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sdas [7]
2 years ago
10

A community hall is in the shape of a cuboid the hall is 40m long 15m high and 3m wide. 10 litre paint covers 25m squared costs

?10. 1m squared floor tiles costs ?3. Work out the total costs of tiles and paints
Mathematics
1 answer:
krek1111 [17]2 years ago
5 0

Answer:

Total cost for tiles and paints is $924.  

Step-by-step explanation:

We have been given that a community hall is in the shape of a cuboid. The hall is 40m long 15m high and 3m wide.

The paint will be required for 4 walls and ceiling.

Let us find area of walls and ceiling.

\text{Area of walls and ceiling}=(2*40*15)+(2*3*15)+(40*3)

\text{Area of walls and ceiling}=1200+90+120

\text{Area of walls and ceiling}=1410

Therefore, the area of walls and ceiling is 1410 square meters.

Given: Cost for 10 litre of paint is $10 and 10 litre paint covers 25 square meter. Therefore,  

\text{ The total painting cost}=10*(\frac{1410}{25})

\text{ The total painting cost}=10*56.4=564

Therefore, the total painting cost is $564.  

Tiles will be required for floor. Let us find the area of floor.

\text{Area of floor} = 40*3\text{ square meters}

\text{Area of floor} =120\text{ square meters}

Given: 1m squared floor tiles costs $3. So,

\text{Total cost for tiles} = 3*120 = 360

Therefore, the total cost for tiles is $360.  

Now let us find combined total cost of tiles and paint.

\text{Combined total cost}= 564+360 = 924

Therefore, the combined total cost of tiles and paint is $924.

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Please answer all of them need this
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First Question

For a better understanding of the solution provided here please find the first attached file which has the diagram of the the isosceles trapezoid.

We dropped perpendiculars from C and D to intersect AB at Q and P respectively.

As can be seen in \Delta BCQ, we can easily find the values of CQ and BQ.

Since, Sin(75^0)=\frac{CQ}{8}

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In a similar manner we can find BQ as:

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All these values can be found in the diagram attached.

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Let us now consider\Delta AQC

We can apply the Pythagorean Theorem here to find the length of the diagonal AC which is the hypotenuse of \Delta AQC.

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Second Question

For this question we can directly apply the formula for the area of a triangle using sines which is as:

Area=\frac{1}{2}(First Side)(Second Side)(Sine of the angle between the two sides)

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Area=\frac{1}{2}\times 218.5\times 224.5\times sin(58.2^0)\approx20845 m^2

Therefore, Option D is the correct option.

Third Question

For this question we will apply the Sine Rule to the \Delta ABC given to us.

Thus, from the triangle we will have:

\frac{AB}{Sin(\angle C)}=\frac{BC}{Sin(\angle A)}

\frac{c}{Sin(\angle C)}=\frac{a}{Sin(\angle A)}

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This gives a to be:

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Fourth Question

Please find the second attachment for a better understanding of the solution provided her.

As can be clearly seen from the attached diagram, we can apply the Cosine Rule here to find the return distance of the plane which is CA.

AC=\sqrt{(AB)^2+(BC)^2-2(AB)(BC)\times Cos(\angle B)}

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Thus, Option D is the answer.





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