Respuesta :

aosera
2a.
Since BC = 10m and M is the midpoint, we can find CM by : 10÷2 = 5m
So, CM = 5m

2b. Now, you have the base ( 5m) & hypotenuse (13m). Pythagoras' Theorem:

[tex]c {}^{2} = {a}^{2} + {b}^{2} [/tex]
Where c is the hypotenuse, a & b are sides.

[tex] {13}^{2} = {5}^{2} + b {}^{2} \\ {5}^{2} + {b}^{2} = 13 {}^{2} \\ {b}^{2} = 13 {}^{2} - {5}^{2} \\ b = \sqrt{144} \\ b = 12[/tex]
b here refers to the height.

2c.
Area of a right-angled triangle =
[tex] \frac{base \times height}{2} [/tex]
So,
[tex] \frac{10 \times 12}{2} \\ = 60 {m}^{2} [/tex]
gmany
a)
[tex]|CM|=\dfrac{|BC|}{2}\to|CM|=\dfrac{10m}{2}=5m[/tex]

b)
[tex]h^2+5^2=13^2\\h^2+25=169\ \ \ |-25\\h^2=144\to h=\sqrt{144}\\h=12m[/tex]

c)
[tex]A_\Delta=\dfrac{|BC|h}{2}\to A\Delta=\dfrac{10\cdot12}{2}=60\ m^2[/tex]

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