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The Triangle and Its Properties Test - 6

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The Triangle and Its Properties Test - 6

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Question 1
1 / -0

In the figure (not drawn to scale), ABC and DEF are two triangles. CA is parallel to FD and CFBE is a straight line. Find the value of x + y.

A
183°

B
134°

C
148°

D
176°

Solution

∠FCA = ∠BFD (corresponding angles)
x = 47°
Now, in triangle ABC:
y = 47° + 89° (Exterior angle property)
y = 136°
x + y = 47° + 136° = 183°
Question 2
1 / -0

In triangle ABC, if AB + BC = 12 cm, BC + CA = 14 cm, CA + AB = 16 cm, then the perimeter of the triangle is __________.

A
21 cm

B
17 cm

C
28 cm

D
22 cm

Solution

It is given that:

AB + BC = 12 cm.....(i)

BC + CA = 14 cm.......(ii)

CA + AB = 16 cm.......(iii)

Adding (i), (ii) and (iii), we get

2(AB + BC + CA) = 12 + 14 + 16

AB + BC + CA = 21 cm
Question 3
1 / -0

In the figure given below (not drawn to scale), ACDF is a rectangle and BDE is a triangle. Find EBD.

A
42°

B
71°

C
128°

D
134°

Solution

ECDB + EBDE = 90° (Angle of a rectangle)
43° + EBDE = 90°
EBDE = 90° - 43° = 47°
In triangle BED:
EEBD + EBDE + EBED = 180° ...(Angle sum property)
62° + 47° + EEBD = 180°
EEBD = 180° - (62° + 47°) = 71°
Question 4
1 / -0

In the given figure, ABCD is a rectangle. Find the value of x.

A
20°

B
30°

C
25°

D
50°

Solution

It is given that ABCD is a rectangle.
ADC = 90°
In triangle ADC:
DAC + ADC + DCA = 180° (Angle sum property)
65° + 90° + x = 180°
x = 25°
Question 5
1 / -0

In the figure given below, ABC and DBE are equilateral triangles. Find the value of x.

A
120°

B
60°

C
80°

D
None of the above

Solution

We can see that BDA is a straight line.

Angle on a straight angle measures 180°.

In BDE:

All the angles of an equilateral triangle are 60°.

∠BDE + ∠ADE = 180^o

So, ∠ADE = x = 180° - 60° = 120°
Question 6
1 / -0

If ABC is a right-angled triangle, right-angled at B and AC is the hypotenuse, then the other two angles of the triangle ABC are _________.

A
straight angles

B
right angles

C
obtuse angles

D
acute angles

Solution

If ABC is a right triangle (as shown above), right-angled at B and AC is the hypotenuse then other two angles of triangle ABC are acute angles.
This is because of the fact that the total sum of all the angles of a triangle is 180^o and if one angle is 90^o then the sum of other 2 angles will be 90^o. Hence, both will be acute angles.
Question 7
1 / -0

In the figure given below, ABCD is a square and LDC is an equilateral triangle. Find ∠ADL.

A
20°

B
35°

C
30°

D
15°

Solution

As triangle LDC is an equilateral triangle, therefore each angle measures 60°.

So, ∠LDM = 60^o.

In a square, all the angles measure 90°.

So:
∠ADM = 90^o

Therefore, ∠ADL = 90° - 60° = 30°
Question 8
1 / -0

In the above figure, the values of y and z, respectively, are

A
85^oand 35^o

B
95^o and 25^o

C
80^o and 40^o

D
95^o and 35^o

Solution

Using opposite angles property:
y = 95^o
As the sum of interior angles of a triangle is 180^o, therefore
180^o = 60^o + y + z
z = 180^o - 60^o- 95^o
z = 25^o
Question 9
1 / -0

Find ∠BPQ.

A
25°

B
30°

C
45°

D
None of these

Solution

∠APS + ∠SPQ + ∠BPQ = 180° (Straight angle)

15° + 120° + ∠BPQ = 180°

∠BPQ = 45°
Question 10
1 / -0

In the figure given below, find ∠ACB.

