Question

18) The length, breadth and height of a
cuboid are 15 m, 16m and 5 dm
respectively. The lateral surface area of
the cuboid is
O a) 45 sq m.
O b) 21 sqm
O c) 201 sqm
O d) 90 Sam​

Answer 1

$\begin{gathered}\begin{gathered}\bf \:Given - \begin{cases} &\sf{length_{(cuboid)} = 16 \: m} \\ &\sf{breadth_{(cuboid)} = 15 \: m}\\ &\sf{height_{(cuboid)} = \: 5 \: dm \: = 0.5 \: m} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To\:find - \begin{cases} &\sf{lateral \: surface \: area_{(cuboid)}}  \end{cases}\end{gathered}\end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

Given

Dimensions of Cuboid

• Length of Cuboid, l = 16 m

• Breadth of Cuboid,b = 15 m

• Height of Cuboid, h = 0.5 m

Now,

We know that

• Lateral surface area of Cuboid is given by

$\rm \: \boxed{ \pink{ \bf \:CSA_{(cuboid)}  \:  =  \tt \: 2(length + breadth) \times height}}$

$\rm :\implies\:CSA_{(cuboid)} \: = \: 2 \times (16 + 15) \times 0.5$

$\rm :\implies\:CSA_{(cuboid)} = 31 \times 1$

$\rm :\implies\: \boxed{ \pink{ \bf \: CSA_{(cuboid)} \:  =  \tt \: 31 \: {m}^{2} }}$

$\underline{ \boxed{ \boxed{ \red{ \bf \: Additional \: Information:}}}}$

Perimeter of rectangle = 2(length× breadth)

Diagonal of rectangle = √(length ²+breadth ²)

Area of square = side²

Perimeter of square = 4× side

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²