Geometry is Awesome!

Chords intersecting at right angles

Let perpendiculars ON and OM to the chords, form a rectangle KNOM. NK = MO = 1, NB = 5

$R^2 = OP^2 = (h+2)^2 + 1^2$ and

$R^2 = OB^2 = h^2 + 5^2$ Solving gives h = 5 and $R^2 = 50$ giving area = $\color{#D61F06} {157}$

• 
• Daigram is not to scale
• As shown above AB CD are perpendicular and 'h' and 'i' are perpendiculars to the chords
• We know AH=4,HB=6,HC=2
• And we know
• $AH \times HB=CH \times HD$
• thus , HD=12
• Now AG = 5,ED=7
• AG-AH=GH
• GH=5-4
• GH=1=EF (since EFGH is a rectangle)
• Now in right angled triangle EDF
• $FE^2+ED^2=FD^2$
• Thus $FD=\sqrt{FE^2+ED^2}=\sqrt{7^2+1^2}=\sqrt{50}$
• AREA = $\pi r^2$ =3.14 x 50= $\boxed{157}$

Solving for $KQ$ , we have

$(KQ)(2) = (4)(6)$ $\implies$ $KQ = 12$

Draw perpendiculars form $M$ to $KQ$ or $KB$ ,

$NQ = \frac{2+12}{2} = 7$

$DB = \frac{4+6}{2} = 5$

$NM = DK = 6-5 = 1$

$DM = KN = 7 - 2 = 5$

By Pythagorean Theorem, we have

$r^2 = 1^2 + 7^2 = 50$ or $r^2 = 5^2 + 5^2 = 50$

Solving for the area of the circle, we have

$A = \pi r^2 = \pi (\sqrt{50})^2 =$ $\large\boxed{\color{#3D99F6}157}$

Calculation can be almost completely replaced by sketching when numbers are this easy. Put the points in the grid as shown above. Connect them. Run perpendicular lines, making sure the slopes are correct. Find where they intersect (point $O$ in the sketch). Points $A$ and $O$ are off by $5$ both in $x$ and $y$ directions, so the distance between them is $R=5\sqrt{2}$ , giving us the area as $\pi\times5^2\times2\approx157.$