4. La figura se muestra un armazón arqueado de 50 metros de longitud con las alturas indicadas. Los "tirantes" verticales están
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*see attachment for the figure described
Answer:
5 units
Step-by-step explanation:
==>Given the figure attached below, let where FH and EG intercepted be K.
Since FH are midpoints of parallel lines, KE = KG = x.
Given that the area of the kite EFGH = 35 square units, and we know the length of one of the diagonals = HF = KF + KH = 2 + 5 = 7, we can solve for x using the formula for the area of a kite.
Area of kite = ½ × d1 × d2
Where d1 = KH = 7
d2 = EG = KE + KG = x + x = 2x
Area of kite EFGH = 35
THUS:
35 = ½ × 7 × 2x
35 = 1 × 7 × x
35 = 7x
Divide both sides by 7
35/7 = x
x = 5
