x \\  \\  =  - 9 {x}^{2}   +  14x + 16

\large\mathfrak{{\pmb{\underline{\orange{Mystique35 }}{\orange{♡}}}}}

3 0
3 years ago
Find the length of the third side. If necessary, round to the nearest tenth.47
IRISSAK [1]

Given a right angle triangle

The length of the legs are 4 and 7

we will find the hypotenuse using the Pythagorean theorem

So,

\begin{gathered} h^2=7^2+4^2=49+16=65 \\ h=\sqrt[]{65}\approx8.062 \end{gathered}

Rounding to the nearest tenth

So, the answer is the length of the third side = 8.1

3 0
11 months ago
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