Area of Semi Circle With an Inscribed Triangle
$begingroup$
Suppose we have a triangle with a right angle at its height, with side a, 10 inches, side b, unknown, and side c, 24 inches; inscribed in a semi-circle. 
Now, I'm asked to find the area of the semi circle only, in terms of $pi$. I was already given the answer, which is $84.5pi -120 in^2$, but I don't understand how to get this conclusion.
I tried finding the area this way, but I get a pretty different answer:
$A=frac{pi r^2}{2}, r=12 inches$
$$A=frac{144pi}{2}$$
$$A=72pi in^2$$
Then I know I have to subtract the triangle because I only want the area of the shaded region (the semi-circle), so
$$10^2+B^2=24^2$$
$$2sqrt{119}$$
So I end up getting an answer of $A=72pi-2sqrt{119} in^2$ which is clearly false. Could anyone explain where I went wrong?
geometry area
asked Dec 17 '18 at 12:46
Lex_iLex_i
807
$endgroup$
add a comment |
$begingroup$
Suppose we have a triangle with a right angle at its height, with side a, 10 inches, side b, unknown, and side c, 24 inches; inscribed in a semi-circle.
Now, I'm asked to find the area of the semi circle only, in terms of $pi$. I was already given the answer, which is $84.5pi -120 in^2$, but I don't understand how to get this conclusion.
I tried finding the area this way, but I get a pretty different answer:
$A=frac{pi r^2}{2}, r=12 inches$
$$A=frac{144pi}{2}$$
$$A=72pi in^2$$
Then I know I have to subtract the triangle because I only want the area of the shaded region (the semi-circle), so
$$10^2+B^2=24^2$$
$$2sqrt{119}$$
So I end up getting an answer of $A=72pi-2sqrt{119} in^2$ which is clearly false. Could anyone explain where I went wrong?
geometry area
asked Dec 17 '18 at 12:46
Lex_iLex_i
807
$endgroup$
add a comment |
$begingroup$
Suppose we have a triangle with a right angle at its height, with side a, 10 inches, side b, unknown, and side c, 24 inches; inscribed in a semi-circle.
Now, I'm asked to find the area of the semi circle only, in terms of $pi$. I was already given the answer, which is $84.5pi -120 in^2$, but I don't understand how to get this conclusion.
I tried finding the area this way, but I get a pretty different answer:
$A=frac{pi r^2}{2}, r=12 inches$
$$A=frac{144pi}{2}$$
$$A=72pi in^2$$
Then I know I have to subtract the triangle because I only want the area of the shaded region (the semi-circle), so
$$10^2+B^2=24^2$$
$$2sqrt{119}$$
So I end up getting an answer of $A=72pi-2sqrt{119} in^2$ which is clearly false. Could anyone explain where I went wrong?
geometry area
asked Dec 17 '18 at 12:46
Lex_iLex_i
807
$endgroup$
Suppose we have a triangle with a right angle at its height, with side a, 10 inches, side b, unknown, and side c, 24 inches; inscribed in a semi-circle.
Now, I'm asked to find the area of the semi circle only, in terms of $pi$. I was already given the answer, which is $84.5pi -120 in^2$, but I don't understand how to get this conclusion.
I tried finding the area this way, but I get a pretty different answer:
$A=frac{pi r^2}{2}, r=12 inches$
$$A=frac{144pi}{2}$$
$$A=72pi in^2$$
Then I know I have to subtract the triangle because I only want the area of the shaded region (the semi-circle), so
$$10^2+B^2=24^2$$
$$2sqrt{119}$$
So I end up getting an answer of $A=72pi-2sqrt{119} in^2$ which is clearly false. Could anyone explain where I went wrong?
geometry area
geometry area
asked Dec 17 '18 at 12:46
Lex_iLex_i
807
asked Dec 17 '18 at 12:46
Lex_iLex_i
807
edited Dec 17 '18 at 13:16
Lex_i
asked Dec 17 '18 at 12:46
Lex_iLex_i
807
asked Dec 17 '18 at 12:46
Lex_iLex_i
807
asked Dec 17 '18 at 12:46
Lex_iLex_i
807
807
add a comment |
add a comment |
1 Answer
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$begingroup$
The question is poorly worded - it is not clear which side of the triangle is the longest (the hypotenuse).
You are assuming that the 24 inch side of the triangle is the longest side and is therefore the diameter of the semi circle.
The book answer is assuming that the unknown side of the triangle is the longest side. In this case the length of this side is $sqrt{10^2+24^2}=26$ inches, the radius of the semicircle is $13$ inches, the area of the triangle is $frac{10 times 24}{2} = 120$ square inches and the area of the semicircle outside of the triangle is
$frac{13^2pi}{2}-120$ square inches.
answered Dec 17 '18 at 13:01
gandalf61gandalf61
8,801725
$endgroup$
$begingroup$
Here is an image of the figure. ibb.co/2M2gwQG
$endgroup$
– Lex_i
Dec 17 '18 at 13:03
$begingroup$
@Lex_i Please include that in the question body rather posting it in a comment, and please use SE's built-in image upload because external links tend to be, in general, unstable.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:11
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
The question is poorly worded - it is not clear which side of the triangle is the longest (the hypotenuse).
