What is the typical rate of seafloor spreading
1 answer:
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Answer:
Line OX = 7.09
Explanation:
The question is incomplete. Find attached the complete question.
SOLUTION
Given:
Line OL is perpendicular to DX
Line DX = 13
Line PO = the hypotenuse of the larger triangle
Line PO = 10
Line BO = hypotenuse of smaller triangle
Line BO = 12
To find length of line OX, we would apply rule of similar and congruent triangles.
∆DPO = ∆XBO
<POD = <BOX
Line DP is parallel to line XB
Since length of side DX = 13
Side DX = side DO + side OX
Let side DO = x
Side OX = 13-x
(Adj of ∆DPO)/(Adj of ∆XBO) = hyp of ∆DPO/hyp of ∆XBO
DO/OX = PO/BO
x/ (13-x) = 10/12
12x = 10(13-x)
12x = 130-10x
12x +10x = 130
22x = 130
x = 130/22 = 65/11
x is approximately = 5.91
Therefore line OX = 13 - 5.91
Line OX = 7.09
