A quadri lateral ABCD is circumscribed
to ellipse whose diagonals passes.
through opposite foci of ellipse
AB=5 BC=6 CD=7 find AD
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A quadrilateral ABCD is circumscribed to ellipse whose diagonals passes.
through opposite foci of ellipse AB=5 BC=6 CD=7 and AD=2√3.
Given,
AB=5
BC=6
CD=7
Let AD = x
Since the quadrilateral ABCD is circumscribed to ellipse, we can divide this quadrilateral into 2 right angles triangles,
quad ABCD = tri ABC + tri ADC
Δ ABC and Δ ADC
In Δ ABC,
AC^2 = AB^2 + BC^2
AC^2 = 5^2 + 6^2
AC^2 = 25 + 36
AC^2 = 61
AC = √61
Now, consider,
In ΔADC,
AC^2 = AD^2 + DC^2
(√61 )^2 = x^2 + 7^2
61 = x^2 + 49
61 - 49 = x^2
12 = x^2
x = √12
∴ AD = √12 = 2√3
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