Question

Me podéis hacer el 36 , 37 , 38 por favor

Answer 1

PROBLEMA 36

$V_{(semiesfera)} = \frac{2}{3} \times \pi \times (3.5 \: cm) {}^{3} \\ V_{(semiesfera)} = \frac{2}{3} \times 3.1416 \times 42.875 \: {cm}^{3} \\ V_{(semiesfera)} = 89.7974 \: {cm}^{3}$

PROBLEMA 37:

PROBLEMA "a":

$P_{(rectangulo)} = 2 \times (3 \: mm + \sqrt{(5 \: mm) {}^{2} - (3 \: mm) {}^{2} } \\ P_{(rectangulo)} = 2 \times (3 \: mm + \sqrt{25 \: {mm}^{2} - 9 \: {mm}^{2} } \\ P_{(rectangulo)} = 2 \times (3 \: mm + \sqrt{16 \: {mm}^{2} } \\ P_{(rectangulo)} = 2 \times (3 \: mm + 4 \: mm) \\P_{(rectangulo)} = 2 \times (7 \: mm) \\ P_{(rectangulo)} = 14 \: mm$

$A_{(rectangulo)} = 3 \: mm \times ( \sqrt{(5 \: mm) {}^{2} - (3 \: mm) {}^{2} } ) \\ A_{(rectangulo)} = 3 \: mm \times ( \sqrt{25 \: {mm}^{2} - 9 \: {mm}^{2} } ) \\ A_{(rectangulo)} = 3 \: mm \times ( \sqrt{16 \: {mm}^{2} } ) \\ A_{(rectangulo)} = 3 \: mm \times 4 \: mm \\ A_{(rectangulo)} = 12 \: mm {}^{2}$

PROBLEMA "b"

$P_{(triangulo)} = 3 \times 10 \: cm \\ P_{(triangulo)} = 30 \: cm$

PROBLEMA "c"

$P_{(triangulo)} = 13 \: cm + 5 \: cm + ( \sqrt{(13 \: cm) {}^{2} - (5 \: cm) {}^{2} } \\ P_{(triangulo)} = 13 \: cm + 5 \: cm + \sqrt{144 \: cm {}^{2} } \\ P_{(triangulo)} = 13 \: cm + 5 \: cm + 12 \: cm \\ P_{(triangulo)} = 30 \: cm$

$A_{(triangulo)} = \frac{5 \: cm \times ( \sqrt{(13 \: cm) {}^{2} - (5 \: cm) {}^{2} } }{2} \\ A_{(triangulo)} = \frac{5 \: cm \times 12 \: cm}{2} \\ A_{(triangulo)} = \frac{60 \: {cm}^{2} }{2} \\ A_{(triangulo)} = 30 \: {cm}^{2}$

PROBLEMA "d"

$P_{(triangulo)} = 90 \: hm \\ 3 \times lado_{(triangulo)} = 90 \: hm \\ lado_{(triangulo)} = 30 \: hm$

PROBLEMA "e"

$P_{(cuadrado)} = 4 \times \frac{50 \: cm}{ \sqrt{2} } \\ P_{(cuadrado)} = \frac{200 \: cm}{ \sqrt{2} } \\ P_{(cuadrado)} = 100 \sqrt{2} \: cm$

PROBLEMA "f"

$P_{(rectangulo)} =2 \times ( 10 \: cm + \sqrt{(116 \: cm) {}^{2} - (10 \: cm) {}^{2} }) \\ P_{(rectangulo)} = 2 \times (10 \: cm + 115.56816 \: cm) \\ P_{(rectangulo)} = 2 \times (125.56816 \: cm) \\ P_{(rectangulo)} = 251.13632 \: cm$

PROBLEMA "g"

$P_{(rectangulo)} = 24 \: m \\ 2 \times (7 \: m + altura) = 24 \: m \\ 7 \: m + altura = 12 \: m \\ altura = 5 \: m$

$A_{(rectangulo)} = base \times altura \\ A_{(rectangulo)} = 7 \: m \times 5 \: m \\ A_{(rectangulo)} = 35 \: {m}^{2}$

PROBLEMA "h"

$P_{(triangulo)} = 3 \times 6 \: m \\ P_{(triangulo)} = 18 \: m$

PROBLEMA "i"

$A_{(triangulo)} = \frac{6 \: cm \times \sqrt{(12 \: cm ) {}^{2} - \frac{(6 \: cm) {}^{2} }{4} } }{2} \\ A_{(triangulo)} = \frac{6 \: cm \times \sqrt{144 \: {cm}^{2} - \frac{36 \: {cm}^{2} }{4} } }{2} \\ A_{(triangulo)} = \frac{6 \: cm \times \sqrt{144 \: {cm}^{2} - 9 \: {cm}^{2} } }{2} \\ A_{(triangulo)} = \frac{6 \: cm \times 11.61895 \: cm}{2} \\ A_{(triangulo)} = \frac{69.7137 \: {cm}^{2} }{2} \\ A_{(triangulo)} = 34.85685 \: {cm}^{2}$

PROBLEMA 38

$A_{(rombo)} = \frac{2 \: cm\times 4 \: cm}{2} \\ A_{(rombo)} = \frac{8 \: {cm}^{2} }{2} \\ A_{(rombo)} = 4 \: {cm}^{2}$

$(lado_{(rombo)} ) {}^{2} = ( \frac{4 \: cm}{2} ) {}^{2} + ( \frac{2 \: cm}{2} ) {}^{2} \\ (lado_{(rombo)} ) {}^{2} = {(2 \: cm)}^{2} + {(1 \: cm)}^{2} \\ (lado_{(rombo)} ) {}^{2} = 4 \: {cm}^{2} + 1 \: cm {}^{2} \\ (lado_{(rombo)} ) {}^{2} = 5 \: {cm}^{2} \\ lado_{(rombo)} = \sqrt{5 \: cm {}^{2} } \\ lado_{(rombo)} = 2.23607 \: cm$