# In a trapezium ABCD , side BC is parallel to AD.Also the lengths of side AB,BC,CD and AD are 8,2,8 and 10 units respectively.Find the radius of the circle that passes through all four points A,B,C and D?

Feb 28, 2018

$\textcolor{b l u e}{R = 2 \sqrt{7}}$

#### Explanation:

The following is a figure for the question. So the whole thing looks like this one below.

Lets make some construction

$B E \text{ perpendicular to " AD , CF " perpendicular to } A D$
$E F = 2 c m$
Rest things are mentioned in figure.
Now applying pythagorean theorem
We get $B E = 4 \sqrt{3}$ (if you want me to include that part too, just let me know in comments)

Before proceeding any further i would like to tell that i'll find circumcentre of $\triangle A B C$ and because of symmetry that becomes circumcentre of trapezium as well.

Since it is a isoceles trapezium
angle A=180°- angleB
angleA=tan^(-1)(sqrt3)=60°
angleB=120°
Ar triangle ABC=1/2 ×sin120° ×8×2=4sqrt3 cm^2 (there's a lot more easier method also)

Now , here's a relation
$\text{area of } \triangle = \frac{a \cdot b \cdot c}{4 R}$ (where a,b,c are sides of triangle and R is radius of circumcircle)
Using cosine rule $A C = 2 \sqrt{21}$

Now again using that relation
$4 \sqrt{3} = \frac{8 \cdot 2 \cdot 2 \sqrt{21}}{4 R}$
$R = 2 \sqrt{7}$ (i tried to show cancelling but preview didn't look any better so no cancelling have been shown).
Hope that helps.