A right cylinder has a radius of 6 m and a surface area of 84 m². Find the surface area of a similar cylinder with a radius of 2
1 answer:
Answer:
The surface area of the second cylinder is equal to
Step-by-step explanation:
we know that
If two figures are similar then
the ratio of their surfaces areas is equal to the scale factor squared
Let
z-------> the scale factor
x------> the surface area of the smaller cylinder (second cylinder)
y-------> the surface area of the original cylinder (first cylinder)
so
Step 1
Find the scale factor
Step 2
Find the surface area of the second cylinder
we have
substitute and solve for x
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See the attached figure.
========================
AB = 10 , FD = 3
∵ D is the midpoint of AB, and F is the mid point of CB
∴ FD // AC , FD = 0.5 AC
∵ Δ ABC is a right triangle at C
∴ FD ⊥ BC
∴ BD = 0.5 AB = 5
∴ in Δ FDB ⇒⇒ BF² = BD² - FD² = 5² - 3² = 16
∴ BF = √16 = 4
∵ F is the mid point of CB
∴ CF = BF = 4 , and CB = 2 BF = 2*4 = 8
∵ D is the midpoint of AB, and E is the mid point of AC
∴ DE // CB , and DE = 0.5 CB = 0.5 * 8 = 4
∴ T<span>he length of line ED is 4
</span>
========================
AB = 10 , FD = 3
∵ D is the midpoint of AB, and F is the mid point of CB
∴ FD // AC , FD = 0.5 AC
∵ Δ ABC is a right triangle at C
∴ FD ⊥ BC
∴ BD = 0.5 AB = 5
∴ in Δ FDB ⇒⇒ BF² = BD² - FD² = 5² - 3² = 16
∴ BF = √16 = 4
∵ F is the mid point of CB
∴ CF = BF = 4 , and CB = 2 BF = 2*4 = 8
∵ D is the midpoint of AB, and E is the mid point of AC
∴ DE // CB , and DE = 0.5 CB = 0.5 * 8 = 4
∴ T<span>he length of line ED is 4
</span>