37.5
Step-by-step explanation:
The line which marks the height 25\sqrt325 Β
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25, square root of, 3, end square root of the large triangle is vertical, and the base of the triangle is horizontal, so their intersection is a right angle. The angles in the large triangle add up to \piΟpi, so the remaining angle is:
\qquad \begin{aligned} \pi - \dfrac\pi2 - \dfrac\pi6 =& \dfrac\pi3 \end{aligned} Β
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The original triangle L U R is shown, but now the length of segment U D is shown to be '25 times the square root of 3', and angle U L R is shown to have a measure of ' pi divided by 3' radians.
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The original triangle L U R is shown, but now the length of segment U D is shown to be '25 times the square root of 3', and angle U L R is shown to have a measure of ' pi divided by 3' radians.
Hint #2
Looking at the left triangle ULDULDU, L, D, the hypotenuse length LULUL, U is:
\qquad \begin{aligned} \dfrac{25\sqrt3}{\sin\!\left({\Large\frac\pi3}\right)} \end{aligned} Β
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Looking at the big triangle LURLURL, U, R, we see that the base length LRLRL, R is:
\qquad \begin{aligned} &\left(LU\right)\dfrac{1}{\sin\!\left({\Large\frac\pi6}\right)} \\ \\ =&\left(\dfrac{25\sqrt3}{\sin\!\left({\Large\frac\pi3}\right)}\right)\dfrac{1}{\sin\!\left({\Large\frac\pi6}\right)} \\ \\ =&\left(\dfrac{25\sqrt3}{ \quad\large\frac{\sqrt{3}}{2}\quad }\right)\dfrac{1}{ \quad\large\frac{1}{2}\quad } \\ \\ =&\left(\dfrac{25}{ \quad\large\frac{1}{2}\quad }\right)\dfrac{1}{ \quad\large\frac{1}{2}\quad } \\ \\ =&100 \\ \end{aligned} Β
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100
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Hint #3
The large triangle has uniform thickness, so the inner triangle's sides are parallel to the outer triangle's sides. Therefore, the inner triangle's angles are congruent to the outer triangle's angles. The two triangles are similar.
Hint #4
Similar triangles have proportional lengths, so we can set up the equation:
\qquad \begin{aligned} \dfrac{h}{w} =& \dfrac{H}{W} \\ \\ \dfrac{h}{50\sqrt{3}} \approx& \dfrac{25\sqrt3}{100} \end{aligned} Β
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Hint #5
\qquad\qquad\, \begin{aligned} \dfrac{h}{50\sqrt{3}} \approx& \dfrac{\sqrt3}{4} \\ \\ {h} \approx& \dfrac{{50\sqrt{3}}\sqrt3}{4} \\ \\ {h} \approx& \dfrac{25\cdot3}{2} \\ \\ h \approx& 37.5 \\ \\ \end{aligned} Β
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Hint #6
To the nearest tenth, the height of the inner triangle is:
\qquad\qquad\, \begin{aligned} h \approx& 37.5 \end{aligned} Β
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37.5
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*note: The answers 37.337.337, point, 3 and 37.437.437, point, 4 may have been obtained by other methods of solving for hhh since this is an approximation. Those answers are also accepted as correct.