I don't understand this:
The plan shows the roads between towns A, B and C. Find the distance:
a) from A to BC;
b) from B to AC;
c) from C to AB.
It's an exercise from an 8th grade book, where trigonometry functions (sin, cos, tg) are two years away, that's why I posted it not into trigonometry section. You can't use triangle similarity at this stage as well. This exercise is given at the end of the "Pythagoras theorem" chapter, where usually mixed exercises are, from earlier (not necessarily about the theorem). And I personally can't see how you can apply Pythagoras theorem here.
Hello, Evaldas!
I'll do part (b).
$\displaystyle \text{The plan shows the roads between towns }A, B, C.$
$\displaystyle \text{Find the distance: (b) from B to AC.}$
Code:B o * |* * | * 15 * | * 8 * |h * * | * * | * * | * A o * * * * * * * o * * * o C : 17-x D x :
We have $\displaystyle \Delta ABC$ with: .$\displaystyle AB = 15,\;BC = 8,\;AC = 17$
Draw $\displaystyle BD \perp AC.\;\text{ Let }h = BD.$
Let $\displaystyle x = DC,\;17-x = AD.$
In right triangle $\displaystyle BDC\!:\;h^2 + x^2 \:=\:8^2 \quad\Rightarrow\quad h^2 \:=\:64 - x^2$ .[1]
In right triangle $\displaystyle BDA\!:\;h^2 + (17-x)^2 \:=\:15^2$ .[2]
Substitute [1] into [2]: .$\displaystyle (64-x^2) + (17-x)^2 \:=\:225$
. . which simplifies to: .$\displaystyle 34x \:=\:128 \quad\Rightarrow\quad x \:=\:\frac{64}{17}$
Substitute into [1]: .$\displaystyle h^2 \:=\:64 - \left(\frac{64}{17}\right)^2 \:=\:\frac{14,\!400}{289} $
. . Therefore: .$\displaystyle h \:=\:\sqrt{\frac{14,\!400}{289}} \:=\:\frac{120}{17} \;=\;7\frac{1}{7}$
Soroban's method does not require ABC to be a right triangle. If you notice that 17^2 = 15^2+ 8^2 ABC is aright triangle
Area = 4x15 = 60 km^2
60 =a1 x17/2 a1=120/17
60 =a2 x8/2 a2=120/8
60 = a3 x15/2 a3 = 120/15 where a's are altitudes to coressponding sides
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