CBSE Class 10 Maths HOTs Similar Triangles

Please refer to CBSE Class 10 Maths HOTs Similar Triangles. Download HOTS questions and answers for Class 10 Mathematics. Read CBSE Class 10 Mathematics HOTs for Chapter 6 Triangles below and download in pdf. High Order Thinking Skills questions come in exams for Mathematics in Class 10 and if prepared properly can help you to score more marks. You can refer to more chapter wise Class 10 Mathematics HOTS Questions with solutions and also get latest topic wise important study material as per NCERT book for Class 10 Mathematics and all other subjects for free on Studiestoday designed as per latest CBSE, NCERT and KVS syllabus and pattern for Class 10

Chapter 6 Triangles Class 10 Mathematics HOTS

Class 10 Mathematics students should refer to the following high order thinking skills questions with answers for Chapter 6 Triangles in Class 10. These HOTS questions with answers for Class 10 Mathematics will come in exams and help you to score good marks

HOTS Questions Chapter 6 Triangles Class 10 Mathematics with Answers

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Please refer to link below to download pdf file of Class 10 Similar Triangles HOTs

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More MCQs for NCERT Class 10 Mathematics Triangles....... 

 

1. In the given figure, AB || CD then the value of ‘x’ is equal to

CBSE Class 10 Maths HOTs Triangles

(A)  44o

(B)  88o

(C)  80o

(D)  100o

Answer : (B)

 

2. In the given figure, AB divides ∠DACin the ratio 1 : 3 and AB = DB. Find the value of x.

CBSE Class 10 Maths HOTs Triangles

(A)  90o

(B)  80o

(C)  70o

(D)  85o

Answer : (A)

 

3. In the given figure, o ∠A =100 and AB = AC, find ∠B and ∠C.

CBSE Class 10 Maths HOTs Triangles

(A)  40o ,40o

(B)  60o ,20o

(C)  45o ,35o

(D)  25o ,55o

Answer : (A)

 

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

(A) SAS

(B) SSA

(C) ASA

(D)SSS

Answer : (B)

 

5. In the given figure, The measure of ∠B'A'C' is

CBSE Class 10 Maths HOTs Triangles

(A) 50o

(B) 60o

(C) 70o

(D) 80o

Answer : (B)

 

6. In the triangles ABC and PQR, three equality relations between some parts are as follows:

AB = PQ, ∠B=∠P , BC = PR Congruence conditions apply:

(A) SAS

(B) ASA

(C) SSS

(D)RHS

Answer : (A)

 

7. If ΔPQR ≅ ΔEFD, then ED = ?

(A) PQ

(B) QR

(C) PR

(D) none of these

Answer : (C)

 

8. In the given figure, AB = AC, AD is the median to base BC. Then, ∠BAD= ?

CBSE Class 10 Maths HOTs Triangles

(A) 550

(B) 700

(C) 350

(D) 1100

Answer : (A)

 

9. In ∠ABC, o ∠B=∠C = 45 . Which is the longest side?

(A) AC

(B) AB

(C) BC

(D) none of these

Answer : (C)

 

10. In ΔABC, if o ∠A = 50 and o ∠B = 60 , determine the shortest and largest sides of the triangle.

(A) BC, AB

(B) AB, BC

(C) AC, BC

(D) none of these

Answer : (A) 

 

11. ABCD is a parallelogram, if the two diagonals are equal, find the measure of ∠ABC.

(A) 500

(B) 600

(C) 900

(D) 1000

Answer : (C)

 

12. In the given figure, AC is the bisector of ∠BAD. Then CD = ?

CBSE Class 10 Maths HOTs Triangles

(A) 2cm

(B) 3 cm

(C) 4 cm

(D) 5cm

Answer : (C)


13. In ΔABCand ΔDEF such that ΔABC ≅ ΔFDE, and AB = 5 cm, ∠B = 40 ,  ∠A = 80 . Which of the following is true?

(A) DF = 5 cm, ∠F = 60

(B) DE = 5 cm,  ∠E = 60

(C) DF = 5 cm, ∠E = 60

(D) DE = 5 cm,  ∠D = 40

Answer : (C)

