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We have, 

ΔABC ~ Δ PQR 

Area (ΔABC) = 25 cm² 

Area (PQR) = 36 cm² 

And AD = 2.4 cm 

And AD and PS are the altitudes 

To find: PS 

Proof: Since, ΔABC ~ ΔPQR 

Then, by area of similar triangle theorem 

Area of ΔABC/Area of ΔPQR = AB² /PQ² 

25/36 = AB²/PQ² 

5/6 = AB/PQ………………..(i)

In ΔABD and Δ PQS

∠B = ∠Q (ΔABC ~ ΔPQR)

∠ADB = ∠PSQ (Each 90°)

Then, ΔABD ~ Δ PQS (By AA similarity) 

So, AB/PS = AD/PS…………(ii) (Corresponding parts of similar Δ are proportional )

 Compare (i) and (ii)

AD/PS = 5/6 

2.4/PS = 5/6 

PS = 2.4 x 6/5 

PS = 2.88cm

Given: in ΔPSR, Q is a point on the side SR such that PQ = PR.

In ΔPRQ,

PR = PQ (given)

⇒ ∠PRQ = ∠PQR (opposite angles to equal sides are equal)

But ∠PQR > ∠PSR (exterior angle of a triangle is greater than each of opposite interior angle)

⇒ ∠PRQ > ∠PSR

⇒ PS > PR (opposite sides to greater angle is greater)

⇒ PS > PQ (as PR = PQ)

∆ABC ~ ∆DEF (given)

2AB = DE, BC = 6 cm (given)

∠E = ∠B and ∠D = ∠A and ∠F = ∠C

2AB = DE

=> AB/DE = 1/2

Therefore,

AB/DE = BC/EF

1/2 = 6/EF

or EF = 12 cm

∵ ∆ABC ~ ∆ DEF 

∴ AB/DE = BC/EF 

⇒ 1/2 = 6/EF 

⇒  EF = 12 cm

When two triangles are similar, then the ratios of the lengths of their corresponding sides are equal. 

Here, ∆ABC ~∆DEF 

∴ AB/DE = BC/EF 

⇒ AB/2AB = 6/EF 

⇒ EF = 12 cm

In ∆PQS, 

PQ + QS > PS …..(i) [Sum of any two sides of a triangle is greater than the third side] 

Similarly, in ∆PSR, 

PR + SR > PS …(ii) [Sum of any two sides of a triangle is greater than the third side] 

∴ PQ + QS + PR + SR > PS + PS 

∴ PQ + QS + SR + PR > 2PS 

∴ PQ + QR + PR > 2PS [Q-S-R]

Correct answer is (a) 70°

D) All the above

Correct option is B) ⌊D = ⌊R

D) ΔLMN ≅ ΔFGH

Correct answer is (c) PQ – QR< PR

In ΔABC, AB = 3.5 cm, AC = 5 cm, BC = 6 cm and in ΔPQR, PR= 3.5 cm, PQ = 5 cm, RQ = 6 cm. Then ΔABC ≅ ΔPQR.

False

Correct answer is (d) 90°

(i) Let OA be X cm. 

∵ ∆ OAB - ∆ OCD 

∴ OA/ OC = AB/CD 

⇒ X/3.5 = 8/5 

⇒ X = 8×3.5/5 = 5.6 

Hence, OA = 5.6 cm 

(ii) Let OD be Y cm 

∵ ∆ OAB - ∆ OCD 

∴ AB/CD = OB/OD 

⇒ 8/5 = 6.4/Y 

⇒ Y = 6.4 ×5/8 = 4 

Hence, DO = 4 cm

False

Sum of any two sides of a triangle is not less than the third side.

Correct option is D) 80°

(d) 145^o

From the question it is given that, one of the angles of triangle is 110^o

We know that, sum of all angles of triangle is equal to 180^o.

So, sum of other 2 angles is 180^o – 110^o = 70^o

Then, both angled get halved 70^o/2 = 35^o

Sum of bisected angles will be half of sum of angle of triangle.

Then, the third angle will be = 180^o – 35^o

= 145^o

True

The measure of any exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.

Correct option is D) 120°

(b) 65^o

Let us assume the interior opposite angles are Q and Q.

Then, 130^o = Q + Q … [from exterior angle property]

2Q = 130^o

Q = 130^o/2

Q = 65^o

Therefore, the measure of each interior opposite angle is 65^o.

Let the other interi6r opposite angle be = x° 

Give interior opposite angle be = 60° 

Now sum of the interior opposite angle = exterior angle 

x° + 60° = 130°

 x° = 130° – 60° = 70° 

∴ The other interior opposite angle = 70°

The line segment joining a vertex of a triangle to the mid-point of its opposite side is called its median.

