From the given figure, the value represented by ‘x’ is ………�

1.	From the given figure, the value represented by ‘x’ is …………(A) 12 cm (B) 11 cm (C) 13 cm (D) 16 cm
Answer» Correct option is (C) 13 cm We have AD = 3 cm & CD = 4 cm, \(\angle ADC=90^\circ\) By using Pythagoras theorem in \(\triangle ADC,\) we obtain \(AC^2=AD^2+CD^2\) \(=3^2+4^2\) \(=9+16=25\) \(\Rightarrow\) \(AC=\sqrt{25}=5\,cm\) Now in right \(\triangle ACB,\) we have \(BC=12\,cm,AC=5\,cm\;and\;AB=x\) \(\therefore\) \(AB^2=AC^2+BC^2\) (By using Pythagoras theorem in right \(\triangle ACB)\) \(=5^2+12^2\) \(=25+144=169\) \(\therefore\) \(AB=\sqrt{169}=13\,cm\) Correct option is: (C) 13 cm

From the given figure, the value represented by ‘x’ is …………(A) 12 cm (B) 11 cm (C) 13 cm (D) 16 cm

Answer»

Correct option is (C) 13 cm

We have AD = 3 cm & CD = 4 cm, \(\angle ADC=90^\circ\)

By using Pythagoras theorem in \(\triangle ADC,\) we obtain

\(AC^2=AD^2+CD^2\)

\(=3^2+4^2\)

\(=9+16=25\)

\(\Rightarrow\) \(AC=\sqrt{25}=5\,cm\)

Now in right \(\triangle ACB,\) we have \(BC=12\,cm,AC=5\,cm\;and\;AB=x\)

\(\therefore\) \(AB^2=AC^2+BC^2\) (By using Pythagoras theorem in right \(\triangle ACB)\)

\(=5^2+12^2\)

\(=25+144=169\)

\(\therefore\) \(AB=\sqrt{169}=13\,cm\)

Correct option is: (C) 13 cm