Consecutive 46761

The block lengths are made up of three consecutive GP members. The sum of the lengths of all edges is 84 cm, and the volume
block is 64 cm³. Determine the surface of the block.

Correct answer:

S =  672 cm²

Step-by-step explanation:

$$\begin{gathered} a+b+c=84cm \\ V=64{\text{cm}}^3 \\ b=qa \\ c=q^{2}a \\ V=abc=a^3\cdot q^3=(a\cdot q{)}^3=b^3 \\ b=\sqrt[3]{V}=\sqrt[3]{64}=4\text{cm} \\ a+c=84-b \\ ac=V/b \\ a(84-b-a)=V/b \\ a(84-b-a)=V/b \\ a(84-4-a)=64/4 \\ -a^2+80a-16=0 \\ {a}^2-80a+16=0 \\ p=1;q=-80;r=16 \\ D=q^2-4pr=80^2-4\cdot 1\cdot 16=6336 \\ D>0 \\ {a}_{1,2}=\frac{-q\pm \sqrt{D}}{2p}=\frac{80\pm \sqrt{6336}}{2}=\frac{80\pm 24\sqrt{11}}{2} \\ {a}_{1,2}=40\pm 39.799497 \\ {a}_1=79.799497484 \\ {a}_2=0.200502516 \\ a=a_2=0.2005\doteq 0.2005 \\ c=V/(b\cdot a)=64/(4\cdot 0.2005)\doteq 79.7995\text{cm} \\ S=2\cdot (a\cdot b+b\cdot c+a\cdot c)=2\cdot (0.2005\cdot 4+4\cdot 79.7995+0.2005\cdot 79.7995)=672=672{\text{cm}}^2 \\ \text{ Verifying Solution: } \\ {V}_2=a\cdot b\cdot c=0.2005\cdot 4\cdot 79.7995=64{\text{cm}}^3 \\ {s}_2=a+b+c=0.2005+4+79.7995=84\text{cm} \\ {q}_1=b/a=4/0.2005\doteq 19.9499 \\ {q}_2=c/b=79.7995/4\doteq 19.9499 \end{gathered}$$

Our quadratic equation calculator calculates it.

Try another example