Triangles Test

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Triangles Test - 7

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Question 1
1 / -0

If in the given figure, ΔABC ~ ΔCDA, then = ?

A

B

C

D

Solution

Since ΔABC and ΔCDA are similar,
=
Question 2
1 / -0

In ΔBAC, BA = BC and CD bisects ACB. Also if CA = CD, find BAC - ABC.

A
24°

B
30°

C
36°

D
42°

Solution

In ΔBAC,
AB = BC
So, CAB = BCA (If opposite sides are equal, then opposite angles are also equal.)
Let CAB = BCA = x
So, ABC = 180° – 2x

In ΔADC, we have
AC = CD (given)
So, CAB = CDA (If opposite sides are equal, then opposite angles are also equal)
So, CAD = CDA = x
ACD = 180° – 2x

Since CD bisects ACB,
180° – 2x =
Solving, x = 72°
BAC = BCA = 72°
BAC + BCA + ABC = 180° (angle sum property)
72° + 72° + ABC = 180°
ABC = 180° - 144°
ABC = 36°
BAC - ABC = 72° - 36° = 36°
Question 3
1 / -0

In rectangle ABCD, AB² + BC² + CD² + DA² = ?

A
OA²+ OB²

B
OB² + OC²

C
OC² + OD²

D
AC² + BD²

Solution

In ΔABC,
B = 90°
So, by Pythagoreous theorem,
AB² + BC² = AC² --- (1)
And in triangle ADC,
AD² + DC²= AC² --- (2)
AB² + BC² + AD² + DC² = AC² + AC²
And, we know that the diagonals of a triangle are equal.
So, AC = BD
AB² + BC² + CD² + DA² = AC² + BD²
Question 4
1 / -0

In the given figure, area of ΔCQP is 32 cm², which is equal to the area of ΔCQR, and AQ = 6 cm. What is the value of AB?

A
2.3 cm

B
2.5 cm

C
2.7 cm

D
2.9 cm

Solution

Area of ΔCQP = Area of ΔCQR = 32 cm²

Area of ΔCQR = = 2CQ

32 = 2CQ
CQ = 16 cm --- (i)
CA = CQ - AQ = 16 - 6 = 10 cm (As it is given that AQ = 6 cm) ---(ii)
Now,
In ΔCQR and ΔCAB,
CAB = CQR (each 90°)
C = C (Common)

Thus, ΔCQR and ΔCAB are similar by AA similarity criteria.

Therefore,

... (because CQ = 16 cm, from (i), CA = 10 cm and from (ii), QR = 4 cm (given))

AB = = = 2.5 cm
Question 5
1 / -0

In the given adjacent figure, ABC = AED = 90°, AE = 8 cm, AB = 12 cm , AD = (2x + 6) cm, and AC = (4x + 2) cm. Find AD and AC.

A
AD = 15 cm, AC = 25 cm

B
AD = 20 cm, AC = 30 cm

C
AD = 25 cm, AC = 35 cm

D
AD = 30 cm, AC = 40 cm

Solution

In ΔAED and ΔABC,
ABC = AED = 90°
A = A (common)

So, ΔAED ~ ΔABC (by AA similarity)
=
=

32x + 16 = 24x + 72
8x = 56
x = 7
So, AD = 2 × 7 + 6 = 20 cm
AC = 4 × 7 + 2 = 30 cm
Question 6
1 / -0

In the given figure, DEB = ACB. What is equal to?

A

B

C

D

Solution

In ΔABC and ΔDBE,
DEB = ACB (given)
ABC = DBE (common)
So, ΔABC ~ ΔDBE (By AA similarity criteria)

Hence, .
Question 7
1 / -0

In the given figure, RN and QM are perpendicular to PQ and PR, respectively. If PR = 12 cm, PQ = 10 cm and QM = 8 cm, then what is the value of 10RN?

A
95 cm

B
96 cm

C
97 cm

D
98 cm

Solution

In ΔPNR and ΔPMQ,
∠QPM = ∠RPN (Common)
∠QMP = ∠RNP (Each 90°)
Thus, ΔPNR and ΔPMQ are similar to each other by AA similarity rule.

In right-angled ΔPMQ,

= = = 6 cm

PM = 6 cm

As, ΔPMQ is similar to ΔPNR,

10RN = 96 cm
Question 8
1 / -0

In the given figure, AD = DC and ADB = CDB.

Which of the following criteria will make ΔADB congruent to ΔCDB?

A
ASA

B
SSS

C
SAS

D
SSA

Solution

In ΔADB and ΔCDB,
AD = DC (Given)
ADB = CDB (Given)
DB = DB (Common)
∴ ΔADB ≌ ΔCDB(by SAS criterion)
Question 9
1 / -0

In the given figure, DE is parallel to BC, FE is parallel to DC, BC = 10 cm, and DE = 6 cm. Find 6FD.

