# 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:Â **Which of the following can’t be the ratio of angles of an acute angled triangle?

a)Â 1:9:9

b)Â 2:3:4

c)Â 3:7:8

d)Â 1:1:1

e)Â 1:2:3

**Question 2:Â **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)Â 8 cm

b)Â 6 cm

c)Â 10 cm

d)Â 5 cm

e)Â 7 cm

**Question 3:Â **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)Â 768 $m^2$

b)Â 534 $m^2$

c)Â 696.5 $m^2$

d)Â 684 $m^2$

**Question 4:Â **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)Â 5

b)Â 21

c)Â 10

d)Â 15

e)Â 14

**Question 5:Â **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)Â 5

b)Â 21

c)Â 10

d)Â 15

e)Â 14

**Question 6:Â **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)Â 11

b)Â 15

c)Â 12

d)Â 13

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

a)Â 6

b)Â 5

c)Â 4

d)Â 3

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

a)Â 3240$^{\circ}$

b)Â 360$^{\circ}$

c)Â 720$^{\circ}$

d)Â 900$^{\circ}$

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

a)Â 157.5$^{\circ}$

b)Â 108$^{\circ}$

c)Â 137.5$^{\circ}$

d)Â 120$^{\circ}$

**Question 10:Â **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)Â 180 degrees

b)Â 100 degrees

c)Â 260 degrees

d)Â None of the above

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**Answers & Solutions:**

**1)Â AnswerÂ (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)Â AnswerÂ (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)Â AnswerÂ (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)Â AnswerÂ (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)Â AnswerÂ (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)Â AnswerÂ (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)Â AnswerÂ (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)Â AnswerÂ (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 =Â (20-2)*180 = 3240

**9)Â AnswerÂ (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)Â AnswerÂ (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.