 

14. In an isosceles triangle, if the vertex angle is twice the sum of the base angles, then the measure of the vertex angle of the triangle is

(A)  100

(B)  120

(C)  110

(D)  130

Answer : (A)

 

15. In the given figure, AB = AC and CD || BA. The value of x is

CBSE Class 10 Maths HOTs Triangles

(A)  52o

(B)  76o

(C)  156o

(D)  104o

Answer : (D)

 

16. If the two polygons are similar then find the value of x.

CBSE Class 10 Maths HOTs Triangles

(A) 6

(B) 16/3

(C) 17/3

(D) 8

Answer : (B)

 

17. If ΔABCis a isosceles triangle where AB = BC and DE || BC, so if AD = 1.8 cm and CE = 5. 2 cm then AC is :

(A) 5.2 cm

(B) 6 cm

(C) 8 cm

(D) 7cm

Answer : (D)

 

18. A ship started from the base of a light house. Height of the ship is 25 ft. The ship is travelling with a speed of 10 feet/sec. So after 5 seconds what is the length if the ship’s shadow on water if height of the light house is 50 ft.

(A) 60 ft

(B) 50 ft

(C) 25 ft

(D)100 ft

Answer : (B)

 

19. If DE || BC then

CBSE Class 10 Maths HOTs Triangles

(A) AB/AC = BC/BE

(B) AB/BC = DE/BE

(C) AC/BE = AB/AD

(D) none of these

Answer : (C)

 

20. In a ΔABC,DE intersects AB and AC at point D and E respectively. If AB = 9, AD = 6, AE = 4, AC = 6 then line DE and BC are

(A) same

(B) parallel

(C) perpendicular

(D) none of these

Answer : (B)

 

21. Find x and if DE || BC.

CBSE Class 10 Maths HOTs Triangles

(A) 1

(B) 2

(C) 3 

(D) 4

Answer : (B)

 

22. Find y if DE || BC.

(A)  250

(B)  300

(C)  400

(D)  450

Answer : (C)

 

23. Let A by any point inside the rectangle KLMN. Then

(A) KL2 + LM2 = NM2 + MA2

(B) KA+ LM2 = KN2 + AL2

(C) AK2 + AM2 = AL2 + AN2

(D) none of these

Answer : (C)

 

24. If DG is the bisector of the angle ∠D in the ΔDEF and DE = 6, DF = 8, EF = 10, then EG =

(A) 15/7

(B) 34/7

(C) 40/7

(D) 30/7

Answer : (D)

 

25. If A, B and C are mid-points of DE, EF and ED of ΔDEF then find the ratio of area of ΔABC and ΔDEF .

(A) 1 : 4

(B) 4 : 1

(C) 3 : 2

(D) 2 : 3

Answer : (B)

 

26. Which of the following is correct?

CBSE Class 10 Maths HOTs Triangles

(A) AC = AB

(B) AC = BC

(C) DC = BD

(D) BC = AC

Answer : (C)

 

27. If two lines DE and LM bisects each other and if DM = 8 cm then EL is:

(A) 8/3 cm

(B) 4 cm

(C) 6 cm

(D) 8cm

Answer : (D)

 

28. Here DR || HI || GF, HJ = 6 cm, JF = 9 cm, GI = 18 cm and GF = 12 cm then HI is:

CBSE Class 10 Maths HOTs Triangles

(A) 9 cm

(B) 8 cm

(C) 6 cm

(D) 10 cm

Answer : (D)

 

29. If ΔABC ≅ ΔDEF. If ∠A = 3x - 60and ∠D = x + 20 then ∠A is:

(A)  60o

(B)  90o

(C)  100o

(D) 

Answer : (A)

 

30. Find x:

CBSE Class 10 Maths HOTs Triangles

(A)  30

(B)  25

(C)  35

(D)  20

Answer : (B)

 

31. Find x.

CBSE Class 10 Maths HOTs Triangles

(A) 11.4

(B) 10.9

(C) 11.6

(D) 12.2

Answer : (A)

 

