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ICSE Class X Math 2013 Paper

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In the given figure, ABA B and DED E are perpendicular to BCB C.
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Question : 14 of 46
Marks: +1, -0
If AB=6 cm,DE=4 cmA B = 6\ \text{cm}, D E = 4\ \text{cm} and AC=15A C = 15 cm\text{cm} . Calculate CDC D .
Solution:  
    In  △ABC  and  △DEC  ,  \;\; \text{In}\; \triangle ABC \;\text{and}\; \triangle DEC\; \text{,}\;
  △ABC∼△DEC  \;\triangle ABC \sim \triangle DEC\;     (proved in (i) part)  \;\; \text{(proved in (i) part)}\;
  ∴    ABDE=ACCD\; \therefore \;\; \frac{AB}{DE} = \frac{AC}{CD}
    Given :  AB=6 cm,DE=4 cm,AC=15\;\; \text{Given :}\; AB = 6\ \text{cm}, DE = 4\ \text{cm}, AC = 15   cm,\; \text{cm},
  ∴    64=15CD\; \therefore \;\; \frac{6}{4} = \frac{15}{CD}
  ⇒    CD=15×46\; \Rightarrow \;\; CD = \frac{15 \times 4}{6}
  ⇒    CD=10 cm.\; \Rightarrow \;\; CD = 10\ \text{cm} .
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