Triangles

∴ The third side is proportional to the corresponding third side.

i.e., The two triangles will be similar by SSS criterion.

∴ The given triangle is not a right triangle.

= 180° – (55° + 25°) = 100° = ∠M

∠Q = ∠S = 25° (Given)

∆QPR ~ ∆STM

i.e., . ∆QPR is not similar to ∆TSM.

∴ ∠A = ∠D = 47°

∠B = ∠E = 63°

∴ ∠C = 180° – (∠A + ∠B) = 180° – (47° + 63°) = 70°

∴ Given statement is true.

∴ AB² = AC² + BC² [By Pythagoras theorem]

⇒ AB² = AC² + AC²

[∵ AC = BC]

⇒ AB² = 2AC²

DE || BC,

∴ $\frac{AD}{DB}=\frac{AE}{EC}$ [By Basic Proportionality Theorem]

$\Rightarrow \frac{x}{x-2}=\frac{x+2}{x-1}$

⇒ x(x – 1) = (x – 2) (x + 2)

⇒ x² – x = x² – 4

⇒ x = 4

We have, PQ = 1.28 cm, PR = 2.56 cm

PE = 0.18 cm, PF = 0.36 cm

Now, EQ = PQ-PE = 1.28 – 0.18 = 1.10 cm and

FR = PR – PF = 2.56 – 0.36 = 2.20 cm

Therefore, EF || QR [By the converse of Basic Proportionality Theorem]

Now, in ∆ABC and ∆DEF, we have

h = 42 Hence, height of tower, DE = 42m

Firstly, in ∆ABC, we have

LM || CB (Given)

Therefore, by Basic Proportionality Theorem, we have

DE || OQ (Given)

[Applying the converse of Basic Proportionality Theorem in ∆PQR]

Given: ∆ABC in which D and E are the mid-points of sides AB and AC respectively.

To prove: DE || BC

Proof: Since D and E are the mid-points of AB and AC respectively

∴ AD = DB and AE = EC

DB EC Therefore, DE || BC (By the converse of Basic Proportionality Theorem)

∆NML is not similar to ∆PQR.

⇒ DC = 10 cm.

In ∆ABE and ∆CFB, we have

∠AEB = ∠CBF (Alternate angles)

∠A = ∠C (Opposite angles of a parallelogram)

∴ ∆ABE ~ ∆CFB (By AA criterion of similarity)

In ∆RPQ and ∆RTS, we have

∠RPQ = ∠RTS (Given)

∠PRQ = ∠TRS = ∠R (Common)

∴ ∆RPQ ~ ∆RTS (By AA criterion of similarity)

Given: A ∆ABC in which D is the mid-point of AB and DE is drawn parallel to BC, which meets AC at E.

To prove: AE = EC

Proof: In ∆ABC, DE || BC

∴ By Basic Proportionality Theorem, we have

$\frac{AD}{DB}=\frac{AAND}{ANDC}$ … (i)

Now, since D is the mid-point of AB

⇒ AD = BD … (ii)

From (i) and (ii), we have

$\frac{AD}{DB}$ = $\frac{AAND}{ANDC}$

$\Rightarrow 1=\frac{AAND}{ANDC}$

Hence, E is the mid-point of AC.

∠B = ∠Q (∵ ∆ABC ~ ∆PQR) … (i)

∠1 = ∠2 [Alternate angles]

∠3 = ∠4 [Alternate angles]

∠CED = ∠AEB [Vertically opposite angles]

∴ ∆EDC ~ ∆EBA [By AA criterion of similarity]

Given: A ∆ABC in which ∠ABC = 90° and AB = BC.

∆ABD and ΔCAE are equilateral triangles.

To Prove: ar(∆ABD) = $\frac{1}{2}$ × ar(∆CAE)

Proof: Let AB = BC = x units.

∴ hyp. CA = √x² + √x² = x√2 units.

Each of the ABD and ∆CAE being equilateral has each angle equal to 60°.

∴ ∆АВD ~ ∆CAE

But, the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Given: Two triangles ABC and DEF, such that

∆ABC ~ ∆DEF and area (∆ABC) = area (∆DEF)

To prove: ∆ABC ≅ ∆DEF

Proof: ∆ABC ~ ∆DEF

⇒ ∠A = ∠D, ∠B = ∠E, ∠C = ∠F

AB = DE, BC = EF, AC = DF

∆ABC ≅ ∆DEF (By SSS criterion of congruency)

Assertion Reason Answer-