ABCD IS A KITE. CA IS PERPENDICULAR TO DB. BC IS 6 CM. CD IS 3 CM. DA IS 5 CM AND AB IS X. FIND X
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Answer:
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Step-by-step explanation:
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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.