Test Level 1: Geometry - 1 - CAT MCQ

Let the diameter of circle I and II be 4x and 3x, respectively.
ΔOSQ is similar to ΔORP by AA criteria.

Let x denote the radius of the smaller circle. The figure shows the lengths marked to the right as follows.

Here,

BO² = AB² + AO²

BO² = x² + x²

BO = √2x

BO + OC = 1

Hence, this is the correct answer.

Let area of the first square be a₁, whose side is s₁.
Let area of the second square be a₂, whose side is s₂.

Then,

So, required ratio of perimeter

PM is a tangent and MQ is the radius.
⇒ QM perpendicular to PM
⇒ PMQ is a right-angled Δ.
PQ² = PM² + MQ² (By Pythagoras Theorem)
PQ² = (7)² + (7)² = 2(7)²
PQ² = 98

PQ = 7√2 cm

 In ΔADC and ΔAFE:
∠A = ∠A {Common}
AD = DF {Given}
∠ADC = ∠AFE
{Corresponding angles between two || lines}

Hence, ΔADC is similar to ΔAFE.

We know: AF = AD + DF = 3 + 3 = 6 units

Since all small triangles are congruent, AC = CE = EG = GI = IK = KM and 6AC = AM.

Now, as ΔANM ~ ΔABC,

Required Ratio = 36 : 1

Since AT = AP, (Radii)
∴ ∠APT = ∠ATP = 40°
ΔAPT is an isosceles Δ, i.e. with AP = AT and the angles subtended by them will be the same.

Angle subtended by the minor arc ADC at centre = 2 × (∠ABC) = 160°
Angle subtended by the complementary major arc ABC at centre = (360 - 160)° = 200°

RS = 6 units, SC = 5 units = CT

RP = 6/2 = 3 units = PS, as perpendicular from the centre to chord divides the chord into 2 equal parts.

Put MN = 10m + n
AB = 10n + m

As OP is rational, so 11 must divide (m + n)(m - n) and m and n are single digit numbers.
Hence, m = 6, n = 5.
Therefore, diameter MN = 10(6) + 5 = 65 units