__RRB Quant Practice Questions – Geometry (17 – 04 – 2018)__**1. In ΔABC,**

**∠**

**B = 5**

**∠**

**C and**

**∠**

**A = 3**

**∠**

**C, then the measure of**

**∠**

**C is**

a) 45°

b) 30°

c) 20°

d) 5°

**Explanation:**We know that in a triangle,

Sum
of angles = 180°

⇒ ∠A + ∠B + ∠C = 180°

⇒ 3∠C + 5∠C + ∠C = 180°

⇒ 9∠C = 180°

⇒ ∠C = 180/9 =
20°

∴ ∠C = 20°

**2. The sum of two angles of a triangle is 116° and their difference is 24°. The measure of smallest angle of the triangle is:**

a) 38°

b) 28°

c) 46°

d) 64°

**Answer: C)**

**Explanation:**Let the angles are x° and y° .

The
sum of two angles of a triangle is 116

^{o}
x°
+ y° = 116° ....(i)

The
difference of two angles of a triangle is 24°.

x°
- y° = 24° .....(ii)

Adding
eq.(i) and eq.(ii), we will get,

2x°
= 116° + 24°

⇒ x° = 140°/2 = 70°

Then,
y° = 116° - 70° = 46°.

Another
third angle of the triangle = 180° - 116° = 64°

∴ The
smallest angle of the triangle is 46°.

**3. In the given figure, x = ?**

a) α + β –
γ

b) α - β +
γ

c) α + β +
γ

d) α + γ –
β

**Answer: C)**

**Explanation:**Let ∠BOC = t,

∴∠AOC = β – t

External
angle of triangle = Sum of internal opposite angle

∴ t + α = x

_{1}and β –t + γ = x_{2}
Where,
x

_{1}+ x_{2}= x
Adding
the two equations we get,

α
+ β + γ = x

**4. If the angles of a right angled triangle are in Arithmetic progression, then the angles are in the ratio**

a) 3 : 5 :
7

b) 0.5 : 1
: 2

c) 1 : 2 :
3

d) 2 : 3 :
4

**Answer: C)**

**Explanation:**Let the angles of the triangle be (a - d), (a) and (a + d).

Now
sum of all the interior angles of a triangle = 180°

⇒ (a - d) +
(a) + (a + d) = 180°

⇒ 3a = 180

⇒ a = 60°

We
know in a right angle triangle the largest angle is 90°

⇒ a + d = 90

⇒ d = 30°

So,
the other angle is a – d = 60 – 30 = 30°

∴ the ratio
of the angles is 1 : 2 : 3

**5. In the figure, ΔODC ~ ΔOBA,**

**∠**

**BOC = 125**

**°**

**and**

**∠**

**CDO = 70**

**°**

**. Value of**

**∠**

**DOC,**

**∠**

**DCO and**

**∠**

**OAB respectively are: Given DC || AB**

a) 65°, 65°
and 65°

b) 45°, 55°
and 55°

c) 55°, 45°
and 65°

d) 55°, 55°
and 55°

**Answer: D)**

**Explanation:**DOB is a straight line.

Therefore,
∠DOC + ∠COB = 180°

⇒ ∠DOC = 180° - 125°

=
55°

In
ΔDOC,

∠DCO + ∠CDO + ∠DOC = 180°

(Sum
of the measures of the angles of a triangle is 180º.)

⇒ ∠DCO + 70° + 55° = 180°

⇒ ∠DCO = 55°

It
is given that ΔODC ~ ΔOBA.

∴ ∠OAB = ∠OCD
[Corresponding angles are equal in similar triangles.]

⇒ ∠OAB = 55°

∴ ∠OAB = ∠OCD
[Corresponding angles are equal in similar triangles.]

⇒ ∠OAB = 55°

**6. The equation of the line if its slope is – 3/7 and it passes through the point (5, – 2) is**

a) 3X + 7y
= 29

b) 3X – 7y
= 1

c) 3X + 7y
= 1

d) 3X – 7y
= 29

**Answer: C)**

**Explanation:**We know that equation of line passed through point (x, y)

⇒ y = mx + c

Given
: m = – 3/7, (x, y) = (5, -2)

⇒ – 2 = - (3/7) × 5 + c

⇒ – 2 = -15/7 + c

⇒ c = 15/7 – 2 = 1/7

⇒ y = -3x/7
+ 1/7

⇒ 3X + 7Y =
1

Hence,
3X + 7y = 1 will be the equation of the line whose slope is – 3/7 and it passes
through the point (5, – 2)

**7. Equation of the straight line parallel to x– axis and also 6 units below x–axis is:**

a) x = -6

b) x = 6

c) y = -6

d) y = 6

**Answer: C)**

**Explanation:**We know that the equation of a straight line parallel to x-axis at a distance b from it is y = b.

Therefore,
the equation of a straight line parallel to x-axis at a distance 6 units below
the x-axis is y = -6

**8. The graphs of 5x + 3y = 12 and 2x + 4y = 0 intersect at the point**

a) (12/7, –24/7)

b) (–12/7,
–24/7)

c) (–12/7,
24/7)

d) None of
these

**Answer: D)**

**Explanation:**Given,

5x
+ 3y = 12 and 2x + 4y = 0

To
find their point of intersection we need to solve both the equations.

2x
+ 4y = 0

⇒ x = – 2y

Substitute
the above value in 5x + 3y = 12

⇒ 5(– 2y) + 3y = 12

⇒ – 10y + 3y = 12

⇒ – 7y = 12

⇒ y = – 12/7

⇒ x = (– 2)(– 12/7) = 24/7

∴ The point
of intersection of the lines 5x + 3y = 12 and 2x + 4y = 0 is (24/7, – 12/7)

**9. The coordinates of one end point of a diameter of a circle are (2, 4) and the coordinates of its centre are (1, 2). Find the co-ordinates of other end of the diameter.**

a) (2, 1)

b) (1, 0)

c) (3, 1)

d) (0, 0)

**Answer: D)**

**Explanation:**We know that, if line segment formed by joining points A (x

_{1}, y

_{1}) and B (x

_{2}, y

_{2}) is divided by point P (x

_{3}, y

_{3}) to a ratio of m: n then:

x

_{3}= (mx_{2}+ nx_{1})/(m + n) and y_{3}= (my_{2}+ ny_{1})/(m + n)
The
center divides the diameter into two equal parts.

∴ m : n = 1
: 1

Given,
co-ordinates of center is (1,2) and co-ordinates of one end of diameter is (2,4).

Let
the coordinates of other end of the diameter is (a, b)

=>
1 = (1*a + 1*2)/(1 + 1) and 2 = (1*b + 1*4)/(1 + 1)

=>
a = 0 and b = 0

∴
co-ordinates of the other end = (0, 0)

**10. Find the area of the triangle formed by the line x + 4y = 3 and the coordinate axes.**

a) 1/3 sq.
unit

b) 4/9 sq.
unit

c) 4/5 sq.
unit

d) 9/8 sq.
unit

**Answer: D)**

**Explanation:**By rearranging the equation and plotting the graph we get, x = 3 – 4y

Thus
the triangle formed by the points are (0,0), (3,0) and (0, ¾)

⟹ Area of
triangle = ½ × base × height

⟹ Area of
triangle = ½ × 3 × ¾

⟹ Area of
triangle = (9/8) sq. Unit