MAT Exam 2022 Paper

Concept:Use the radius of the circle and the right triangle formed by joining the centre to one end of a perpendicular chord.Explanation:Diameter $AB = MN = 3$ cm, so radius $OA = OB = OD = 1.5$ cm.Given $AE:EB = 1:2$.Let $AE = x$ and $EB = 2x$. Since $AB = 3$, we get $x + 2x = 3$, so $x = 1$.Thus $AE = 1$ cm, and $EB = 2$ cm.Now $AO = 1.5$ cm, so $EO = AO - AE = 1.5 - 1 = 0.5$ cm.Similarly, $NL:LM = 1:2$ on diameter $MN = 3$ cm, so the point $L$ is $1$ cm from $N$.Since $ON = 1.5$ cm, we get $OL = 1.5 - 1 = 0.5$ cm.Therefore $EO = OL = 0.5$ cm.The line $DF$ is a chord perpendicular to $MN$, so $L$ is its midpoint. Join $O$ to $D$.In right triangle $OLD$, $OD = 1.5$ cm and $OL = 0.5$ cm.Using Pythagoras, $DL^2 = OD^2 - OL^2 = (1.5)^2 - (0.5)^2 = 2$So $DL = \sqrt{2}$ cm.Also, $HL = EO = 0.5$ cm.Thus $DH = DL - HL = \sqrt{2} - \frac{1}{2} = \frac{2\sqrt{2}-1}{2}$ cm.Answer:$\frac{2\sqrt{2}-1}{2}$ cm, i.e. option B.