Right-angled triangle ABC

Question

(I'm only a Year 7 so please explain clearly how you found the solution)

In the right angled triangle $ABC$, a point $M$ on the hypotenuse $BC$ is such that $AM$ is perpendicular to $BC$. Also, $MC$ is $8$ cm longer than $BM$, and the ratio $AB:AC=3:5$ How many centimetres is the hypotenuse?

I started by calling the length of $BM=y$, and $MC=y+8$ and then using similar triangle ratios created fractions $\frac{AM}{MC} = \frac{BM}{AM}$ which cancelled out to make $\frac{BM}{MC}$ which substitutes into $\frac{y}{y+8}$ but then I got stuck. Can anyone help me?. By the way, the answer is $17$.

Answer 1

This is how I did it:

Since $ \angle AMC = \angle AMB = 90°$ , apply the Pythagoras Theorem in $ \triangle AMC $ and $ \triangle AMB $ $$\implies (y+8)^2 + x^2 = (5k)^2$$ $$\implies y^2 + 16y + 64 + x^2 = 25k^2 ...(1)$$

Also, $$x^2 + y^2 = (3k)^2$$ $$\implies x^2 + y^2 = 9k^2 ....(2)$$

Now, substituting $(2)$ into $(1)$ , $$9k^2 + 64 + 16y = 25k^2$$ $$\implies 16y + 64 = 16k^2$$ $$\implies y + 4 = k^2$$ (dividing both sides by $16$ )

$$\implies BC = 2y + 8 = 2k^2$$

Now, on applying Pythagoras Theorem in $\triangle BAC$ , we get

$$ 9k^2 + 25k^2 = 4k^4$$ $$\implies 34 = 4k^2$$ (dividing both sides by $ k^2 $ )

$$\implies BC = 2k^2 = 17 cm $$ (dividing both sides by $ 2 $ )

NOTE: To clear your confusion regarding $3k$ and $5k$, the $k$ has been written because $AB$ and $AC$ may not necessarily be $3cm$ and $5cm$ respectively (and they are not in this case also). In order to maintain the ratio, we have expressed $AB$ as $3k$ and $AC$ as $5k$, where $k$ can be any positive real number. I hope it's clear. If it's not, then please write in the comments below.

Answer 2

Call $BM=y$. We know that $AB=0.6AC$. Time to invoke Euclid: $$(0.6AC)^2=y\cdot(2y+8)\text{ and } AC^2=(y+8)\cdot (2y+8).$$ Now divide the first equation by the second to solve $$0.36=\frac{y}{y+8}.$$

EDIT: If you replace $8$ by $a$ and $0.6$ by $t$, the hypotenuse will be $$a\cdot\frac{1+t^2}{1-t^2}.$$

(Provided that $t\neq1$; in this case there is no solution unless $a=0$.)