ChandlerBong wrote:
What is the ratio of the area of the smaller triangle to the larger triangle if x = 4 and y = 2?
A. 1:4
B. 1:3
C. 1:2
D. 2:3
E. 3:4
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Question: \(\frac{\text{Area of} \triangle CDE }{ \text{Area of} \triangle AHG}\)
\(\triangle CDE\)Given- CE = \(x\sqrt{2}\)
- DE = x
- \(\angle CDE = 90\)
\(CE^2 = CD^2 + DE^2\)
CD = x
Therefore \(\angle DEC = \angle CED = 45\)
Drop a perpendicular from point C to AH at point B. BH will be parallel to HG.
\(\angle BCD = 90\), therefore \(\angle ACB = \angle BAC = 45\)
Drop a perpendicular from point E to HG at point F. EF will be parallel to BG.
\(\angle DEF = 90\), therefore \(\angle FEG = \angle EGF = 45\)
\(\triangle ABC\) (45 - 45-90)
\(BC = y\)
\(AC = \sqrt{2}y\)
\(AB = y\)
\(\triangle EFG\) (45 - 45-90)
\(EF = y\)
\(EG = \sqrt{2}y\)
\(FG = y\)
\(\triangle AHG\)AH = x + y + y = 2y + x = 8 units
HG = x + y + y = 2y + x = 8 units
Area = \(\frac{1}{2} AH * HG = 32\) units
\(\triangle CDE\)CD = x = 4 units
DE = x = 4 units
Area = \(\frac{1}{2} CD * DE = 8\) units
Required Ratio = \(\frac{\text{Area of} \triangle CDE }{ \text{Area of} \triangle AHG}\) =\(\frac{ 8}{32}\) = \(\frac{1}{4}\)
Option A