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The square of the hypotenuse is equal to the sum of the squares of the other two sides. Here, the hypotenuse is the longest side and it’s always opposite the right angle. 

Correct option is (B) 9 : 64

\(\because DE\bot AC\;\&\;BC\bot AC\)

\(\therefore DE\,||\,BC\)

Now in triangles \(\triangle ADE\;\&\;\triangle ABC,\)

\(\angle ADE=\angle ABC,\)   (Corresponding angles as DE || BC)

\(\angle AED=\angle ACB\) \(=90^\circ\)

\(\angle DAE=\angle BAC\)      (Common angle)

\(\therefore\) \(\triangle ADE\sim\triangle ABC\)        (By AAA similarity rule)

\(\therefore\frac{ar(\triangle ADE)}{ar(\triangle ABC)}=(\frac{AD}{AB})^2\)

\(=(\frac{DE}{BC})^2=(\frac{AE}{EC})^2\)

\(\Rightarrow\frac{ar(\triangle ADE)}{ar(\triangle ABC)}=(\frac{AE}{AE+EC})^2\)

\(=(\frac{3}{3+5})^2=(\frac{3}{8})^2\)

\(=\frac{3^2}{8^2}=\frac9{64}\)

\(\therefore\) \(ar(\triangle ADE):ar(\triangle ABC)\) = 9:64

Correct option is: (B) 9 : 64

In two triangles, if three sides of the one are proportional to the corresponding sides of the other, the triangles are similar.

If the corresponding sides of two triangles are proportional then their corresponding angles are equal, and hence the two triangles are similar.

SSS similarity criterion: If in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.

SAS similarity criterion: If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

In two triangles, if two sides of the one are proportional to the corresponding sides of the other and their included angles are equal, the two triangles are similar.

If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional then the two triangles are similar.

Correct option is (C) 13 cm

We have AD = 3 cm & CD = 4 cm, \(\angle ADC=90^\circ\)

By using Pythagoras theorem in \(\triangle ADC,\) we obtain

\(AC^2=AD^2+CD^2\)

\(=3^2+4^2\)

\(=9+16=25\)

\(\Rightarrow\) \(AC=\sqrt{25}=5\,cm\)

Now in right \(\triangle ACB,\) we have \(BC=12\,cm,AC=5\,cm\;and\;AB=x\)

\(\therefore\) \(AB^2=AC^2+BC^2\)          (By using Pythagoras theorem in right \(\triangle ACB)\) 

\(=5^2+12^2\)

\(=25+144=169\)

\(\therefore\) \(AB=\sqrt{169}=13\,cm\)

Correct option is: (C) 13 cm

If the corresponding sides of two triangles are proportional then their corresponding angles are equal, and hence the two triangles are similar.

If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional then the two triangles are similar.

In a right angled triangle, the square on the hypotenuse is equal to the sum of squares on the other two sides.

The square of the hypotenuse is equal to the sum of the squares of the other two sides. Here, the hypotenuse is the longest side and it’s always opposite the right angle.

Answer: 

(i) ΔABC ≈ ΔPQR 

(ii) ΔDEF ≈ ΔNML 

(iii) Not congruent 

(iv) ΔADB ≈ ΔADC

Correct answer is (b) a scalene triangle only

Data: In the quadrilateral ABCD. the diagonals intersect at 0’ such that \(\frac{AO}{BO} = \frac{CO}{DO}\)

To Prove: ABCD is a trapezium. 

In the qudrilateral ABCD, \(\frac{AO}{BO} = \frac{CO}{DO}\)

∴ \(\frac{AO}{Co} = \frac{OB}{OD}\)

It means sides of ∆AOB and ∆DOC divides proportionately. 

∴ ∆AOB ||| ∆DOC. 

Similarly. ∆AOD ||| ∆BOC. 

Now, ∆AOB + ∆AOD = ∆BOC + ∆DOC 

∆ABD = ∆ABC. 

Both triangles are on the same base AB and between two pair of lines and equal in area. 

∴ AB || DC.

