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PLEASE assist me in this geometry problem !!!

PLEASE assist me in this geometry problem !!!-example-1

2 Answers

2 votes
15.27in^2

Using the Pythagorean theorem, we can find the hypotenuse of the triangle, or base, which is also convieniently the diameter of the semicircle. Since 3, 4, 5, is a pythagorean triple, and 6 and 8 are multiples of that triple, then we can find the third multiple which is 10. (You can still use the Pythagorean theorem, but this method is faster for this case.)

Since the base/diameter is 10 in, the radius of the semicircle is 5 in (radius is 1/2 diameter)

Using the semicircle area formula, A = 1/2pir^2, we can substitute r and find the area.

Thus, A = 1/2(3.141)(5)^2 = 39.27

Since we have to find the shaded area, we will subtract the area of the triangle from the semicircle. Finding the area of the triangle, where A = 1/2 x b x h and b = 8 and h = 6,
A = 24in^2

Thus, the area of the shaded region is the area of the semicircle subtracted by the area of the triangle, or

39.27in^2 - 24in^2 = 15.27 in^2
User Selia
by
4.9k points
3 votes

Answer:

Solution given:

p=6in

b=8in

By using Pythagoras law

diameter=
√(6²+8²)=10in

r=5in

Now

Area of shaded region=area of semi circle- area of triangle

=½(πr²)-½p*b

=½(π×5²)-½(6*8)

=15.269

=15.3in² is a required answer.

User Grimgrom
by
5.5k points