Question

In the given figure, ABCD is a quadrilateral. $\angle B+\angle D$ is equal to

• A. ${120}^0$
• B. ${190}^0$
• C. ${180}^0$
• D. ${200}^0$

Answer 1

The correct option is C ${180}^0$

Given: AB = CD = 3 cm and AD = BC = 4 cm
Consider, $A{B}^2+A{D}^2=3^2+4^2$
$=9+16$
$=5^2$
$=B{D}^2$
$\therefore$ ABD is a right-angled triangle such that $\angle A=90°$ (Converse of Pythagoras theorem)
Similarly, $\angle BCD=90°$.
In the quadrilateral ABCD, by Angle sum property,
$\angle A+\angle B+\angle C+\angle D=360°$
$\Rightarrow 90°+\angle B+90°+\angle D=360°$
$\Rightarrow \angle B+\angle D=180°$
Hence, the correct answer is option (c).