32. A kite got stuck on top of a 20 feet wall. A ladder is used by person to get the kite. It should be placed in such a manner that the top of ladder should rest on top of the wall and bottom of the ladder should be 15 feet away from the bottom of wall. Height of the ladder is:

(A) 22 feet

(B) 20 feet

(C) 25 feet

(D)14 feet

Answer : (C)

 

33. If ΔABChas  ∠B = 90o and D and E are points on BC where when connected to A, AD and AE trisects the angle A. Then

(A) AE2 = 3AC2/8 + 5AD2/8

(B) AC2 = 3AE2/8 + 5AD2/8

(C) AC2 = AE2 + AD2

(D) none of these

Answer : (A)

 

34. If XYZ is a triangle where o ∠Z = 90 . If L is the mid-point of YZ then

(A) XY2 = 4XL2 = 3XZ2

(B) XY2 + 3XZ2 = 4XL2

(C) XY2 + XZ2 = XL2

(D) none of these

Answer : (B)

 

35. If hypotenuse LM is common for both the triangles i.e., ΔKLMand ΔLMN then

CBSE Class 10 Maths HOTs Triangles

(A) KX x XM = LX x LM

(B) KX x KL = LM x MX

(C) KX x XM = LX x XN

(D) none of these

Answer : (C)

 

36. Here BG || CD and FG || DE when which of the following are correct?

CBSE Class 10 Maths HOTs Triangles

(A) AC/BG = AE/DE 

(B) AB/AC = AF/ AE

(C) AD/AG = AC/AE

(D) none of these

Answer : (B)  


37. Here QN || LM and QO || LN, so which of the following is correct?

CBSE Class 10 Maths HOTs Triangles

(A) PO x PN = NM x ON

(B) POxMN = PNxON

(C) ONxNL =OQ

(D) none of these

Answer : (B)



38. If AB || CD then ΔABE and ΔDCEwill be

CBSE Class 10 Maths HOTs Triangles

(A) similar

(B) concreate

(C) equilateral

(D) cannot say

Answer : (A)


39. If ∠BAE = ∠ECDthen ∠ABD and ∠CDEwill be

CBSE Class 10 Maths HOTs Triangles

(A) congruent

(B) similar

(C) right angle triangle

(D) cannot say

Answer : (B)

 

40. If ΔAEB and ΔDCAboth are right angled triangle then which of the following is correct?

CBSE Class 10 Maths HOTs Triangles

(A) CD/EB = DA/ AC

(B) AB/AC = AD/DC

(C) EA/AC = EB/CD

(D) none of these

Answer : (C)

 

Please click the below link to access CBSE Class 10 Triangles HOTs

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MORE QUESTION

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Please refer to link below for CBSE Class 10 Mathematics HOTs Triangles Set A

SIMILAR TRIANGLES 

 
Geometry is the right foundation of all painting, I have decided to teach its rudiments and principles to all youngsters eager for art. 
 
1.  ABC is a right-angled triangle, right-angled at A. A circle is inscribed in it. The lengths of the two sides containing the right angle are 6cm and 8 cm. Find the radius of the in circle.
(Ans: r=2)
CBSE_Class_10_maths_Similar_Triangles_2
 
 
 
 
 
 
 
 
 
Ans:    BC = 10cm
y + z = 8cm
x + z = 6cm
x + y = 10
  x + y + z = 12
z = 12 – 10
z = 2 cm
∴radius = 2cm
 
2.  ABC is a triangle. PQ is the line segment intersecting AB in P and AC in Q such that PQ parallel to BC and divides triangle ABC into two parts equal in area. Find BP: AB. 
Ans: Refer example problem of text book.
 
