Rule for any two right triangles ABC and DEF:
If \(\frac{AB}{DE} = \frac{BC}{EF}\), then \(\frac{AB}{DE} =\) \(\frac{BC}{EF} = \frac{AC}{DF}\).
As a result, the two triangles are SIMILAR.
In other words:
If two sets of corresponding sides are in the same ratio, then the third set of corresponding sides must also be in that ratio, with the result that the two triangles are similar.
Rule for two similar triangles:
If each side of the larger triangle is \(x\) times the corresponding side in the smaller triangle, then the area of the larger triangle is \(x^2\) times the area of the smaller triangle.NandishSS wrote:
(2)The width of the 18''-screen is 20% greater than that of the 15''-screen. We are not aware of the length?
So I marked as A
Prompt:
Ratio of the diagonals \(= \frac{greater-dagonal}{smaller-diagonal} = \frac{18}{15} = \frac{6}{5}\)
Statement 2:
Ratios of the widths \(= \frac{greater-width}{smaller-width} = \frac{1.2}{1} = \frac{6}{5}\)
The two diagonals and the two widths are in the same ratio.
As a result, the two lengths must also be in this ratio:
\(\frac{greater-length}{smaller-length} = \frac{6}{5}\)
Since all of the sides are in the same ratio, the two triangles are SIMILAR.
Since the length of each side in the larger side is \(\frac{6}{5}\) the length of the corresponding side in the smaller triangle, the area of the larger triangle is \((\frac{6}{5})^2\) the area of the smaller triangle.
By extension, the area of the larger screen is \((\frac{6}{5})^2\) the area of the smaller screen.
SUFFICIENT.
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