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ryzh [129]
3 years ago
7

What is the area of the figure? Enter your answer in the box. 400 in²

Mathematics
2 answers:
lesya692 [45]3 years ago
7 0

Answer:

should be 112 cm^2

Step-by-step explanation:

10cm*10cm+(1/2(4*6)

dem82 [27]3 years ago
5 0

Answer:

112

Step-by-step explanation:

To make this easier we can pull apart the two know shapes which is the square and triangle

the formula to find the area square is length x width

so 10 x 10=100 this is the area of the triangle

the picture doesn't show the measurements of the triangle but we can still find it

16 in goes from the top of the triangle to the bottom of the square and we know the right side of the square is 10 in so to find the length of the left side of the triangle we can subtract 10 from 16 which is 6.

now we need to find the bottom part of the triangle we can see that it measures 6 in to the right of the triangle and we know that the side of the square is 10 in so all we do is subtract 6 from 10 and you get 4

the formula to find the area of a triangle is height x base divided by 2

so 6 x 4 divided by 2 is 12

now we know the area of the triangle and square is 12 and 100 so we add these together to get 112

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Orbet used 24 square tiles to make a rectangle. The perimeter of the rectangle is 20 units. What would be the perimeters of the
Dmitriy789 [7]

Answer:

The perimeter of other rectangles are 50, 28, 22 units

The area of all possible rectangles are same and equals to 24 unit²

Step-by-step explanation:

Given - Orbet used 24 square tiles to make a rectangle. The perimeter of the rectangle is 20 units.

To find -  What would be the perimeters of the other rectangles Orbet could make using only these 24 tiles? How do the areas of the rectangles compare?

Proof -

We know that

Are of Rectangle = Length × Breadth

And Perimeter of Rectangle = 2 (Length + Breadth )

Now,

Given that,

Orbet used 24 square tiles

And perimeter of the rectangle made = 20

So,

Possible Length of rectangle = 6

Possible breadth of Rectangle = 4

Or Vice-versa.

Total number of Rectangles possible = 4

Possibilities are -  1 × 24, 2 × 12, 3 × 8, 4 × 6

Case I :

Length of rectangle = 1

Breadth of rectangle = 24

∴ Perimeter of rectangle = 2(1 + 24) = 2(25) = 50 units

Area of rectangle = 1 × 24 = 24 unit²

Case II :

Length of rectangle = 2

Breadth of rectangle = 12

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Area of rectangle = 2 × 12 = 24 unit²

Case III :

Length of rectangle = 3

Breadth of rectangle = 8

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Area of rectangle = 3 × 8 = 24 unit²

Case IV :

Length of rectangle = 4

Breadth of rectangle = 6

∴ Perimeter of rectangle = 2(4 + 6) = 2(10) = 20 units

Area of rectangle = 4 × 6 = 24 unit²

∴ we get

The perimeter of other rectangles are 50, 28, 22 units

The area of all possible rectangles are same and equals to 24 unit²

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Answer:

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