*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.
