A point is taken on the hypotenuse at a distance of 2 ft. from the vertex adjacent to the 4 ft. side. Find the distance from this point to the vertex of the right angle.
I tried equating the area of bigger right angled triangle to the sum of two smaller ones using Heron's formula and I got the answer too. But too much calculations were needed to be performed to achieve the same. Is there any smarter way to achieve the answer?
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As in the other answer, draw the triangle so that its vertices have coordinates $A(0,0)$, $B(3,0)$, and $C(3,4)$. Let $P$ be the point on the hypotenuse $2$ units away from $C$. This means that $\overline{AP}$ has length $3$. Draw the vertical line down from $P$ and let $Q$ be the point at which this line intersects $\overline{AB}$.
Notice that $\triangle AQP$ and $\triangle ABC$ are similar triangles. This means that the ratio of their hypotenuses equals the ratios of their corresponding bases:
$$\frac{AP}{AC}=\frac{AQ}{AB}=\frac{QP}{BC}$$
$$\frac{3}{5}=\frac{AQ}{3}=\frac{QP}{4}$$
This means $AQ=9/5$ and $QP=12/5$. But these are the same as the $x$ and $y$ coordinates of $P$, respectively.
The distance we want to find is $BP$. $\overline{BP}$ is the hypotenuse of the right triangle with bases $\overline{QB}$ and $\overline{QP}$. We know $QP=12/5$, and we can find
$$QB = AB - AQ = 3 - \frac{9}{5} = \frac{6}{5}$$
A right triangle with bases $1$ and $2$ has hypotenuse $\sqrt{5}$, so our triangle with bases $6/5$ and $12/5$ will have hypotenuse
$$BP = \frac{6\sqrt{5}}{5}$$
Draw the triangle so that its vertices have coordinates $(0,0)$, $(3,0)$ and $(3,4)$.
Then the point along the hypotenuse that is $2$ ft. away from $(3,4)$ has coordinates $(3\cos\theta, 3\sin\theta)=(3\cdot\frac{3}{5}, 3\cdot\frac{4}{5})=(\frac{9}{5},\frac{12}{5})$.
Now use the distance formula to get the distance from this point to $(3,0)$