Question : A circle touches all four sides of a quadrilateral ABCD. If AB = 18 cm, BC = 21 cm, and AD = 15 cm, then length CD is:
Option 1: 16 cm
Option 2: 14 cm
Option 3: 12 cm
Option 4: 18 cm
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Correct Answer: 18 cm
Solution : Given: A circle touches all four sides of a quadrilateral ABCD. AB=18 cm, BC=21 cm and AD=15 cm. We know that the lengths of tangents drawn from an external point are equal. Using this we get, AB + CD = AD + BC So, CD = (AD + BC) – AB = (15 + 21) – 18 = 36 – 18 = 18 cm Hence, the correct answer is 18 cm.
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Question : A circle is inscribed in a ΔABC having sides AB = 16 cm, BC = 20 cm, and AC = 24 cm, and sides AB, BC, and AC touch circle at D, E, and F, respectively. The measure of AD is:
Question : The quadrilateral ABCD is circumscribed by a circle. If the lengths of AB, BC, and CD are 7 cm, 8.5 cm, and 9.2 cm, respectively, then the length (in cm) of DA is:
Question : In the figure, chords AB and CD of a circle intersect externally at P. If AB = 4 cm, CD = 11 cm and PD =15 cm, then the length of PB is:
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