Test Level 2: Geometry - 1 - CAT MCQ

In ΔABC, AE and CD are the medians of the triangle.
So, AE = 6 and CD = 8

hence, BE = EC = 1/2BC and AD  = DB= 1/2AB

Adding (l) and (2), we get 

∠PQR = 50° (Alternate Segment Theorem)

∠PQO + ∠ OQR = 50°
30° + ∠OQR = 50°
∠OQR = 20°

Since PO = OQ, (Radii of the same circle)
∠OPQ = ∠OQP = 30°
∠OQR = ∠ORQ = 20°

As shown in the figure, join PO and QO, where O is the centre.
Arc PMQ = 150°
(∵ Given ∠POQ = 150°)
Now, join PN and QN as shown in the figure.

∠PNQ = 1/2 ∠POQ = 75° [∵ The angle subtended by an arc or a chord of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle]
Now, ∠PQS = ∠PNQ [∵ Angles in alternate segments are always equal]
∠PQS = 75°

PM ⊥ AB
Any perpendicular from the centre to the chord bisects the chord.

In ΔAMP
AP² = MP² + AM²
AP² = 82 + 152 = 64 + 225 = 289

In ΔCNP
CP² = NP² + CN²
17² = NP² + 8²
17² - 8² = NP²
NP = 15

Opposite sides of a parallelogram are parallel and equal.
Therefore, PQ = SR and PS = QR.
Suppose the length of SC is x units and that of CR is y units.
∵ SR = PQ
∴ x + y = 8 ... (1)
SC and SD are tangents to the circle, so SC = SD = x units.
CR and RB are tangents to the circle, so CR = BR = y units.
SP = (x + 3) = RQ = (5 + y)
x - y = 2 ... (2)
Solving the equations (1) and (2), we get
x = 5 units and y = 3 units
∴ PS = 3 + x = 3 + 5 = 8 units

Given; PS || RQ, ∠1 = ∠2, ∠3 = ∠4
Let ∠STR = θ
∠PSR + ∠SRQ = 180° {Sum of co-interior angles is 180°}
∠1 + ∠2 + ∠3 + ∠4 = 180°
2∠2 + 2∠3 = 180°
∠2 + ∠3 = 90° ...(1)
Now, ∠2 + ∠θ + ∠3 = 180° {Sum of the angles in a triangle is 180°}
∠θ + 90° = 180° {from (1)}
∠θ = 180° - 90°
∠θ = 90°
∴ ∠STR = ∠θ = 90°

Sum of interior angles of a polygon = (n - 2) x 180°, where n is the number of sides of polygon.
Now, (n - 2) x 180° = 2520°
n - 2 = 14
n = 16

Given, AB || CD.
70° = 30° +∠ECD
∠ECD = 40° 
∠ECD + ∠CEF = 140° + 40° = 180° 
So, EF || CD.
∠EFG = ∠FGB
∠BGF = 50°

In triangle AOB and AOC, two angles are correspondingly equal.
⇒ ∠AOB = ∠AOC = z
Now, ∠AOB + ∠AOC + ∠BOC = 360°
z + z + 100° = 360°
Or z = 130°
Hence, option (2) is correct.

Draw a triangle running through the centres of the circles as shown above. Since the circles are of equal radii, ΔABC will be an equilateral triangle with side 4r.
Area bounded by but not included in the circles (A) = Area of the equilateral triangle – (Area of 3 semicircles + Area of 3 sectors of angle 60° each)