3.  In a right triangle ABC, right angled at C, P and Q are points of the sides CA and CB respectively, which divide these sides in the ratio 2: 1.
Prove that 9AQ2= 9AC2 +4BC2
9BP2= 9BC2 + 4AC2
9 (AQ2+BP2) = 13AB2
 
Ans: Since P divides AC in the ratio 2 : 1
 CBSE_Class_10_maths_Similar_Triangles_3
 
 
 
 
 
 
 
 
 
 
 
 
CP = 2/3 AC
QC = 2/3 Bc  
AQ2 = QC2 + AC2 
AQ2 =
4/9  BC2 + AC2 
9 AQ2 = 4 BC2  + 9AC                                …………………. (1) Similarly we get 9 BP2 = 9BC2 + 4AC2 ……..…………(2) Adding (1) and (2) we get 9(AQ2 + BP2) = 13AB2
 
4.  P and Q are the mid points on the sides CA and CB respectively of triangle ABC right angled at C. Prove that 4(AQ2 +BP2) = 5AB2
 
Self Practice
 
5.  In an equilateral triangle ABC, the side BC is trisected at D.
Prove that 9AD2 = 7AB2
 
Self Practice
 
6.  There is a staircase as shown in figure connecting points A and B. Measurements of steps are marked in the figure. Find the straight distance between A and B. (Ans:10) 
CBSE_Class_10_maths_Similar_Triangles_4 
Ans: Apply Pythagoras theorem for each right triangle add to get length of AB. 
 
7.  Find the length of the second diagonal of a rhombus, whose side is 5cm and one of the diagonals is 6cm. (Ans: 8cm) 
 
Ans: Length of the other diagonal = 2(BO)
where BO = 4cm
∴ BD = 8cm.
 
8.  Prove that three times the sum of the squares of the sides of a triangle is equal to four times the sum of the squares of the medians of the triangle.
 
Ans: To prove 3(AB2 + BC2 + CA2) = 4 (AD2+ BE2 + CF2)
In any triangle sum of squares of any two sides is equal to twice the square of half of third side, together with twice the square of median bisecting it. 
 
If AD is the median
CBSE_Class_10_maths_Similar_Triangles_5
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
AB2 + AC2 = 2 {AD2 + BC2/4}
 
2(AB2 + AC2) = 4AD2 + BC2
Similarly by taking BE & CF as medians we get
  2 (AB2 + BC2) = 4BE2 + AC2
& 2 (AC2 + BC2) = 4CF2 + AB2
Adding we get
  3(AB2 + BC2 + AC2) = 4 (AD2+ BE2 + CF2) D
 
9.  ABC  is  an  isosceles triangle is  which  AB=AC=10cm. BC=12. PQRS  is  a rectangle inside the isosceles triangle.  Given PQ=SR= y cm, PS=QR=2x.  Prove that x = 6 −3 y/4 
CBSE_Class_10_maths_Similar_Triangles_6
 
 
 
 
 
 
 
 
 
 
 
 
 
Ans: AL = 8 cm
∆BPQ ∼ ∆BAL
  BQ/PQ =  BL/AL 
 => 6 − x/y = 6/8 
 =>x = 6 −3y/4 
Hence proved
 
10. If ABC is an obtuse angled triangle, obtuse angled at B and if AD⊥CB Prove that AC2 =AB2 + BC2+2BCxBD
Ans: AC2 = AD2 + CD2
= AD2 + (BC + BD)2
= AD2 + BC2 + 2BC.BD+BD2
= AB2 + BC2 + 2BC.BD
CBSE_Class_10_maths_Similar_Triangles_7
 
 
 
 
 
 
 
 
 
11.  If ABC is an acute angled triangle , acute angled at B and AD⊥BC
prove that AC2 =AB2 + BC2 −2BCxBD 
 
Ans: Proceed as sum no. 10. 
 
12.  Prove that in any triangle the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median, which bisects the third side.
 
Ans: To prove AB2 + AC2 = 2AD2 + 2 (1/2 BC )2
CBSE_Class_10_maths_Similar_Triangles_8
 
 
 
 
 
 
 
 
 
Draw AE ⊥ BC 
Apply property of Q. No.10 & 11. In ∆ ABD since ∠D > 900
∴AB2 = AD2 + BD2 + 2BD x DE ….(1)
∆ ACD  since ∠ D < 90o
AC2 = AD2 + DC2 - 2DC x DE ….(2)
Adding (1) & (2)
AB2 + AC2 = 2(AD2 + BD2)
= 2(AD2 + (1/2BC )2)
 
