Question

In a quadrilateral ABCD, if AB = 5 m, BC = 5 m, CD = 6 m, AD = 6 m and diagonal AC = 6 m, what is its area?

• A. $2(4+3\sqrt{3})m^2$
• B. $3(4+3\sqrt{3})m^2$
• C. $5(4+3\sqrt{3})m^2$
• D. $7(4+3\sqrt{3})m^2$

Answer 1

The correct option is B

$3(4+3\sqrt{3})m^2$

Given: AB = 5m, BC = 5m, CD = 6m, DA = 6m, AC = 6m

For this quadrilateral ABCD, we can obtain its area as follows:

Area (ABCD) = area ($\Delta$ABC) + area ( $\Delta$ ACD)

Also, $\Delta$ ABC is an isosceles triangle, while $\Delta$ ACD is an equilateral triangle.

(AB = BC = 5m) (AC = CD = DA = 6m)

So, area ( $\Delta ACD)=\left(\sqrt{\frac{3}{4}}\right)\times (side{)}^2=\left(\sqrt{\frac{3}{4}}\right)\times (6{)}^2$

$=9\sqrt{3}sq.m$

And, height of the isosceles $\Delta$ABC = h = BE

(BE is the altitude from vertex B on the side AC)

Applying Pythagoras theorem:

$(BE{)}^2+(EC{)}^2=(BC{)}^2$

$\Rightarrow h^2+3^2=5^2$

(for an isosceles triangle, altitude from the vertex containing the equal sides is also the bisector of the third side: $EC=\frac{\left(AC\right)}{2}=\left(\frac{6}{2}\right)=3m$ )

$\Rightarrow h^2$ = (25 - 9) = 16

$\Rightarrow$ h = 4 m

$Area\left(\Delta ABC\right)=\left(\frac{1}{2}\right)\times h\times \left(AC\right)=\left(\frac{1}{2}\right)\times 4\times 6$

= 12 sq. m

So, area (ABCD) = (12 + 9 $\sqrt{3}$)

= 3 (4 + 3 $\sqrt{3}$ ) sq. m.