Re: In a quadrilateral ABCD, AB=AD=5, BC=6 and CD=1.
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09 Jan 2022, 10:08
It all depends on the triangle that can be constructed by the two sides AB and AD, and the diagonal BD.
But BD will get restricted by the two sides of triangle BCD. BD cannot be greater than sum of other two sides so <6+1 or <7, and cannot be lesser than the difference of other two sides, that is >6-1 or >5.
Now let’s get back to the triangle ABD.
1) Least value of angle A is when we take least value of opposite side BD.
So the triangle becomes 5,5,>5.
Almost an equilateral triangle but with the side BD slightly more, so the opposite angle A will be slightly more than 60.
2) Maximum value of A.
Say A is 90, then BD becomes hypotenuse and would be equal to \(\sqrt{5^2+5^2}=5\sqrt{2}=5*1.414=7\)
But BD <6, so A has to be less than 90.
Range : 60<A<90
Only 75 possible
C