• # question_answer Which one of the following is not true for the square ABCD given below? A)  $\text{AB}+\text{BC+CD}=3\times \text{AD}$ B)  $\angle \text{ABC}+\angle \text{ADC}=\angle \text{BAD+}\angle \text{BCD}$ C)  $\text{AB}+\text{BC}=\text{AD}+\text{CD}$ D)  $\angle \text{ABC}+\angle \text{BCD}+\angle \text{ADC}=360{}^\circ$ E) None of these

Explanation: Option (d) is correct. Since, ABCD is a square, therefore, $\angle \text{ABC}=\angle \text{BCD}=\angle \text{ADC}=\angle \text{BAD}=90{}^\circ$(each) And, AB= BC = CD = AD = 1 unit (say) Now, in option (a); AB + BC + CD = ($1+1+1$) units = 3 units = $3\times 1$ unit = $3\times \text{AD}$ In option (b)$\angle \text{ABC}+\angle \text{ADC}=90{}^\circ +90{}^\circ =180{}^\circ$; $\angle \text{BAD}+\angle \text{BCD}=90{}^\circ +90{}^\circ =180{}^\circ$ In option (c); AB + BC = 1 unit + 1 unit = 2 units; AD + CD = 1 unit + 1 unit = 2 units In option (d); $\angle \text{ABC}+\angle \text{BCD}+\angle \text{ADC}=$ $90{}^\circ +90{}^\circ +90{}^\circ =270{}^\circ$ (not $360{}^\circ$)