Correct answer is 55°

As we know that , sum of angle of a triangle is 180^{0 .} 

^{So, 90+35+ third angle =180}

Third angle = 180-125

                     =55

Yes, Two triangles having equal corresponding sides are congruent and all congruent Δs have equal angles, hence they are similar too.

DE  || BC

:. AD / AB = AE / EC

or,   3 / 4 = 6 / EC

    :.       EC = 8 cm

Correct answer is (d)

The Obtuse triangle always has altitude outside itself.

 The sum of an exterior angle of a triangle and its adjacent angle is always a right angle.

To prove

(c² – a² + b²) tan A = (a² – b² + c²) tan B = (b² – c² + a²⁾ tan C

We know

tan A = \(\frac{abc}R\)\(\frac{1}{b^2+c^2-a^2}\).......(a)

Similarity, tan B = \(\frac{abc}R\)\(\frac{1}{b^2+c^2-a^2}\) and tanC = \(\frac{abc}R\)\(\frac{1}{b^2+c^2-a^2}\)

Therefore,

(c² - a² + b²) tanA = (a² - b² + c²) tanB = (b² - c² +a²) tanc = \(\frac{abc}R\)

Hence we can conclude comparing above equations.

(c² – a² + b²) tanA = (a² – b² + c²) tanB = (b² – c² + a²) tanC 

[Proved]

To prove: a sin A – b sin B = c sin (A – B) 

Left hand side, 

= a sin A – b sin B = (b cosC + c cosB) sinA – (c cosA + a cosC) sinB 

= b cosC sinA + c cosB sinA – c cosA sinB – a cosC sinB 

= c(sinA cosB – cosA sinB) + cosC(b sinA – a sinB)

We know a/sinA = b/sinB = c/sinC = 2R that, where R is the circumradius.

Therefore, 

= c(sinA cosB – cosA sinB) + cosC(2R sinB sinA – 2R sinA sinB) 

= c(si nA cosB – cosA sinB) 

= c sin (A – B) 

= Right hand side. [Proved]

To prove: a² sin ( B – C ) = (b² – c²) sin A 

We know that, a/sinA = b/sinB = c/sinC = 2R where R is the circumradius.

Therefore, 

a = 2R sinA ---- (a) 

Similarly, b = 2R sinB and c = 2R sinC 

From Right hand side,

= (b² – c²) sin A 

= {(2R sinB)² – (2R sinC)² } sinA 

= 4R2 ( sin²B – sin²C )sinA 

We know, sin²B – sin²C = sin(B + C)sin(B – C) 

So,

= 4R² (sin(B + C)sin(B – C))sinA 

= 4R² (sin(π – A)sin(B – C))sinA [ As, A + B + C = π] 

= 4R² (sinAsin(B – C))sinA [ As, sin(π −  θ) = sin θ ] 

= 4R²sin²A sin(B – C) 

= a 2sin(B – C) [From (a)] 

= Left hand side. [Proved]

Given 

DE || BC, BC = 8 cm, AB = 6 cm and DA = 1.5 cm. 

So, 

PR = 6 cm. 

In ΔABC and ΔADE, 

∠ABC = ∠ADE [corresponding angles] 

∠ACB = ∠AED [corresponding angles] 

∠A = ∠A [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. 

∴ ΔABC ~ ΔADE 

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

⇒ \(\frac{BC}{DE}\) = \(\frac{AB}{DA}\)

⇒ \(\frac{8}{DE}\) = \(\frac{6}{1.5}\)

∴ DE = 2 cm

Correct option is (C) All quadrilaterals

All circles, all squares & all equilateral triangles are similar but all quadrilaterals are not similar.

A parallelogram is not similar to a trapezium.

Correct option is: (C) All quadrilaterals

Correct option is  B) A.S.A.

Correct option is   A) S.S.S.

Correct option is  C) SAS

Correct option is (C) In a right angled triangle, the hypotenuse is the smallest side

In any triangle, sum of all three angles is \(180^\circ.\)

\(\therefore\) In \(\triangle ABC,\) \(\angle A+\angle B+\angle C\) \(=180^\circ\)

In a triangle angle opposite to longer side is greater.

\(\therefore\) In \(\triangle ABC,\) if AB > BC, then \(\angle C>\angle A\)

(Also include angle sign in option (B))

In a right angled triangle, the hypotenuse is the longest side.

In a triangle, angles opposite to equal sides are equal.

\(\therefore\) In \(\triangle ABC,\) if AB = BC, then \(\angle C=\angle A\)

C) In a right angled triangle, the hypotenuse is the smallest side