All categories

2 years ago

Helllp please Find the area of The following figures. Dimensions are in centimeters.

Mathematics

2 answers:

lukranit [14]2 years ago

5 0

Answer:

$\huge\mathfrak\red{Answer...} \\ \\ area \: of \: square \: = side {}^{2} \\ \\ side = \: 6 \: cm \\ \\ area \: of \: square \: = 6 \times 6 \\ \\ area \: of \: square \: = 36 \: cm.sq \\ \\ area \: of \: semicircle \: = (1 \div 2)\pi \: r {}^{2} \\ \\ radius = 2 \: cm \\ \\ area \: of \: semicircle = 6.28 \: cm.sq \\ \\ area \: of \: whole \: figure \: = 36 + 6.28 \\ \\ area = 42.28 \: cm.sq$

AleksAgata [21]2 years ago

3 0

Answer:

Seperate the shapes!

Step-by-step explanation:

CIRCLE

First we want to find the area of the circle which would be:

\pi r^{2}

3.14159265359 x 2^{2}

3.14159265359 x 4cm

= 12.5663706144cm^{2} /2 (must divide by two because it is half of a circle)

= 6.28318530718cm^{2}

RECTANGLE/SQUARE

Find the area of the square.

AREA = L x W

Length = 2cm + ?

Width = 6cm

To find the length we must use the diameter of the circle (which is the radius but times it by 2)

To find diameter: r x 2

= 2cm x 2

= 4cm

Length = 4 + 2 = 6cm

Width = 6 cm

Area (Length x Width) = 6cm x 6cm

= 36cm^{2}

TOTAL AREA:

Just add the area of the circle and the area of the square together.

Area total = 36cm^{2} + 6.28318530718cm^{2}

= 42.2831853072cm^{2}

* Also round your total answer to the nearest tenth or hundreth if needed to do so.

You might be interested in

What is the area of this figure? Select from the drop-down menu to correctly complete the statement. The area of the figure is c

melomori [17]

Answer:

The answer is 146

Step-by-step explanation:

5 0

3 years ago

Which line is the best model for the data in the scatter plot?

brilliants [131]

Yeah the first one (A) seems like the best to me :)

4 0

3 years ago

Read 2 more answers

Graph the image of the figure after a dilation with a scale factor of 2 centered at (−2,−2). Use the Polygon tool to graph the q

kolbaska11 [484]

Answer:

see below

Step-by-step explanation:

Every vertex moves twice as far from the center of dilation as it is in the pre-image.

Perhaps the easiest image point to find is the one at lower left. In the pre-image it is 2 units left of the center of dilation, so the image point will be 2×2 = 4 units left of the center of dilation. It will be located at (-6, -2).

Other points on the image can be found either by reference to the center of dilation, or by reference to known image points. For example, the next point clockwise is 1 left and 4 up in the pre-image, so will be 2 left and 8 up from (-6, -2) in the image. That is, the pre-image point (-5, 2) becomes image point (-8, 6). You will note that (-5, 2) is 3 left and 4 up from the center of dilation, and that (-8, 6) is 6 left and 8 up from the center of dilation (twice as far away).