The Triangle and Its Properties Test

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The Triangle and Its Properties Test - 7

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Question 1
1 / -0

What are the measures of angles 'a' and 'b' in the given figure?

A
a = 70°, b = 40°

B
a = 60°, b = 50°

C
a = 60°, b = 70°

D
a = 70°, b = 50°

Solution

According to the given figure, angle 'a' is the vertically opposite angle to 60° and hence, it is equal to 60°.
Therefore, a = 60°
So,
60° + 70° + b = 180° (sum of angles of a triangle is 180°)
130° + b = 180°
Therefore, b = 50°
Question 2
1 / -0

Find the value of x, if DC = CF.

A
62°

B
65°

C
85°

D
86°

Solution

Given, CF = DC
∠FDC = ∠CFD (As CF = DC)
∠CFD = 30°
∠CFD = ∠EFA (Vertically opposite angles)
∠EFA = 30°
In triangle AFE,
30° + x = 95° (Sum of two interior angles of the triangle is equal to the opposite exterior angle)
x = 65°
Question 3
1 / -0

What are the measures of 'x' and 'y' in the isosceles triangle ABC given below?

A
x = 96°, y = 192°

B
x = 42°, y = 42°

C
x = 42°, y = 84°

D
x = 42°, y = 138°

Solution

In the given figure:
exterior angle at C = y

ABC = x
ACB = x (Angles opposite to equal sides are equal)
Also, ∠BAC = 96° (Vertically opposite angle theorem)
y = 96° + x (Exterior angle property) .........(1)

In triangle ABC,
x + ACB + 96° = 180°
x + x + 96° = 180°(Angle sum property of a triangle)
2x = 180° - 96°
2x = 84°
x = = 42°
From (1),
y = 96° + x
y = 96° + 42° = 138°
Question 4
1 / -0

Directions: Triangles PQR and RQS are isosceles triangles. Find ∠SQR.

A
40°

B
50°

C
60°

D
65°

Solution

In isosceles triangle RSQ, ∠SQR = ∠SRQ = x.
Sum of interior angles of triangle = 180°
∠SQR + ∠QSR + ∠SRQ = 180°
2x + ∠QSR = 180°
2x = 180° - 80° = 100°
x = 50°
Question 5
1 / -0

Directions: Find from right angled .

A
80°

B
60°

C
50°

D
40°

Solution

In PQR,

PQR = PQS + SQR

90° = PQS + 25°
PQS = 65°

InPQS,
SPQ + PQS + QSP = 180°

50° + 65° + QSP = 180°
QSP = 180° - 115°
QSP = 65°

Now, QSR + QSP = 180°
QSR + 65° = 180°
QSR = 115°

In QSR,
RQS + QSR + SRQ = 180°

25° + 115° + SRQ = 180°
SRQ = 180° - 140°
SRQ = 40°
Question 6
1 / -0

PQR is an equilateral triangle with its sides PQ = QR = RP. What is the value of ∠PQR in the triangle?

A
60°

B
110°

C
120°

D
30°

Solution

All the angles of an equilateral triangle measure 60°.
Question 7
1 / -0

What are the values of 'x' and 'y' in the given figure? (Note: AB = BC)

A
x = 45°, y = 90°

B
x = 60°, y = 120°

C
x = 45°, y = 120°

D
x = 45°, y = 135°

Solution

In the given figure,
B = 90°
y is the exterior angle.
BCA = x
Since AB = BC, BCA = CAB = x (because angles opposite to equal sides are equal)
Exterior angle of a triangle = Sum of interior opposite angles of a triangle
y = 90° + x ... (i)
Also, in ABC,
x + x + 90° = 180°
2x = 180° - 90°
2x = 90°
x = 45° ... (ii)
Put the value of x in (i).
y = 90° + 45° = 135°
So, x = 45°, y = 135°
Question 8
1 / -0

Find the missing angle in the following figure.

A
50°

B
63°

C
43°

D
53°

Solution

Sum of all the angles of a triangle is 180°.
So,
55° + 82° + ?° = 180°
?° = 180° - 137° = 43°
So, the missing angle is 43°.
Question 9
1 / -0

Find the value of angle x° from the given group of equilateral triangles.

A
60°

B
80°

C
90°

D
120°

Solution

The combination also makes an equilateral triangle. So, each individual angle will also be 60°.
So, x = 60°
Question 10
1 / -0

Directions: In the diagram given below, triangle BAC is a right angled triangle and quadrilateral ADOE is a rhombus. Using these properties, find the value of 5x - y.

