If $A : B = 1 : 2, B : C = 3 : 4$ and $C : D = 5 : 6$ , find $D : A$ .

6 : 5

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5 : 16

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5 : 6

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16 : 5

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Solution

The correct option is D $16 : 5$
Given, $A : B = 1 : 2, B : C = 3 : 4$ and $C : D = 5 : 6$

Therefore, $\frac{D}{C} = \frac{6}{5}, \frac{C}{B} = \frac{4}{3}, \frac{B}{A} = \frac{2}{1}$

$\Rightarrow \frac{D}{A} = \frac{D}{C} \times \frac{C}{B} \times \frac{B}{A} = \frac{6}{5} \times \frac{4}{3} \times \frac{2}{1} = \frac{16}{5}$

$\Rightarrow D : A = 16 : 5$ .

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Find the area of the quadrilateral ABCD in which AB = 42 cm, BC = 12 cm, CD = 29 cm, DA = 34 cm and diagonal BD = 20 cm.

In the figure, prove that:

$(i) C D + D A + A B + B C > 2 A C$

$(i i) C D + D A + A B > B C$

Let ABCD be a quadrilateral with diagonals AC and BD. Prove the following statements (Compare these with the previous problem);

(a) AB + BC + CD > AD;

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