Or AB2 + AC2 = 2 (AD2 + BD2) Hence proved 
 
13. If A be the area of a right triangle and b one of the sides containing the right angle, prove that the length of the altitude on the hypotenuse is
2 Ab √b4  + 4 A2 
CBSE_Class_10_maths_Similar_Triangles_1 
  ∠ABC = 60o
Hence proved
 
15. ABCD is a rectangle. ∆ ADE and ∆ ABF are two triangles such that ∠E=∠F as shown in the figure. Prove that AD x AF=AE x AB.
Ans: Consider ∆ ADE and ∆ ABF
∠D = ∠B = 90o
∠E = ∠F (given)
∴∆ ADE ≅  ∆ ABF
AD/AB = AE/AF
=>AD x AF = AB x AE Proved
CBSE_Class_10_maths_Similar_Triangles_9
 
 
 
 
 
 
 
 
 
16. In the given figure, ∠AEF=∠AFE and E is the mid-point of CA. Prove that
 BD/CD =  BF/ CE
CBSE_Class_10_maths_Similar_Triangles_10
 
 
 
 
 
 
 
 
 
 
 
 
 
Ans: Draw CG IIDF
In ∆ BDF
CG II DF
∴ BD/CD = BF/ GF ………….(1) BPT
In ∆AFE
∠AEF=∠AFE
 =>AF=AE
 =>AF=AE=CE…………..(2)
In ∆ ACG
E is the mid point of AC
=> FG = AF
∴ From (1) & (2)
BD/CD = BF /CE
Hence proved 
 
17. ABCD is a parallelogram in the given figure, AB is divided at P and CD and Q so that AP:PB=3:2 and CQ:QD=4:1. If PQ meets AC at R, prove that AR= 3/7 AC.
  CBSE_Class_10_maths_Similar_Triangles_11
 
 
 
 
 
 
 
 
 
 
 
 
Ans: ∆APR ∼ ∆CQR (AA)
   AP/CQ = PR/QR = AR/CR
   AP/CQ  = AR/CR & AP= 3/ 5 AB 
  3AB/5CQ = AR/CR & CQ= 4/  5  CD = 4/5AB
CBSE_Class_10_maths_Similar_Triangles_12
 
 
 
 
 
 
 
 
 
 
 
 
 
 AR/CR = 3/4
  CR/AR = 4/3 
(CR + AR)/AR = 4/3+1
  AC/AR = 7/ 3
  AR = 3/7 AC
Hence proved 
 
18. Prove that the area of a rhombus on the hypotenuse of a right-angled triangle, with one of the angles as 60o, is equal to the sum of the areas of rhombuses with one of their angles as 60o drawn on the other two sides.
 CBSE_Class_10_maths_Similar_Triangles_13
 
 
 
 
 
 
 
 
Ans: Hint: Area of Rhombus of side a & one angle of 60o 
= √3/2 x a x a =√ 3/2 a2
 
19. An aeroplane leaves an airport and flies due north at a speed of 1000 km/h.  At the same time, another plane leaves the same airport and flies due west at a speed of 1200
km/h. How far apart will be the two planes after 1½ hours. (Ans:300√61Km)
N
Ans: ON = 1500km (dist = s x t) OW = 1800 km 
NW =
CBSE_Class_10_maths_Similar_Triangles_14
 
 
 
 
 
 
 
 
 
 
 
 
 
√ (15002 + 18002)
=   √5490000
=300 √61 km 
 
20. ABC is a right-angled isosceles triangle, right-angled at B.   AP, the bisector of
∠BAC, intersects BC at P. Prove that AC2 = AP2 + 2(1+√2)BP2 
Ans: AC = √2 AB (Since AB = BC)
AB/AC = BP/ CP (Bisector Theorem)
CBSE_Class_10_maths_Similar_Triangles_15
 
 
 
 
 
 
 
 
 
 
 
 
 
=>CP =√2 BP
AC2 – AP2 = AC2 – (AB2 + BP2)
= AC2 – AB2 - BP2
= BC2 - BP2
= (BP + PC)2 - BP2
= (BP + √2 BP)2 – BP2
= 2BP2 + 2 √2  BP
 
ADDITIONAL QUESTIONS
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