A
5AF

B
10AF

C
15AF

D
20AF

Solution

Given that DE is parallel to BC.
Using basic proportionality theorem in ΔABC and ΔADE,
= --- (1)

Also, FE is parallel to DC.
Using basic proportionality theorem in ΔAFE and ΔADC,
= --- (2)

From (1) and (2),
= --- (3)

In ΔABC and ΔADE,
A = A (Common)
ADE = ABC (Since DE is parallel to BC, ADE and ABC are corresponding angles.)
Thus, ΔABC and ΔADE are the similar triangles by AA similarity criteria.

So,

Substituting this in equation (3),

6FD = 10AF
Question 10
1 / -0

In the following figure, AB and CD are perpendicular to BD. Find the length of side AC.

A
14 cm

B
13 cm

C
12 cm

D
11 cm

Solution

AB = PD = 13 cm
So, CP = CD - PD
CP = 18 - 13 = 5 cm
AP = BD = 12 cm
ΔAPC is a right angled triangle.
Using Pythagoras theorem,
AC² = AP² + PC²
= 12² + 5²
= 144 + 25 = 169
AC = 13 cm
Question 11
1 / -0

In the given figure, ST || PR and SR || PA. What will be the value of QA – QR, if QT = 3 cm and TR = 9 cm?

A
40 cm

B
31 cm

C
13 cm

D
36 cm

Solution

QA – QR = RA.
So, we need to find the value of RA.
In ΔQPR, ST || PR.
= (By Basic proportionality theorem) --- (I)

In ΔQPA, SR || PA.
= (By Basic proportionality theorem) --- (ii)

From (i) and (ii),
=

=

=

=

RA = (12 × 9) ÷ 3
RA = 36 cm
Question 12
1 / -0

In the given figure, MNO and OPM are isosceles right triangles. What is the length of OP?

A
8 units

B
10 units

C
5units

D
7units

Solution

In ΔONM,
ON² + NM² = OM²........(Pythagoras theorem)
5² + 5² = OM²
OM = 5units

In ΔOMP,
OM = MP = 5units
Now, OM² + PM² = OP²(5)² + (5)² = OP²50 + 50 = OP²
OP² = 100
OP = 10 units
Question 13
1 / -0

In ΔABC, right-angled at B, AD = AE and CF = CE. If DEF is equal to x, find the value of x.

A
35°

B
45°

C
62°

D
72°

Solution

Let AED = y = ADE (∵ AD = AE)
So, A = 180° - 2y
Similarly, C = 180° - 2z (If CEF = CFE = z)

So, A + C + 90° = 180° (Angle sum property of triangle)
180° - 2y + 180° - 2z + 90° = 180°
So, y + z = 135°
Hence, x = 180° - 135° = 45° (Linear pair)
Question 14
1 / -0

Sima is sliding down a slide in a park, which is in the shape of a right-angled triangle. She slides down with a length of A m and then returns back to the slide by climbing up with a perpendicular length of B m. If A - B = 2 m, find out the length of the base of the slide in terms of A.

A
(2A + 2) metres

B
() metres

C
metres

D
(4A) metres

Solution

Let the length of the base of the slide be C m.
Given that A – B = 2 m --- (i)
By Pythagoras theorem,
A² = B² + C²
C² = A² - B²
= (A - B)(A + B)
= 2(B + A) --- from (i)
= 2(A - 2 + A)
= 2(2A - 2)
C²= 4A - 4 m
C = metres
Question 15
1 / -0

In a ΔMQR, QT ⊥ MR, PQ = QR, and QP is a line that touches MR. Which of the following option is correct?

A
QM² = PM² + 2PM.PT + 2RQ²

B
QM² = 2PM.PT + RQ²

C
QM² = PM² + 2PM.PT + RQ²

D
QM² = QT² + 2PM.PT + RQ²

Solution

Given,
Q = 90° and QT ⊥ PR
Using Pythagoras theorem in ΔMQR,
RM² = QM² + QR² ...(1)
Using Pythagoras theorem in ΔQTM,
QM² = MT² + QT²
Because MT = PM + PT,
QM² = (PM + PT)² + QT²
QM² = PM² + PT² + 2PM.PT + QT² ...(2)In ΔPTQ (by Pythagoras theorem),
PQ² = PT² + QT² ...(3)

From (3), we get equation (2) as
QM² = PM² + 2PM.PT + PQ² ...(4)

Putting the value of QM² in (1) equation,
RM² = QM² + QR²
RM² = PM² + 2PM.PT + PQ² + QR²
RM² - QR² = PM² + 2PM.PT + PQ²

In ΔMQR,
RM² - QR² = QM²
QM² = PM² + 2PM.PT + PQ²
PQ = QR (Given)
So,
QM² = PM² + 2PM.PT + RQ²
Question 16
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Arpan starts moving from the foot of the tower (height 8 m) on the ground, covers 4 m and stops at point C for some time. Then, he again starts moving in the same direction, further covers 11 m, and marks that point as D. What is the approximate difference between the distance of the top of the tower from point D and from point C?

