4. W trójkącie prostokątnym wysokość poprowadzona z wierzchołka kąta prostego podzieliła przeciwprostokątną na odcinki o długości 2 i 8. Oblicz:
a) Pole trójkąta.
b) długości odcinków, na jakie dwusieczna kąta prostego podzieliła przeciwległy bok.
c) stosunek pola koła wpisanego w trójkąt do opisanego na trójkącie.
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a)
Rysunek w załączniku.
z = 8+2=10
![a^2 = 8^2 + h^2 a^2 = 8^2 + h^2](https://tex.z-dn.net/?f=a%5E2+%3D+8%5E2+%2B+h%5E2)
![\\a^2=10^2 - b^2 \\a^2=10^2 - b^2](https://tex.z-dn.net/?f=%5C%5Ca%5E2%3D10%5E2+-+b%5E2)
![\\b^2= h^2+2^2 \\b^2= h^2+2^2](https://tex.z-dn.net/?f=%5C%5Cb%5E2%3D+h%5E2%2B2%5E2)
![\\a^2 = 10^2-(h^2+2^2) \\a^2 = 10^2-(h^2+2^2)](https://tex.z-dn.net/?f=%5C%5Ca%5E2+%3D+10%5E2-%28h%5E2%2B2%5E2%29)
![\\a^2=100-h^2-4 \\a^2=100-h^2-4](https://tex.z-dn.net/?f=%5C%5Ca%5E2%3D100-h%5E2-4)
![\\100-h^2-4=64+h^2 \\100-h^2-4=64+h^2](https://tex.z-dn.net/?f=%5C%5C100-h%5E2-4%3D64%2Bh%5E2)
![\\2h^2=32 \\2h^2=32](https://tex.z-dn.net/?f=%5C%5C2h%5E2%3D32)
![\\h^2 =16 \\h^2 =16](https://tex.z-dn.net/?f=%5C%5Ch%5E2+%3D16)
![\\h=4 \\h=4](https://tex.z-dn.net/?f=%5C%5Ch%3D4)
![\\z=10 \\z=10](https://tex.z-dn.net/?f=%5C%5Cz%3D10)
![\\P= \frac{1}{2} \cdot h \cdot z \\P= \frac{1}{2} \cdot h \cdot z](https://tex.z-dn.net/?f=%5C%5CP%3D+%5Cfrac%7B1%7D%7B2%7D+%5Ccdot+h+%5Ccdot+z)
![\\ P = \frac{1}{2} \cdot 10 \cdot 4 = 20 \\ P = \frac{1}{2} \cdot 10 \cdot 4 = 20](https://tex.z-dn.net/?f=%5C%5C+P+%3D+%5Cfrac%7B1%7D%7B2%7D+%5Ccdot+10+%5Ccdot+4+%3D+20)
![\frac{BD}{DC} = \frac{b}{a} \frac{BD}{DC} = \frac{b}{a}](https://tex.z-dn.net/?f=%5Cfrac%7BBD%7D%7BDC%7D+%3D+%5Cfrac%7Bb%7D%7Ba%7D)
![\\ b^2 = 16+4 = 20 \rightarrow b= \sqrt{20} \\ b^2 = 16+4 = 20 \rightarrow b= \sqrt{20}](https://tex.z-dn.net/?f=%5C%5C+b%5E2+%3D+16%2B4+%3D+20+%5Crightarrow+b%3D+%5Csqrt%7B20%7D)
![\\a^2 = 64+16 = 80 \rightarrow a = \sqrt{80} \\a^2 = 64+16 = 80 \rightarrow a = \sqrt{80}](https://tex.z-dn.net/?f=%5C%5Ca%5E2+%3D+64%2B16+%3D+80+%5Crightarrow+a+%3D+%5Csqrt%7B80%7D)
![\\ BD + DC = 10 \rightarrow BD= 10-DC \\ BD + DC = 10 \rightarrow BD= 10-DC](https://tex.z-dn.net/?f=%5C%5C+BD+%2B+DC+%3D+10+%5Crightarrow+BD%3D+10-DC)
![\\\frac{10-DC}{DC}=\frac{\sqrt{20}}{ \sqrt{80}} \\\frac{10-DC}{DC}=\frac{\sqrt{20}}{ \sqrt{80}}](https://tex.z-dn.net/?f=%5C%5C%5Cfrac%7B10-DC%7D%7BDC%7D%3D%5Cfrac%7B%5Csqrt%7B20%7D%7D%7B+%5Csqrt%7B80%7D%7D)
![\\\frac{10-DC}{DC}= \frac{1}{2} \\\frac{10-DC}{DC}= \frac{1}{2}](https://tex.z-dn.net/?f=%5C%5C%5Cfrac%7B10-DC%7D%7BDC%7D%3D+%5Cfrac%7B1%7D%7B2%7D)
![\\20-2DC=DC \rightarrow 3DC=20\rightarrow DC = \frac{20}{3} \rightarrow BD = \frac{10}{3} \\20-2DC=DC \rightarrow 3DC=20\rightarrow DC = \frac{20}{3} \rightarrow BD = \frac{10}{3}](https://tex.z-dn.net/?f=%5C%5C20-2DC%3DDC+%5Crightarrow+3DC%3D20%5Crightarrow+DC+%3D+%5Cfrac%7B20%7D%7B3%7D+%5Crightarrow+BD+%3D+%5Cfrac%7B10%7D%7B3%7D)
![P = \pi \cdot r^2 P = \pi \cdot r^2](https://tex.z-dn.net/?f=P+%3D+%5Cpi+%5Ccdot+r%5E2)
![\\R_{1} = \frac{1}{2} \cdot z = 5 \\R_{1} = \frac{1}{2} \cdot z = 5](https://tex.z-dn.net/?f=%5C%5CR_%7B1%7D+%3D+%5Cfrac%7B1%7D%7B2%7D+%5Ccdot+z+%3D+5)
![\\P_{1} = 25\pi \\P_{1} = 25\pi](https://tex.z-dn.net/?f=%5C%5CP_%7B1%7D+%3D+25%5Cpi)
b)
c)
Sory, że do tego momentu, ale już dzisiaj myśleć nie mogę. :