Question

Find the volume and area for the objects shown and answer Question

Answer 1

Explanation:

You must write formulas regarding the volume and surface area of ​​the given solids.

`\bold{\#1\ Rectangular\ prism:}\\\\V=lwh\\SA=2lw+2lh+2wh=2(lw+lh+wh)\\\\\bold{\#2\ Cylinder:}\\\\V=\pi r^2h\\SA=2\pi r^2+2\pi rh=2\pir(r+h)\\\\\bold{\#3\ Sphere:}\\\\V=(4)/(3)\pi r^3\\SA=4\pi r^2`

`\bold{\#4\ Cone:}\\\\V=(1)/(3)\pi r^2h\\\\\text{we need calculate the length of a slant length}\ l\\\text{use the Pythagorean theorem:}\\\\l^2=r^2+h^2\to l=√(r^2+h^2)\\\\SA=\pi r^2+\pi rl=\pi r^2+\pi r√(r^2+h^2)=\pi r(r+√(r^2+h^2))\\\\\bold{\#5\ Rectangular\ Pyramid:}\\\\V=(1)/(3)lwh\\\\`

`\\\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((l)/(2)\right)^2+h^2\to h_1=\sqrt{\left((l)/(2)\right)^2+h^2}=\sqrt{(l^2)/(4)+h^2}=\sqrt{(l^2)/(4)+(4h^2)/(4)}\\\\h_1=\sqrt{(l^2+4h^2)/(4)}=(√(l^2+4h^2))/(\sqrt4)=(√(l^2+4h^2))/(2)`

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

`SA=lw+2\cdot(lh_1)/(2)+2\cdot(wh_2)/(2)\\\\SA=lw+2\!\!\!\!\diagup\cdot(l\cdot(√(l^2+4h^2))/(2))/(2\!\!\!\!\diagup)+2\!\!\!\!\diagup\cdot(w\cdot(√(w^2+4h^2))/(2))/(2\!\!\!\!\diagup)\\\\SA=lw+(l√(l^2+4h^2))/(2)+(w√(w^2+4h^2))/(2)\\\\SA=(2lw)/(2)+(l√(l^2+4h^2))/(2)+(w√(w^2+4h^2))/(2)\\\\SA=(2lw+l√(l^2+4h^2)+w√(w^2+4h^2))/(2)`