Two small circles are drawn that touch each other, and both circles touch the large circle. 7 in. 7 in. Which is closest to the area of the shaded region? *the photo is much better*
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308 inches

Step-by-step explanation:

since the two radiuses of the small circles become the diameter of the big circle, you can substitute 14 into the equation A= pi r^2 to get A=pi(14)^2, which gets you 615 inches. But then you have to subtract the area of two of the small circles, which if you add them two up, you get 307.87 inches. Then, you subtract the area of the big circle with the two small circles to get 307.87, which is closest to 308 inches.

Area of shaded region = 307.88 inch ^2

= 308 inch^2

Step-by-step explanation:

Radius of smaller circle = 7 inch radius

Area of SC = π * 7^2

= 153.93804

= 153.93 inch^2

2 x 7 = 14   (for main radius as two circles touching each other inscribed

would translate as R = 2r.

Area of two circles = 2 * 153.93804 = 307.87608

= 307.88 inch^2

Area of whole large circle

Area LC  = π * 14^2

= 615.75216

Area of shaded region

Area of Sh Region = 615.75216 - 307.87608 =307.87608

= 307.88 inch ^2

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

now, we can cross multiply

and we get

so, this could be step to prove it