Determine the area of each composite figure.
The figure is a circle inside a semicircle just like the picture in this website below, the r in the picture in the website is not given.
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There is no space between the h and ttps:
The diameter of the circle is 10 cm and the diameter (the base) is 35 cm.
My answer:
A=(πD^2)/4
A=(π(35)^2)/4
A=962.11cm
A=(πD^2)/4
A=(π(10)^2)/4
A=78.54
962.11 - 78.54=883.57
my final answer is 883.57.
My textbook says 402.5cm^2
Please help me!
Usually the area of a circle is given in terms of its radius, not its diameter.
However your areas are correct,
BUT, don't you have a semicircle?
so it should be (1/2)(962.11cm^2) - 78.54 cm^2
= 402.52 cm^2
It is a circle inside a semicircle
To determine the area of the composite figure, we need to break it down into simpler shapes and then find the individual areas. In this case, we have a circle inside a semicircle.
First, let's find the area of the circle. The formula for the area of a circle is A = πr^2, where r is the radius. Given the diameter of the circle is 10 cm, we can find the radius by dividing the diameter by 2: r = 10 cm / 2 = 5 cm. Now, the area of the circle is A = π(5 cm)^2 = 25π cm^2.
Next, let's find the area of the semicircle. The formula for the area of a semicircle is A = (πr^2) / 2. Since the diameter of the semicircle is given as 35 cm, the radius is 35 cm / 2 = 17.5 cm. Now, the area of the semicircle is A = (π(17.5 cm)^2) / 2 = 306.25π cm^2.
To find the area of the composite figure, we need to subtract the area of the circle from the area of the semicircle. So, the total area is:
A = 306.25π cm^2 - 25π cm^2
A = 281.25π cm^2
To get the numerical value of the area, we need to substitute π with the approximation of 3.14159. Therefore:
A ≈ 281.25 * 3.14159 cm^2
A ≈ 883.6 cm^2
So, the correct answer is approximately 883.6 cm^2.
If your textbook states the answer as 402.5 cm^2, there may be some discrepancy or mistake in the problem statement or the solution provided. It is worth double-checking the given dimensions and the formula used in the textbook solution.