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In the adjoining figure, ABCD is a trapezium in which AB || DC; AB = 7cm; AD = BC = 5cm and the distance between AB and DC is 4cm. Find the length of DC and hence, find the area of trap. ABCD. |
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Answer» Consider △ ALD Based on the Pythagoras theorem AL2 + DL2 = AD2 By substituting the values 42 + DL2 = 52 So we get DL2 = 52 – 42 DL2 = 25 – 16 By subtraction DL2 = 9 By taking square root DL = √ 9 So we get DL = 3 cm Consider △ BMC Based on the Pythagoras theorem MC2 + MB2 = CB2 By substituting the values MC2 + 42 = 52 So we get MC2 = 52 – 42 MC2 = 25 – 16 By subtraction MC2 = 9 By taking square root MC = √ 9 So we get MC = 3 cm From the figure we know that LM = AB = 7cm So we know that CD = DL + LM + MC By substituting the values CD = 3 + 7 + 3 By addition CD = 13cm Area of Trapezium ABCD = ½ (sum of parallel sides × distance between them) So we get Area of Trapezium ABCD = ½ × (CD + AB) × AL By substituting the values Area of Trapezium ABCD = ½ × (13 + 7) × 4 On further calculation Area of Trapezium ABCD = 20 × 2 By multiplication Area of Trapezium ABCD = 40 cm2 Therefore, length of DC = 13cm and area of trapezium ABCD = 40 cm2. |
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