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In ΔABC and ΔCDA,

∠BAC = ∠DCA (Alternate interior angles, as p || q)

AC = CA (Common)

∠BCA = ∠DAC (Alternate interior angles, as l || m)

∴ ΔABC ≅ ΔCDA (By ASA congruence rule)

The correct option is: (B) If a line is drawn parallel to one side of the triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio.

If a line is draw parallel to one side of a triangle intersect the other two sides, then it divides the other two sides in the same ratio.

(i) (AAA Similarity) If two triangles are equiangular, then they are similar.

(ii) (SSS Similarity) If the corresponding sides of two triangles are proportional, then they are similar.

(iii) (SAS Similarity) If in two triangle’s one pair of corresponding sides are proportional and the included angles are equal then the two triangles are similar.

Triangles are the most well-known idea whose applications are seen oftentimes in everyday life. Solve the MCQ Questions of Triangles with Answer here and become familiar with triangles. Get significant questions for class 10 Maths here too.

Practice Class 10 Maths MCQ Question of Triangles

Check out the Multiple Choice Questions of Triangles with Answers which are given below: -

1. In a \(\Delta ABC\) , AD is the bisector of ∠BAC . If AB = 8 cm, BD = 6 cm and DC = 3 cm. Find AC

(a) 4 cm
(b) 6 cm
(c) 3 cm
(d) 8 cm

2. A square and a rhombus are always

(a) similar
(b) congruent
(c) similar but not congruent
(d) neither similar nor congruent

3. Which geometric figures are always similar?​

(a) Circles
(b) Circles and all regular polygons
(c) Circles and triangles
(d) Regular

4. D and E are respectively the midpoints on the sides AB and AC of a triangle ABC and BC = 6 cm. If DE || BC, then the length of DE (in cm) is

(a) 2.5
(b) 3
(c) 5
(d) 6

5. The lengths of the diagonals of a rhombus are 16 cm and 12cm. Then, the length of the side of the rhombus is

(a) 9 cm
(b) 10 cm
(c) 8 cm
(d) 20 cm

6. The ratio of the areas of two similar triangles is equal to the:​

(a) square of the ratio of their corresponding sides.
(b) the ratio of their corresponding sides
(c) square of the ratio of their corresponding angles
(d) None of the above

7. Two congruent triangles are actually similar triangles with the ratio of corresponding sides as.​

(a) 1:2
(b) 1:1
(c) 1:3
(d) 2:1

8. Which of the following triangles have the same side lengths?

(a)Scalene
(b)Isosceles
(c)Equilateral
(d)None of these

9. Corresponding sides of two similar triangles are in the ratio of 2:3. If the area of small triangle is 48 sq.cm, then the area of large triangle is:

(a)230 sq.cm.
(b)106 sq.cm
(c)107 sq.cm.
(d)108 sq.cm

10. If perimeter of a triangle is 100cm and the length of two sides are 30cm and 40cm, the length of third side will be:

(a)30cm
(b)40cm
(c)50cm
(d)60cm

11. If triangles ABC and DEF are similar and AB=4cm, DE=6cm, EF=9cm and FD=12cm, the perimeter of triangle is:

(a)22cm
(b)20cm
(c)21cm
(d)18cm

12. The height of an equilateral triangle of side 5cm is:

(a)4.33
(b)3.9
(c)5
(d)4

13. Areas of two similar triangles are 36 \(cm^2\) and 100 \(cm^2\). If the length of a side of the larger triangle is 20 cm, then the length of the corresponding side of the smaller triangle is:

(a) 12cm
(b) 13cm
(c) 14cm
(d) 15cm

14. If ABC and DEF are two triangles and AB/DE=BC/FD, then the two triangles are similar if

(a)∠A=∠F
(b)∠B=∠D
(c)∠A=∠D
(d)∠B=∠E

15. Sides of two similar triangles are in the ratio 4: 9. Areas of these triangles are in the ratio

(a)2: 3
(b)4: 9
(c)81: 16
(d)16: 81

16. The length of each side of a rhombus whose diagonals are of lengths 10 cm and 24 cm is

(a) 25 cm
(b) 13 cm
(c) 26 cm
(d) 34 cm

17. ΔABC ~ ΔDEF, ar (ΔABC) = 9 cm\(^2\), ar (ΔDEF) = 16 cm\(^2\). If BC = 2.1 cm, then the measure of EF is

