Triangles Test

Result

Triangles Test - 1

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Question 1
1 / -0

In triangle ABC, BC is parallel to PQ. Find the value of x.

A
2

B
1

C

D

E
None of these

Solution

In triangle ABC, BC || PQ.

∴By BPT (Basic Proportionality Theorem),

=

So, =
4x(5x - 3) - 3(5x - 3) = 8x(3x - 1) - 7(3x - 1)
20x² - 12x - 15x + 9 = 24x² - 8x - 21x + 7
0 = 4x² - 2x - 2
2x² - x - 1 = 0
2x² - 2x + x - 1 = 0
2x(x - 1) + 1(x - 1) = 0

x = 1, -
Negative value of x is not possible as it would make the sides negative.
So, x = 1 is the correct answer.
Question 2
1 / -0

In PQR, S and T are points on sides PQ and PR, respectively. ST is parallel to QR. If S divides PQ in the ratio of 4 : 3 and length QR is 4.9 cm, then find ST.

A
5.6 cm

B
7.2 cm

C
3.6 cm

D
2.8 cm

E
Can't be determined

Solution

PST is similar to PQR.

Therefore,

y = 2.8 cm
Therefore, ST = 2.8 cm.
Question 3
1 / -0

Triangle PAB and square ABCD lie in perpendicular planes. If PA = 3 units, PB = 4 units and AB = 5 units, then find the length of PD.

A
units

B
units

C
units

D
units

E
units

Solution

Since the line segment AD is perpendicular to the plane of PAB, therefore angle PAD is a right angle.
In right triangle PAD:
PA = 3 units and AD = AB = 5 units
By Pythagoras theorem:
PD = = units
PB = 4 units was not needed.
Question 4
1 / -0

All the five angles marked in the given figure are equal. Each angle is equal to

A
30°

B
35°

C
36°

D
45°

E
52°

Solution

Let measure of each angle be x.
x + x + 3x = 180°
⇒ x = 36°
Question 5
1 / -0

In the following figure, OA bisects A, ABO = ACO and BOC = 100°. Find the measure of AOB.

A
120°

B
130°

C
140°

D
150°

E
160°

Solution

In triangle AOB and AOC, two angles are correspondingly equal.
⇒ AOB = AOC = z
Now, AOB + AOC + BOC = 360°
z + z + 100° = 360°Or z = 130°Hence, option (2) is correct.
Question 6
1 / -0

A triangle ABC is right-angled at A. The measure of angle B is 22° with M as the mid-point of side BC. What is the measure of AMC?

A
33°

B
40°

C
44°

D
56°

E
60°

Solution

BCA = 180° - (22° + 90°) = 180° - 112° = 68°
AMC = ?
ΔCMA is isosceles triangle with AM = MC.
∴ CAM = 68°
CMA = 180° - 2 68° = 180° - 136° = 44°
Question 7
1 / -0

What is the value of y in the given figure?

A
16 units

B
13 units

C
10 units

D
8 units

E
6 units

Solution

y = ? Property to be used - both smaller s are similar to the bigger triangle and hence each other
Question 8
1 / -0

In the figure below (not drawn to scale), if AD = CD = BC and BCE = 96°, then what is the measure of DBC?

A
32°

B
84°

C
64°

D
48°

E
90°

Solution

Given, AD = CD = BC and BCE = 96°

CAD = ACD ...(i)
CDB = CBD ...(ii)
Now, CDB = CAD + ACD = 2CAD ...(iii)
Also, CAD + CBD = 96°
CAD + 2CAD = 96° ...from (iii)
CAD = 32°
CDB = 64° ...from (iii)
DBC = 64° ...from (ii)
Question 9
1 / -0

In a right triangle ABC, AD = AE and CF = CE. If angle DEF is equal to x, find the value of x.

A
35°

B
45°

C
62°

D
72°

E
60°

Solution

Let AED = y = ADE
So, angle A = 180° - 2y
Similarly, angle C = 180° - 2z (if angle CEF = angle CFE = z)
So, A + C + 90° = 180°180° - 2y + 180° - 2z + 90° = 180°
So, y + z = 135°Hence, x = 180° - 135° = 45°
Question 10
1 / -0

In the figure above, QRS is an equilateral triangle and QTS is an isosceles triangle. If a = 13°, then what is the value of b?

A
13°

B
23°

C
30°

D
47°

E
53°

Solution

ΔQRS is an equilateral triangle.
R = Q = S = 60°
ΔQTS is an isosceles triangle.
TQS = TSQ = b°
RSQ = a° + b°
⇒ a° + b° = 60°
⇒ 13° + b° = 60°
b° = 47°
Question 11
1 / -0

ΔABC, ΔCDE, ΔEFG, ΔGHI, ΔIJK and ΔKLM are congruent to one another and similar to ΔANM. What is the ratio of the area of ΔANM to the area of ABC?

A
18 : 1

B
20 : 1

C
24 : 1

D
36 : 1

E
54 : 1

Solution

Since all small triangles are congruent, AC = CE = EG = GI = IK = KM and 6AC = AM.
Now, as ΔANM ~ ΔABC,
Required Ratio = 36 : 1
Question 12
1 / -0

The given figure is formed by joining 18 equilateral triangles, each of perimeter 6. What is the perimeter of the figure?

A
20

B
24

C
28

D
36

E
48

Solution

Side of each equilateral triangle = 2
Perimeter of the given figure = 12 2 = 24
Question 13
1 / -0

In the triangle ABC, BM and CM are the respective angle bisectors of the exterior angles CBD and BCE, meeting at point M. What is the relation between BMC and A?

A
A = 90° - M

B
M = 90° - A

C
A = 90° + M

D
M = 90° + A

E
M = 90° - A

Solution

In ABC,
A + 180° - 2x + 180° - 2y = 180° (Sum of angles of triangle = 180°)
A + 180° = 2(x + y)
x + y = 90° + A …… (1)
In BMC,
x + y + M = 180°x + y = 180° - M …… (2)
From (1) and (2):
90° + A = 180° - M
M = 90° - A
Question 14
1 / -0

In the given figure, length (AC) = 3, length (BC) = 4, length(AE) = 6

Column A Column B

length (DE) 8

Compare the two quantities to decide whether

A
the quantity in Column A is greater

B
the quantity in Column B is greater

C
the two quantities are equal

D
the relationship cannot be determined from the information given

Solution

Since we don`t know about similarity of ΔABC and ΔADE, we can`t find l(DE).

Question 15

1 / -0

Directions: Compare the two quantities in column A and column B.

Column A	Column B
The ratio of the area of a right angled triangle ABC and of another triangle whose sides containing the right angle, are three times and four times the sides of DABC, respectively.	The ratio of the areas of rectangle PQRS and another rectangle whose sides are 4 times the side of the rectangle PQRS.