A
45°

B
55°

C
40°

D
None of these

Solution

Let ∠ABC = x.

x + 15° = 90°

x = 75°

Now, in ABC:

60° + 75° + ∠ACB = 180° (Angle sum property)

∠ACB = 45°
Question 11
1 / -0

In the given figure (not drawn to scale), AB is parallel to CD, ∠EOH = 85° and ∠DMF is given to be 50°. Find the value of x + y - z.

A
55°

B
45°

C
50°

D
60°

Solution

85° + ∠y = 180° (Linear Pair)

∠y = 180° - 85° = 95°

∠z = 85° (Vertically opposite angles)

∠DMF = ∠OMN = 50° (Vertically opposite angles)

In OMN,

∠OMN + ∠ONM + ∠NOM = 180^o

50^o+ ∠ONM + 85^o= 180^o

∠ONM = 45^o

∠ONM = ∠x = 45° (alternate interior angles)

So, x + y - z = 45° + 95° - 85°

= 55°
Question 12
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 13
1 / -0

In WXY, ∠YXW = 3y°, ∠YWX = (4y - 4)° and ∠WYZ = 52°. Find the value of y.

A
4

B
5

C
6

D
8

Solution

We know that:
Exterior angle = Sum of interior opposite angles
So, in XYW:
∠ZYW = (4y - 4)° + (3y)°
52 = 7y - 4
7y = 56y = 8
Question 14
1 / -0

What is the measure of x in the figure below? (Given that AB = AC)

A
120°

B
50°

C
150°

D
100°

Solution

AB = AC
∴ABC = ACB
So, ACB = 50°
Now, x = ACB + ABC [∵ Exterior angle property]
x = 50° + 50° = 100°
Question 15
1 / -0

In triangle ABC given above, B = 75°, ACB = 70°. Find the value of x - y.

A
75°

B
35°

C
110°

D
90°

Solution

x = 180° - 70° = 110° (Linear Pair)
y = 180° - (75° + 70°) = 35° (Sum of angles of a triangle)
x - y = 110° - 35° = 75°
Question 16
1 / -0

A 35-foot tall tree is leaning against a wall and is positioned such that the base of the tree is 21 feet from the base of the wall. How far above the ground is the point where the tree touches the wall?

A
28 feet

B
26 feet

C
24 feet

D
22 feet

Solution

The above figure is the representation of the situation given in the question.

In this case, a tree is leaning against a wall making a right-angled triangle.

Now,
(Hypotenuse)²= (Base)²+ (Perpendicular)²35²= 21²+ (Perpendicular)²
(Perpendicular)²= 1225 - 441 = 784.

Perpendicular = = 28 feet.

The point where the tree touches the wall is 28 feet above the ground.
Question 17
1 / -0

There is a pipe which has broken at a height of 8 m from the ground. The top end of the broken pipe touches the ground at a distance of 15 m from the base of the pipe. What is original length of the pipe?

A
10 m

B
20 m

C
25 m

D
30 m

Solution

Let PN be the original length of the pipe.

In MNO:

MO² = MN² + NO² MO² = 8² + 15²= 64 + 225
MO² = 289
MO = 17 m

Now,
PN = PM + MN
PN = MO + MN
PN = 17 + 8
PN = 25 m (Original length of the pipe)
Question 18
1 / -0

There is a park where children play in the evening. The park is in the shape of a rhombus with diagonals equal to 60 m and 80 m. What is the perimeter of the park?

A
120 m

B
140 m

C
160 m

D
200 m

Solution

We know that the diagonals of a rhombus bisect each other at 90°.
In the given figure, PQRS represents a rhombus park.
Given: SQ = 80 m, PR = 60 m

SO = = = 40 m

OR = = = 30 m

In SOR,
SR²= OR² + SO² SR² = 30² + 40²
SR² = 900 + 1600 = 2500
SR = 50 m
Perimeter = 4(SR)
= 4(50)
= 200 m
Question 19
1 / -0

There is a 41 m long ladder, which is placed against the balcony of a building. The balcony is at a height of 9 m from the ground. Calculate the distance between the foot of the ladder and the building's base.