(a) 2.8 cm
(b) 4.2 cm
(c) 2.5 cm
(d) 4.1 cm

18. In an isosceles triangle ABC, if AB = AC = 25 cm and BC = 14 cm, then the measure of altitude from A on BC is

(a) 20 cm
(b) 22 cm
(c) 18 cm
(d) 24 cm

19. If in triangles ABC and DEF,\(\frac{AB}{DE}=\frac{BC}{FD}\) , then they will be similar, if

(a) ∠B = ∠E
(b) ∠A = ∠D
(c) ∠B = ∠D
(d) ∠A = ∠F

20. If ΔABC ~ ΔDEF and ΔABC is not similar to ΔDEF then which of the following is not true?

(a) BC.EF = AC.FD
(b) AB.ED = AC.DE
(c) BC.DE = AB.EE
(d) BC.DE = AB.FD

Answers & Explanations

1. Answer: (a) 4 cm

​​Explanation: Given AD is the bisector of ∠BAC. AB = 8 cm, DC = 3 cm and BD = 6 cm. 

We know that the internal bisector of angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

\(\Rightarrow \frac{AB}{AC}=\frac{BD}{DC}\)

\(\Rightarrow \frac{8}{AC}=\frac{6}{3}\)

∴ AC = 4 cm

2. Answer: (d) neither similar nor congruent

3. Answer: (b) Circles and all regular polygons

4. Answer: (b) 3

Explanation: By midpoint theorem,

DE=\(\frac{1}{2}\) BC

DE =\(\frac{1}{2}\) of 6

DE=3cm

5. Answer: (b) 10 cm

Explanation: The diagonals of rhombus bisect each other at right angle, so side of rhombus is the hypotenuse for the triangles formed.

Therefore,

By Pythagoras theorem

\((\frac{16}{2})^2 + (\frac{12}{2})^2 = \mathrm{Side}^2\)

⇒ \(8^2 + 6^2 = \mathrm{Side}^2\)

⇒ \(64 + 36 = \mathrm{Side}^2\)

⇒ Side = 10 cm

6. Answer: (a) square of the ratio of their corresponding sides.

7. Answer: (b) 1:1

8. Answer: (c)Equilateral

Explanation: Equilateral triangles have all its sides and all angles equal.

9. Answer: (d)108 sq.cm

Explanation: Let \(A_1\) and \(A_2\) are areas of the small and large triangle.

Then,

\(\frac{A_2}{A_1}\)=(side of large triangle/side of small triangle)

\(\frac{A_2}{48}=(\frac{3}{2})^2\)

\(A_2\)=108 sq.cm.

10. Answer: (a)30cm

Explanation: Perimeter of triangle = sum of all its sides

P = 30+40+x

100=70+x

x=30cm

11. Answer: (d)18cm

Explanation: ABC ~ DEF

AB=4cm, DE=6cm, EF=9cm and FD=12cm

AB/DE = BC/EF = AC/DF

4/6 = BC/9 = AC/12

BC = (4.9)/6 = 6cm

AC = (12.4)/6 = 8cm

Perimeter = AB+BC+AC

= 4+6+8

=18cm

12. Answer: (a)4.33

Explanation: The height of the equilateral triangle ABC divides the base into two equal parts at point D.

Therefore,

BD=DC= 2.5cm

In triangle ABD, using Pythagoras theorem,

\(AB^2=AD^2+BD^2\)

\(5^2=AD^2+2.5^2\)

\(AD^2 = 25-6.25\)

\(AD^2=18.75\)

AD=4.33 cm

13. Answer: (a) 12cm

Explanation: Let the side of smaller triangle be x cm.

ar(Larger Triangle)/ar(Smaller Triangle) = (side of larger triangle/side of smaller triangle)\(^2\)

\(\frac{100}{36} = (\frac{20}{x})^2\)

x = √144

X = 12 cm

14. Answer: (b) ∠B=∠D

15. Answer: (d)16: 81

Explanation: Let ABC and DEF are two similar triangles, such that,

ΔABC ~ ΔDEF

And AB/DE = AC/DF = BC/EF = 4/9

As the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides,

∴ Area(ΔABC)/Area(ΔDEF) = \(\frac{{AB}^2}{{DE}^2}\)

∴ Area(ΔABC)/Area(ΔDEF) = \((\frac{4}{9})^2 = \frac{16}{81} = 16: 81\)

16. Answer: (b) 13 cm

Explanation: Given that

Diagonals of the rhombus

Let \(d_1\) =10cm and \(d_2\)= 24cm

Diagonals meet at the centre and forms right-angled triangles.

