Answer:
3. 150.72 in²
4. 535.2cm²
Step-by-step Explanation:
3. The solid formed by the net given in problem 3 is the net of a cylinder.
The cylinder bases are the 2 circles, while the curved surface of the cylinder is the rectangle.
The surface area = Area of the 2 circles + area of the rectangle
Take π as 3.14
radius of circle = ½ of 4 = 2 in
Area of the 2 circles = 2(πr²) = 2*3.14*2²
Area of the 2 circles = 25.12 in²
Area of the rectangle = L*W
width is given as 10 in.
Length (L) = the circumference or perimeter of the circle = πd = 3.14*4 = 12.56 in
Area of rectangle = L*W = 12.56*10 = 125.6 in²
Surface area of net = Area of the 2 circles + area of the rectangle
= 25.12 + 125.6 = 150.72 in²
4. Surface area of the net (S.A) = 2(area of triangle) + 3(area of rectangle)
= 
Where,
b = 8 cm
h = ![\sqrt{8^2 - 4^2} = \sqrt{48} = 6.9 cm} (Pythagorean theorem)w = 8 cm[tex]S.A = 2(0.5*8*6.9) + 3(20*8)](https://tex.z-dn.net/?f=%20%5Csqrt%7B8%5E2%20-%204%5E2%7D%20%3D%20%5Csqrt%7B48%7D%20%3D%206.9%20cm%7D%20%28Pythagorean%20theorem%29%3C%2Fp%3E%3Cp%3Ew%20%3D%208%20cm%3C%2Fp%3E%3Cp%3E%5Btex%5DS.A%20%3D%20%202%280.5%2A8%2A6.9%29%20%2B%203%2820%2A8%29)



12 + 36 = 48
your answer is 48 inch
hope this helps
The volume would be 201.06
Answer:
A.
Step-by-step explanation:
<span>19.8 m
We have 2 similar triangles. For the 1st triangle, we have two legs, one being the man's height of 1.8m and the second being the length of his shadow at 6m. So we have the ratio 1.8/6. The other triangle is comprised of the pole's height and length of the pole's shadow which will be 60m + 6m = 66meters. So we have another ratio of x/66 which has the same value as the first ratio. So let's solve for x.
1.8/6 = x/66
66*1.8/6 = x
11*1.8 = x
19.8 = x
So the light pole is 19.8 meters tall.</span>