Plis do 17 mam to pani wysłać z obliczeniami Plis dam naj

[tex]Zad. 35\\a)\\h=9cm\\a=\frac{h}3=\frac{9cm}3=3cm\\\\P=\frac{ah}2 \to P=\frac{3cm*9cm}2=13.5cm^2[/tex]
[tex]b)\\a=20cm\\b=\frac{a}4=\frac{20cm}4=5cm\\\\P=\frac{ab}2 \to P=\frac{20cm*5cm}2=10cm*5cm=50cm^2[/tex]
[tex]c)\\\\P=49dm^2\\a=7dm\\\\\frac{7dm*h}2=49dm^2 /*2\\7dm*h=98dm^2 /:7cm\\h=14dm[/tex]
[tex]d)\\P=150cm^2\\|AD| = h=20cm\\|BC|=a\\\\\frac{a*20cm}2=150cm^2 /*2\\a*20cm=300cm^2 /:20cm\\a=15cm[/tex]
a) Odp. Pole tego trójkąta wynosi 13,5[tex]cm^{2}[/tex].
b) Odp. Pole tego trójkąta wynosi 50[tex]cm^{2}[/tex].
c) Odp. Wysokość opuszczona na ten bok ma 14 dm.
d) Odp. Bok BC ma 15 cm.
Szczegółowe wyjaśnienie:
a) 9cm : 3 = 3 cm P = a*h/2 P = 9cm * 3cm/2 = 27[tex]cm^{2}[/tex]/2 = 13,5[tex]cm^{2}[/tex]
b) 20cm : 4 = 5 cm P = a*h/2 P = 20cm * 5cm/2 = 100[tex]cm^{2}[/tex]/2 = 50[tex]cm^{2}[/tex]
c) P = 49[tex]dm^{2}[/tex]
49[tex]dm^{2}[/tex] = 7dm* x/2 //*2
98[tex]dm^{2}[/tex] = 7dm * x //:7 dm
x = 14 dm
d) P = 150[tex]cm^{2}[/tex]
150[tex]cm^{2}[/tex] = a * 20cm/2 /*2
300[tex]cm^{2}[/tex] = a*20cm //:20cm
a = 15 cm