So by using pythagoras theorem

Length of the base = 10/2 = 5cm
Length of the height = 24/2 = 12cm

Hypotenuse\(^2\) = side\(^2\)+ side\(^2\)

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

Hypotenuse\(^2\) = 25 + 144

Hypotenuse\(^2\) = 169

On taking square root we get,

Hypotenuse = 13{ 13 X 13=169 => √169=13}

Hence the side of the rhombus is 13cm.

17. Answer: (a) 2.8 cm

Explanation: Given Ar (ΔABC) = 9 cm\(^2\), ar (ΔDEF) = 16 cm\(^2\) and BC = 2.1 cm
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

\(\Rightarrow \frac{ar(\Delta ABC)}{ar(\Delta DEF)}=\frac{{BC}^2}{{EF}^2}\)

\(\Rightarrow \frac{9}{16}=\frac{{2.1}^2}{{EF}^2}\)

\(\Rightarrow \frac{3}{4}=\frac{2.1}{EF}\)

\(\therefore EF=2.8\)cm

18. Answer: (d) 24 cm

Explanation: Given in an isosceles ΔABC, AB = AC = 25 cm and BC = 14 cm 

Here altitude from A to BC is AD. 

We know in isosceles triangle altitude on non-equal sides is also median. 

⇒ BD = CD = BC/2 = 7 cm 

Applying Pythagoras Theorem, 

⇒ AB² = BD² + AD² 

⇒ 25² = 7² + AD² 

⇒ AD² = 625 – 49 = 576 

⇒ AD = 24 

∴ Measure of altitude from A to BC is 24 cm

19. Answer: (c) ∠B = ∠D

20. Answer: (c) BC.DE = AB.EE

Click here to practice more MCQ Questions from Chapter Triangles Class 10 Maths

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

Applying Thales’ theorem, we get: 

AD/DB = AE/EC 

∵ AD = 3.6 cm , AB = 10 cm, AE = 4.5cm 

∴ DB = 10 − 3.6 = 6.4cm 

Or, 3.6/6.4 = 4.5/EC 

Or, EC = 6.4×4.5/3.6 

Or, EC= 8 cm 

Thus, AC = AE + EC 

= 4.5 + 8 = 12.5 cm 

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

Applying Thales’ Theorem, we get : 

AD/DB = AE/EC 

Adding 1 to both sides, we get :

AD/DB + 1 = AE/ EC +1 

⇒ AB/DB = AC/EC

⇒ 13.3/DB = 11.9/5.1 

⇒ DB = 13.3×5.1/11.9 = 5.7 cm 

Therefore, AD=AB-DB=13.5-5.7=7.6 cm 

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

Applying Thales’ theorem, we get : 

AD/DB = AE/EC 

⇒ 4/7 = AE/EC 

Adding 1 to both the sides, we get : 

11/7 = AC/EC 

⇒ EC = 6.6×7/11 = 4.2 cm 

Therefore,

AE = AC – EC = 6.6 – 4.2 = 2.4 cm 

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

Applying Thales’ theorem, we get: 

AD/AB = AE/AC 

⇒ 8/15 = AE/ AE+EC 

⇒ 8/15 = AE/ AE+3.5 

⇒ 8AE + 28 = 15AE 

⇒ 7AE = 28 

⇒ AE = 4cm

If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

In a triangle, if a line parallel to one side is drawn, it will divide the other two sides proportionally.

If a line divides any two sides of a triangle in the same ratio. Then, the line must be parallel to the third side.

∠BAD ≅ ∠DAC ….(i) [Seg AD bisects ∠BAC] 

∠ADB is the exterior angle of ∆ADC. 

