Question

13. In the given figure, ABCD is a kite. The value of angle xº is
(A) 86°
(B) 100°
(C) 1040
(D) none of these
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Answer 1

Answer:

Step-by-step explanation:

Since ABCD is a kite,

AD = CD (adjacent sides are equal in a kite)

and the diagonals intersect at right angles

Mark the intersecting point of diagonals as O.

Now, in ∆ADC,

AD = CD

Hence ∆ is isosceles triangle.

So, angle DCA = angle DAC = 50°

Now, in ∆AOB,

angle AOB = 90° (diagonal intersect at right angles)

and angle OBA = 36°

So, by angle sum property,

angle OAB + 36 + 90 = 180°

=> angle OAB = 180 - 90 - 36

=> angle OAB = 54°

now, x = angle OAB + angle DAC

=> x = 54 + 50

=> x = 104°