The block

The block, the edges formed by three consecutive GP members, has a surface area of 112 cm². The sum of the edges that pass through one vertex is 14 cm. Calculate the volume of this block.

Correct answer:

V =  64 cm³

Step-by-step explanation:

$$\begin{gathered} b=qa \\ c=q{a}^2 \\ S=112{\text{cm}}^2 \\ S=2(ab+bc+ca)=2(q\cdot a^2+q^3{a}^2+q^2{a}^2) \\ S=2a^2(q+q^2+q^3) \\ a+b+c=14 \\ a+aq+a{q}^2=14 \\ a(1+q+q^2)=14 \\ a(q+q^2+q^3)=14q \\ S=2a^2(q+q^2+q^3)=2a^{2}14q\mathrm{/}a=2\cdot 14\cdot a\cdot q \\ b=aq \\ b={\displaystyle \frac{S}{2\cdot 14}}={\displaystyle \frac{112}{2\cdot 14}}=4\text{ cm } \\ S=2(b\mathrm{/}q\cdot b+b\cdot q\cdot b+b\cdot q\cdot b\mathrm{/}q) \\ S=2(b^2\mathrm{/}q+b^2\cdot q+b^2) \\ S\mathrm{/}2\mathrm{/}{b}^2=1\mathrm{/}q+q+1 \\ q\ast S\mathrm{/}2\mathrm{/}{b}^2=1+q^2+q \\ q\cdot 112\mathrm{/}2\mathrm{/}{4}^2=1+q^2+q \\ -q^2+2.5q-1=0 \\ {q}^2-2.5q+1=0 \\ a=1;b=-2.5;c=1 \\ D=b^2-4ac=2.5^2-4\cdot 1\cdot 1=2.25 \\ D>0 \\ {q}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{2.5\pm \sqrt{2.25}}{2}} \\ {q}_{1,2}=1.25\pm 0.75 \\ {q}_1=2 \\ {q}_2=0.5 \\ \text{ Factored form of the equation: } \\ (q-2)(q-0.5)=0 \\ q=2 \\ a=b\mathrm{/}q=4\mathrm{/}2=2 \\ c=q\cdot b=2\cdot 4=8 \\ V=a\cdot b\cdot c=2\cdot 4\cdot 8=64=64{\text{cm}}^3 \\ \text{ Verifying Solution: } \\ {S}_1=2\cdot (a\cdot b+b\cdot c+c\cdot a)=2\cdot (2\cdot 4+4\cdot 8+8\cdot 2)=112{\text{cm}}^2 \\ {s}_1=a+b+c=2+4+8=14\text{ cm} \end{gathered}$$

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