∴ ∠ADB > ∠DAC ….(ii) [Property of exterior angle] 

∴ ∠ADB > ∠BAD ….(iii) [From (i) and (ii)] 

In AABD,

 ∠ADB > ∠BAD [From (iii)] 

∴ AB > BD [Side opposite to greater angle is greater]

In two triangles, if three angles of the one triangle are equal to the three angles of the other, the triangles are similar.

The line joining the midpoints of two sides of a triangle, is parallel to the third side.

Yes, a median always lie in the interior of the triangle.

If two angles are correspondingly equal to the two angles of another triangle, then the two triangles are similar.

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

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

i) 3 + 4 > 5 ; 4 + 5 > 3 ; 3 5 > 4, 

∴ Yes, a triangle can be formed with these sides.

 ii)  ∴Yes. A triangle can be formed with these sides.

iii) 4 + 4 > 8

∴ With these sides, a triangle cannot he formed. 

iv) 3 + 5 > 7; 5 + 7 > 3; 3 + 7 > 5 

∴ Yes. A triangle can he formed with these sides.

AAA similarity criterion: In two triangles, if corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.

C) Circumcircle

Correct option is  D) 120°

(i) (a) AR = PE 

(b) RT = EN 

(c) AT = PN

ii) (a) RT = EN and 

(b) PN = AT

iii) (a) ∠A = ∠P 

b) ∠T = ∠N

Correct option is  B) 5 cm

Correct option is  D) A.A.A

(c) SAS

Given, AD = EC

⇒ AD + DE = DE + EC 

⇒ AE = DC 

Also, AB = BC 

⇒ ∠BCA = ∠BAC       (isos. Δproperty)

⇒ ∠BCD = ∠BAE 

∴ In Δs ABE and CBD,

AB = CB            (Given) 

AE = DC           (Proved above) 

∠BAE =∠BCD   (Proved above)

∴ Δ ABE ≅ Δ CBD    (SAS)

Correct answer is ∠PON = 90°, ∠NPO = 20°

Correct answer is ∠x = 75°, ∠y = 135°

Correct option is (C) ≅

Symbol of “is congruent to” is \(\cong.\)

i.e., \(\triangle ABC\cong\triangle DEF\) is said as triangle ABC is congruent to triangle DEF.

Correct option is: (C) ≅

Correct option is (A) 11 : 4

We have DE || BC

and \(\frac{AE}{EC}=\frac47\)

\(\Rightarrow\) \(EC=\frac74AE\)

\(\because AC=AE+EC\)

\(=AE+\frac74AE\)

\(=\frac{11}4AE\)

\(\frac{AC}{AE}=\frac{11}4\)                   ________________(1)

In triangles \(\triangle ABC\;\&\;\triangle ADE,\)

\(\angle ABC=\angle ADE,\)

\(\angle ACB=\angle AED\)      (Corresponding angles as DE || BC)

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

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

\(\therefore\) \(\frac{BC}{DE}=\frac{AC}{AE}=\frac{11}4\)                         (From (1))

\(\therefore\) BC : DE = 11 : 4

Correct option is: (A) 11 : 4

In Fig. 6.22, ΔPQR ≅ ΔXZY

We are providing Class 9 Maths MCQ Questions of Triangles that covers topics of triangles orthocenter, circumcenter of the triangle, and center of the triangle etc. Students can practice objective types questions to score good marks in the upcoming exam. MCQ Questions are provided here with answers. 

Students can tackle Class 9 Maths MCQ Questions with Answers to access their preparation level. Let's start practice of given MCQ Questions bleow: -

Practice MCQ Questions for Class 9 Maths

1. If △ABC≅△PQR, then which of the following is not true?

(a) BC=PQ
(b) AC=PR
(c) QR=BC
(d) AB=PQ

2. In triangle ABC, if AB=BC and ∠B = 70°, ∠A will be:

(a) 70°
(b) 110°
(c) 55°
(d) 130°

3. In two triangles DEF and PQR, if DE = QR, EF = PR and FD = PQ, then

(a) ∆DEF ≅ ∆PQR
(b) ∆FED ≅ ∆PRQ
(c) ∆EDF ≅ ∆RPQ
(d) ∆PQR ≅ ∆EFD

4. In ∆ABC, BC = AB and ∠B = 80°. Then ∠A is equal to:

(a) 80°

(b) 40°

(c) 50°

(d) 100°

5. For two triangles, if two angles and the included side of one triangle are equal to two angles and the included side of another triangle. Then the congruency rule is:

