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ale4655 [162]
3 years ago
6

△HKO∼△FGO Two right triangles. The first triangle is labeled triangle HKO. Angle H is the right angle. Side KO which is the hypo

tenuse has a length of 10 units. The other right triangle is labeled triangle FGO. In this triangle, angle F is the right angle and side GO is the hypotenuse with a length of 15 units. The triangles meet at angle O forming a vertical angle. Angles H and F are both right angles. Points H comma O comma and G are on the same line. Points F comma O comma and K are on the same line.
What is the ratio of the area of △HKO to the area of △FGO ? Express your answer as a fraction. Enter your answer in the box.
Mathematics
1 answer:
Leya [2.2K]3 years ago
5 0
<h2>the area is the 74 that is the answer </h2>
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Surface Area of Cylinder=2πrh+2πr2
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  • Find the surface area when r is 8 inches and h is 8 inches.

\qquad A. 160π in²

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\qquad C. 288π in²

\qquad D. 256π in² ☑

We are given –

\qquad ⇢Radius of cylinder , r = 8 inches

\qquad⇢ Height of cylinder, h = 8 inches.

We are asked to find surface area of the given cylinder.

Formula to find the surface cylinder given by –

\bf \star \pink{ Surface\: Area_{(Cylinder)} = 2\pi rh +2\pi r^2 }

Now, Substitute given values –

\sf  \twoheadrightarrow  Surface\: Area_{(Cylinder)} = 2\pi rh +2\pi r^2

\sf  \twoheadrightarrow Surface\: Area_{(Cylinder)}  = 2 \pi \times 8 \times 8 + 2\pi \times 8^2

\sf  \twoheadrightarrow Surface\: Area_{(Cylinder)} = 2 \pi \times 8^2 + 2\pi \times 8^2

\sf  \twoheadrightarrow Surface\: Area_{(Cylinder)} = 2\pi \times 64 +2\pi \times 64

\sf  \twoheadrightarrow Surface\: Area_{(Cylinder)}  =128 \pi + 128\pi

\purple{\bf  \twoheadrightarrow Surface\: Area_{(Cylinder)}  = 256\pi \: in^2 }

  • Henceforth,Option D is the correct answer.

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