Hello!

Kite is a quadrilateral which has two adjacent pairs of equal sides.

It is known that diagonals of a kite have an intersection (denote it O) and are perpendicular. Also, the diagonal which separates two pairs of equal sides bisects the other diagonal.

We don't know whether AB=AD and CB=CD, or BA=BC and DA=DC.

Please refer to the picture uploaded.

Let's prove that the first option is impossible. Consider the triangle ABC. It has a height from B of length 11.6/2=5.8cm. Then AB>=5.8 and BC>=5.8. Also, because `/_` B=116°>90°,

`AB^2+BC^2 lt AC^2 = 6.4^2`

(by the law of cosine, cos(116°)<0).

But `AB^2+BC^2gt=2*5.8^2,` so we get `2*5.8^2 lt 6.4^2,`

which is false (the left is 67.28, the right is 40.96). This contradiction proves that AB=AD and CB=CD is impossible.

Now consider the second option, BA=BC and DA=DC. Then BD bisects AC and AO=AC/2=3.2cm.

Triangle AOB is right (angle AOB is right), and BO bisects `/_`B, so `/_`ABO=116°/2=58°.

Then AO/AB=sin58°, `AB=(AO)/sin(58°)=3.2/sin(58°).`

Also AO/BO=tan58°, `BO=(AO)/tan58°=3.2/tan58°.`

So `DO=BD-BO=11.6-3.2/tan58°,`

and `AD=sqrt(AO^2+DO^2)=sqrt(3.2^2+(11.6-3.2/tan58°)^2).`

Now compute these lengths apporoximately:

`AB=3.2/sin58° approx3.77cm,`

`BO=3.2/tan58° approx 2.00cm,`

`DO approx 11.6-2.00 = 9.6cm,`

`AD approx sqrt(3.2^2+9.6^2) approx 10.12cm.`

The answer: AB=BC approx 3.77.cm, AD=DC approx 10.12cm.

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