(a) SSS
(b) ASA
(c) SAS
(d) None of the above

6. A triangle in which two sides are equal is called:

(a) Scalene triangle
(b) Equilateral triangle
(c) Isosceles triangle
(d) None of the above

7. The angles opposite to equal sides of a triangle are:

(a) Equal
(b) Unequal
(c) supplementary angles
(d) Complementary angles

8. If E and F are the midpoints of equal sides AB and AC of a triangle ABC. Then:

(a) BF=AC
(b) BF=AF
(c) CE=AB
(d) BF = CE

9. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Then:

(a) BE>CF
(b) BE<CF
(c) BE=CF
(d) None of the above

10. If ABC and DBC are two isosceles triangles on the same base BC. Then:

(a) ∠ABD = ∠ACD
(b) ∠ABD > ∠ACD
(c) ∠ABD < ∠ACD
(d) None of the above

11. If ABC is an equilateral triangle, then each angle equals to:

(a) 90°
(B)180°
(c) 120°
(d) 60°

12. If AD is an altitude of an isosceles triangle ABC in which AB = AC. Then:

(a) BD=CD
(b) BD>CD
(c) BD<CD
(d) None of the above

13. In a right triangle, the longest side is:

(a) Perpendicular
(b) Hypotenuse
(c) Base
(d) None of the above

14. In ∆PQR, if ∠R > ∠Q, then

(a) QR > PR
(b) PQ > PR
(c) PQ < PR
(d) QR < PR

15. D is a point on the side BC of a ΔABC such that AD bisects ∠BAC. Then

(a) BD : DC = AB : AC
(b) CD > CA
(c) BD > BA
(d) BA > BD

16. All the medians of a triangle are equal in case of a:

(a) Scalene triangle
(b) Right angled triangle
(c) Equilateral triangle
(d) Isosceles triangle

17. Which of the following is not a criterion for congruence of triangles?

(a) SAS
(b) ASA
(c) SSA
(d) SSS

18. In triangles ABC and PQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are

(a) Isosceles and congruent
(b) Isosceles but not congruent
(c) Congruent but not isosceles
(d) Neither congruent nor isosceles

19. In ∆ PQR, ∠R = ∠P and QR = 4 cm and PR = 5 cm. Then the length of PQ is

(a) 2 cm
(b) 2.5 cm
(c) 4 cm
(d) 5 cm

20. If AB = QR, BC = PR and CA = PQ, then

(a) ∆ PQR ≅ ∆ BCA
(b) ∆ BAC ≅ ∆ RPQ
(c) ∆ CBA ≅ ∆ PRQ
(d) ∆ ABC ≅ ∆ PQR

21. In triangle PQR if ∠Q = 90°, then:

(a) PQ is the longest side
(b) QR is the longest side
(c) PR is the longest side
(d) None of these

22. Two sides of a triangle are of lengths 5 cm and 1.5 cm. The length of the third side of the triangle cannot be

(a) 3.4 cm
(b) 3.6 cm
(c) 3.8 cm
(d) 4.1 cm

Click here: - Practice MCQ Questions for Class 9 Maths

Answer: 

1. Answer: (a) BC=PQ

Explanation: Given, ABC≅PQR
Thus, corresponding sides are equal.
Hence, AB=PQ 
AC=PR
BC=QR
Hence, BC=PQ is not true for the triangles.

2. Answer: (c) 55°

Explanation: Given,

AB = BC

Hence, ∠A=∠C

And ∠B = 70°

By angle sum property of triangle we know:

∠A+∠B+∠C = 180°

2∠A+∠B=180°

2∠A = 180-∠B = 180-70 = 110°

∠A = 55°

3. Answer: (b) ∆FED ≅ ∆PRQ

4. Answer: (c) 50°

5. Answer: (b) ASA

6. Answer: (c) Isosceles triangle

7. Answer: (a) Equal

8. Answer: (d) BF = CE

Explanation: AB and AC are equal sides.

