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Question

The given figure shows a parallelogram ABCD. E is a point in AD and CE produced meets BA produced at point F. If $$AE=4 cm, AF= 8$$ cm and AB $$=12$$ cm. find the perimeter of the parallelogram ABCD.

179361_3c2c07a0a6db459b8a44226785987e0f.png


A
42 cm
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B
44 cm
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C
46 cm
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D
48 cm
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E
40 cm
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F
38 cm
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G
36 cm
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Solution

The correct option is B $$44$$ cm
In $$\triangle AFE$$ and $$\triangle BFC$$,
$$\angle AFE = \angle BFC$$ (Common angle)
$$\angle AEF = \angle BCF$$ (Corresponding angles of parallel lines)
$$\angle FAE = \angle CBF$$ (Corresponding angles of parallel lines)
Thus, $$\triangle AFE \sim \triangle CFB$$ ($$AAA$$ rule)
Hence, $$\dfrac{AF}{BF} = \dfrac{AE}{BC}$$
$$\dfrac{AF}{AF + AB} = \dfrac{AE}{BC}$$
$$\dfrac{8}{8 + 12} = \dfrac{4}{BC}$$
$$BC = \dfrac{4 \times 20}{8}$$
$$BC = 10$$ cm
Perimeter of parallelogram = $$AB + BC + BC+ AD$$ 
Perimeter of parallelogram = $$2 (AB + BC)$$ 
Perimeter of parallelogram = $$2  (12 + 10)$$  (Opposite sides are equal) 
Perimeter of parallelogram = $$44$$ cm

Mathematics

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