Question

In parallelogram $ABCD$ the length $AB$ and $CD$ are both $4$, the length of diagonal $AC=4$ and the length of diagonal $BD=6$. The length $AD$ equal to

• A. $\sqrt{10}$
• B. $\sqrt{12}$
• C. $\sqrt{15}$
• D. $\sqrt{20}$

Answer 1

Answer: (A)

$\sqrt{10}$

Let $AC$ and $BD$ intersect at $O$.

$\therefore DO$ is the median of triangle $ADC$

Using Apollonius' theorem in$\Delta ADC$, we get
$\Rightarrow A{D}^2+C{D}^2=2\left(\right(AO{)}^2+\left(DO{)}^2\right)=A{D}^2+(4{)}^2=2\left(\right(2{)}^2+\left(3{)}^2\right)$
$\Rightarrow A{D}^2+16=2(4+9)=26$
$\Rightarrow A{D}^2+16=2(4+9)=26$
$AD=\sqrt{10}=BC$

Hence, option A.