Question

Raman wants a new carpet for his dining room. The dimensions of the dining room are 10.5 m by 8 m. What will be the area of his carpet that covers the entire dining room?


18.5 sq. m
185 sq. m
80 sq. m
84 sq. m

Please Answer​

Answer 1

Required Answer :

Given -

• Dining room with dimensions 10.5 m and 8 m

To Find -

• Area of his carpet that covers the entire dining room

Formula Used -

• Area of Rectangle = Length × Breadth

Solution -

To find the area of his carpet that covers the entire dining room, we will be using the formula, Area of Rectangle = Length × Breadth, so let's do it$!$

Area of Rectangle = Length × Breadth

Area of Rectangle = 10.5 × 8

Area of Rectangle = 84 m²

∴ Area of his carpet that covers the entire dining room is 84 m²

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Additional Information :

$\boxed{\begin {minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {minipage}}$

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Answer 2

Given:-

The dimensions of the dining room are 10.5 m by 8 m.

To Find:-

The area of his carpet that covers the entire dining room.

Using the formula:-

Area of a rectangle = length × breadth

Solution:-

Area of his carpet = side × side

  ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ = 10.5 × 8m = 84 sq.m

$⠀\;\;\;\;⠀\;\;⠀⠀\;\;\;\;\;\;⠀⠀ = \tt \sqrt{84} \: m$

More Information:-

$\begin{gathered}\boxed{\begin{array}{c}\sf Area~ of ~a ~square~ = ~side × side~ (side)² \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:  \:  \:  \:  \:   \\ \\ {\sf Area \:  of ~a ~rectangle~ = length~ × ~breadth \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }  \\ \\ {\sf Area~ of ~a ~circle~ = ~πr² \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: }  \\ \\ {\sf Area~of~a~triangle~ =  \dfrac{1}{2}  × b× h \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  } \\  \\ {\sf Surface ~area ~ of~a~cylinder~=~2π ×~r×h \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \\{\sf Surface~area~of~a~sphere~=~4πr² \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: } \: \end{array}}\end{gathered}$