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A quadrilateral ABCD is drawn to circumscribe a circle. If AB =12 cm, BC = 15 cm and CD = 14 cm, find AD.​

Answer»

<P>Answer:

Given: Let ABCD is a quadrilateral which circumscribe the circle with CENTER 'O'. The quadrilateral ABCD touches circle at point P , Q , R and S.

To proof: AB+CD = AD+BC

Proof:

AP=AS…(i)

BP=BQ…(ii)

CR=CQ…(III)

DR=DS…(iv)

Adding equation (i),(ii),(iii) and (iv) we get…

AP+BP+CR+DR=AS+BQ+CQ+DS

According to FIGURE [AB=AP+PB, AD=AS+SD, CD=DR+CR, BC=BQ+CQ]

Or, AB+CD=AD+BC (Proved).

In this question a quadrilateral ABCD is drawn to circumscribe a circle.

Hence,

AB+CD=AD+BC

Or, 12+14=AD+15

Or, AD = 11 cm



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