An Isosceles right triangle $ABC$ , $AB=BC= 4 cm$
Point $p$ is a midpoint of $BC$ , points $q$, $s$ lies on $AC$,$AB$ respectively , such that the triangle $pqs$ is an equilateral triangle ;
what is the area of triangle $pqs$
[Math] Area of equilateral triangle inscribed in right triangle
geometry
Related Solutions
Hint:
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The red solid line is the height dropped onto the hypotenuse, i.e. $h = \frac{ab}{c}$ and the red dotted lines are of the same length. The green parallel lines are unnecessary, but might get you some intuitions.
Good luck! ;-)
The fewer steps the better; I found I had to use the geometric mean construction and I wonder whether this step might be avoidable.
Something like a geometric mean is unavoidable since the problem is a quadratic equation for the side length of the equilateral triangle.
Here is a relatively efficient construction using a geometric mean. The savings is in re-using one side of ABC as the base of the equilateral triangle, and in using a non-perpendicular line to measure altitude, and allowing the semicircle construction of geometric mean to be applied.
On one of the sides, say AB, build an equilateral triangle ABD. Extend line CD to intersect AB at P. Find a length $g$ equal to the geometric mean of PC and PD, and take a point H on PD with PH = $g$. The equilateral triangle whose "height" measured along line PCD (from P) is $g$ has the same area as the given triangle.
The construction of the geometric mean is made easier by extending line CD to E so that P is the midpoint of EC, then using ED as diameter of a circle and taking $g=|PG|$ for PG a perpendicular to CD with G on the circle. Then draw the parallel to AB through H and intersect it with AD (at point K) to cut off the correct side length (AK) of equilateral triangle.
Best Answer
Rotate $\Delta Bps$ clockwise $60^\circ$ about $p$ as shown, to form $\Delta B'pq$.
Using $\Delta B'Cp$, find that $B'C=2\sqrt3~\mathrm{cm}$.
Using Sine Law on $\Delta B'Cq$, find that $\dfrac{B'q}{\sin 15^\circ}=\dfrac{B'C}{\sin 105^\circ}$, so $B'q=(4\sqrt3-6)~\mathrm{cm}$.
Use Pythagoras' theorem on $\Delta B'pq$ to find that $pq=2\sqrt{22-12\sqrt3}~\mathrm{cm}$.
Therefore the area is $\dfrac12a^2\sin60^\circ=(22\sqrt3-36)~\mathrm{cm}^2$.