Question

if each edge of a cube is increased by 50% . Find the percentage increase in the surface area

Answer 1

Answer:125%

Step-by-step explanation:

Let the edge = a cm

So increase by 50 % = a + a/2 = 3a/2

Total surface Area of original cube = 6a2

TSA of new cube = 6(3a2)2 =6(9a24)= 13.5a2

Increase in area = 13.5a2−6a2 =7.5a2

7.5a2 Increase % =7.5a26a2×100 = 125%

Answer 2

AnswEr:

Let the length of the each edge of the cube be l units. Then,

• $\sf{S}_{1}$ = TSA = 6l²

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• Length of each edge when each edge is increased by 50% .

$\qquad \sf \: = l + \frac{50}{100} \times l = \frac{3l}{2}$

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$\sf{S}_{2}$ = $\text{TSA\:of\:new\:cube}$

$\qquad \sf = 6 \times ( \dfrac{3l}{2} ) {}^{2} = \dfrac{27}{2} \: {l}^{2}$

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$\therefore$ $\sf\underline{Percentage\:increase\:in\:surface\:area}$

$\qquad \sf = \frac{s_2 - s_1}{s_1} \times 100 \\ \\ \qquad \sf = \frac{ \frac{27}{2} \: {l}^{2} - 6 {l}^{2} }{6 {l}^{2} } \times 100 \\ \\ \qquad \sf = \frac{15}{12} \times 100 = 125$

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