Chapter 6 Triangles NCERT Exemplar Solutions Exercise 6.1 Class 10 Maths
Chapter Name  NCERT Maths Exemplar Solutions for Chapter 6 Triangles Exercise 6.1 
Book Name  NCERT Exemplar for Class 10 Maths 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 6.1 Solutions
Multiple Choice Questions
Choose the correct answer from the given four options:
1. In figure, if ∠BAC = 90° and AD ⊥ BC . Then,
If the lengths of the diagonals of rhombus are 16 cm and 12 cm. Then, the length of the sides of the rhombus is
(A) BD.CD = BC^{2}
(B) AB.AC = BC^{2}
(C) BD.CD = AD^{2}
(D) AB.AC = AD^{2}
Solution
(C) BD.CD = AD^{2}
In ∆ADB and ∆ADC,
We have,
∠D = ∠D = 90° (∵ AD ⊥ BC)
∠DBA = ∠DAC [each angle = 90° ∠C]
From AAA similarity rule,
∆ADB ∼ ∆ADC
Therefore,
BD/AD = AD/CD
⇒ BD.CD = AD^{2}
2. If the lengths of the diagonals of rhombus are 16 cm and 12 cm. Then, the length of the sides of the rhombus is
(A) 9 cm
(B) 10 cm
(C) 8 cm
(D) 20 cm
Solution
(B) 10 cm
We have,
A rhombus is a simple quadrilateral whose four sides are of same length and diagonals are perpendicular bisector of each other.
Now,
AC = 16cm and
BD = 12 cm
∠AOB = 90°
AC and BD bisects each other
AO = 1/2 AC
BO = 1/2 BD
So,
AO = 8cm
BO = 6cm
In right angled ∆AOB,
By Pythagoras theorem,
We have,
AB^{2} = AO^{2} + OB^{2}
⇒ AB^{2} = 8^{2} + 6^{2}
= 64 36 = 100
⇒ AB = √100 = 10 cm
As the four sides of a rhombus are equal.
So, one side of rhombus = 10 cm.
3. If ∆ABC ∼ ∆EDF and ∆ABC is not similar to ∆DEF, then which of the following is not true ?
(A) BC.EF = AC.FD
(B) AB.EF = AC.DE
(C) BC.DE = AB.EF
(D) BC.DE = AB.FD
Solution
(C) BC.DE = AB.EF
If sides of one triangle are proportional to the side of the other triangle, and the corresponding angles are also equal, then the triangles are similar by SSS similarity.
So, ∆ABC ∼ ∆EDF
By similarity rule,
AB/ED = BC/DF = AC/EF
At first we take,
AB/ED = BC/DF
⇒ AB.DF = ED.BC
Hence, option (D) BC · DE = AB.FD is true
Now taking,
BC/DF = AC/EF , we get
⇒ BC.EF = AC.DF
Hence, option (A) BC.EF = AC.FD is true
Now if,
AB/ED = AC/EF, we get ,
⇒ AB.EF = ED.AC
Hence, option (B) AB.EF = AC.DE is true.
(A) ∆PQR ~ ∆CAB
(B) ∆PQR ~ ∆ABC
(C) ∆CBA ~ ∆PQR
(D) ∆BCA ~∆PQR
AB/QR = BC/PR = CA/PQ
∆PQR ~ ∆CAB
(A) 50°
(B) 30°
(C) 60°
(D) 100°
In ∆APB and ∆CPD,
∠APB = ∠CPD = 50° (vertically opposite angles)
AP/PD = 6/5 ...(i)
And,
BP/CP = 3/2.5
⇒ BP/CP = 6/5 ...(ii)
From equations (i) and (ii),
AP/PD = BP/CP
Therefore,
∆APB ～∆DPC [using SAS similarity rule]
From ∆APB,
(B) DE/PQ = EF/RP
(C) DE/QR = DF/PQ
(D) EF/RP = DE/QR
We have,
In ∆DEF and ∆PQR,
∠D = ∠Q,
∠R = ∠E
∆DEF ～ ∆QRP
∠F = ∠P
DF/QP = ED/RQ = FE/PR
(A) Congruent but not similar
In ∆ABC and ∆DEF,
(A) 9
(B) 3
(C) 1/3
(D) 1/9
We have,
∆ABC ～ ∆PQR
BC/QR = 1/3
We know that, the ratio of the areas of two similar triangles is equal to square of the ratio of their corresponding sides.
ar(PRQ)/ar(BCA) = QR^{2}/BC^{2}
QR^{2} /BC^{2} = 3^{2} /1^{2} = 9
(A) DE = 12 cm, ∠F = 50°
(B) DE = 12 cm, ∠F = 100°
(C) EF = 12 cm, ∠D = 100°
(D) EF = 12 cm, ∠D = 30°
∠B = ∠F
180°  (50° + 30°)
= 100°
Now,
AB/DF = AC/DE
⇒ 5/7.5 = 8/DE
⇒ DE = 12 cm
(A) ∠B = ∠E
(B) ∠A = ∠D
(C) ∠B = ∠D
(D) ∠A = ∠F
Given, in ∆ABC and ∆EDF,
AB/DE = BC/FD
Therefore,
∆ABC ～ ∆EDF if, ∠B = ∠D [By SAS similarity criterion]
(A) 10 cm
(B) 12 cm
(C) 20/3 cm
(D) 8 cm
(A) PR . QR = RS^{2}
(B) QS^{2} + RS^{2} = QR^{2}
(C) PR^{2} + QR^{2} = PQ^{2}
(D) PS^{2} + RS^{2} = PR^{2}