Question

2. The surface area of a cuboid is 720 m². Its length and breadth are in the ratio 4:3 and
height is 6 m. Find the:
(i) Length of the cuboid (ii) Breadth of the cuboid​

Answer 1

$\sf\small\underline\purple{Let:-}$

$\tt{\implies Cuboid\:_{(length\:and\:wide)}=x}$

$\tt{\implies Length\:_{(cuboid)}=4x}$

$\tt{\implies Breadth\:_{(cuboid)}=3x}$

$\sf\small\underline\purple{Given:-}$

$\tt{\implies Surface\: Area\:_{(cuboid)}=720m^2}$

$\tt{\implies Cuboid\:_{(length: breadth)}=4:3}$

$\tt{\implies Cuboid\:_{(height)}=6m}$

$\sf\small\underline\purple{To\: Find:-}$

$\tt{\implies Cuboid\:_{(length\: and\: breadth)}=?}$

$\sf\small\underline\purple{Solution:-}$

To calculate the length and breadth of cuboid , at first we have to assume the length and breadth of cuboid be "x" . After that we have to apply formula TSA of cuboid to find the length and breadth of cuboid:-]

$\tt{\implies T.S.A\:_{(cuboid)}=2(lb+bh+lh)}$

$\tt{\implies T.S.A\:_{(cuboid)}=2(4x*3x+3x*6+4x*6)}$

$\tt{\implies T.S.A\:_{(cuboid)}=2(12x^2+18x+24x)}$

$\tt{\implies T.S.A\:_{(cuboid)}=2(12x^2+42x)}$

$\tt{\implies 24x^2+84x=720}$

$\tt{\implies 24x^2+84x-720=0}$

• Dividing by 12 on both sides:-]

$\tt{\implies 2x^2+7x-60=0}$

$\tt{\implies 2x^2-8x+15x-60=0}$

$\tt{\implies 2x(x-4)+15(x-4)=0}$

$\tt{\implies (2x+15)(x-4)=0}$

$\tt{\implies \therefore x=-15/2\:or\:4}$

Here negative value can't be length or breadth of cuboid so, we take value for length or breadth for cuboid is x=4m

$\sf\large{Hence'}$

$\bf{\implies Length\:_{(cuboid)}=4x=4*4=16m}$

$\bf{\implies Breadth\:_{(cuboid)}=3x=3*4=12m}$

Answer 2

Given :-

TSA of cuboid = 720 m²

Ratio of length and breadth = 4:3

Height = 6 m

To Find :-

(i) Length of the cuboid (ii) Breadth of the cuboid

Solution :-

Let,

Length = 4x

Breadth = 3x

We know that

TSA = 2(lb + bh + lh).

$\sf TSA = 2(4x \times 3x + 3x \times 6 + 4x \times 6)$

$\sf 720 = 2(12x^{2} + 18x + 24x)$

$\sf 720 = 2(12x^{2} +42x)$

$\sf 24x^{2} +84x - 720 = 0$

$\sf 2x^{2} +7x - 60 = 0$

$\sf 2x^{2} - 8x + 15x - 60 = 0$

Taking x as common

$\sf 2x(x - 4) + 15(x - 4) = 0$

$\sf (2x + 15),(x - 4) = 0$

Either

x = -15/2

or

x = 4

Since length can't be negative. So neglect the -15/2

${\textsf{\textbf{\red{\underline{Length = 4(4) = 16 m}}}}}$

${\textsf{\textbf{\red{\underline{Breadth = 3(4) = 12 m}}}}}$