You are assuming that the 24 inch side of the triangle is the longest side and is therefore the diameter of the semi circle.
The book answer is assuming that the unknown side of the triangle is the longest side. In this case the length of this side is $sqrt{10^2+24^2}=26$ inches, the radius of the semicircle is $13$ inches, the area of the triangle is $frac{10 times 24}{2} = 120$ square inches and the area of the semicircle outside of the triangle is
$frac{13^2pi}{2}-120$ square inches.
answered Dec 17 '18 at 13:01
gandalf61gandalf61
8,801725
$endgroup$
$begingroup$
Here is an image of the figure. ibb.co/2M2gwQG
$endgroup$
– Lex_i
Dec 17 '18 at 13:03
$begingroup$
@Lex_i Please include that in the question body rather posting it in a comment, and please use SE's built-in image upload because external links tend to be, in general, unstable.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:11
add a comment |
$begingroup$
The question is poorly worded - it is not clear which side of the triangle is the longest (the hypotenuse).
You are assuming that the 24 inch side of the triangle is the longest side and is therefore the diameter of the semi circle.
The book answer is assuming that the unknown side of the triangle is the longest side. In this case the length of this side is $sqrt{10^2+24^2}=26$ inches, the radius of the semicircle is $13$ inches, the area of the triangle is $frac{10 times 24}{2} = 120$ square inches and the area of the semicircle outside of the triangle is
$frac{13^2pi}{2}-120$ square inches.
answered Dec 17 '18 at 13:01
gandalf61gandalf61
8,801725
$endgroup$
$begingroup$
Here is an image of the figure. ibb.co/2M2gwQG
$endgroup$
– Lex_i
Dec 17 '18 at 13:03
$begingroup$
@Lex_i Please include that in the question body rather posting it in a comment, and please use SE's built-in image upload because external links tend to be, in general, unstable.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:11
add a comment |
$begingroup$
The question is poorly worded - it is not clear which side of the triangle is the longest (the hypotenuse).
You are assuming that the 24 inch side of the triangle is the longest side and is therefore the diameter of the semi circle.
The book answer is assuming that the unknown side of the triangle is the longest side. In this case the length of this side is $sqrt{10^2+24^2}=26$ inches, the radius of the semicircle is $13$ inches, the area of the triangle is $frac{10 times 24}{2} = 120$ square inches and the area of the semicircle outside of the triangle is
$frac{13^2pi}{2}-120$ square inches.
answered Dec 17 '18 at 13:01
gandalf61gandalf61
8,801725
$endgroup$
The question is poorly worded - it is not clear which side of the triangle is the longest (the hypotenuse).
You are assuming that the 24 inch side of the triangle is the longest side and is therefore the diameter of the semi circle.
The book answer is assuming that the unknown side of the triangle is the longest side. In this case the length of this side is $sqrt{10^2+24^2}=26$ inches, the radius of the semicircle is $13$ inches, the area of the triangle is $frac{10 times 24}{2} = 120$ square inches and the area of the semicircle outside of the triangle is
$frac{13^2pi}{2}-120$ square inches.
answered Dec 17 '18 at 13:01
gandalf61gandalf61
8,801725
answered Dec 17 '18 at 13:01
gandalf61gandalf61
8,801725
answered Dec 17 '18 at 13:01
gandalf61gandalf61
8,801725
answered Dec 17 '18 at 13:01
gandalf61gandalf61
8,801725
8,801725
$begingroup$
Here is an image of the figure. ibb.co/2M2gwQG
$endgroup$
– Lex_i
Dec 17 '18 at 13:03
$begingroup$
@Lex_i Please include that in the question body rather posting it in a comment, and please use SE's built-in image upload because external links tend to be, in general, unstable.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:11
add a comment |
$begingroup$
Here is an image of the figure. ibb.co/2M2gwQG
$endgroup$
– Lex_i
Dec 17 '18 at 13:03
$begingroup$
@Lex_i Please include that in the question body rather posting it in a comment, and please use SE's built-in image upload because external links tend to be, in general, unstable.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:11
$begingroup$
Here is an image of the figure. ibb.co/2M2gwQG
$endgroup$
– Lex_i
Dec 17 '18 at 13:03
$begingroup$
Here is an image of the figure. ibb.co/2M2gwQG
$endgroup$
– Lex_i
Dec 17 '18 at 13:03
$begingroup$
@Lex_i Please include that in the question body rather posting it in a comment, and please use SE's built-in image upload because external links tend to be, in general, unstable.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:11
$begingroup$
@Lex_i Please include that in the question body rather posting it in a comment, and please use SE's built-in image upload because external links tend to be, in general, unstable.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:11
add a comment |
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