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Correct option is: (B) DE || BC

Correct option is  C) 12.5

Correct option is A) 58°

We have, D and E as the midpoint of AB and AC 

So, according to the midpoint therom 

DE||BC and DE = 1/2 BC……………..(i) 

In ΔADE and ΔABC

∠A = ∠A (Common)

∠ADE = ∠B (Corresponding angles)

Then, ΔADE ~ ΔABC (By AA similarity) 

By area of similar triangle theorem 

Area ΔADE/ Area ΔABC = DE²/BC² 

Or, (1/2BC)²/(BC)² 

Or, 1/4

In Δ ANC and Δ ABC

∠C = ∠c (Common)

∠ANC = ∠BAC (Each 90°)

Then, Δ ANC ~ Δ ABC (By AA similarity) 

By area of similarity triangle theorem. 

Area of ΔABC/Area of ΔPQR = AC² /BC² 

Or, 5²/12²

Or, 25/144

Correct option is C) b + c > a

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

\(\therefore\) b + c > a is true.

Correct option is B) 4 cm, 5 cm, 7 cm

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

(A) 2 + 3 = 5 not strictly greater

Thus, these sides do not form any triangle

(B) 4 + 5 = 9 > 7,

4 + 7 = 11 > 5 and

5 + 7 = 12 > 4

Thus, these sides will form a triangle.

(C) 1 + 2 = 3 not strictly greater

Thus, these sides do not form any triangle

(D) 5 + 6 = 11 < 13

Thus, these sides do not form any triangle.

∠Q = 180° – (95° + 65°) = 20° 

∴ ∠Q < ∠T < ∠P

 ∴ PT < PQ < TQ 

Correct option is (B) PQ < TQ 

Correct option is D) 3 cm, 4 cm, 7 cm

\(\because\) 3 + 4 = 7(not strictly greater)

Thus, sides 3 cm, 4 cm, 7 cm do not form any triangle because sum of any two sides is always greater than the third side.

Correct option is  A) PQ = 5 cm, ⌊P = 105°, ⌊Q = 95°

Correct option is  B) 11 cm, 7 cm, 18 cm

Correct option is  A) 3

i. PQ = 12, QR = 5 [Given] 

In APQR, ∠Q = 90° [Given] 

∴ PR² = QR² + PQ² [Pythagoras theorem]

 = 25 + 144 

∴ PR² =169 

∴ PR = 13 units [Taking square root of both sides]

ii. In right angled APQR, seg QS is the median on hypotenuse PR.

∴ QS = \(\frac{1}2\)PR [In a right angled triangle, the length of the median on the hypotenuse is half the length of the hypotenuse] 

= \(\frac{1}2\) x 13 

∴ l(QS) = 6.5 units

Here AB² = 576, BC² = 100 and AC² = 676. 

So, AC² = AB² + BC² 

Hence, the given triangle is a right triangle.

False

We know that,

Corresponding angles are equal in similar triangles.

So, we get,

∠D = ∠R

∠E = ∠P

∠F = ∠Q

False

According to the question,

Let us assume that,

A = 25 cm

B = 5 cm

C = 24 cm

Now, Using Pythagoras Theorem,

We have,

A² = B² + C²

B² + C² = (5)² + (24)²

B² + C² = 25 + 576

B² + C² = 601

A² = 600

600 ≠ 601

A² ≠ B² + C²

Since the sides does not satisfy the property of Pythagoras theorem, triangle with sides 25cm, 5cm and 24cm is not a right triangle.

False

As per question,

Let suppose,

a = 25 cm,

b = 5 cm and

c = 24 cm

Now,

b² + c² = (5)² + (24)²

b² + c² = 25 + 576

→ b² + c² = 601 ≤ (25)²

According to the Pythagoras theorem, in a right-angle triangle the sum of square of two sides must be equal to square of third side. But Given sides do not make a right triangle because it does not satisfy the property of Pythagoras theorem so the triangle with sides 25cm, 5cm and 24cm is not a right triangle.

