In the given figure, $A B C D$ is a trapezium in which $A D ∥ B C$ ,
$∠ A B C$ , $A D = 16 c m$ , $A C = 41 c m$ and $B C = 40 c m .$ Find the length of
$A B$

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Solution

Give: $∠ A B C = 90^{\circ}$
$B C = 40 c m$
$A C = 41 c m$
By Pythagoras theorem, we have
$A C^{2} = A B^{2} + B C^{2}$
$\Rightarrow A B^{2} = 41^{2} - 40^{2}$
$\Rightarrow A B^{2} = 1681 - 1600$
$\Rightarrow A B^{2} = 9$
$\Rightarrow A B = \sqrt{9}$
$∴ A B = 3 c m$

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In the given figure, ABCD is a trapezium in which AD∥BC,∠ABC, AD=16 cm, AC=41 cm and BC=40 cm. Find the length ofAB

Give: ∠ABC=90∘BC=40 cmAC=41 cmBy Pythagoras theorem, we have AC2=AB2+BC2⇒AB2=412−402⇒AB2=1681−1600⇒AB2=9⇒AB=√9∴AB=3 cm

In the given figure, $A B C D$ is a trapezium in which $A D ∥ B C$ ,
$∠ A B C$ , $A D = 16 c m$ , $A C = 41 c m$ and $B C = 40 c m .$ Find the length of
$A B$

Give: $∠ A B C = 90^{\circ}$
$B C = 40 c m$
$A C = 41 c m$
By Pythagoras theorem, we have
$A C^{2} = A B^{2} + B C^{2}$
$\Rightarrow A B^{2} = 41^{2} - 40^{2}$
$\Rightarrow A B^{2} = 1681 - 1600$
$\Rightarrow A B^{2} = 9$
$\Rightarrow A B = \sqrt{9}$
$∴ A B = 3 c m$