A
8.1 m

B
8.3 m

C
8.9 m

D
9.5 m

Solution

Height of the tower, i.e. AB = 8 m
Distance of point C from the foot of the tower, i.e. CB = 4 m
Distance of point C from the top of the tower, i.e. AC² = AB² + CB²
AC² = 8² + 4²
AC² = 64 + 16
AC² = 80
AC = = 8.9 m --- (1)

Now,
Distance of point D from the foot of the tower, i.e. BD = 11 m + 4 m = 15 m
Distance of point D from the top of the tower is AD.

Applying Pythagoras Theorem,
AD² = BD² + AB²
AD² = 225 + 64
AD² = 289
AD = 17 m --- (2)
Difference = 17 m – 8.9 m = 8.1 m
Question 17
1 / -0

In the sextant experiment, the distance between the top of the two buildings on the plane ground comes out to be 5 m. If the height of the two buildings is 16 m and 13 m, what is the distance between the two buildings?

A
3 m

B
4 m

C
5 m

D
6 m

Solution

Let AB = 16 m and DC = 13 m be the heights of two buildings, and AD = 5 m be the distance between the top of the two buildings.
Draw a straight line DE from point D.
In triangle ADE,
AD = 5 m
AE = AB – DC
AE = 16 m – 13 m
AE = 3 m
AD² = AE² + DE²
5² = 3² + DE²
DE² = 25 – 9
DE² = 16
DE = 4 m
Question 18
1 / -0

In the given figure, DF || AG, DE || AB, AB = 15, CD = 8, AD = a, DE = 10, FG = b and CG = 6. The ratio a : b is equal to

A
1 : 2

B
2 : 1

C
1 : 3

D
2 : 3

Solution

ΔCDE ~ ΔCAB (AA Similarity)

Now, in ΔCAG, DF || AG
By BPT;

Or or 3b = 6 or b = 2
Question 19
1 / -0

In a triangular field, three players A, B and C are standing in the corners of the field. The distance of player A from player C is same as the distance of player A from B. The distance of player B from player C is 30 m and the perpendicular distance of player A from his front side CB is 30 m. If the distance between the points D and E marked on the triangular field is 14 m and the line from D to E is parallel to the line CB, then what is the distance between player A and the midpoint of the line DE?

A
12 m

B
14 m

C
15 m

D
17 m

Solution

Drawing AG ⊥ CB,
In ΔACG and ΔADF,
∠G = ∠F (right angle)
∠A = ∠A (common angle)
So, ΔACG is similar to ΔADF (by AA rule).
=
=
AF = × 30
AF = 14 m
Question 20
1 / -0

There are two buildings, A and B, standing parallel to each other. Height of building A is 6 m and that of building B is 17 m. A ribbon is used to connect the foot of building B to the top of building A, whose length is 10 m. Another ribbon is used to connect the top of building B to the foot of building A. Both ribbons cross each other at some point.

(i) What is the distance between the feet of building A and building B?
(ii) What is the perpendicular distance from the ground to the point where ribbons cross, if the distance of that point on ground from the foot of building B is 6 m?

A
8 m, 4 m

B
8 m, 4.5 m

C
7.5 m, 5 m

D
6 m, 4.5 m

Solution

(i) The distance between the foot of building A and building B is DC.
AC = 10 m
AD = 6 m

Using Pythagoras theorem,
AC² = AD² + DC²
10² = 6² + DC²
100 = 36 + DC²
DC² = 100 – 36
DC² = 64
DC = 8 m

(ii)

Let E be the point where the ribbons cross and draw a perpendicular EF on DC.
From ΔADC and ΔEFC,
∠D = ∠F (right angle)
∠C = ∠C (common angle)
ΔADC and ΔEFC are congruent.

