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3 years ago

13

How do the surface areas of similar prisms compare when dimensions are doubled?

1 answer:

zhuklara [117]3 years ago

6 0

Answer:

The surface area of the larger prism is 4 times the surface area of the smaller one.

(which agrees with answer B in your list)

Step-by-step explanation:

Notice that the lateral surface area of prisms always consists of parallelograms, whose areas are given by the product base times height. = B x H

Therefore if the dimensions of the prism double, then the base and height of the parallelograms will double as well, making such individual ares of the lateral faces go from:

B x H to --> 2 B x 2 H = 4 (B x H)

that is 4 times the original lateral face's area.

with the faces for the top and bottom bases of the prism, something similar happens, so we conclude that the surface area of the larger prism is 4 times the surface area of the smaller one.

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In a parallelogram ABCD point K belongs to diagonal BD so that BK:DK=1:4. If the extension of AK meets BC at point E, what is th

olya-2409 [2.1K]

Answer:

$\frac{BE}{EC} =\frac{1}{3}$

Step-by-step explanation:

In the diagram below we have

ABCD is a parallelogram. K is the point on diagonal BD, such that

$\frac{BK}{CK} =\frac{1}{4}$

And AK meets BC at E

now in Δ AKD and Δ BKE

∠AKD =∠BKE ( vertically opposite angles are equal)

since BC ║ AD and BD is transversal

∠ADK = ∠KBE ( alternate interior angles are equal )

By angle angle (AA) similarity theorem

Δ ADK and Δ EBK are similar

so we have

$\frac{AD}{BE} =\frac{DK}{BK}$

$\frac{AD}{BE} =\frac{4}{1}$

$\frac{BC}{BE}=\frac{4}{1}$ ( ABCD is parallelogram so AD=BC)

$\frac{BE+EC}{BE}=\frac{4}{1}$ ( BC= BE+EC)

$\frac{BE}{BE} +\frac{EC}{BE}=\frac{4}{1}$

$1+\frac{EC}{BE}=4$

$\frac{EC}{BE}=3$ ( subtracting 1 from both side )

$\frac{EC}{BE}=\frac{3}{1}$

taking reciprocal both side

$\frac{BE}{EC} =\frac{1}{3}$

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