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The answer 7.7 provided by my teacher I need help with the work

The answer 7.7 provided by my teacher I need help with the work-example-1
User Creak
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1 Answer

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22 votes

The side length of the square, a=4.

An equilateral triangle has equal sides. Since the base of the triangle is one side of the square, we get


AB=BC=AC=4

Since BQ⊥AC, BP is perpendicular to AC, and hence, BP is the altitude the equilateral triangle ABC.

The altitude BP of the equilateral triangle ABC is,


\begin{gathered} BP=\frac{\sqrt[]{3}}{2}a \\ BP=\frac{\sqrt[]{3}}{2}*4 \\ BP=3.464 \\ \end{gathered}

Also, since BQ⊥ED, BQ⊥AC and P is midpoint of AC, Q is midpoint of ED.


\begin{gathered} EQ=(1)/(2)ED=(1)/(2)a \\ EQ=(1)/(2)*4 \\ EQ=2\text{ } \end{gathered}


\begin{gathered} BQ=BP+PQ \\ BQ=3.464+4 \\ BQ=7.464 \end{gathered}

Now using Pythagoras theorem in triangle BQE,


\begin{gathered} BE=\sqrt[]{BQ^2+EQ^2} \\ BE=\sqrt[]{(7.464)^2+2^2} \\ BE=7.7 \end{gathered}

Therefore, BE=7.7.

User Keshava GN
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