CBSE Class 10 Maths HOTs Similar Triangles Set 01

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

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

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

(A)  44^o

(B)  88^o

(C)  80^o

(D)  100^o

Answer : (B)

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

(A)  90^o

(B)  80^o

(C)  70^o

(D)  85^o

Answer : (A)

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

(A)  40^o ,40^o

(B)  60^o ,20^o

(C)  45^o ,35^o

(D)  25^o ,55^o

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

(A) 50^o

(B) 60^o

(C) 70^o

(D) 80^o

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= ?

(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 = ?

(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^o  ,  ∠A = 80^o  . Which of the following is true?

(A) DF = 5 cm, ∠F = 60^o 

(B) DE = 5 cm,  ∠E = 60^o 

(C) DF = 5 cm, ∠E = 60^o 

(D) DE = 5 cm,  ∠D = 40^o 

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^o 

(B)  120^o 

(C)  110^o 

(D)  130^o 

Answer : (A)

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

(A)  52^o

(B)  76^o

(C)  156^o

(D)  104^o

Answer : (D)

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

(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

(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.

(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) KA² + 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?

(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:

(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)  60^o

(B)  90^o

(C)  100^o

(D) 

Answer : (A)

30. Find x:

(A)  30

(B)  25

(C)  35

(D)  20

Answer : (B)

31. Find x.

(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 = 90^o 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

(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?

(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?

(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

(A) similar

(B) concreate

(C) equilateral

(D) cannot say

Answer : (A)

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

(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?

(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

MORE QUESTION

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

SIMILAR TRIANGLES 

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

 

 

 

 

 

 

 

 

 

 

 

 

 

CP = 2/3 AC

QC = 2/3 Bc  

AQ² = QC² + AC² 

AQ² =

4/9  BC² + AC² 

9 AQ² = 4 BC²  + 9AC²                                 …………………. (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) 

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

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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 

 

 

 

 

 

 

 

 

 

 

 

 

 

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

 

 

 

 

 

 

 

 

 

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

 

 

 

 

 

 

 

 

 

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 < 90^o

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 

  ∠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 = 90^o

∠E = ∠F (given)

∴∆ ADE ≅  ∆ ABF

AD/AB = AE/AF

=>AD x AF = AB x AE Proved

 

 

 

 

 

 

 

 

 

16. In the given figure, ∠AEF=∠AFE and E is the mid-point of CA. Prove that

 BD/CD =  BF/ CE

 

 

 

 

 

 

 

 

 

 

 

 

 

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.

  

 

 

 

 

 

 

 

 

 

 

 

 

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

 

 

 

 

 

 

 

 

 

 

 

 

 

 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 60^o, is equal to the sum of the areas of rhombuses with one of their angles as 60^o drawn on the other two sides.

 

 

 

 

 

 

 

 

 

Ans: Hint: Area of Rhombus of side a & one angle of 60^o 

= √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 =

 

 

 

 

 

 

 

 

 

 

 

 

 

√(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)

 

 

 

 

 

 

 

 

 

 

 

 

 

=>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  BP²