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Correct option is  D) XY > XZ + ZY

Sum of any two sides is always greater than the third side.

\(\therefore\)  XY > XZ +ZY is not a situation under which ΔABC is constructed.

We need 

HG = TS 

HJ = TR

Correct answer is (c) right

From ΔADC, we have

AC² = AD² + CD² (Pythagoras Theorem)  .... (1)

From ΔADB, we have

AB² = AD² + BD²  (Pythagoras Theorem) ......(2)

Subtracting (1) from (2), we have

AB² – AC² = BD² – CD²

or, AB² + CD² = BD² + AC²

The Obtuse triangle always has altitude outside itself.

Correct option is: (C) EF || QR

Correct option is: (D) 2.4

Correct option is: (A) 4

A triangle is said to be Equilateral if each one of its sides has the same length.

Correct answer is y = 30°

Two angles are said to be Congruent, if they have equal measures.

Median is also called Altitude in an equilateral triangle. 

If one angle of a triangle is equal to the sum of other two, then the measure of that angle is 90°.

Measures of each of the angles of an equilateral triangle is 60°

In an isosceles triangle, angles opposite to equal sides are Equal.

In an isosceles triangle, two angles are always Equal.

Correct answer is 20°, 80°, 80°

Correct answer is (b) 65°

Correct answer is 36°, 54°, 90°

Correct answer is 360°

According to the question,

∆ NSQ ≅ ∆MTR

∠1 = ∠2

Since,

∆NSQ = ∆MTR

So,

SQ = TR ….(i)

Also,

∠1 = ∠2 ⇒ PT = PS….(ii)

[Since, sides opposite to equal angles are also equal]

From Equation (i) and (ii).

PS/SQ = PT/TR

⇒ ST || QR

By converse of basic proportionality theorem, If a line is drawn parallel to one side of a triangle to intersect the other sides in distinct points, the other two sides are divided in the same ratio.

∴ ∠1 = PQR

And

∠2 = ∠PRQ

In ∆PTS and ∆PRQ.

∠P = ∠P [Common angles]

∠1 = ∠PQR (proved)

∠2 = ∠PRQ (proved)

∴ ∆PTS – ∆PRQ

[By AAA similarity criteria]

Hence proved.

Correct option is (C) 3 cm

(A) \(\because\) 4+1 = 5 < 6

\(\therefore\) 4 cm, 1 cm and 6 cm do not form any triangle.

(B) \(\because\) 4+2 = 6 (not strictly greater than 6)

\(\therefore\) 4 cm, 2 cm and 6 cm do not form any triangle.

(C) \(\because\) 4+3 = 7 > 6, 4+6 = 10 > 3 and 3+6 = 9 > 4

\(\therefore\) 3 cm, 4 cm and 6 cm do not form any triangle.

(D) \(\because\) 4+1.5 = 5.5 < 6

\(\therefore\) 4 cm, 1.5 cm and 6 cm do not form any triangle.

Correct option is  C) 3 cm

Correct option is (D) 5 cm, 8 cm, 1 cm

Since, sum of any two sides in a triangle is always greater than the third side.

\(\because\) 5 cm + 1 cm = 6 cm < 8 cm

\(\therefore\) Sides of 5 cm, 1 cm and 8 cm do not form any triangle.

D) 5 cm, 8 cm, 1 cm

Correct option is  C) ∠C > ∠B

B) RHS congruency rule

By exterior angle theorem,

∠ACD = ∠A + ∠B

∠ACD = 68° + ∠B

\(\frac{1}{2}\)∠ACD = 34° + \(\frac{1}{2}\)∠B

34° = \(\frac{1}{2}\)∠ACD - ∠EBC (i)

Now,

In ΔBEC

∠ECD = ∠EBC + ∠E

∠E = ∠ECD - ∠EBC

∠E = \(\frac{1}{2}\)∠ACD - ∠EBC (ii)

From (i) and (ii), we get

∠E = 34°

(B) AC = DE

Given, in ΔABC and ΔDEF, AB = DF and ∠A = ∠D

We know that, two triangles will be congruent by ASA rule, if two angles and the included side of one triangle are equal to the two angles and the included side of other triangle.

∴ AC = DE

seg PT ⊥ ray ST, seg PR ⊥ ray SR [Given] 

seg PR ≅ seg PT 

∴ Point P lies on the bisector of ∠TSR [Any point equidistant from the sides of an angle is on the bisector of the angle] 

∴ Ray SP is the bisector of ∠RST.

∠RSP = 56° [Given] 

∴ ∠RSP = (1/2) ∠RST 

= (1/2) x 56° 

∴ ∠RSP = 28°

Correct answer is (C) ∠B = ∠D

Correct answer is (B) similar but not congruent

Correct option is (A) AAS

Given \(\angle ACB=\angle PRQ\)

& \(\angle ABC=\angle PQR\)

\(\therefore\) \(\triangle ABC\sim\triangle PQR\)    (By AA similarity criterion)

Correct option is: (A) AAS

Correct option is (D) No similarity

Given that : In \(\triangle PQR,\) PQ = 6, QR = 10 & \(\angle PQR=90^\circ\)

& In \(\triangle LMN,\) LM = 3, MN = 4 & \(\angle LMN=90^\circ\)

\(\because\) \(\frac{PQ}{LM}=\frac63=\frac21\)

& \(\frac{QR}{MN}=\frac{10}4=2.5\)

\(\because2\neq2.5\)

\(\therefore\) \(\frac{PQ}{LM}\neq\frac{QR}{MN}\)

\(\therefore\) \(\triangle PQR\) is not similar to \(\triangle LMN.\)

Hence, there is no similarity between both triangles.