(c)Two triangles are similar if their corresponding sides are proportional. According to the statement: 

∆ABC~ ∆DEF

if AB/DE = AC/DF = BC/EF 

We have, 

PA ⏊ AC, and RC ⏊ AC 

Let AB = a and BC = b 

In ΔCQB and ΔCPA

∠QCB = ∠PCA(common)

∠QBC = ∠PCA(Each 90º)

Then, ΔCQB ~ ΔCPA (By AA similarity)

So,\(\frac{QB}{PA}\) =\(\frac{CB}{CA}\)  (Corresponding parts of similar triangle area proportion)

Or,\(\frac{y}x\) = \(\frac{b}{{a}+{b}}\) -----------(i) 

In ΔAQB and ΔARC

∠QAB = ∠RAC(common)

∠ABQ = ∠AQC(Each 90º)

Then, ΔAQB ~ ΔARC (By AA similarity)

So,\(\frac{QB}{RC}\) = \(\frac{AB}{CA}\) (Corresponding parts of similar triangle area proportion)

Or,\(\frac{y}x\) =\(\frac{b}{{a}+{b}}\)  -----------(ii) 

Adding equation i & ii

\(\frac{y}{x}+\frac{y}{x} \) = \(\frac{b}{{a}+{b}}\) = \(\frac{a}{{a}+{b}}\)

or,\({y}(\frac{1}{x}+\frac{1}{z})\) = \(\frac{b+a}{a+b}\)

or,\({y}(\frac{1}{x}+\frac{1}{z})\) = 1

or \(\frac{1}{x}+\frac{1}{z} \) = \(\frac{1}y\)

Intersecting point: P 

Concurrent point: Q

Given: PQ || RS

To Prove: ΔPOQ ~ ΔSOR

Let us take ΔPOQ and ΔSOR

∠OPQ = ∠OSR (as PQ || RS, Alternate angles)

∠POQ = ∠ROS (vertically opposite angles)

∠OQP = ∠ORS (as PQ || RS, Alternate angles)

∴ ΔPOQ ~ ΔSOR (by AAA similarity criterion)

Hence Proved

OA . OB = OC . OD (Given)

So, OA/OC=OD/OB ....  (1)

Also, we have ∠AOD =∠COB (Vertically opposite angles) ...... (2)

Therefore, from (1) and (2), ΔAOD ~  ΔCOB (SAS similarity criterion)

So, ∠A = ∠C and ∠D =∠B

(Corresponding angles of similar triangles)

Correct option is B) ΔWON

Correct option is (C) 9

\(\because\) LM || AB

\(\therefore\) \(\angle CLM=\angle CAB\)     (Corresponding angles as LM || AB)

& \(\angle CML=\angle CBA\)     (Corresponding angles as LM || AB)

\(\therefore\) \(\triangle CLM\sim\triangle CAB\)    (By AA similarity rule)

\(\therefore\) \(\frac{LC}{AC}=\frac{MC}{BC}\)                (By properties of similar triangles)

\(\Rightarrow\) \(\frac{AC-AL}{AC}=\frac{BC-BM}{BC}\)

\(\Rightarrow\) \(\frac{2x-(x-3)}{2x}=\frac{2x+3-(x-2)}{2x+3}\)               (Given)

\(\Rightarrow\) \(\frac{x+3}{2x}=\frac{x+5}{2x+3}\)

\(\Rightarrow(x+3)(2x+3)=2x(x+5)\)

\(\Rightarrow2x^2+9x+9=2x^2+10x\)

\(\Rightarrow10x-9x=9\)

\(\Rightarrow x=9\)

Correct option is: (C) 9

The answer is ΔRAT ≅ ΔWON

Given 

∠M = ∠N = 46° 

It forms a pair of corresponding angles, hence LM || PN. 

In ΔLMK and ΔPNK, 

∠LMK = ∠PNK [corresponding angles] 

∠MLK = ∠NPK [corresponding angles] 

∠K = ∠K [common angle] 

We know that AAA similarity criterion states that in two triangles, if corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar. 

∴ ΔLMK ~ ΔPNK 

We know that two triangles are similar if their corresponding sides are proportional.

⇒ \(\frac{ML}{NP}\) = \(\frac{MK}{NK}\)

⇒ \(\frac{a}{x}\) = \(\frac{b+c}{c}\)

∴ x = \(\frac{ac}{b+c}\)

i) ΔABC ≅ ΔDEF 

∵ ∠B = ∠E (∵ Angle sum property Z E = 180° – (70° + 60°) = 50°) 

BC = EF 

∠C = ∠F 

∴ By SAS congruence 

ΔABC ≅ ΔDEF

ii) In ΔMNL and ΔTSR 

MN = ST 

NL = TR 

∠ M = ∠ T (∵ ΔMNL is flipped to get ΔTSR) 

∴ ΔMNL ≅ ΔTSR

Given that P bisects AB and DC. 

