Math, asked by Anonymous, 5 months ago

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Answers

Answered by Anonymous

$\Large{\underbrace{\underline{\sf{Understanding\; the\; Concept}}}}$

Here in this question, concept of perimeter and area of rectangle as well as of square is used. To find the area and perimeter we have special formula for each type of quadrilateral. Sometimes perimeter and area seems to be a difficult concept. So let's understanding what they actually are!

Perimeter→ The sum of all sides of the given figure is called perimeter. It is also termed as the boundary of the figure.

Area→ The place or space occupied by given figure is called it's area.

So let's try to solve it by using formulas :)

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Given:-

• Length of building=178m

• Breadth of building=68m

• Height of building=28m

• Length of front door=3m

• Breadth of front door=1.5m

• Each side of window=1.5m

• Length of school compound=862m

• Breadth of school compound=629m

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To find:-

• Perimeter of the building

• Area of roof of the building

• Area of the front door

• Area of a window

• Perimeter of school compound

• Area of school compound

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Formula used:-

• Perimeter of rectangle=2(L+B)

[Here L refers to length and B refers to Breadth]

• Perimeter of square=4×Side

• Area of square=Side²

• Area of rectangle=Length×Breadth

• Perimeter of cuboid=4(L+B+H)

[Here, L=Length, B=Breadth, H=Height]

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Solution:-

In the 1st question we have to find perimeter of the building. The building is a 3rd dimensional figure and is in the shape of a cuboid.

So:-

★ Perimeter of the building=4(L+B+H)

Perimeter of the building=4(178m+68m+28m)

Perimeter of the building=4(274m)

Perimeter of the building=1096m

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In the 2nd question, we have to find area of the roof.

We can see that roof is in the shape of a rectangle.

So:-

Area of roof=Length×Breadth

Area of roof=178m×68m

Area of roof=12,104m²

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In the 3rd question, we have to find area of front door. The front door is in the shape of rectangle.

So:-

Area of front door=Length×Breadth

Area of front door=3m×1.5m

Area of front door=4.5m²

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In the 4th question, we have to find the area of a window. Window is in the shape of a square.

So:-

Area of a window=Side²

Area of a window=1.5m×1.5m

Area of a window=2.25m²

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In the 5th question, we have to find the perimeter of school compound. The school compound is in the shape of a rectangle.

So:-

Perimeter of school compound=2(L+B)

Perimeter of school compound=2(862m+629m)

Perimeter of school compound=2(1491m)

Perimeter of school compound=2,982m

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In the 6th question, we have to find area of school compound. School compound is in the shape of rectangle.

So:-

Area of school compound=Length×Breadth

Area of school compound=862m×629m

Area of school compound=5,42,198m²

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So the answers are:-

1.) 1096m

2.) 12,104m²

3.) 4.5m²

4.) 2.25m²

5.) 2982m

6.) 5,42,198m²

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Note→ The length and breadth which I used is different for each figure.

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Learn more with brainly:

https://brainly.in/question/15741392

https://brainly.in/question/19428360

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Answered by CɛƖɛxtríα

Required answers:-

The perimeter of the building is 1,096 m.
The area of the roof of the building is 12,104 m².
The area of the front door is 4.5 m².
The area of a window is 2.25 m².
The perimeter of the school compound is 2,982 m.
The area of the school compound is 5,42,198 m².

Step-by-step explanation:-

${\underline{\underline{\bf{1.\:Perimeter\:of\:the\: building:}}}}$

Given that, the shape of the building is cuboid and it's length is 178, breadth is 68 m and height is 28 m. It's perimeter can be found by adding the measures of all the edges. The edges can be grouped as- 4 lengths, 4 breadths and 4 heights. So, we can find the answer by inserting the given measures in:

$\leadsto{\sf{\purple{4(l+b+h)\:units}}}$

$\\ \:\:\:\:\::\implies{\sf{4\times (178+68+28)}}$

$\\ \:\:\:\:\::\implies{\sf{4\times (246+28)}}$

$\\ \:\:\:\:\::\implies{\sf{4\times 274}}$

$\\ \:\:\:\:\::\implies{\boxed{\frak{\red{1,096\:m}}}}$

$\:$

${\underline{\underline{\bf{2.\:Area\:of\:the\:roof:}}}}$

The shape of the roof is rectangle. And it's length and breadth are 178 m and 68 m, respectively. So, it's area can be found by applying the formula:

$\leadsto{\sf{\purple{lb\:sq.units}}}$

$\\ \:\:\:\:\::\implies{\sf{178\times 68}}$

$\\ \:\:\:\:\::\implies{\boxed{\frak{\red{12,104\:m^2}}}}$

$\:$

${\underline{\underline{\bf{3.\:Area\:of\:the\: front\:door:}}}}$

The shape of the front door is rectangle where its length is 3 m and breadth is 1.5 m. It's area equals on substituting the measures in the formula:

$\leadsto{\sf{\purple{lb\:sq.units}}}$

$\\ \:\:\:\:\::\implies{\sf{3\times 1.5}}$

$\\ \:\:\:\:\::\implies{\boxed{\frak{\red{4.5\:m^2}}}}$

$\:$

${\underline{\underline{\bf{4.\:Area\:of\:a\:window:}}}}$

The windows in the building are in the shape of square. And it's side is in the measure: 1.5 m. We're asked to find the area of a window. So, we can find it by using the formula:

$\leadsto{\sf{\purple{a^2\:sq.units}}}$

$\\ \:\:\:\:\::\implies{\sf{(1.5)^2}}$

$\\ \:\:\:\:\::\implies{\sf{1.5\times 1.5}}$

$\\ \:\:\:\:\::\implies{\boxed{\frak{\red{2.25\:m^2}}}}$

$\:$

${\underline{\underline{\bf{5.\:Perimeter\:of\:the\: school\: compound:}}}}$

The school compound is in rectangular shape and its length and breadth are 862 m and 629 m, respectively. What we're asked to find? It's perimeter! The answer can be obtained by inserting the measures of length and breadth in the formula:

$\leadsto{\sf{\purple{2(l+b)\:units}}}$