=
EF =
EF = 4.5 m

Question 21

1 / -0

Match the following:

Column I	Column II
(A) In ΔPQR and ΔMNO, if P = M, Q = N, then ΔPQR ~ ΔMNO.	(I) Basic Proportionality Theorem (BPT)
(B) If a line is parallel to a side of a △ which intersects the other two sides into two distinct points, then the line divides those sides in proportion.	(II) SAS Similarity Criterion
(C ) If in two triangles, one pair of corresponding sides are proportional and the included angles are equal, then the two triangles are similar.	(III) AA Similarity Criteria
(D) In ΔXYZ and ΔLMN, XZ = 6 cm, YZ = 7 cm, XY = 8 cm, LM = 24 cm, LN = 18 cm, and MN = 21 cm, then ΔXYZ ~ ΔLMN.	(IV) SSS Similarity Criteria

(A) --- (II), (B) --- (III), (C) --- (I), (D) --- (IV)

(A) --- (III), (B) --- (I), (C) --- (II), (D) --- (IV)

(A) --- (I), (B) --- (III), (C) --- (II), (D) --- (IV)

(A) --- (III), (B) --- (II), (C) --- (I), (D) --- (IV)

Solution

(A) – (III) The AA criterion for triangle similarity states that if two triangles have two pairs of congruent angles, then the triangles are similar.
So, if in ΔPQR and ΔMNO, P = M, Q = N, then ΔPQR ~ ΔMNO, by AA similarity criteria.

(B) – (I) According to Basic Proportionality Theorem, if a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.

(C ) - (II) According to SAS similarity criteria: If in two triangles, one pair of corresponding sides is proportional and the included angles are equal, then the two triangles are similar.

(D) – (IV) The SSS criterion for triangle similarity states that if three sides of one triangle are proportional to three sides of another triangle, then the triangles are similar.

In ΔXYZ and ΔLMN,

=

So by SSS criteria, ΔXYZ and ΔLMN are similar.

Question 22
1 / -0

Which of the following statements is correct?

(A) If all the corresponding sides of ΔABC and ΔPQR are equal, the triangles are congruent.
(B)

In the above picture, only shape I and II are similar because the ratio of corresponding sides is equal.

(C) A pair of similar triangles has three equal angles and proportional sides.
(D)
The triangles are congruent, according to SAS.

A
(A) only

B
(A) and (C)

C
(B) only

D
(B) and (C)

Solution

(A) - This is correct because if all 3 corresponding sides of and are equal, they are congruent by SSS.
(B) - This is incorrect because the ratio of the corresponding angles is equal in shape I and III. So, shape I and III are similar.
(C) - This statement is correct because if two triangles are similar, they have three equal angles and proportional sides.
(D) - This statement is incorrect because the rule applied here is RHS as both are right-angled triangles. The measure of hypotenuse of both triangles is same, and measure of corresponding sides is equal.
Hence, option 2 is the correct answer.
Question 23
1 / -0

In the given figure, PS is a median of ΔPQR, and PM is perpendicular to QR. Find:
(i) PR²
(ii) PQ²

A
(i) + (2MS.QR) (ii) - (MS.QR)

B
(i) + (MS.QR) (ii) - (MS.QR)

C
(i) + (MS.QR) (ii) - (MS.QR)

D
(i) + (MS.QR) (ii) - (MS.QR)

Solution

(i) In △PMS, by Pythagoras theorem:
PM² + MS² = PS² --- (i)

In ΔPMR, by Pythagoras theorem:
PM² + MR² = PR²
PM² + ( MS + SR)² = PR² (because MR = MS + SR)
PM² + MS² + SR² + 2 × MS × SR = PR²
(PM² + MS²) + SR²+ 2(MS.SR) = PR²PS² + SR²+ 2MS.SR = PR² --- (from (i))
+ 2 MS. = PR² (because PS is the median of triangle PQR)
So, PR² = PS² + + (MS.QR)

(ii) In ΔPMQ,
PQ² = PM² + MQ²
PQ² = (PS² - MS²) + (QS - MS)^{2 ...}(from (i) and MQ = QS - MS)
PQ² = PS² - MS² + QS² + MS² –2MS.QS
PQ² = PS² + QS² –2MS.QS
(because PS is the median of triangle PQR)
So, - (MS.QR)
Question 24
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°

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 25
1 / -0

In triangle ABC, ∠C and in triangle PQR, ∠R are 90°. If A = P, DQ = 2.5 cm and AD is the perpendicular bisector of QP, then find the length of AB.

A
20 cm

B
25 cm

C
30 cm

D
35 cm

Solution

In triangle ABC and triangle PQR,
QRP = ACB = 90° (Each 90°)
CAB = RPQ (given)

Triangle ABC and triangle PQR are similar to each other by AA similarity criteria,

(PQ = QD + DP, i.e. PQ = 2.5 + 2.5 = 5 cm (Because AD is the perpendicular bisector of PQ)

AB = 20 cm