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In a parallelogram ABCD, AB = 20 cm and AD = 12 cm. The bisector of angle A meets DC at E and BC produced at F. Find the length of CF. - Mathematics

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प्रश्न

In a parallelogram ABCD, AB = 20 cm and AD = 12 cm. The bisector of angle A meets DC at E and BC produced at F.

Find the length of CF.

योग

उत्तर


Given AB = 20 cm and AD = 12 cm.

∠DAE = ∠BAF = x and

∠DAE = ∠AFB = x

In ΔABF

∠BAF = ∠BFA

Hence, ABF is an isosceles triangle

So AB = BF = 20

BF = 20

∠ADE = ∠ABF

∴ AD = BC = 12

BC + CF = 20

12 + CF = 20

CF = 20 - 12

CF = 8 cm

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Names of Polygons
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अध्याय 14: Rectilinear Figures [Quadrilaterals: Parallelogram, Rectangle, Rhombus, Square and Trapezium] - Exercise 14 (B) [पृष्ठ १७६]

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सेलिना Concise Mathematics [English] Class 9 ICSE
अध्याय 14 Rectilinear Figures [Quadrilaterals: Parallelogram, Rectangle, Rhombus, Square and Trapezium]
Exercise 14 (B) | Q 9 | पृष्ठ १७६
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