A right angled triangle has the sides 3cm, 4 cm and 5 cm. find out the area of the greatest square that can be inscribed in it with one of the vertices of the square on the hypotenuse??
[Math] Maximum square that can be incribed in a right angle with one vertex on the hypotenuse
geometry
Related Solutions
Your approach looks fine to me. Indeed, the most natural way to show that a line bisects the area of a square is to show that it passes through the center of that square. Here this follows from the fact that $O$ is the midpoint of the arc $AB$ of the circumcircle of $\triangle APB$ and the fact that the inner angle bisector in a triangle passes through the midpoint of the opposite arc of the circumcircle.
If triangle has its vertices on a circle and one side is a diameter of the circle, then the angle opposite that side is a right angle.
The center of the circle is $C=\left(-\frac{3}{2},2\right)$ and the radius is $\frac{\sqrt{29}}{2}$ so its equation can be found. The slope of the perpendicular bisector of the segment $AB$ has slope $-\frac{5}{2}$ and contains $C$. The line intersects the circle at $E$ and at $D$ but all that is wanted is the slope of $EA$ and $BE$ which can be found by finding the coordinates of $E$. The slope of $DA$ is the same as $BE$ and the slope of $DB$ is the same as the slope of $EA$.
However, that way is considerably messy. It is probably easier to use trigonometry.
The line $AB$ makes an angle $\theta=\arctan\left(\frac{2}{5}\right)$ with the horizontal. The side $AD$ makes an angle $45^\circ+\arctan\left(\frac{2}{5}\right)$ so its slope is $\tan\left(45^\circ+\arctan\left(\frac{2}{5}\right)\right)=\frac{7}{3}$. Since $EA$ is perpendicular to $AD$ it has slope $-\frac{3}{7}$.
Note: This uses the identity $\tan(X+Y)=\dfrac{\tan X+\tan Y}{1-\tan X\tan Y}$.
The equation of the lines containing the four sides satisfying the conditions of the problem are
- AB: $y-3=\frac{7}{3}(x-1)$
- BD: $y-1=\frac{7}{3}(x+4)$
- AD: $y-3=-\frac{3}{7}(x-1)$
- BE: $y-1=-\frac{3}{7}(x+4)$
These simplify as follows:
- AB: $\quad7x-3y+2=0$
- BD: $\quad7x-3y+31=0$
- AD: $\quad3x+7y-24=0$
- BE: $\quad3x+7y+5=0$
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Best Answer
In the left triangle $$ s=\frac3{\frac54+\frac35}=\frac{60}{37} $$ In the right triangle $$ s=\frac3{\frac34+1}=\frac{12}7 $$