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hichkok12 [17]
2 years ago
9

4. Find area of irregular shape 5. Find area of BLUE portion of shape

Mathematics
1 answer:
lisov135 [29]2 years ago
7 0

Answer:

4. Option 3: 65.1 cm^2

5. Option 3: 3.4 inches^2

Step-by-step explanation:

<u>Question 4 :</u>

We can see that there is a rectangle and a half circle in the picture.

So we will find the area of both separately

For the area of rectangle

Area=length*width

=8*5

=40 cm^2

We know that the line which is representing the diameter has length 8 cm

To find the radius

r=d/2

=8/2

=4 cm

As we cannot find the area of half circle, we will find the area of full circle and will divide it in half.

So,

area= πr^2

=3.14*(4)

=3.14*16

=50.24 cm^2

Area of half circle=(Area of full circle)/2

=50.24/2

=25.12 cm^2

So the area of shape will be

Area of rectangle+area of half circle

=40 cm^2+25.12 cm^2

=65.12 cm^2

Rounding off to nearest 10 will give 65.1 cm^2

So, option 3 is the correct answer ..

<u>Question 5 :</u>

To find the area of shaded region we have to find the area of circle and area of square in which it is inscribed

So,

area of circle= πr^2

=3.14*(2)^2

=3.14*4

=12.56 inches^2

And to find the area of square we need one of its side, we can clearly see that the diameter of circle will be equal to the side of square

So,

Side of square=s=r*2

=2*2

=4 inches

Area of square=s^2

=4^2

=16 inches^2

To find the area of shaded region we have to subtract the area of circle from the area of square

So,

Area of Shaded Region=Area of Square-Area of Circle

=16-12.56

=3.44 inches^2

Rounding off to the nearest 10 will give 3.4 inches^2 ..

So, option 3 is the correct answer ..

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Write a translation rule that maps point D ( 7 , − 3 ) onto point D ' ( 2 , 5 ). I NEED HELPPPP!
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Answer:

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Step-by-step explanation:

T _ { ? ,?}

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the coordinates of D can be represented as (x1,y1), and the coordinates of D' can be represented as (x,y).

you can simply take the difference in the x values and difference in the y values from the preimage to image.

like this:

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