MCQ
In Fig. ABCD is an isosceles trapezium, its perimeter is
- A$(8+4 \sqrt{2})$ units
- B$(10+2 \sqrt{2})$ units
- ✓$(10+4 \sqrt{2})$ units
- D$(11+4 \sqrt{2})$ units
✓
Answer
Correct option: C.
$(10+4 \sqrt{2})$ units
(C)$(10+4 \sqrt{2})$ units
In $\triangle A E D$, we obtain
$
\tan 45^{\circ}=\frac{A E}{D E} \Rightarrow 1=\frac{2}{D E} \Rightarrow E D=2 \Rightarrow C F=2
$
Again in $\triangle A E D$, we obtain
$
\sin 45^{\circ}=\frac{A E}{A D} \Rightarrow \frac{1}{\sqrt{2}}=\frac{2}{A D} \Rightarrow A D=2 \sqrt{2}
$
$\begin{array}{l}C D=C l+I l+C D=2+3+2=7 \\ \text { Perimeter }=A B+B C+C D+A D=3+2 \sqrt{2}+7+2 \sqrt{2}=(10+4 \sqrt{2}) \text { units }\end{array}$
In $\triangle A E D$, we obtain
$
\tan 45^{\circ}=\frac{A E}{D E} \Rightarrow 1=\frac{2}{D E} \Rightarrow E D=2 \Rightarrow C F=2
$
Again in $\triangle A E D$, we obtain
$
\sin 45^{\circ}=\frac{A E}{A D} \Rightarrow \frac{1}{\sqrt{2}}=\frac{2}{A D} \Rightarrow A D=2 \sqrt{2}
$
$\begin{array}{l}C D=C l+I l+C D=2+3+2=7 \\ \text { Perimeter }=A B+B C+C D+A D=3+2 \sqrt{2}+7+2 \sqrt{2}=(10+4 \sqrt{2}) \text { units }\end{array}$
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