Below is the graph of a trigonometric function. It has a minimum point at \left(-5.4,1.25\right)(−5.4,1.25)left parenthesis, min
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Answer:
5 units
Step-by-step explanation:
Let point O be the point of intersection of the kite diagonals.
|OF| = 2, |OH| = 5
|FH| = |OF| + |OH| = 2 + 5 = 7
FH and EG are the diagonals of the kite. Hence the area of thee kite is:
Area of kite EFGH = (FH * EG) / 2
Substituting:
35 = (7 * |EG|) / 2
|EG| * 7 = 70
|EG| = 10 units
The longer diagonal of a kite bisects the shorter one, therefore |GO| = |EO| = 10 / 2 = 5 units
x = |GO| = |EO| = 5 units
