Question

rectangular room 5.6 metres into 4.8 m is to be carpeted with a carpet 80 cm wide find the length of a carpet required​

Answer 1

❏ Question:-

Rectangular room 5.6 metres into 4.8 m is to be carpeted with a carpet 80 cm wide find the length of a carpet required.

❏ Solution:-

✦Given:-

➝For the Room Floor:

• Length (l)=5.6 metres
• breadth (b)=4.8 metres.

➝For the Room carpet:

• breadth (b')=80 cm= 0.8 metres.

✦To Find:-

• length of the carpet=?=l' (let)

✦Ans:-

Now, the Area of the Room floor is,

$\sf\longrightarrow Area_{room\:floor}=(5.6\times4.8)\:m{}^{2}$

Now , the area of the Room carpet

$\sf\longrightarrow Area_{carpet}=(0.8\times l')\:m{}^{2}$

Now, the floor needs to be covered by the carpet so,

$\bf\therefore$ the Area of the Room floor = the Area of the Room carpet.

$\sf\longrightarrow(5.6\times4.8)=0.8\times l'$

$\sf\longrightarrow\frac{(\cancel{5.6}\times4.8)}{\cancel{0.8}}=l'$

$\sf\longrightarrow 7\times 4.8=l'$

$\sf\longrightarrow\boxed {\large{\red{l'=33.6\:\:m}}}$

$\bf\therefore$ Length of the Carpet=33.6 m.

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Some Useful Formulas

✦CIRCLE✦

❚ For a Circle of diameter r ,

(1)Area is given by,

$\sf\longrightarrow \boxed{Area=\pi r{}^{2}}$

(2) Circumference is given by,

$\sf\longrightarrow\boxed{ Circumference=2\pi r=\pi d}$

✦RECTANGLE✦

For a rectangle of length l and breadth b,

$\sf\longrightarrow \boxed{Area=length\times breadth}$

$\sf\longrightarrow \boxed{Perimeter=2\times(length+Breadth)}$

$\sf\longrightarrow \boxed{Diagonal=\sqrt{length{}^{2}+Breadth{}^{2}}}$

✦SQUARE✦

For a square of side a ,

$\sf\longrightarrow \boxed{Area=Side{}^{2}}$

$\sf\longrightarrow \boxed{Perimeter=4\times side}$

$\sf\longrightarrow \boxed{Diagonal=\sqrt{2}\times side}$

✦TRIANGLE✦

Triangle of sides a, b and c

$\sf\longrightarrow \boxed{Perimeter=a+b+c}[/tex</p><p>[tex]\sf\longrightarrow \boxed{S=\frac{a+b+c}{2}}$

[Where, S= Half of Perimeter]

$\sf\longrightarrow \boxed{Area=\sqrt{S(S-a)(S-b)(S-c)}}$

[ using Heron's Formula]

✰For a Right angled triangle of base b and height h,

$\sf\longrightarrow \boxed{Area=\frac{1}{2}\times base\times height}$

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

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