∆OBA ~ ∆ODC (data) 

In ∆OBA and ∆ODC, we have 

∠ODC = ∠OBA = 70° (. Alternate angle) 

∠AOB = 180°- 125° = 55° ( Adjacent angle) 

∴ ∠AOB = ∠DOC = 55° (: VertIcally opposite angles) 

∠OAB = ∠OCD = 55° (. Alternate angles) 

∴ ∠DOC=55° 

∠DCO = 55% 

∠OAB= 55%.

D) An acute angled triangle

C) An equilateral triangle

Correct option is  A) 110°

Given,

ABC is a triangle

∠A = 72° and internal bisectors of B and C meet O.

In ΔABC

∠A + ∠B + ∠C = 180°

72° + ∠B + ∠C = 180°

∠B + ∠C = 180° – 72°

∠B + ∠C = 108°

Divide both sides by 2, we get

\(\frac{∠B}{2}\) + \(\frac{∠C}{2}\) = \(\frac{108}{2}\)

\(\frac{∠B}{2}\) + \(\frac{∠C}{2}\) = 54°

∠OBC + ∠OCB = 54° (i)

Now, in ΔBOC

∠OBC + ∠OCB + ∠BOC = 180°

54° + ∠BOC = 180°[Using (i)]

∠BOC = 180° – 54°

= 126°

Correct option is  A) 110°

(i) It is given that DB is a straight line. 

Therefore, 

∠DOC + ∠COB = 180°

∠DOC = 180° − 115° = 65°

(ii) In ∆ DOC, we have: 

∠ODC + ∠DCO + ∠DOC = 180°

Therefore, 

70° + ∠DCO + 65° = 180°

⇒ ∠DCO = 180 − 70 − 65 = 45°

(iii) It is given that ∆ ODC - ∆ OBA 

Therefore, ∠OAB = ∠OCD = 45°

(iv) Again, ∆ ODC- ∆ OBA 

Therefore, 

∠OBA = ∠ODC = 70°

From given statement:

In Δ ADC

EO || AB || DC

By thales theorem: AE/ED = AO/OC …(1)

In Δ DAB,

EO || AB

So, By thales theorem: DE/EA = DO/OB …(2)

From (1) and (2)

AO/OC = DO/OB

(5x – 7) / (2x + 1) = (7x-5) / (7x+1)

(5x – 7)(7x + 1) = (7x – 5)(2x + 1)

35x² + 5x – 49x – 7 = 14x² – 10x + 7x – 5

35x² – 14x² – 44x + 3x – 7 + 5 = 0

21x² – 42x + x – 2 = 0

21(x – 2) + (x – 2) = 0

(21x + 1)(x – 2) = 0

Either (21x + 1) = 0 or (x – 2) = 0

x = -1/21 (does not satisfy) or x = 2

=> x = 2.

Here ∆ODC ~ ∆OBA, so

∠D = ∠B = 70°

∠C = ∠A

∠COD = ∠AOB

(i) But ∠DOC + ∠BOC = 180° (Linear pair)

∠DOC + 115°= 180°

∠DOC = 180° – 115° = 65°

(ii) ∠DOC + ∠CDO + ∠DCO = 180° (Angles of a triangle)

65° + 70° + ∠DCO = 180°

135° + ∠DCO = 180°

∠DCO = 180° – 135° = 45°

(iii) ∠AOB = ∠DOC = 65° (vertically opposite angles)

∠OAB = ∠DCO = 45° (Since ∆ODC ~ ∆OBA)

(iv) ∠OBA = ∠CDO = 70° (Since ∆ODC ~ ∆OBA)

Correct answer is (c) 18 m

As the sum (6 cm) of two sides 5 cm and 1 cm is less than third side. It is not possible to construct a triangle with the given measures.

In ∆ABC, M and N are points on the sides AB and AC respectively such that BM = CN and if ∠B = ∠C.

We know that, sides opposite to equal angles are equal.

AB = AC

BM = CN ( given)

AB – BM = AC – CN

=> AM = An

From ∆ABC

AM/MB = AN/NC

Therefore, MN ||BN

In trapezium ABCD, AB ‖ CD and the diagonals AC and BD intersect at O. 

