<h3>
Answer:</h3>
- x = 13; NP = 2 18/23 ≈ 2.8; NL = 5 5/23 ≈ 5.2
- D. x = 17, y = 5
<h3>
Step-by-step explanation:</h3>
1. In order for this problem to be workable, we need to assume PQ ║ ML. Then ΔPNQ ~ ΔLNM and ∠L = ∠P = 60°.
... ∠L = 60°
... (3x+21)° = 60°
... 3x = 39 . . . . . . . divide by °, subtract 21
... x = 13 . . . . . . . . .divide by 3
The side lengths of similar triangles are proportional, so we have ...
... (3.2 cm)/y = (6 cm)/(8-y)
Multiplying by the product of the denominators, and dividing by (cm), we have ...
... 3.2(8 -y) = 6y
... 3.2×8 = 9.2y . . . . . add 3.2y
... y = 3.2×8/9.2 = 64/23 = 2 18/23 ≈ 2.78 = NP
Then ...
... 8-y = NL ≈ 5.22
In summary: x = 13; NP ≈ 2.8; NL ≈ 5.2
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2. If we assume figures ABCD and PSRQ are similar, we can find the values of the variables. We assume ∠C ≅ ∠R, so ...
... 4x +27° = 95°
... 4x = 68° . . . . . subtract 27°
... x = 17° . . . . . . . divide by 4
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AB/AD = PS/PQ . . . . . corresponding sides of similar figures are proportional
... 4y/(3y-5) = 10/5 . . . . . units of ft cancel
... 5×4y = 10(3y -5) . . . . . multiply by 5(3y-5)
... 20y = 30y -50 . . . . . . simplify
... 50 = 10y . . . . . . . . . . . add 50-20y
... 5 = y . . . . . . . . . . . . . . divide by 10
Using this value of y, we have ...
AB = 4y ft = 20 ft; AD = (3y-5) ft = 10 ft. Both these values are double the corresponding lengths on PSRQ.
In summary, x = 17°, y = 5. . . . . . (note that the ° symbol is appropriate for x)
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You should have your teacher show you how to work these problems <em>using only the given information</em>.
In the second problem, you cannot start with the assumption that the figures are similar, as that is what you're being asked to prove. Please note, too, that two sides and one angle of a quadrilateral are insufficient to show similarity.
