Question

Find the area of an equilateral triangle whose side, $a=2+\sqrt{3}$ units.

[Hint: Area of an equilateral triangle =
$\frac{\left(\sqrt{3}\right)}{4}$${\left(side\right)}^2$]

• A. <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $2\times \sqrt{3}$ sq. units
• B. $\frac{\sqrt{3}}{2}(7+4\sqrt{3})$ sq. units
• C. <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\frac{7\sqrt{3}+12}{4}$ sq. units
• D. <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\frac{3}{4}$ sq. units

Answer 1

The correct option is C <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\frac{7\sqrt{3}+12}{4}$ sq. units

Area of an equilateral triangle = $\frac{\left(\sqrt{3}\right)}{4}$ × <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> ${\left(side\right)}^2$

= $\frac{\left(\sqrt{3}\right)}{4}$${(2+\sqrt{3})}^2$

= $\frac{\left(\sqrt{3}\right)}{4}(4+3+4\sqrt{3})$

=$\frac{\left(\sqrt{3}\right)}{4}(7+4\sqrt{3})$

=$\frac{7\sqrt{3}+12}{4}$ sq. units
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->