Question

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Answer 1

Respuesta:

3) x = 32, r = 8

4) x = 12, r = 2

5) x = 27.2, r = 5.4

6) x = 7, r = 3

Explicación paso a paso:

Para hallar x deberas utilizar el Teorema de Pitágoras

$h^{2} =(c_{1})^{2} +(c_{2})^{2}$

y para hallar el r deberas utilizar la siguiente formula:

$r=\frac{a*b}{a+b+c}$

3) x = $c_{2}$

   $h^{2} = (c_{1})^{2} + (c_{2})^{2} \\40^{2} = 24^{2} + (c_{2})^{2}\\1600 = 576 + (c_{2})^{2}\\1600 - 576 = (c_{2})^{2}\\1024 = (c_{2})^{2}\\\sqrt{1024}=c_2\\ 32=c_2$                 $r=\frac{AB*BC}{AB+BC+AC}\\\\r=\frac{32*24}{24+32+40}\\\\r=\frac{768}{96}\\\\r=8$

4) x = $c_{2}$

$h^{2} = (c_{1})^{2} + (c_{2})^{2} \\13^{2} = 5^{2} + (c_{2})^{2}\\ 169= 25 + (c_{2})^{2}\\169 - 25 = (c_{2})^{2}\\144 = (c_{2})^{2}\\\sqrt{144}=c_2\\ 12=c_2$                     $r=\frac{AB*BC}{AB+BC+AC}\\\\r=\frac{12*5}{12+5+13}\\\\r=\frac{60}{30}\\\\r=2$

5) x = h

$h^{2} = (c_{1})^{2} + (c_{2})^{2} \\h^{2} = 16^{2} + 22^{2}\\ h^{2}= 256 + 484\\h^{2} = 740\\h = \sqrt{740}\\h= 27.20$                    $r=\frac{AB*BC}{AB+BC+AC}\\\\r=\frac{16*22}{16+22+27.20}\\\\r=\frac{352}{65.2}\\\\r=5.4$

6) x = $c_{1}$

$h^{2} = (c_{1})^{2} + (c_{2})^{2} \\25^{2} = (c_{1})^{2} + 24^{2}\\ 625= (c_{1})^{2}+576\\625 - 576 = (c_{1})^{2}\\49 = (c_{1})^{2}\\\sqrt{49}=c_1\\ 7=c_1$                   $r=\frac{AB*BC}{AB+BC+AC}\\\\r=\frac{7*24}{7+24+25}\\\\r=\frac{168}{56}\\\\r=3$