Question

(iii) If each edge of a cube is increased by 50%, then how much the total surface area of
the cube will be increased in percent - let us write by calculating it.
​

Answer 1

GivEn:-

• Each edge of cube is increased by 50%.

To find:-

• How much the total surface area of the cube will be increased in percent?

SoluTion:-

Lets original edge of cube be x units.

⋆ CUBE

$\setlength{\unitlength}{0.65cm}\begin{picture}(2,3)\thicklines\put(2,6){\line(1,0){3.3}}\put(2,9){\line(1,0){3.3}}\put(5.3,9){\line(0,-1){3}}\put(2,6){\line(0,1){3}}\put(0,7.3){\line(1,0){3.3}}\put(0,10.3){\line(1,0){3.3}}\put(0,10.3){\line(0,-1){3}}\put(3.3,7.3){\line(0,1){3}}\put(2,6){\line(-3,2){2}}\put(2,9){\line(-3,2){2}}\put(5.3,9){\line(-3,2){2}}\put(5.3,6){\line(-3,2){2}}\put(3.4,5.5){\sf x units}\put(0,6.3){\sf x units}\put(5.5,7.5){\sf x units}\end{picture}$

As we know that,

$\dag\;{\underline{\boxed{\bf{\purple{Total\;Surface\;Area\;of\;cube = 6a^2}}}}}$

Therefore,

✇ Original Surface area of cube = 6x² units.

★ If edge of cube is increased by 50%,

Then, New edge of cube = x + $\sf \bigg( \dfrac{50}{100} \bigg) x$

$:\implies\sf x + \bigg( \dfrac{1}{2} \bigg) x$

$:\implies\sf x + \bigg( \dfrac{1}{2} \bigg) x$

$:\implies\sf x + 0.5x$

$:\implies\sf 1.5x\; units$ ★

Therefore,

✇ New surface area of cube = 6 × (1.5x)²

$:\implies\bf 13.5x^2\; units$ ★

━━━━━━━━━━━━━━━

✩ Persentage increase in Surface area -

$:\implies\sf \dfrac{(new\;surface\;area - original\;surface\;area)}{original\;surface\;area} \times 100$

$:\implies\sf \dfrac{13.5x^2 - 6x^2}{6x^2} \times 100$

$:\implies\sf \dfrac{7.5x^2}{6x^2} \times 100$

$:\implies\sf{\underline{\boxed{\bf{\pink{125 \% }}}}}$ ★

$\therefore\sf \underline{The\; total\; surface\; area \;of\; the\; cube\; will \;be\; increased\; by\; \bf{125 \% }.}$

━━━━━━━━━━━━━━━

Additional Information:-

★ Formula related to cube:-

⠀⠀⠀⠀✩ Lateral surface area of cube = 4a²

⠀⠀⠀⠀✩ Total surface area of cube = 6a²

⠀⠀⠀⠀✩ Volume of cube = a³

⠀⠀⠀⠀✩ Diagonal of cube = $\sf \sqrt{3a}$

Answer 2

GIVEN :

• Edge of a cube is increased by 50%.

TO FIND :

• The total surface area of the cube will be increased in percent = ?

SOLUTION :

Let the edge be a cm.

Surface area = 6a²

Edge of cube is increased by 50% then,

$\dashrightarrow\:$$\sf a + \dfrac{50}{100}a$

$\dashrightarrow\:$$\sf a + \dfrac{a}{2}$

$\dashrightarrow\:$$\sf \dfrac{3a}{2}$

Then,

$\dashrightarrow\:$ $\sf Surface\: area = 6 \left ( \dfrac{3a}{2}\right)^2$

$\dashrightarrow\:$ $\sf 6 \times \dfrac{9a^2}{4}$

$\dashrightarrow\:$ $\sf \dfrac{27x^2}{2}$

➨ Increase in Surface area = 27a²/2 - 6a²

➨ 27a² - 12a²/2

➨ 15a²/2

Hence,

★$\bf \dfrac{Increased\:in\:surface\:area}{Original\: surface\:area}\times 100$

➨ 15a²/2 /6a² × 100

➨ 15a²/2 × 1/6a² × 100

➨ 125 %

Therefore, the total surface area of the cube will be increased in percent = 125%.