SSC CGL Previous Year Questions: Geometry- 4 - SSC CGL MCQ

Let the three sides are x, (x + 1) and (x + 2)
In a  right angle triangle, sum of square of base and
height is equal to the square of hypotenuse
∴ (x + 2)² = x² + (x + 1)²
4x + 4 = x2 + 2x + 1
x² – 2x – 3 = 0
(x – 3) (x + 1) = 0 ⇒ x  = 3
Thus, three sides are x = 3 unit
x + 1 = 3 + 1 = 4 unit
x + 2 = 3 + 2 = 5 unit
Smallest side = x = 3 unit.

ΔABC ~ ΔPQR (given)

  (Corresponding sides are proportional)

Since DE is parallel to BC
ΔADE ~ ΔABC

Sum of interior angles of polygon = (n – 2) × 180°
(n – 2) × 180° = 1440
n – 2 = 1440/180 = 8
n = 10
Hence, the number of sides is 10.

In ΔABC, AB ⊥ AC,
BC² = AB² + AC²
BC² = (5)² + (12)²
BC² = 25 + 144
BC² = 169
BC = √169 = 13 cm

Let two circles with center O and O' and radius
OA = 25 cm and O' B = 9 cm.
draw BC || OO'
then O' B = OC = 9 cm
AC = 25 – 9 = 16 cm
From ΔABC,
BC = OO' = 25 + 9 = 34 cm
AC = 16 cm

In a triangle sum of two sides are always greater than third sides and point of intersection of medians of a triangle is called centroid of that triangle, which divides the median into 2 : 1 ratio.
∴ BG : GE = 2 : 1; CG : GF = 2 : 1
AG : GD = 2 : 1

Now, In ΔBGC, BG + GC > BC

Let ABC be the triangle and D, E and F are midpoints of BC, CA and AB respectively.
In ΔABC, AD is median
AB + AC > 2 AD
Similarly, we get
BC + AC > 2 CF
BC + AB > 2 BE
On adding the above inequations, we get
(AB + AC + BC + AC + BC + AB) > 2(AD + BE + CF)
2 (AB + AC + BC) > 2 (AD + BE + CF)
∴ AB + BC + BC > AD + BE + CF
Thus, the perimeter of triangle is greater than the sum of the medians.

In ΔABC, If 
then by the converse of internal angle bisector theorem, we get that AD bisects angle A.

Again, we know that, sum of angles in a triangle = 180°
∴ ∠A + ∠B + ∠C = 180°
∠A + 70° + 50° = 180° ⇒ ∠A = 60°

∠AOB = 90° ; OA = OB = r
∴ ∠BAO = ∠ABO = 45°
∴ ∠AOC = 110°; OA = OC = r
∴ ∠OAC = ∠OCA = 70/2 = 35°
∴ ∠BAC = 45° + 35° = 80°

From question, DE  || AC

OA = OC = 10 cm (radius)
AB = 12 cm
AP = PB = 6 cm
CD = 16 cm
CQ = QD = 8 cm

Let, AB = BC = CA = 2a cm,
AD ⊥ BC

∠B + ∠D = 180°
∠A + ∠C = 180°
∠DAC = θ {given}
∠AED = 90° {given}
In ΔAED,
∴ ∠ADE = 90° – θ = ∠CDE
∴ ∠ABC = 180° – 2 (90° – θ) = 2θ

Here OB = 4 cm (radius)
OA = 5 cm
As AB is a tangent to the circle,
∴ ∠OBA = 90°
OA = 5, OB = 4

∠A + ∠B = 65°
∴ ∠C = 180° – 65° = 115°
∠B + ∠C = 140°
∴ ∠B = 140° – 115° = 25°

We know that Altitude from a point on a line is the shortest distance of that point from the line.
AP <  AB
BQ < BC
CR <  AC
∴ AP + BQ + CR < AB + BC + AC

Area of triangle = Inradius × Semi-perimeter
= 6 ×16 = 96  sq. cm.

As D and E are mid point so,
DE is parallel to BC
So ∠AED = ∠C = 35°
Since ∠A= 80°
Then ∠ADE = 65°
∠EDB is supplement to ∠ADE
So, ∠EDB = 180° - ∠ADE
= 180° - 65° =115° 

Number of common tengent = 3

Let two circles with centre O and O' touches each other at point P. From point T tangent TQ and TR are drawn on two circles equal radius.
As we know two tangent drawn from a external points are always equal
So, TQ = TP ...(i)
and TP = TR ...(ii)
from (i) and (ii), TQ = TP = TR

In ΔABC, ∠ACB = 90°
∴ ∠ACB + ∠ACD = 
90°+ 50° = 140°
As angle mode by triangle
in semicircle is equal to 90°
∴ In quad. ABCD, ∠BAD + ∠BCD = 180° (angle of opp. pair of quad is equal to 180°)
∠BAD =180° - 140° = 40°

AP is a tangent and OA is a radius.
Therefore, OA is ⊥ at AP.
So, In ΔOAP
OP² = OA² + AP² = 5² + 12²
OP² = 25 + 144 = 169
OP = 13 cm.

∴ ∠ACB = 180º – 140º = 40º
In  ΔABC, ∠BAC + ∠ABC + ∠ACB = 180º
∠BAC + 3 ∠BAC + 40º = 180º
4∠BAC = 180º – 40º
∠BAC = 140/4 = 35º

In ΔOBP.
OB = OP (radius)

∴ ∠OBP = ∠OPB = 35º
In ΔAOP
OA = OP (radius)
∴  ∠OAP = ∠OPA = 25º
Now, ∠APB = ∠OPA + ∠OPB = 25º + 35º = 60º
Hence, ∠AOB = 2∠APB
(Since, Angle be substended by arc at centre is twice the angle subtend at the perimeter)
= 2 × 60º = 120º

Let ΔABC is a equilateral triangle of side AB = BC = AC = 2a unit.
AD, BE and CF are three altitudes in the DABC,
In equilateral triangle ∠A = ∠B = ∠C = 60°

Similarly, In ΔACF,

Here, we get AD = BE = CF
Hence, three altitude are equal
Thus, triangle must be a equilateral triangle.

∴ ∠BEC = 130°
∴ ∠DEC = 180° – 130° = 50°
∴ ∠EDC = 180° – 50° – 20° = 110°
∴ ∠BAC = ∠EDC = 110°
(Angles on the same arc)

In ΔOPR, OP = OR = radius
∠OPR = ∠ORP = 25°
In ΔOQR OQ = OR = radius
∠OQR= ∠ORQ

∴ ∠PRQ = ∠PRO + ∠ORQ = 25° + 35° = 60°

Here, AC and BD are chords of the circle.
∴ AP . CP = BP . DP