Correct option is: (D) No similarity

Data: S and T are points on sides PR and QR of ∆PQR such that 

∠P = ∠RTS. 

To Prove: ∆RPQ ~ ∆RTS. 

In ∆RPQ and ∆RTS, 

∠P = ∠RTS (Data) 

∠PRQ = ∠SRT (Common) 

∴ 3rd angle ∠PRQ = ∠SRT 

∴ These are equiangular angular triangles. 

∴ Here A.A.A. similarity criterion. 

∴ ∆RPQ ~ ∆RTS.

Angle opposite to side LM is ∠N.

Correct answer is Yes, SSS, ∠ N = 40°

Correct option is: (C) AAA

Data: Here \(\frac{QR}{QS} = \frac{QT}{PR}\)  and ∠1 = ∠2

To prove: ∆PQS ~ ∆TQR. 

In ∆PQS and ∆TQR, we have

\(\frac{QR}{QS} = \frac{QT}{PR}\)

∠PQR = ∠PRQ (∵ ∠1 = ∠2) 

Here, Similarity criterion used here is side, angle, side (SAS). 

∴ ∆PQS ~ ∆TQR.

Yes,

(i) In ΔPQR and ΔPSR,

PQ = PS  [given]

∠1 = ∠2  [given]

PR = PR [common side]

By SAS congruence criterion, ΔPQR ≅ ΔPSR

(ii) Yes, QR = SR

Correct option is C) congruent

(i) False

(ii) True

(iii) False

(iv) False

(v) True

(vi) True

Proof: 

In ∆ADB and ∆BEA, seg BD ≅ seg AE [Given] 

∠ADB ≅ ∠BEA = 90° [Given] 

seg AB ≅ seg BA [Common side] 

∴ ∆ADB ≅ ∆BEA [Hypotenuse-side test] 

∴ seg AD ≅ seg BE [c. s. c. t.]

From given:

∠ABC = 90°, BD ⊥ AC.

AB = 5.7 cm, BD = 3.8 cm, CD = 5.4 cm

In ∆ABC and ∆BDC,

∠ABC = ∠BDC (each 90°)

∠BCA = ∠BCD (common angles)

∆ABC ~ ∆BDC (AA axiom)

So, corresponding sides are proportional

AB/BD = BC/CD

=> 5.7/3.8 = BC/5.4

=> BC = 8.1

Correct option is: (C) \(\frac{AM}{MB}=\frac{AN}{ND}\)

In ∆LMN and ∆PNM, 

seg LM ≅ seg PN 

seg LN ≅ seg PM [Given] 

seg MN ≅ seg NM [Common side] 

∴ ∆LMN ≅ ∆PNM [SSS test]

∴ ∠LMN ≅ ∠PNM, 

∴ ∠MLN ≅ ∠NPM, and ∠LNM ≅ ∠PMN[c.a.c.t.]

We have,

∠ABC = 90º and BD perpendicular AC

In ΔABY and ΔBDC

∠C = ∠C (Common)

∠ABC = ∠BDC(each 90º angles)

then, Δ ABC ~ΔBDC (By AA Similarity) 

So,\(\frac{AB}{BD}\) = \(\frac{BC}{DC}\) (Corresponding parts of similar triangle area proportion) 

Or \(\frac{5.7}{3.8}\) = \(\frac{BC}{5.4}\)

Or BC = 5.7/3.8 x 8.1 

Or BC = 12.15 cm

Given: \(\frac{QT}{PR}\)= \(\frac{QR}{QS}\)

∠1 = ∠2 

R.T.P : △PQS ~ △TQR 

Proof: In △PQR; ∠1 = ∠2 

Thus, PQ = PR [∵ sides opp. to equal angles are equal]

\(\frac{QT}{PR}\)=\(\frac{QR}{QS}\) ⇒ \(\frac{QT}{PQ}\)= \(\frac{QR}{QS}\)

i.e., the line PS divides the two sides QT and QR of △TQR in the same ratio. 

Hence, PS // TR. 

[∵ If a line join of any two points on any two sides of triangle divides the two sides in the same ratio, then the line is parallel to the third side] 

Hence, PS // TR (converse of B.P.T)

Now in △PQS and △TQR 

∠QPS = ∠QTR 

[∵ ∠P, ∠T are corresponding angles for PS // TR] 

∠QSP = ∠QRT 

[∵ ∠S, ∠R are corresponding angles for PS // TR] 

∠Q = ∠Q (common) 

∴ △PQS ~ △TQR (by AAA similarity)

i) False

ii) True

ii) False