Question

One dimension of a cube is increased by 1 inch to form a rectangular block. Suppose that the volume of the new block is 150 cubic inches. Find the length of an edge of the original cube. (Area of cube = s 3 )

Answer 1

Given:-

• one dimension of a cube is increased by 1 inches to form a rectangular block.
• Volume of new block means = 150 inches³

To find:-

• edge of the original cube.

Supposition:-

• let the edge of the block be x inches.
• length of the block = (x+1) inches
• breadth of the block = x inches
• Height of rectangular block = x inches.

Formula of it's volume:-

$\underline{\boxed{\sf Volume = l×b×h}}$

here:-

• l = length
• B = breath
• h = height

Solution:-

$\longrightarrow$ 150 = l×b×h

$\longrightarrow$ 150 = (x+1)(x)(x)

$\longrightarrow$ 150 = x²(x+1)

$\longrightarrow$ 150 = x³+x²

$\longrightarrow$ 0 = x³-5x+6x²-30x+30x-150

$\longrightarrow$ x²(x-5)+6x(x-5)+30(x-5) = 0

$\longrightarrow$ (x-5)(x²+6x+30) = 0

$\longrightarrow$ x = 5 inches

and for x²+6x+30

$\longrightarrow$ D = b²-4ac

$\longrightarrow$ D = (6)²-4(1)(30) = -84

$\longrightarrow$ D = -84

$\longrightarrow$ D < 0

hence, this equation has no real roots.

So, we will go for the first value we got i.e 5 inches.

hence, length of this cube is 5 inches.

Answer 2

${\large{\rm{\underbrace{\underline{Correct \; Question}}}}}$

${\sf{Correct \: question}}$

One dimension of a cube is increased by 1 inch to form a rectangular block. Suppose that the volume of the new block is 150 cubic inches. Find the length of an edge of the original cube. (Area of cube = s³).

${\sf{Let's \: understand \: the \: concept}}$

This question says that one dimension of cube by 1 inch to form a rectangular block let's suppose that the volume of the new block is 150 inches³. We have to find the length of an edge of the original cube. And the question says that we have to take area of cube as s³. The biggest thing is that the question is from a very interesting chapter of mathematics named Quadratic equation. So work according to the question and the chapter rules.

${\large{\rm{\underline{Given \; that}}}}$

${\bullet{\leadsto}}$ Dimension of cube by 1 inch to form a rectangular block

${\bullet{\leadsto}}$ The volume of the new block is 150 inches³.

${\bullet{\leadsto}}$ Take area of cube as s³

${\large{\rm{\underline{To \; find}}}}$

${\bullet{\leadsto}}$ The length of an edge of the original cube.

${\large{\rm{\underline{Solution}}}}$

${\bullet{\leadsto}}$ The length of an edge of the original cube = 5 inches²

${\large{\rm{\underline{Full \; Solution}}}}$

✽ According to the question, let the side of cube be s.

✽ As we already know that the formula to find volume of cube is a³

But according to the question volume be s³

Where,

${\rightarrow}$ a / s denotes sides.

✽ According to the given data,

${\longmapsto}$ Rectangle block volume = (s+1)s² = 150

Henceforth,

${\longmapsto}$ s³ + s² - 150 = 0

${\longmapsto}$ s³ - 5s² + 6s² - 30s + 30s + 30s - 150 = 0

Middle term splitting method,

${\longmapsto}$ s²(s-5) + 6s(s-5) + 30(s-5) = 0

${\longmapsto}$ (s-5) (s²+6s+30) = 0

And as we see that the trinomes has none real root.

${\longmapsto}$ 5 inches³

✽ Henceforth, the answer is 5 inches ³

${\large{\rm{\underline{Let's \: verify}}}}$

${\longmapsto}$ (5+1) × 5 × 5 = 150

${\longmapsto}$ 6 × 5 × 5 = 150

${\longmapsto}$ 30 × 5 = 150

${\longmapsto}$ 150 = 150

${\longmapsto}$ LHS = RHS

Hence, verified !!

${\large{\rm{\underline{Know \; more}}}}$

Cube :

$\setlength{\unitlength}{4mm}\begin{picture}(10,6)\thicklines\put(0,1){\line(0,1){10}}\put(0,1){\line(1,0){10}}\put(10,1){\line(0,1){10}}\put(0,11){\line(1,0){10}}\put(0,11){\line(1,1){5}}\put(10,11){\line(1,1){5}}\put(10,1){\line(1,1){5}}\put(0,1){\line(1,1){5}}\put(5,6){\line(1,0){10}}\put(5,6){\line(0,1){10}}\put(5,16){\line(1,0){10}}\put(15,6){\line(0,1){10}}\put(4.6,-0.5){\bf\large y m}\put(13.5,3){\bf\large z m}\put(-4,5.8){\bf\large x m}\end{picture}$

Rectangle :

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large x cm}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large y cm}\put(-0.5,-0.4){\bf A}\put(-0.5,3.2){\bf D}\put(5.3,-0.4){\bf B}\put(5.3,3.2){\bf C}\end{picture}$

Formulas related to SA & Volume :

$\begin{array}{|c|c|c|}\cline{1-3}\bf Shape&\bf Volume\ formula&\bf Surface\ area formula\\\cline{1-3}\sf Cube&\tt l^3}&\tt 6l^2\\\cline{1-3}\sf Cuboid&\tt lbh&\tt 2(lb+bh+lh)\\\cline{1-3}\sf Cylinder&\tt {\pi}r^2h&\tt 2\pi{r}(r+h)\\\cline{1-3}\sf Hollow\ cylinder&\tt \pi{h}(R^2-r^2)&\tt 2\pi{rh}+2\pi{Rh}+2\pi(R^2-r^2)\\\cline{1-3}\sf Cone&\tt 1/3\ \pi{r^2}h&\tt \pi{r}(r+s)\\\cline{1-3}\sf Sphere&\tt 4/3\ \pi{r}^3&\tt 4\pi{r}^2\\\cline{1-3}\sf Hemisphere&\tt 2/3\ \pi{r^3}&\tt 3\pi{r}^2\\\cline{1-3}\end{array}$