Therefore, 

AO/ OC = BO/OD 

⇒ 5x−7/2x+1 = 7x−5/7x+1 

⇒ (5x-7) (7x+1) = (7x-5) (2x+1) 

⇒35x² + 5x – 49x – 7 = 14² – 10x + 7x – 5 

⇒ 21x² – 41x – 2 = 0 

⇒ 21x² – 42x + x -2 =0 

⇒ 21x (x-2) +1 (x-2) =0 

⇒ (x-2) (21x + 1) =0 

⇒ x = 2, - 1/21 

∵ x ≠ - 1/21 

∴ x = 2

Given: ABCD is a parallelogram

To prove: 

(i) DM / MN = DC/BN

(ii) DN/DM =AN/ DC

Proof: In ∆ DMC and ∆ NMB 

∠DMC = ∠NMB (Vertically opposite angle) 

∠DCM = ∠NBM (Alternate angles) 

By AAA- Similarity 

∆DMC ~ ∆NMB 

∴DM /MN = DC/ BN

NOW, MN/DM = BN/DC

Adding 1 to both sides, we get 

MN/DM + 1 = BN/DC + 1 

⇒ MN+DM/DM = BN+DC/DC

⇒ MN+DM/DM = BN+ AB/DC [∵ ABCD is a parallelogram] 

⇒ DN/DM = AN/DC

(i) It is give that AD bisects ∠A. 

Applying angle – bisector theorem in ∆ ABC, we get: 

BD/DC = AB/AC 

⇒ 5.6/DC = 6.4/8 

⇒ DC = 8×5.6/6.4 = 7 cm

(ii) It is given that AD bisects ∠A. 

Applying angle – bisector theorem in ∆ ABC, we get: 

BD/DC = AB/AC 

Let BD be x cm. 

Therefore, DC = (6- x) cm 

⇒ x/ 6−x = 10/14 

⇒ 14x = 60-10x 

⇒ 24x = 60 

⇒ x = 2.5 cm 

Thus, BD = 2.5 cm 

DC = 6-2.5 = 3.5 cm 

(iii) It is given that AD bisector ∠A. 

Applying angle – bisector theorem in ∆ ABC, we get: 

BD/DC = AB/AC 

BD = 3.2 cm, BC = 6 cm 

Therefore, DC = 6- 3.2 = 2.8 cm 

⇒ 3.2/2.8 = 5.6/AC 

⇒ AC = 5.6×2.8/3.2 = 4.9 cm

(iv) It is given that AD bisects ∠A. 

Applying angle – bisector theorem in ∆ ABC, we get: 

BD/DC = AB/AC 

⇒ BD/3 = 5.6/4 

⇒ BD = 5.6×3/4 

⇒ BD = 4.2 cm 

Hence, BC = 3+ 4.2 = 7.2 cm

(i) We have: 

AD/DE = 5.7/9.5 = 0.6  

AE/EC = 4.8/8 = 0.6  

Hence, AD/DB = AE/EC 

Applying the converse of Thales’ theorem, 

We conclude that DE || BC. 

(ii) We have: 

AB = 11.7 cm, DB = 6.5 cm 

Therefore, 

AD = 11.7 -6.5 = 5.2 cm 

Similarly, 

AC = 11.2 cm, AE = 4.2 cm 

Therefore, 

EC = 11.2 – 4.2 = 7 cm 

Now, 

AD/DB = 5.2/6.5 = 4/5 

AE/EC = 4.2/7 

Thus, AD/DB ≠ AE/EC 

Applying the converse of Thales’ theorem, 

We conclude that DE is not parallel to BC. 

(iii) We have: 

AB = 10.8 cm, AD = 6.3 cm 

Therefore,

DB = 10.8 – 6.3 = 4.5 cm 

Similarly, 

AC = 9.6 cm, EC = 4cm 

Therefore, 

AE = 9.6 – 4 = 5.6 cm 

Now, 

AD/DB = 6.3/4.5 = 7/5 

AE/EC = 5.6/4 = 7/5 

⟹ AD/DB = AE/EC 

Applying the converse of Thales’ theorem, 

We conclude that DE ‖ BC. 

(iv) We have : 

AD = 7.2 cm, AB = 12 cm 

Therefore, 

DB = 12 – 7.2 = 4.8 cm 

Similarly, 

AE = 6.4 cm, AC = 10 cm 

Therefore, 

EC = 10 – 6.4 = 3.6 cm 

Now, 

AD/DB = 7.2/4.8 = 3/2 

AE/EC = 6.4/3.6 = 16/9 

This, AD/DB ≠ AE/EC 

Applying the converse of Thales’ theorem, 

We conclude that DE is not parallel to BC.

