Angles Questions for NMAT – Download [PDF]

Download Angles Questions for NMAT PDF – NMAT Fill in the blanks questions pdf by Cracku. Top 10 very important Angles Questions for NMAT based on asked questions in previous exam papers.

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Question 1: Which of the following can’t be the ratio of angles of an acute angled triangle?

a) 1:9:9

b) 2:3:4

c) 3:7:8

d) 1:1:1

e) 1:2:3

Question 2: Two right angled triangles ABC and DCB are drawn on the same side of BC. If BC = 30, AB = 10 and CD = 15, and AC and BD intersect at P, find the distance of P from BC.

a) 8 cm

b) 6 cm

c) 10 cm

d) 5 cm

e) 7 cm

Question 3: Two sides of a plot measure 32 m and 24 m and the angle between them is a perfect right angle. The other two sides measure 25 m each and the other three angles are not right angles. What is the area of the plot?
[CAT 2001]

a) 768 $m^2$

b) 534 $m^2$

c) 696.5 $m^2$

d) 684 $m^2$

Question 4: Consider obtuse-angled triangles with sides 8 cm, 15 cm and x cm. If x is an integer then how many such triangles exist?

[CAT 2008]

a) 5

b) 21

c) 10

d) 15

e) 14

Question 5: Consider obtuse-angled triangles with sides 8 cm, 15 cm and x cm. If x is an integer then how many such triangles exist?

a) 5

b) 21

c) 10

d) 15

e) 14

Question 6: 2 polygons with (n-1) and (n+2) sides respectively have their exterior angles in such a way that the difference between the exterior angles is 6 degrees. What is the value of n?

a) 11

b) 15

c) 12

d) 13

Question 7: What is the number of distinct triangles with integral valued sides and perimeter 14?[CAT 2000]

a) 6

b) 5

c) 4

d) 3

Question 8: The sum of all interior angles of an isosagon ?

a) 3240$^{\circ}$

b) 360$^{\circ}$

c) 720$^{\circ}$

d) 900$^{\circ}$

Question 9: If two complementary angles are in the ratio 1: 4, find the supplement of larger angle ?

a) 157.5$^{\circ}$

b) 108$^{\circ}$

c) 137.5$^{\circ}$

d) 120$^{\circ}$

Question 10: A pair of parallel lines are cut by a transversal and the sum of a pair of alternate interior angles A and B is 100 degrees. What is the sum of the pair of corresponding angles that has A as one of its angles?

a) 180 degrees

b) 100 degrees

c) 260 degrees

d) None of the above

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Answers & Solutions:

1) Answer (E)

In an acute angled triangle, the sum of the smallest two angles is greater than the third angle. If the ratio is 1:2:3, the angles are 30, 60 and 90 which is a right angled triangle.

2) Answer (B)

Let PT be the perpendicular drawn from P to BC.
Let PT be ‘y’. Let CT be ‘x’. This means BT = 30-x

Triangles ABC and PTC are similar.
=> $\frac{PT}{AB}=\frac{CT}{BC} => \frac{y}{10}=\frac{x}{30}$ —– (1)

Triangles DBC and PBT are similar.
=> $\frac{PT}{CD}=\frac{BT}{BC} => \frac{y}{15}=\frac{30-x}{30}$ —– (2)

Solving equations (1) and (2) we get y = 6 => PT = 6 cm.

3) Answer (D)

Length of the diagonal of the right triangle is 40. The height of the isosceles triangle formed, with 40 as its base is 15.
So, area = $\frac{1}{2}* 32 * 24 + \frac{1}{2} * 40 * 15 = 384 + 300 = 684 m^2$

4) Answer (C)

For obtuse-angles triangle, $c^2 > a^2 + b^2$ and c < a+b

If 15 is the greatest side, 8+x > 15 => x > 7 and 225 > 64 + x^2 => x^2 < 161 => x <= 12

So, x = 8, 9, 10, 11, 12

If x is the greatest side, then 8 + 15 > x => x < 23

x^2 > 225 + 64 = 289 => x > 17

So, x = 18, 19, 20, 21, 22

So, the number of possibilities is 10

5) Answer (C)

For obtuse-angles triangle, $c^2 > a^2 + b^2$ and c < a+b
If 15 is the greatest side, 8+x > 15 => x > 7 and $225 > 64 + x^2 => x^2 < 161 => x <= 12$
So, x = 8, 9, 10, 11, 12
If x is the greatest side, then 8 + 15 > x => x < 23
$x^2$ > 225 + 64 = 289 => x > 17
So, x = 18, 19, 20, 21, 22
So, the number of possibilities is 10

6) Answer (D)

The exterior angles are given by 360/(n-1) and 360/(n+2)
So, 360/(n-1) – 360/(n+2) = 6 => 720 + 360 = 6(n^2 -3n – 2) => (n-1)(n+2) = 180.
15*12 = 180, so, n-1 =12 and n+2 = 15 and n = 13

7) Answer (C)

Let the sides be x, y and 14-(x+y)
x+y > 14-(x+y) => x+y > 7
x+14-x-y > y => y < 7
Similarly, x < 7
If x = 1, y = 7 (not possible)
So, if x = 2, y = 6
if x = 3, y = 5
if x = 4, y = 4, 5
The cases for x = 5 and 6 are already taken care of by y.
Number of possible cases = 4

8) Answer (A)

Isosagon is a regular polygon with 20 sides.

Sum of interior angles = (n-2)*180; where n is the number of sides.

Sum of angles = (20-2)*180 = 3240

9) Answer (B)

Let the two angles be X and 90-X.
Hence, X = 4(90-X) or X = 72 $^{\circ}$
Hence, the supplement of the larger angle is
180$^{\circ}$ – 4/5(90$^{\circ}$)=108$^{\circ}$

10) Answer (B)

Since A and B are alternate interior angles, they are equal and each is equal to 50 degrees. Since corresponding angles are equal, the sum of A and its corresponding angle, say C is 50+50 = 100 degrees.

We hope these Angles Questions for NMAT pdf for the NMAT exam will be highly useful for your Preparation.