A balalaika is a Russian stringed instrument. Show that the triangular parts of the two balalaikas are congruent for x = 6.
1 answer:
*the diagram of the Russian stringed instrument is attached below.
Answer/Step-by-step explanation:
To show that the traingular parts of the two balalaikas instruments are congruent, substitute x = 6, to find the missing measurements that is given in both ∆s.
Parts of the first ∆:
WY = (2x - 2) in = 2(6) - 2 = 12 - 2 = 10 in
m<Y = 9x = 9(6) = 54°.
XY = 12 in
Parts of the second ∆:
m<F = 72°
HG = (x + 6) in = 6 + 6 = 12 in
HF = 10 in
m<G = 54°
m<H = 180 - (72° + 54°)
m<H = 180 - 126
m<H = 54°
From the information we have, let's match the parts that are congruent to each other in both ∆s:
WY ≅ FH (both are 10 in)
XY ≅ GH (both are 12 in)
<Y ≅ <G (both are 54°)
Thus, since two sides (WY and XY) and an included angle (<Y) of ∆WXY is congruent to two corresponding sides (FH and GH) and an included angle (<G) in ∆FGH, therefore, ∆WXY ≅ ∆FGH by the Side-Angle-Side (SAS) Congruence Theorem.
This is enough proof to show that the triangular parts of the two balalaikas are congruent for x = 6.
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Answer: |p-72% |≤ 4%
Step-by-step explanation:
Let p be the population proportion.
The absolute inequality about p using an absolute value inequality.:
, where E = margin of error,
= sample proportion
Given: A poll result of 72% with a margin of error of 4% indicates that p is most likely to be between 68% and 76% .
|p-72% |≤ 4%
⇒ 72% - 4% ≤ p ≤ 72% +4%
⇒ 68% ≤ p ≤ 76%.
i.e. p is most likely to be between 68% and 76% (.