A
35 m

B
40 m

C
45 m

D
50 m

Solution

Let MP = Length of ladder, NM = Height of balcony

In MNP:

MP² = MN² + NP²
41² = 9² + NP²
NP² = 1681 - 81

NP² = 1600

NP = 40 m

The distance between the foot of the ladder and the building's base is 40 m.
Question 20
1 / -0

A rectangle has a diagonal of 50 cm and length of 40 cm.Find the sum of all sides of a rectangle.

A
120 cm

B
130 cm

C
140 cm

D
150 cm

Solution

In PQR:

PQ²+ QR² = PR²
PR²- PQ²= QR²
50² - 40² = QR²
2500 - 1600 = QR²
QR² = 900

QR = 30 cm

Now,
Sum = 30 + 40 + 30 + 40 = 140 cm
Question 21
1 / -0

Calculate the missing angle from the figure above.

A
47°

B
53°

C
55°

D
57°

Solution

∠ABC = 180° (straight angle)

∠ABD + ∠DBC = 180°

124° + ∠DBC = 180°

∠DBC = 180° - 124° = 56°

Now, sum of all the angles of a triangle is 180°.

So, ∠BCD + ∠CDB + ∠DBC = 180°

77° + ∠CDB + 56° = 180°

∠CDB = 180° - 56° - 77° = 47°
Question 22
1 / -0

Which of the following statements is/are true?

Statement-I: An isosceles triangle has two sides of equal length, and all different angles.
Statement-II:The reflex angle is the larger angle. It is more than 180° but less than 360°

A
Only Statement-I is true

B
Only statement-II is true

C
Both the statements are true

D
Both the statements are false.

Solution

Statement-I is false because an isosceles triangle has two sides of equal length and two equal angles.

Statement -II is true as the reflex angle is the larger angle. It is more than 180° but less than 360°
Question 23
1 / -0

Fill in the blanks:

(i) The polygon with least number of sides is a __________ with three sides.
(ii) The longest side of a right-angled triangle is called the ____________, and it is always found opposite the right angle.
(iii) A scalene triangle has _______ sides of equal length and __________ equal angles.

A
i - line, ii - hypotenuse, iii - no, no

B
i - triangle, ii - base, iii - two, no

C
i - triangle, ii - hypotenuse, iii - no, no

D
None of the above

Solution

(i) The polygon with least number of sides is a triangle, with three sides.
(ii) The longest side of a right-angled triangle is called the hypotenuse, and it is always found opposite the right angle.
(iii) A scalene triangle has no sides of equal length and no equal angles.
Question 24
1 / -0

Which of the following statements is/are TRUE?

Statement 1- The sum of two similar sides of an isosceles triangle is greater than the third side.
Statement 2- In a PQR, a random point A is drawn on QR. Is it true that PQ + QR + PR > 2PA?

A
Both 1 and 2

B
Only 1

C
Only 2

D
Neither 1 nor 2

Solution

(1)
Statement 1 is true.

(2) We have PQR, as shown below.

In PQA:
PQ + QA > PA ---(1)

In PAQ:
PQ + QA > PA

Similarly,
In PAR,
AR + PR > PA

On adding these two equations, we get

PQ + QA + AR + PR > PA + PA

PQ + QR + PR > 2PA

Statement 2 is true.
Question 25
1 / -0

Find the measures of ∠d and ∠o. Given that ∠XWU = 44°; ∠XUW = 60°; ∠XVU = 30°.

A
d = 104°, o = 46°

B
d = 110°, o = 56°

C
d = 120°, o = 66°

D
None of the above

Solution

In △XUW
∠WXU + ∠XUW + ∠XWU = 180° (Sum of all the angles of a triangle is 180⁰)
∠e + 60° + 44° = 180°
∠e = 180° - 104° = 76°

and, ∠e + ∠d = 180° (Straight angle)
So, ∠d = 180° - 76° = 104°

Similarly, In △UXV
∠XVU + ∠VUX + ∠UXV = 180° (Sum of all the angles of a triangle is 180⁰)
∠c + ∠o + ∠d = 180°
30° + ∠o + 104° = 180°
∠o = 46°

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