Find the area of the shape shown below HELP PLSSSSSSSSSSSS
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Answer:
D) 3.8 cm
Step-by-step explanation:
There are several ways this problem can be solved. Maybe the easiest is to use the Law of Cosines to find angle BAC. Then trig functions can be used to find the length of the chord.
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In triangle BAC, the Law of Cosines tells us ...
a² = b² +c² -2bc·cos(A)
A = arccos((b² +c² -a²)/(2bc)) = arccos((8² +6² -3²)/(2·8·6)) = arccos(91/96)
A ≈ 18.573°
The measure of half the chord is AB times the sine of this angle:
BD = 2(AB·sin(A)) ≈ 3.82222
The length of the common chord is about 3.8 cm.
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<em>Additional comment</em>
Another solution can be found using Heron's formula to find the area of triangle ABC. From that, its altitude can be found.
Area ABC = √(s(s-a)(s-b)(s-c)) . . . . where s=(a+b+c)/2
s=(3+8+6)/2 = 8.5
A = √(8.5(8.5 -3)(8.5 -8)(8.5 -6)) = √54.4375 ≈ 7.64444
The altitude of triangle ABC to segment AC is given by ...
A = 1/2bh
h = 2A/b = 2(7.64444)/8 = 1.911111
BD = 2h = 3.822222