(i) In ∆ ABC, it is given that DE || BC. 

Applying Thales’ theorem, we have : 

AD/DB = AE/EC 

⇒ x/x−2 = x+2/x−1 

⇒X(x-1) = (x-2) (x+2) 

⇒x² – x = x² − 4 

⇒x= 4 cm 

(ii) In ∆ ABC, it is given that DE ‖ BC. 

Applying Thales’ theorem, we have : 

AD/DB = AE/EC 

⇒ 4/x−4 = 8/3x−19 

⇒ 4 (3x-19) = 8 (x-4) 

⇒12x – 76 = 8x – 32 

⇒4x = 44 

⇒ x = 11 cm 

(iii) In ∆ ABC, it is given that DE || BC. 

Applying Thales’ theorem, we have : 

AD/DB = AE/EC 

⇒ 7x−4/3x+4 = 5x−2/3x 

⇒3x (7x-4) = (5x-2) (3x+4)

⇒21x² – 12x = 15x² + 14x-8 

⇒6x² – 26x + 8 = 0 

⇒(x-4) (6x-2) =0 

⇒ x = 4, 1/3 

∵ x ≠ 1/3 (as if x = 1/3 then AE will be negative) 

∴ x = 4 cm

Data: altitudes AD and CE of ABC Intersect each other at the point P. 

To Prove: 

(i) ∆AEP ~ ∆CDP 

(ii) ∆ABD ~ ∆CBE 

(iii) ∆AEP ~ ∆ADB 

(iv) ∆PDC ~ ∆BEC 

(i) In ∆AEP and ∆CDP. 

∠AEP = ∠CDP = 90° (data) 

∠APE = ∠CPD (Vertically opposite angle) 

∴ ∠PAE = ∠PCD These are equiangular triangles. 

Similarity criterion for ∆ is A.A.A 

∴ ∆AEP ~ ∆CDP 

(ii) In ∆ABD and ∆CBE. 

∠ADB = ∠CEB = 90° (data) 

∠ABD = ∠CBE (common) 

∴ ∠DAB = ∠BCE 

These are equiangular triangles. 

∴ Similarity criterion for is A.A.A. 

∴ ∆ABD ~ ∆CBE 

(iii) In ∆AEP and ∆ADB. 

∠AEB = ∠ADB = 90° (data) 

∠PAE = ∠DAB (common) 

∴ ∠APE = ∠ABD. 

∴ These are equiangular triangles. 

∴ Similarity criterion for is A.A.A. 

∴ ∆AEP ~ ∆ADB 

(iv) In ∆PDC and ∆BEC. 

∠PDC = ∠BEC = 90° (data) 

∠PCD = ∠BCE (common) 

∴ ∠CPD = ∠CBE 

∴ These are equiangular angular triangles. 

Similarity criterion for ∆ is A.A.A. 

∴ ∆PDC ~ ∆BEC.

Mean of first ‘n’ even numbers = 2x

Correct option is (A) 2 : 3

The ratio of perimeters of two similar triangle is equal to the ratio of their corresponding sides/Altitudes/Medians.

\(\because\) Given that the ratio of corresponding medians of two similar triangles is 2:3.

\(\therefore\) The ratio of perimeter of similar triangle = 2:3

Correct option is: (A) 2 : 3

(c) 3 : 2

Median of an equilateral triangle = \(\frac{\sqrt2}{2}\) x side

Let the sides of the two equilateral triangles be a₁ and a₂ respected. Then,

\(\frac{\frac{\sqrt3}{2}a_1}{\frac{\sqrt3}{2}a_2}\) = \(\frac32\) ⇒ \(\frac{a_1}{a_2}\) = \(\frac32\).

(c) 3 : 2

Equilateral triangles are similar triangles. In similar triangles, the ratio of their corresponding sides is the same as the ratio of their medians.

Correct option is B) Centroid

C) ∆ ABC is isosceles

In a triangle, sum of squares of two sides is equal to the square of the third side.

False