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In ∆ABC, seg BD bisects ∠ABC. [Given] 

∴ AB/BC = AD/CD [Property of angle bisector of a triangle] 

∴ x/(x + 5) = (x - 2)/(x + 2) 

∴ x(x + 2) = (x – 2)(x + 5) 

∴ x² + 2x = x² + 5x – 2x – 10 

∴ 2x = 3x – 10 

∴ 10 = 3x – 2x 

∴ x = 10

In ∆ABC and ∆EDC,

∠ABC ≅ ∠EDC [Each angle is of measure 75°]

∠ACB ≅ ∠ECD [Common angle]

∴ ∆ABC ~ ∆EDC [AA test of similarity]

One to one correspondence is

ABC ↔ EDC

Proof:

In ∆ABP and ∆CDP, 

AP/CP = BP/DP [Given] 

∠APB ≅ ∠CPD [Vertically opposite angles] 

∴ ∆ABP ~ ∆CDP [SAS test of similarity]

Correct answer is

(D) 4√2

Correct answer is

(B) The triangles are similar but not congruent.

BC = \(\cfrac{\sqrt3}2\) x AC----[Side opposite to 60°]

= \(\cfrac{\sqrt3}2\) x 24

\(\therefore\) BC = 12√3 units

Note that,

6² + 8² = 10² ,

\(\therefore\) By converse of Pythagoras theorem, the given triangle is a right angled triangle.

9 × A(∆PQR) = 16 × A(∆LMN) [Given]

∴ (ΔLMN)/A(PQR) = 9/16 (i)

Now, ∆LMN ~ ∆PQR [Given] 

∴ (ΔLMN)/A(PQR) = MN²/QR² (ii) [Theorem of areas of similar triangles]

∴ MN²/QR² = 9/16 [From (i) and (ii)] 

∴MN = QR = 3/4  [Taking square root of both sides] 

 ∴ MN/20  = 3/4 

∴ MN = (20 x 3)/4

 ∴ MN = 15 units

A(∆ABC) = A(∆LMN)

\(\therefore\) \(\cfrac12\) x BC x AD = 1 2 x MN x LP

\(\therefore\) \(\cfrac12\) x 5 x 8 = \(\cfrac12\) x MN x 4

\(\therefore\) MN = \(\cfrac{5\times8}4\)

\(\therefore\) MN = 10 cm

Given: In ∆ABC, D is mid-point of BC since AD is median AM ⊥ BC and AC > AB.

To prove:

(i) AC² = AD² + BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\)

(ii) AB² = AD² – BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\)

(iii) AC² + AB² = 2AD² + \(\frac { 1 }{ 2 }\) BC²

Proof : (i) In right angled ∆AMD

AD² = AM² + DM²

AM² = AD² – DM²

in right ∆AMC

AC² = AM² + MC² …..(ii)

From equation (i) and (ii),

AC² = (AD² – DM²) + MC²

⇒ AC² = (AD² – DM²) + (DM + DC)² (∵ MC = DM + DC)

⇒ AC² = AD² – DM² + DM² + DC² + 2DM.DC

AC² = AD² + DC² + 2DM.DC

AC² = AD² + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\) + 2DM·\(\frac { BC }{ 2 }\)

Thus, AC² = AD² + BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\) …..(iii)

(ii) In right ∆AMB

AB² = AM² + BM²

= (AD² – DM²) + BM² [using equation (i)]

= (AD² – DM²) + (BD – DM)²

= AD² – DM² + BD² + DM² – 2 BD.DM

AD² + BD² – 2BD.DM

= AD² + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\) – 2 × \(\frac { 1 }{ 2 }\) BC. DM

∴ AB² = AD – BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\)

Thus, AB² = AD – BC.DM + \({ \left( \frac { BC }{ 2 } \right) }^{ 2 }\) ……(iv)

Adding equation (iii) and (iv)

AB² + AC² = 2AD² + 2 × \(\frac { 1 }{ 4 }\) BC²

= 2AD² + \(\frac { 1 }{ 2 }\) BC²

Thus, AB² + AC² = 2AD² + \(\frac { 1 }{ 2 }\) BC²

Given

AB ⊥ BC, DC ⊥ BC and DE ⊥ AC

To prove : ∆CED ~ ∆ABC

Proof : AB ⊥ BC and DC ⊥ BC

∴ AB || DC

When AB || DC cuts by transversal AC

∠BAC = ∠ACD

or ∠BAC = ∠ECD

∠B = ∠E = 90° (given) …(ii)

∴ ∠ACB = ∠EDC [due to (i) and (ii)]

⇒ ∆ABC ~ ∆CED.

Answer is (A) 6 cm

In ∆ABC

DE || BC

∴ AD/BD = AE/CE (Basic Prop. Theorem)

AD/(AB - AD) = AE/CE

⇒ 8/(12 - 8) = 12/CE

⇒ 8/4 = 12/CE

⇒ 8 × CE = 12 × 4

⇒ CE = (12 x 4)/8

⇒ CE = 6 cm

Answer is True

Given : In ∆ABC, AB = AC and BC² = AC × DC,

where D is any point on side AC.

To prove : BD = BC

Proof : Given that BC² = AC × DC

⇒ BC/DC = AC/BC …..(i)

Now in ∆ABC and ∆BDC

∠C = ∠C (common angle)

and BC/DC = AC/BC [from (i)]

Thus, by SAS similarity criterion

∆ABC ~ ∆BDC

⇒ AC/BC = AB/BD …..(ii)

Given, AB = AC

Thus, AC/BC = AC/BD

⇒ BC = BD.

Answer is True

Answer is True

In ∆ADC

GE || DC

∴ AG/AD = AE/AC (By Basic Prop. Theorem)

Thus, in ∆ABC

∵ FE || BC

∴ AE/AC = AF/AB

From equation (i) and (ii)

AG/AD = AF/AB

The ratio between the areas of two similar triangles is same as the square of the ratio between their corresponding sides.

So, the ratio between the sides of the two triangles = 4 : 5

(i) The ratio between the perimeters of two similar triangles is same as the ratio between their sides.

∴ Required ratio = 4 : 5

(ii) The ratio between the altitudes of two similar triangles is same as the ratio between their sides.

∴ Required ratio = 4 : 5

(iii) The ratio between the medians of two similar triangles is same as the ratio between their sides.

∴ Required ratio = 4 : 5

Answer is (B) DE/PQ = EF/RP

side PQ || side SR [Given]

and seg SQ is their transversal. 

∴ ∠QSR = ∠SQP [Altemate angles] 

∴ ∠ASR = ∠AQP (i) [Q – A – S] 

In ∆ASR and ∆AQP, 

∠ASR = ∠AQP [From (i)] 

∠SAR ≅ ∠QAP [Vertically opposite angles] 

∆ASR ~ ∆AQP [AA test of similarity]

∴ AS/AQ = SR/PQ (ii) [Corresponding sides of similar trianges]

But, AS = 5 AQ  [Given]

∴ AS/AQ = 5/1  (iii)

∴ SR/PQ = 5/1 [Form (ii) and (iii)]

∴ SR =5 PQ

Proof 

∠APQ = ∠ABC = 60° ...[Given] 

∴ ∠APQ ≅ ∠ABC 

∴ side PQ || side BC ...(i)  [Corresponding angles test] 

In ∆ABC, 

sidePQ || sideBC ...[From (i)] 

∴ AP/PB = AQ/QC  ....[Basic proportionality theorem]