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DiKsa [7]
2 years ago
6

Find the height of a triangle with base 4 cm and area 20 cm2. cm

Mathematics
2 answers:
Illusion [34]2 years ago
8 0

Answer:

are of traingle is 1/2XBXH

1/2X4XH=20cm²

H=20X2÷4

H= 10 cm

hope this helps u

Mice21 [21]2 years ago
5 0
<h3><u>Given</u>: </h3>

  • Base of traingle = 4cm
  • Area of Traingle = 20cm^2

~

<h3><u>Need to Find?</u></h3>

  • Height of Traingle

~

<h2><u>Solution: </u></h2>

  • Let Height (H) of traingle be <u>H cm</u>

~

<u>Using formula:</u>

  • \underline{ \boxed { \textbf{ \textsf{Area  \: of  \: Traingle = 1/2 × B × H   }}}}

~

<u>Putting values in the formula:</u>

\longrightarrow \sf 20 = 1/2 × 4 × H

\longrightarrow \sf 20= 2 × H

\longrightarrow \sf 20/2 = H

\longrightarrow \sf H = 10cm

~

\therefore <u>Height of Traingle is 10cm</u>

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x_{1} = 3.5 and y_{1} = 6.4

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Step-by-step explanation:

Mathematically speaking, the statement is equivalent to this 2-variable non-linear system:

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x^{2} + y^{2} = 53.21

First, x is cleared in the first equation:

x = 9.9 - y

Now, the variable is substituted in the second one:

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And some algebra is done in order to simplify the expression:

98.01-19.8\cdot y +2\cdot y^{2} = 53.21

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Roots are found by means of the General Equation for Second-Order Polynomials:

y_{1} \approx \frac{32}{5} and y_{2} \approx \frac{7}{2}

There are two different values for x:

y = y_{1}

x_{1} = 9.9-6.4

x_{1} = 3.5

y = y_{2}

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There are two possibilities:

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The climb distance is:200 * Sec [ArcTan [10/200]] = 10√ 401 km

and the decent distance is:300 * Sec [ArcTan [10/300]] = 10√901 km

So add the given results above with 500 km and this will be the additional distance that plane moves through the air.

The answer would be in the unit meters.

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Hope this is the answer that you are looking for.

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