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GrogVix [38]
2 years ago
5

Help me please!

Mathematics
1 answer:
uranmaximum [27]2 years ago
7 0
90 + 2x + 104= 360,
360-194 = 2x
166 = 2x, x =83.

X+67 = 150, X+37= 120.

-> 150, 120
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\\\text{we need to calculate the height of two different side walls}\ h_1\ \text{and}\ h_2\\\text{use the Pythagorean theorem:}\\\\h_1^2=\left(\dfrac{l}{2}\right)^2+h^2\to h_1=\sqrt{\left(\dfrac{l}{2}\right)^2+h^2}=\sqrt{\dfrac{l^2}{4}+h^2}=\sqrt{\dfrac{l^2}{4}+\dfrac{4h^2}{4}}\\\\h_1=\sqrt{\dfrac{l^2+4h^2}{4}}=\dfrac{\sqrt{l^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{l^2+4h^2}}{2}

\\\\h_2^2=\left(\dfrac{w}{2}\right)^2+h^2\to h_2=\sqrt{\left(\dfrac{w}{2}\right)^2+h^2}=\sqrt{\dfrac{w^2}{4}+h^2}=\sqrt{\dfrac{w^2}{4}+\dfrac{4h^2}{4}}\\\\h_2=\sqrt{\dfrac{w^2+4h^2}{4}}=\dfrac{\sqrt{w^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{w^2+4h^2}}{2}

SA=lw+2\cdot\dfrac{lh_1}{2}+2\cdot\dfrac{wh_2}{2}\\\\SA=lw+2\!\!\!\!\diagup\cdot\dfrac{l\cdot\frac{\sqrt{l^2+4h^2}}{2}}{2\!\!\!\!\diagup}+2\!\!\!\!\diagup\cdot\dfrac{w\cdot\frac{\sqrt{w^2+4h^2}}{2}}{2\!\!\!\!\diagup}\\\\SA=lw+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw}{2}+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw+l\sqrt{l^2+4h^2}+w\sqrt{w^2+4h^2}}{2}

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