In given figure XY and X'Y' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X'Y' at B. Prove that ∠ AOB = 90°.
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Join OC
In Δ OPA and Δ OCA
OP = OC (radii of same circle)
PA = CA (length of two tangents)
AO = AO (Common)
Therefore, Δ OPA ≅ Δ OCA (By SSS congruency criterion)
Hence, ∠ 1 = ∠ 2 (CPCT)
Similarly ∠ 3 = ∠ 4
Now, ∠ PAB + ∠ QBA = 180°
2∠2 + 2∠4 = 180°
∠2 + ∠4 = 90°
∠AOB = 90° (Angle sum property
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Joining BD, there are two triangles.
Area of quad ABCD = Ar △ABD + Ar △BCD
Ar △ABD = ½| (-5)(-5 - 5) + (-4)(5 - 7) + (4)(7 + 5) |
= 53 sq units
Ar △BCD = ½| (-4)(-6 - 5) + (-1)(5 + 5) + (4)(-5 + 6) |
= 19 sq units
Hence, area of quad ABCD = 53 + 19 = 72 sq units
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