Question

Find the area of the quadrilateral ABCD in which AD=24 cm,in triangle BAD=90 and BCD froms an equilateral triangle whose each side is 26 cm.

Answer 1

ar(ABCD)= ar(ABD)+ ar(BDC)
= 90+ 1/2× 26× 22.52
=90+ 292.76
=382.76 cm.cm

Answer 2

Answer:

Step-by-step explanation:

From triangle ABD, using Pythagoras theorem,

$(BD)^{2}=(AB)^{2}+(AD)^{2}$

$(26)^{2}=(24)^{2}+(AB)^{2}$

$676-576=(AB)^{2}$

$AB=10 cm$

Now, Area of triangle BAD=$\frac{1}{2}{\times}BA{\times}AD=\frac{1}{2}{\times}10{\times}24=120 cm^{2}$

Area triangle BCD=$\frac{\sqrt{3}}{4}{\times}(26)^{2}=292.37 cm^{2}$

Now, Area of Quadrilateral ABCD= Area of Triangle BAD+ area of triangle BCD

=$120+292.37=412.37 cm^{2}$