Question

$AB=1$

In the figure above, $\Delta ABC$ is equilateral, D & E are mid points of AB & BC respectively. What is the perimeter of $\Delta AEB$?

• A. $1.5+0.5\sqrt{3}$
• B. $3+\sqrt{3}$
• C. $1.5+\sqrt{3}$
• D. $2.5$

Answer 1

The correct option is A $1.5+0.5\sqrt{3}$

Property:line joining mid points of three sides of a triangle is parallel to the third side and its length is equal to half.

$thelengthofthethirdside.$

$=>DE=\frac{1}{2}AC$

$AB=BC-=AC=1(given,equilateral)$

$=>DE=\frac{1}{2}$

$In△ACE:$

$\angle C={60}^{\circ }$

$usingcosinerule,$

$\cos C=\frac{{AC}^2+{CE}^2-{AE}^2}{2AC.CE}$

$\cos {60}^{\circ }=\frac{1^2+{\left(\frac{1}{2}\right)}^2-{AE}^2}{2\left(1\right)\left(\frac{1}{2}\right)}$

$=>\frac{1}{2}=\frac{1+\frac{1}{4}-{AE}^2}{1}$

$=>{AE}^2=1+\frac{1}{4}-\frac{1}{2}$

$=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$

$=>AE=\frac{\sqrt{3}}{2}$

$puttingin1$

$=>perimeter=\frac{\sqrt{3}}{2}+\frac{1}{2}+1$

$=\frac{\sqrt{3}}{2}+\frac{3}{2}$

$=0.5\sqrt{3}+1.5$