Question

Ac is the diameter of circle and is a diagonal of the quadrilateral abcd inscribed in it whose sides are ab=30,cd=10,bc=40 units. Then the area of that quadrilateral is

Answer 1

Answer: 1089.88 sq units.

Step-by-step explanation: ABCD is a quadrilateral inscribed in the circle with diameter AC. So, ΔABC and ΔADC are right angled triangles at ∠B=∠D=90°.

Now, area of ΔABC is given by

$area1= \dfrac{1}{2}\times base\times altitude=\dfrac{1}{2}\times AB\times BC=600 ~sq units.$

From Pythagoras theorem, we have

$AC^2=AB^2+BC^2\\\RightarrowAC=\sqrt{AB^2+BC^2}=\sqrt{40^2+30^2}=50.$

Again, in ΔACD

$AC^2=AD^2+CD^2\\\Rightarrow AD=\sqrt{AC^2-CD^2}=\sqrt{50^2-10^2}=20\sqrt{6}.$

Therefore, area of ΔACD  is

$area2=\dfrac{1}{2}\times AD\times CD=200\sqrt{6} ~squnits$.

Hence, the area of the quadrilateral ABCD is

$area1+area2=600+200\sqrt{6}=1089.88$ sq units.

Answer 2

Answer:

Bro,area 2is wrong

Step-by-step explanation:

1/2*20√6*10=100√6