Question

pls solve 27 and 26 no useless answers i will talk to a moderator and remove you from brainly, you wont be able to join again if you send useless answer solve it and send in notebook with the help of screenshot

Answer 1

$\mathfrak{\huge{\pink{\underline{Solution \: 1:-}}}}$

$\large{\underline{\rm{\purple{Given:-}}}}$

Length of the carpet = $\sf 6\dfrac{2}{3} \: m$

Breadth of the carpet = 4 m

Length of the room = $\sf 5\dfrac{1}{3} \: m$

Breadth of the room = $\sf 3\dfrac{1}{3} \: m$

$\large{\underline{\rm{\purple{To \: Find:-}}}}$

The fraction of area that have to be cut off to fit the carpet into the floor.

$\large{\underline{\rm{\purple{Analysis:-}}}}$

Find the area of the carpet and the room.

Subtract the area of the room from the area of the carpet to get the fraction of area that have to be cut off to fit the carpet into the floor.

$\large{\underline{\rm{\purple{Answer:-}}}}$

Area of the carpet = Length × Breadth

Area of the carpet = $\sf 6\dfrac{2}{3} \times 4$

$\implies \sf \dfrac{20}{3} \times 4$

$\implies \sf \dfrac{80}{3}\: m^{2}$

Next,

Area of the room = Length × Breadth

Area of the room = $\sf 5\dfrac{1}{3} \times \sf 3\dfrac{1}{3}$

$\implies \sf \dfrac{16}{3} \times \dfrac{10}{3}$

$\implies \sf \dfrac{160}{9} \: m^{2}$

Now, finding the area to be cut off,

Area of carpet to be cut off = Area of carpet − Area of the room

Area of carpet to be cut off = $\sf \dfrac{80}{3} -\dfrac{160}{9}$

Finding the LCM of their denominators,

LCM = 9

Making the denominators equal to subtract,

$\sf \dfrac{80}{3} \times 3=\dfrac{240}{9}$

∴ $\sf \dfrac{240}{9}-\dfrac{160}{9}$

$\sf = \dfrac{80}{9}$

$= \underline{\underline{\sf 8\dfrac{8}{9} \: m^{2} }}$

Therefore $\sf 8\dfrac{8}{9} \: m^{2}$ of carpet should be cut off.

$\mathfrak{\huge{\pink{\underline{Solution \: 2:-}}}}$

$\large{\underline{\rm{\purple{Given:-}}}}$

Length of the park = 80 m

Breadth of the park = 60 m

$\large{\underline{\rm{\purple{To \: Find:-}}}}$

Distance shorter the route across the park than the route around its edges.

$\large{\underline{\rm{\purple{Analysis:-}}}}$

Find the longer route and the shorter route.

Subtract the shorter route from the longer route to get the distance the route across the park than the route around it's edge.

$\large{\underline{\rm{\purple{Answer:-}}}}$

Using Pythagoras theorem,

Diagonal² = Length² + Breadth²

$\implies \sf 60^{2}+80^{2}$

$\implies \sf 3600+6400$

$\implies \sf 10000$

Diagonal = $\sf \sqrt{10000} = 100 \: m$

Therefore,

Shorter route = 100 metres

Longer route = Length + Breadth

$\implies \sf 60+80$

$\implies \sf 140 \: m$

Distance the route shorter = Longer route − Shorter route

Distance the route shorter = $\sf 140-100$

$\implies \sf \underline{\underline{40 \: m}}$

Hence, the route is shorter by 40 metres.