Question

De la acest exercițiu subpunctele a și c. Mulțumesc anticipat!​

Answer 1

Explicație pas cu pas:

a)

$l = 6 \\ m = 6$

$a_{b} = \frac{l}{2} = 3$

$h = \sqrt{ {m}^{2} - \left( \frac{l \sqrt{2} }{2} \right)^{2} } = \sqrt{ {6}^{2} - {(3 \sqrt{2})}^{2} } \\ = \sqrt{36 - 18} = \sqrt{18} = 3 \sqrt{2}$

$A_{b} = {l}^{2} = {6}^{2} = 36$

$A_{l} = \frac{P_{b} \times a_{p}}{2} = \frac{4 \times l \times \sqrt{ {h}^{2} + a_{b}^{2} } }{2} \\ = 36 \sqrt{3}$

$A_{t} = A_{b} + A_{l} = 36 + 36 \sqrt{3} = 36( \sqrt{3} + 1)$

$V = \frac{A_{b} \times h}{3} = \frac{36 \times 3 \sqrt{2} }{3} = 36 \sqrt{2}$

b)

$l = 2 \times a_{b} = 2 \times 20 = 40$

$m = \sqrt{ {h}^{2} + {\left( \frac{l \sqrt{2} }{2} \right)}^{2} } = \sqrt{ {12}^{2} + {\left( \frac{40 \sqrt{2} }{2} \right)}^{2}} \\ = \sqrt{144 + 800} = \sqrt{944} = 4 \sqrt{59}$

$a_{b} = 20$

$h = 12$

$A_{b} = {l}^{2} = {40}^{2} = 1600$

$A_{l} = \frac{P_{b} \times a_{p}}{2} = \frac{4 \times l \times \sqrt{ {h}^{2} + a_{b}^{2} } }{2} = \frac{4 \times 40 \sqrt{ {12}^{2} + {20}^{2} } }{2} \\ = 80 \sqrt{544} = 320 \sqrt{34}$

$A_{t} = A_{b} + A_{l} = 1600 + 320 \sqrt{34} \\ = 320(5 + \sqrt{34})$

$V = \frac{A_{b} \times h}{3} = \frac{1600 \times 12}{3} = 6400$

c)

$l = 10$

$m = \sqrt{ {h}^{2} + {\left( \frac{l \sqrt{2} }{2} \right)}^{2} } = \sqrt{ {12}^{2} + {\left( \frac{10 \sqrt{2} }{2} \right)}^{2}} \\ = \sqrt{144 + 50} = \sqrt{194}$

$a_{b} = \frac{l}{2} = \frac{10}{2} = 5$

$h = \frac{3 \times V}{A_{b}} = \frac{3 \times 400}{100} = 12$

$A_{b} = {l}^{2} = {10}^{2} = 100$

$A_{l} = \frac{P_{b} \times a_{p}}{2} = \frac{4 \times l \times \sqrt{ {h}^{2} + a_{b}^{2} } }{2} =\frac{4 \times 10 \times \sqrt{ {12}^{2} + 5^{2} } } {2} \\ = 20 \sqrt{144 + 25} = 20 \times 13 = 260$

$A_{t} = A_{b} + A_{l} = 100 + 260 = 360$

$V = 400$