Question

Calculate the area of the quadrilateral ABCD in which
AB = BD = AD = 10 cm, angel BCD = 90° and CD = 8 cm.
(Take √3 = 1.732).

Answer 1

Step-by-step explanation:

area of abd

$\frac{ \sqrt{3} }{4} \times {10}^{2}$

$\frac{1.732}{4} \times 100$

=43.3

${cm}^{2}$

for finding BC we can use Pythagoras thm

${10}^{2} - {8}^{2} = {bc}^{2}$

$\sqrt{36} = 6$

now area of another triangle

$\frac{1}{2} \times 8 \times 6$

24

total area =67.3

${cm}^{2}$