AB = AC (Given)

∠A = ∠A (Common angle)

AE = AF (Halves of equal sides)

∆ ABF ≅ ∆ ACE (By SAS rule)

Hence, BF = CE (CPCT)

9. Answer: (c) BE=CF

Explanation: ∠A = ∠A (common arm)

∠AEB = ∠AFC (Right angles)

AB = AC (Given)

∴ ΔAEB ≅ ΔAFC

Hence, BE = CF (by CPCT)

10. Answer: (a) ∠ABD = ∠ACD

Explanation: AD = AD (Common arm)

AB = AC (Sides of isosceles triangle)

BD = CD (Sides of isosceles triangle)

So, ΔABD ≅ ΔACD.

∴ ∠ABD = ∠ACD (By CPCT)

11. Answer: (d) 60°

Explanation: Equilateral triangle has all its sides equal and each angle measures 60°.

AB= BC = AC (All sides are equal)

Hence, ∠A = ∠B = ∠C (Opposite angles of equal sides)

Also, we know that,

∠A + ∠B + ∠C = 180°

⇒ 3∠A = 180°

⇒ ∠A = 60°

∴ ∠A = ∠B = ∠C = 60°

12. Answer: (c) BD<CD

Explanation: In ΔABD and ΔACD,

∠ADB = ∠ADC = 90°

AB = AC (Given)

AD = AD (Common)

∴ ΔABD ≅ ΔACD (By RHS congruence condition)

BD = CD (By CPCT)

13. Answer: (b) Hypotenuse

Explanation: In triangle ABC, right-angled at B.

∠B = 90

By angle sum property, we know:

∠A + ∠B + ∠C = 180

Hence, ∠A + ∠C = 90

So, ∠B is the largest angle.

Therefore, the side (hypotenuse) opposite to largest angle will be longest one.

14. Answer: (b) PQ > PR

15. Answer: (a) BD : DC = AB : AC

16. Answer: (c) Equilateral triangle

17. Answer: (c) SSA

Explanation: SSA is not a criterion for the congruence of triangles. Whereas SAS, ASA and SSS are the criteria for the congruence of triangles. 

18. Answer: (b) Isosceles but not congruent

Explanation: Consider two triangles, ABC and PQR. If the sides AB = AC and ∠C = ∠P and ∠B = ∠Q, then the two triangles are said to be isosceles, but they are not congruent.

19. Answer: (c) 4 cm

Explanation: Given that, in a triangle PQR, ∠R = ∠P.

Since, ∠R = ∠P, the sides opposite to the equal angles are also equal.

Hence, the length of PQ is 4 cm.

20. Answer: (c) ∆ CBA ≅ ∆ PRQ

Explanation: Consider two triangles ABC and PQR.

Given that, AB = QR, BC = PR and CA = PQ.

By using Side-Side-Side (SSS rule),

We can say, ∆ CBA ≅ ∆ PRQ.

21. Answer: (c) PR is the longest side

22. Answer: (a) 3.4 cm

Explanation: If two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of the third side of the triangle cannot be 3.4 cm. Because the difference between the two sides of a triangle should be less than the third side.

Given: Angles of a triangle are in the ratio 1 : 2 : 3

To prove: Its corresponding sides are in the ratio 1:√3:2

Let the angles are x , 2x , 3x 

Therefore, x + 2x + 3x = 180°. 

6x = 180°. 

x = 30°.

So, the angles are 30°. , 60°. , 90°. 

So, the ratio of the corresponding sides are: 

= sin30°. : sin60°. : sin90°.

= 1/2: √3/2:1

= 1:√3:2 [proved]

False

Ratio of interior angles =1:2:3 

Let the angles are x,2x,3x 

x+2x+3x=180

6x = 180

So,

x = 30

2x = 60

3x = 90

let, exterior angle be a, b, c.

a = 360−30

a = 330

b = 360−60

b = 300

c = 360−90

c = 270

a:b:c = 330:300:270

a:b:c = 11:10:9

(i) AB=ST 

AC = SR (given) 

∠A = ∠S 

∴ ΔABC ≅ ΔSTR (S.A.S) 

Also ∠A = ∠S, ∠B = ∠T, ∠C = ∠R

(ii) From the figure, 

FO = RO 

OQ = OS 

∠POQ = ∠ROS 

∴ ΔPOQ ≅ ΔROS (SAS) 

Also ∠P = ∠R, ∠Q = ∠S, and PQ = RS

(iii)  From the figure, IAP Board So IJ 

WD = OR 

∠W= ∠R 

WO = RD 

∴ ΔDWO ≅ ΔORD 

Also ∠EO = ∠OE; ∠WDO = ∠ROD; 

∠WOD = ∠RDO

(iv) Here the two triangles ΔABC and ΔCDA are not congruent.