A
90°

B
120°

C
170°

D
None of the above

Solution

AC is acting as the diagonal of rhombus ADOE. So, the angle x° must be 45°.

In triangle AEO,
x° + 90° + ∠EOA = 180°
45° + 90° + ∠EOA = 180°
So, ∠EOA = 180° - 135° = 45°

And, ∠AOC = 180°
or, ∠EOA + ∠EOC = 180°
45° + y° = 180°
y° = 180° - 45° = 135°
So, 5x - y = (5 × 45) - 135 = 90°
Question 11
1 / -0

In the figure given below (not drawn to scale), ABC and DEF are two triangles, AB is parallel to FD and AFCE is a straight line. Find the value of P + Q.

A
198°

B
199°

C
205°

D
208°

Solution

BAC = DFE (Corresponding angles as AB//FD)
So, P = 61°
Now, in ABC,
Q = 61° + 76° (using Exterior angle property in triangle ABC)
Q = 137°
P + Q = 61° + 137°
= 198°
Question 12
1 / -0

Which of the following CANNOT be the measurements of the sides of a triangle?

A
3 cm, 4 cm, 5 cm

B
2 cm, 4 cm, 6 cm

C
2.5 cm, 3.5 cm, 4.5 cm

D
2.3 cm, 6.4 cm, 5.2 cm

Solution

Since in option 2,
2 + 4 = 6,
it is not a possible measurements of sides of a triangle as the length of the third side must be less than the sum of other two sides.
Question 13
1 / -0

Directions: In the figure, not drawn to scale, MOPR is a rectangle and NPO is a triangle. Find ∠NQO.

A
45°

B
55°

C
65°

D
75°

Solution

MON + NOP = 90° (Angle of rectangle)
MON + 60° = 90°
MON = 90° - 60°
MON = 30°

In QNO,
QNO + NQO + QON = 180° (Angle sum property)

75° + NQO + 30° = 180°
NQO = 180° - 105°
NQO = 75°
Question 14
1 / -0

Directions: In the figure, PQRS is a rectangle and △RTO is an isosceles triangle. Find x.

A
30°

B
40°

C
45°

D
50°

Solution

It is given that PQRS is a rectangle.

So, PSR = 90°

In PSR,
SPR + PSR + PRS = 180° (Angle sum property)

60° + PSR + x = 180°

60° + 90° + x = 180°

x = 180° - 150°

x = 30°
Question 15
1 / -0

Find the measure of ∠CEA.

A
75°

B
85°

C
90°

D
95°

Solution

EDB = CED + ECD (Sum of two interior angles is equal to exterior angle)

130° = 35°+ CED
CED = 130° - 35°
CED = 95°
DEC + CEA = 180° (Linear Pair)

∠CEA = 180° - 95° = 85°
Question 16
1 / -0

There is a building which is 12 m tall. Ram is standing on the ground 9 m away from the building. Find the diagonal distance between roof of the building and feet of Ram.

A
13 m

B
14 m

C
15 m

D
20 m

Solution

In PQR,
PQ² + QR² = PR² (By Pythagoras theorem)
(12)² + (9)²= PR²
144 + 81 = PR²
PR² = 225
PR = 15 m
Question 17
1 / -0

A ladder of length 15 m is placed against a wall. The ground distance between the wall and the lowest point of the ladder is 9 m. Find the height where the ladder touches the wall.

A
10 m

B
11 m

C
12 m

D
15 m

Solution

In △ACB, using Pythagoras Theorem,
AB² = AC² + BC²
15² = h² + 9²
225 = h² + 81
h² = 225 - 81
h² = 144
h = 12 m
Question 18
1 / -0

The diagonal of a rectangle is 10 cm and its length is 8 cm. What is the perimeter of the rectangle?

A
20 cm

B
24 cm

C
28 cm

D
32 cm

Solution

In ABC, using Pythagoras Theorem,
AC² = AB² + BC²
10² = 8² + BC²
100 = 64 + BC²
BC² = 100 - 64
BC² = 36
BC = 6 cm

In the rectangle, BC = AD = 6 cm
and AB = CD = 8 cm
Now, perimeter of the rectangle = AB + BC + CD + DA
= 8 + 6 + 8 + 6
= 28 cm
Question 19
1 / -0

If Tom runs 45 cm in north direction and then 60 cm towards east, find the shortest distance that he needs to cover to get back to the starting point.