Now in ΔAPC and ΔBPD 

AP = BP (∵ P bisects AB) 

CP = DP (∵ P bisects CD)

∠ APC = ∠ BPD 

∴ ΔAPC ≅ ΔRPP (∵ SAS congruence)

Given : In ΔABC , CA – CB and AP x BQ = AC² 

To prove :- ΔAPC ~ BCQ 

Proof:- 

AP X BQ = AC² (Given) 

Or, AP x BC = AC x AC 

Or, AP x BC = AC x BC (AC = BC given) 

Or, AP/BC = AC/PQ ………………(i) 

Since, CA = CB (Given) 

Then,∠CAB = ∠CBA …………….(ii) (Opposite angle to equal sides)

NOW ∠CAB + ∠CAP = 180° …………(iii) (Linear pair of angle)

And ∠CBA + ∠CBQ = 180° …………..(iv) (Linear pair of angle)

Compare equation (ii) (iii) & (iv)

∠CAP = ∠CBQ ……………..(v)

In ΔAPC and ΔBCQ

∠CAP = ∠CBQ (From equation v)

AP/BC = AC/PQ (From equation i) 

Then , ΔAPC ~ ΔBCQ (By SAS similarity)

Given 

AB⏊BC, DC ⏊ BC and DE ⏊AC 

To prove:- 

ΔCED ~ΔABC 

Proof:-

∠BAC + ∠BCA = 90° …………..(i) (By angle sum property)

And, ∠BCA + ∠ECD = 90°……(ii) (DC ⏊ BC given)

Compare equation (i) and (ii)

∠BAC = ∠ECD……………..(iii)

In ΔCED and ΔABC

∠CED = ∠ABC (Each 90°)

∠ECD = ∠BAC (From equation iii)

Then, ΔCED ~ΔABC.

We have,

∠B = ∠M = 90°

And, ∠BAC = ∠MAP

In ΔABC and ΔAMP

∠B = ∠M (each 90°)

∠BAC = ∠MAP (Given)

Then, ΔABC  ~ ΔAMP (By AA similarity) 

So, CA/PM = BC/MP(Corresponding parts of similar triangle are proportional)

In ΔABC, by Pythagoras theorem 

AB² = AC² + BC²

Or, AB² = 5² + 12²

Or, AB² = 25 + 144

Or, AB² = 169

Or AB = 13 (Square root both side) 

In Δ AED and Δ ACB

∠A = ∠A (Common)

∠AED = ∠ACB (Each 90°)

Then, Δ AED ~ Δ ACB(Corresponding parts of similar triangle are proportional) 

So, AE/AC = DE/ CB =AD/ AB 

Or, AE/5 = DE/12 = 3/13 

Or, AE/5 = 3/13 and DE/12 = 3/13 

Or, AE = 15/13cm and DE = 36/13 cm

Let AB be a tower 

CD be a stick , CD = 6m 

Shadow of AB is BE = 28cm 

Shadow of CD is DF = 4m 

At same time light rays from sun will fail on tower and stick at the same angle

So,∠DCF = ∠BAE

And ∠DFC = ∠BEA

∠CDF = ∠ABE (Tower and stick are vertically to ground)

Therefore ΔABE ~ ΔCDF (By AAA similarity)

So, AB/CD = BE/DF 

AB/6 = 28/4 

AB/6 = 7 

AB = 7 x 6 

AB = 42 m 

So, height of tower will be 42 meter.

If two legs of a right triangle are equal to two legs of another right triangle, then the right triangles are congruent.

True

Sum of the lengths of two sides of a triangle > length of the third side

Here, 1.5 cm + 3.4 cm = 4.9 cm < 5 cm

∴ Third side ≠ 3.4 cm

 Correct option is (D) 3.4 cm

Correct answer is (b) AB = EF

In ⊥∆ACB, ∠C = 90°, BC = ? 

AC² + CB² = AB² 

(8)² + CB² = (10)² 

64 + CB² = 100 

CB² = 100 – 64 

CB² = 36 

∴ CB = 6 

∴ Ladder is at a distance of 6m from the base of the wall.

PQ > PR (Given) 

Therefore, ∠R > ∠Q (Angles opposite the longer side is greater) 

So, ∠SRQ > ∠SQR (Half of each angle) 

Therefore, SQ > SR (Side opposite the greater angle will be longer)