From the adjacent figure 

AL // CD and E is the midpoint of BC.

i.e., BE = EC 

from ΔBEL & ΔCED.

∠B = ∠C [ ∵ Alternate interior angles] 

∠DEC = ∠BEL [ ∵ vertically opposite, angles] 

∴ By A.S.A rule 

Δ EBL ≅ Δ ECD.

i) yes. 

∠ACB = ∠DBC 

BC = BC 

∠ABC = ∠DCB (by angle sum property) 

∴ ΔABC ≅ ΔDCB (A.S.A)

ii) yes 

∠A = ∠D 

∠ABO = ∠DCO (angle sum property) 

AB = DC [From (i)]

∴ ΔAOB ≅ ΔDOC (A.S.A) 

Otherwise ΔAOB and ΔDOC are similar by A.A.A. In congruent triangles corresponding parts are equal.

In ΔPQX, ΔPRY 

Sides QX = RY {Given} 

Sides PX = PY {Given} 

∠PXY = ∠PYX 

180 – ∠PXY = 180° – ∠PYX 

⇒ ∠PXQ = ∠PYR {Angle}

By S – A – S congruence criterion. 

ΔPQX ≅ ΔPRY

In ∆ACL, AC = CL 

So, ∆ACL is an isosceles triangle.

In isosceles triangle, the angles which are opposite to equal sides are also equal. 

So, ∠A = ∠L = x° ∠C = 56° (Vertically opposite angle) 

In ∆ACL, we know ∠A + ∠C + ∠L = 180° 

⇒ x + 56° + x = 180° 

⇒ 2x + 56° – 56° = 180° – 56°

 ⇒ 2x =124° 

⇒ 2x/2 = 124°/2

∴ x = 62° 

Exterior angle at A = ∠C + ∠L 

⇒ y = 56 + 62 

⇒ y = 118°

Correct option is (B) \(AB^2 + BC^2\)

Baudhayan theorem is today known as the Pythagoras theorem and according to this theorem in a right angled triangle ABC right angled at B, we have

\((\text{Hypotenuse})^2=(\text{Base})^2+(\text{Altitude})^2\)

i.e., \(AC^2=AB^2+BC^2\)

Hence, according to Baudhayan theorem in a triangle ABC right angled at B, we have \(AC^2=AB^2+BC^2\)

Correct option is: (B) AB² + BC²

Correct option is (A) 4 : 1

In triangles \(\triangle AOB\;and\;\triangle COD,\)

\(\angle OAB=\angle OCD\)        (Alternative interior angles as AB || DC)

\(\angle OBA=\angle ODC\)         (Alternative interior angles as AB || DC)

\(\angle AOB=\angle COD\)         (Vertically opposite angles)

\(\therefore\) \(\triangle AOB\sim\triangle COD\)    (By AAA similarity rule)

\(\therefore\) \(\frac{ar(\triangle AOB)}{ar(\triangle COD)}=(\frac{AB}{CD})^2\)    (By properties of similar triangle)

\(=(\frac21)^2=\frac41\)                 \((\because AB=2CD\Rightarrow\frac{AB}{CD}=\frac21)\)

\(\therefore\) \(ar(\triangle AOB):ar(\triangle COD)=4:1\)

Correct option is: (A) 4 : 1

Correct option is A) 47°

In right angled triangle

One angle = 90°

Other angle = 43°

\(\therefore\) Third angle = 180° - 90° - 43°

 = 90° - 43° = 47°

(b) 3

The perpendicular line segment from a vertex of a triangle to its opposite side is called an altitude of the triangle. A triangle has 3 altitudes.

Correct answer is (a) Median

(a) Median

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

Correct option is A) altitude

Sol. Given angles are 100°, 30°, 45° 

Sum of the angles = 100° + 30° + 45° = 175° < 180° 

So, 100°, 30°, 45° cannot form a triangle.

Given angles are 65°, 45°, 70°. 

Sum of the angles = 65° + 45° + 70° = 180° 

So, 65°, 45°, 70° can form a triangle.