A
57 cm

B
65 cm

C
75 cm

D
80 cm

Solution

In OAB,
(OB)² = (60)² + (45)²
= 15² x (4² + 3²)
= 15² x 5²
= (15 x 5)²
OB = 75 cm
Question 20
1 / -0

A wire is tied from the top of a pole to the ground at a distance 6 m from the base of the pole. If the pole is 8 m high, what is the length of the wire?

A
14 m

B
10 m

C
2 m

D
None of these

Solution

From the given question, we come to know that:

The above diagram represents a triangle.
In this triangle, let the length of the wire be H, the height of the pole be P and the distance between the base of the pole and the base of the wire be B. Therefore, using Pythagoras Theorem, we get

H² = P² + B²

H² = 8² + 6² = 64 + 36

H² = 100

H = 10 m

Therefore, the length of the wire is 10 m.
Question 21
1 / -0

What is the value of (x + y) in the given figure?

A
125°

B
180°

C
200°

D
225°

Solution

BCD = BAC + ABC [Exterior angle property of triangle]
y = 45° + ABC --- (i)
Also, EBC = BAC + ACB [Exterior angle property of triangle]
x = 45° + ACB --- (ii)
Adding equations (i) and (ii), we get
y + x = 45° + (ABC + 45° + ACB)
= 45° + 180° [∵ABC + ACB + 45° = 180°; angle sum property of triangle]
y + x = 225°
Or x + y = 225°
Hence, (4) is the correct option.

Question 22

1 / -0

Directions: State 'T' for true and 'F' for false.

(i) Triangle can have two angles of 90°.
(ii) In a triangle, if there are two acute angles, then third one would be an obtuse angle.
(iii) In a right angle triangle, there are always two acute angles.

	(i)	(ii)	(iii)
A	T	T	T
B	T	F	T
C	F	F	T
D	F	F	F

A

B

C

D

Solution

(i) Triangle cannot have two angles of 90°, so given statement is false. Right triangle has one 90 degree angle and two acute (< 90 degree) angles.

(ii) In a triangle, if there are two acute angles, then third one could be an obtuse angle or it could be an acute angle as well, so given statement is false. An acute triangle is a triangle with all three angles acute (less than 90°). An obtuse triangle is one with one obtuse angle (greater than 90°) and two acute angles.

(iii) In a right angle triangle, there are always two acute angles. So, given statement is true. If a triangle has two acute angles and a single right angle, then the triangle is called a right triangle.

Question 23

1 / -0

In this question, three statements are given with some blanks. Fill in the blanks with the help of the table given below.

(i) A triangle can have maximum A altitudes.
(ii) The angles of a triangle are in the ratio 1 : 2 : 3. The measure of the largest angle is B .
(iii) If the base of a right-angled triangle is 8 cm and its perpendicular is 15 cm, then the length of hypotenuse will be C cm.

Options	(i)	(ii)	(iii)
A	3	90°	15
B	3	90°	17
C	2	105°	15
D	3	105°	17

A

B

C

D

Solution

(i) A triangle can have maximum 3 altitudes.

(ii) Let the angles be x, 2x and 3x.
Now,
x + 2x + 3x = 180° (Angle sum property of triangle)
6x = 180°
x = 30°
Largest angle = 3x = 3 × 30° = 90°

(iii) According to Pythagoras Theorem,
(Hypotenuse)² = (Base)² + (Perpendicular)^2
Here,(Hypotenuse)² = (8)² + (15)²
(Hypotenuse)² = 64 + 225 = 289

Hypotenuse = 17 cm

Question 24
1 / -0

Which of the following is/are not true?

a. A ladder, which has its feet 9 m from the base of a wall, reaches a window on the wall 40 m above the ground. Hence, the ladder is 41 m long.
b. If a ship sails 33 km due south and then 56 km due west, it is 75 km from the starting point.
c. If a ladder, 65 m long, reaches a point on the wall of a house, 63 m above the ground, it is 17 m from the base of the wall.

A
Only a

B
Only a and b

C
Only b and c

D
None of these

Solution

a.

(Here, AC = Length of ladder)
Statement a is correct.

b.

Statement b is incorrect.

c.

(Here, AC is the length of ladder)

Statement c is incorrect.
Question 25
1 / -0

In the figure, if QT PR, then find the measures of x and y.

A
x = 50°, y = 50°

B
x = 50°, y = 80°

C
x = 50°, y = 70°

D
x = 40°, y = 80°

Solution

In QTR,
QTR + TQR + R = 180°90° + 40° + x = 180°
x = 50°
In PSR,
PSR + SPR + SRP = 180°PSR + 30° + 50°= 180°PSR = 100°
y = 180°- PSR = 80°