The given figure shows a quadrilateral ABCD in which $A D = 13 c m, D C = 12 c m, B C = 3$ $c m$ and $∠ A B D = ∠ B C D = 90^{\circ}$ . Calculate the
length of AB.

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Solution

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of of the squares on the remaining two sides.
First, we consider the $△ A D C$ and applying Pathagoras theorem we get,
$D B^{2} = D C^{2} + B C^{2}$
$D B^{2} = 12^{2} + 3^{2}$
$D B^{2} = 144 + 9$
$D B^{2} = 153$
Now, we consider the $△ A B D$ and applying Pathagoras theorem we get,
$D A^{2} = D B^{2} + B A^{2}$
$13^{2} = 153 + B A^{2}$
$B A^{2} = 169 - 153 = 16$
$B A = 4$
The length of $A B$ is $4 c m$

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