Question

The volume of a cuboid is 420 cubic cm. Find the surface area of ​​the cuboid if its breadth and height are 7 cm and 5 cm respectively.

Spammed answer report on the spot...​

Answer 1

Step-by-step explanation:

Volume of Cuboid = 420 cm³

Breadth of Cuboid = 7 cm

Height of Cuboid = 5 cm

To Find :

Surface Area of Cuboid = ?

Solution :

~ Formula Used

Surface used:

Where :

➻ L = Length

➻ B = Breadth or Width

➻ H = Height

➻ TSA = Total Surface Area

~ Calculating the Surface Area

⟹TSA(Cuboid)=2(lb+bh+hl) :

⟹TSA(Cuboid)=2[(12×7)+(7×5)+(5×12)]

⟹TSA(Cuboid)=2(84+35+60) :

⟹TSA(Cuboid)=2×179

Surface Area of the Cuboid =  358 cm² 

Hope it helps

Answer 2

Question-

The volume of a cuboid is 420 cubic cm. Find the surface area of the cuboid if its breadth and height are 7 cm and 5 cm respectively.

$\begin{gathered} \\ {\underline{\rule{200pt}{3pt}}} \end{gathered}$

${\large{\underline{\frak{Required\:Answer}}}}$

${\underline{\bf{358cm^2}}}$

$\begin{gathered} \\ {\underline{\rule{200pt}{3pt}}} \end{gathered}$

Given-

• Volume of cuboid=${\sf{420cm^3}}$
• breadth of cuboid=7 cm
• height of cuboid=5cm

$\begin{gathered} \\ {\underline{\rule{200pt}{3pt}}} \end{gathered}$

To find-

• Surface area of cuboid

$\begin{gathered} \\ {\underline{\rule{200pt}{3pt}}} \end{gathered}$

${\Large{\underline{\underline{\sf{\maltese Formula\:Using}}}}}$

~Volume of cuboid-

${\large{\underline{\boxed{\red{\sf{v_{(cuboid)}=lbh}}}}}}$

~Surface Area of cuboid

${\large{\underline{\boxed{\red{\sf{SA_{(cuboid)}=2(lb+bh+hl)}}}}}}$

Here:

• v=volume
• SA=Surface Area
• l=length
• b=breadth
• h=height

$\begin{gathered} \\ {\underline{\rule{200pt}{3pt}}} \end{gathered}$

${\Large{\underline{\underline{\bf{Solution}}}}}$

$\begin{gathered} \\ \qquad{\rule{150pt}{1pt}} \end{gathered}$

${\Large{\underline{\underline{\frak{~Calculating\:Length\:of\:cuboid}}}}}$

${\pink{\dashrightarrow}{\qquad{\quad{\sf{v_{(cuboid)}=lbh}}}}}$

${\pink{\dashrightarrow}{\qquad{\quad{\sf{420 cm^3_{(cuboid)}=l \times 7 \times 5}}}}}$

${\pink{\dashrightarrow}{\qquad{\quad{\sf{420_{(cuboid)}=l \times 35}}}}}$

${\pink{\dashrightarrow}{\qquad{\quad{\sf{l=\dfrac{420}{35}}}}}}$

${\pink{\dashrightarrow}{\qquad{\quad{\sf{l=\cancel\dfrac{420}{35}=12}}}}}$

${\blue{\rightsquigarrow}{\qquad{\quad{\underline{\red{\sf{Length\:of\:cuboid=12\:cm}}}}}}}$

$\begin{gathered} \\ \qquad{\rule{150pt}{1pt}} \end{gathered}$

${\Large{\underline{\underline{\frak{~Calculating\:Surface\:Area\:of\:cuboid}}}}}$

${\red{\dashrightarrow}{\qquad{\quad{\sf{SA_{(cuboid)}=2(lb+bh+hl)}}}}}$

${\red{\dashrightarrow}{\qquad{\quad{\sf{SA_{(cuboid)}=2[(12 \times 7)+(7 \times 5)+(5 \times 12)])}}}}}$

${\red{\dashrightarrow}{\qquad{\quad{\sf{SA_{(cuboid)}=2(84+35+60)}}}}}$

${\red{\dashrightarrow}{\qquad{\quad{\sf{SA_{(cuboid)}=2\times 179}}}}}$

${\red{\dashrightarrow}{\qquad{\quad{\sf{SA_{(cuboid)}=2\times 179}}}}}$

${\blue{\rightsquigarrow}{\qquad{\quad{\purple{\sf{SA_{(Cuboid)}=358}}}}}}$

$\begin{gathered} \\ \qquad{\rule{150pt}{1pt}} \end{gathered}$

$\begin{gathered} \\ {\underline{\rule{200pt}{3pt}}} \end{gathered}$

Therefore

${\large{\underline{\bf{\bigstar Surface\:Area\:of\:given\:cuboidis\:358cm^2}}}}$

$\begin{gathered} \\ {\underline{\rule{200pt}{3pt}}} \end{gathered}$