Test: Geometry- 1 - CAT MCQ

• When the length of stick = 20 m, then length of shadow = 10 m i.e. in this case length = 2 * shadow
• With the same angle of inclination of the sun, the length of the tower that casts a shadow of 50 m is: 2 * 50m = 100m
  ⇒ Height of tower = 100 m

• The triangle BOC is an isosceles triangle with sides OB and OC both being equal as they are the radii of the circle. Hence, the angle OBC = angle OCB = 30°.
• Hence, the third angle of the triangle BOC i.e. Angle BOC would be equal to 120°.
  ⇒ BOC = AOD = 120° 
• Also, in the isosceles triangle DOA: 
  Angle ODA = Angle DAO = x = 30°

By the rule of tangents, we get:

⇒ 12² = (x + 7)x
⇒ 144 = x² + 7x
⇒ x² + 7x – 144 = 0
⇒ x² +16x – 9x –144 = 0
⇒ x(x + 16) – 9(x + 16) = 0
⇒ x = 9 or –16

–16 can’t be the length, hence this value is discarded. Thus, x = 9

By the rule of chords, cutting externally, we get:

(9 + 6) * 6 = (5 + x) * 5
90 = 25 + 5x
5x = 65
x = 13 cm

• In △PAB

  ⇒  ∠PAB=40o         [ Given ]

  ⇒  ∠BPA=90o      [ angle inscribed in a semi-circle ]

  ⇒  ∠PAB+∠PBA+∠BPA=180o

  ∴   40o+∠PBA+90o=180o

  ∴   ∠PBA=180o−130o

  ∴   ∠PBA=50o

  ⇒  ∠PBA=∠PCA=50o     [ angles inscribed in a same arc PA ] 

  ∴   ∠PCA=50o

• Since the lines, AB and CD are parallel to each other, and the lines RD and AN are parallel, it means that the triangles RBF and NCI are similar to each other. Since the ratio of CN : BR = 1.333, if we take BR as 3, we will get CN as 4.
• This means that the ratio of BF : CI would also be 3 : 4.
  Also, the ratio of BR : RS : ST : TA = BF : FG : GH : HI = 3 : 5 : 2 : 7 (given).

Hence, the correct answer is 3 : 5 : 2 : 7 : 4 

∠AOC of minor sector = 140°
∠AOC of major sector=360° - 140° = 220°
Theorem: The angle subtended at the centre is twice the angle formed at the circumference of the circle.
Hence,

∴ The measure of ∠x = 110°

In ΔQRS, QR = RS
⇒ ㄥRQS = ㄥRSQ (because angles opposite to equal sides are equal).
Thus:

ㄥRQS + ㄥRSQ = 180° - 100° = 80°
ㄥRQS = ㄥRSQ = 40°
ㄥPQS = 180° – 40° = 140°  (sum of angles on a line = 180°)

Then again, ㄥQPS = ㄥQSP (since angles opposite to equal sides are equal)

ㄥQPS + ㄥQSP = 180° – 140° = 40°
ㄥQPS = ㄥQSP = 20° 

COD is a  straight line 
∴ ∠δ + ∠v =180⁰ ⇒ 3v +v =180 ⇒ 4v = 180 ⇒ v =45⁰.

In ∠BOC

⇒ x + y = 80°

⇒ 2x + 2y. = 160°

Also, 2x + 2y + 2z = 180°

⇒ 160° + 2z = 180°

⇒ ∠BAC = 2z = 20°

• Imagine a circle on the corner of the rectangle. 
• 3 quarters of the circle lie outside the rectangle and 1 quarter lies inside. 
• Hence, Required ratio = 3 : 1

Area of △ABP, △APQ and AQCD are in GP.
Let their areas be A, Ar and Ar² 
Now, Area of AQCD = 4 × Area of △ABP
⇒ Ar² = 4 × A
⇒ r = 2

∴ Area of △ABP : Area of △APQ : Area of AQCD = 1 : 2 : 4

From the figure given above
Area of △ABP = 1/2 × BP × AB = 1/2 × BP × 9 = 4.5 × BP
Area of △APQ = 1/2 × PQ × AB = 1/2 × PQ × 9 = 4.5 × PQ
Area of AQCD = 1/2 × (QC + AD) × AB = 1/2 × (QC + 6) × 9 = 4.5 × (QC + 6)

Now, Area of △APQ = 2 × Area of △ABP
⇒ 4.5 × PQ = 2 × 4.5 × BP
⇒ PQ = 2BP

∴ QC = 6 - BP - PQ = 6 - 3BP

Now, Area of AQCD = 4 × Area of △ABP
⇒ 4.5 × (QC + 6) = 4 × 4.5 × BP
⇒ 4.5 × (6 - 3BP + 6) = 4 × 4.5 × BP
⇒ 6 - 3BP + 6 = 4BP
⇒ BP = 12/7

∴ PQ = 24/7 and

∴ QC = 6 - 3 × 12/7 = 6/7

⇒ BP : PQ : QC = 12/7 : 24/7 : 6/7 = 2 : 4 : 1

Hence, option (a).