12 - Question bank - Triangles - Class 10 - Maths Class 10 Notes | EduRev

Crash Course for Class 10 Maths by Let's tute

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Class 10 : 12 - Question bank - Triangles - Class 10 - Maths Class 10 Notes | EduRev

 Page 1


                                                                                                                        
 
A 
C B 
D 
   
  
Triangles 
 
 
 
 
1)  Let  ABC  DEF and their areas be, respectively, 64cm
2 
and 121 cm
2. 
If EF=15.4 cm, find 
BC. 
Solution:           
Given:  ABC    DEF and A( ABC)=64cm
2 
& A( DEF)= 121cm
2 
and EF = 15.4cm. 
To find: BC = ? 
We know that, the ratio of the areas of two similar triangles is equal to the square of the ratio of  
their sides. 
  
       
       
  
    
 
    
 
  
i.e. 
  
   
  
    
 
      
 
 
   BC
2 
 = 
         
   
 = 
        
   
  = 125.44 
  BC = 11.2 cm 
 
2) In an equilateral triangle ABC, AD   BC. Prove that 3AB
2 
= 4AD
2
. 
Solution: 
Given:  ABC is an equilateral triangle and  AD   BC. 
To prove that:  3AB
2 
= 4AD
2 
Proof:  
Since, AD   BC therefore, BD = DC. 
Now, in right triangle  ABD, we have 
AB
2
 = AD
2
 + BD
2 
---
 
[Pythagoras theorem] 
Since BD = DC therefore, BD = 
 
 
BC 
  AB
2
 = AD
2
 +  
 
 
   
 
 
  AB
2
 = AD
2
 + 
 
 
  
 
 
  AB
2
 = AD
2
 + 
  
 
 
 
  AB
2
 = AD
2
 + 
  
 
 
  ------ [Since, AB = BC = AC – sides of an equilateral triangle] 
  AB
2 
- 
    
 
 
  
 = AD
2 
  
   
 
 
 = AD
2
  
  3AB
2 
 =  4AD
2
 
Page 2


                                                                                                                        
 
A 
C B 
D 
   
  
Triangles 
 
 
 
 
1)  Let  ABC  DEF and their areas be, respectively, 64cm
2 
and 121 cm
2. 
If EF=15.4 cm, find 
BC. 
Solution:           
Given:  ABC    DEF and A( ABC)=64cm
2 
& A( DEF)= 121cm
2 
and EF = 15.4cm. 
To find: BC = ? 
We know that, the ratio of the areas of two similar triangles is equal to the square of the ratio of  
their sides. 
  
       
       
  
    
 
    
 
  
i.e. 
  
   
  
    
 
      
 
 
   BC
2 
 = 
         
   
 = 
        
   
  = 125.44 
  BC = 11.2 cm 
 
2) In an equilateral triangle ABC, AD   BC. Prove that 3AB
2 
= 4AD
2
. 
Solution: 
Given:  ABC is an equilateral triangle and  AD   BC. 
To prove that:  3AB
2 
= 4AD
2 
Proof:  
Since, AD   BC therefore, BD = DC. 
Now, in right triangle  ABD, we have 
AB
2
 = AD
2
 + BD
2 
---
 
[Pythagoras theorem] 
Since BD = DC therefore, BD = 
 
 
BC 
  AB
2
 = AD
2
 +  
 
 
   
 
 
  AB
2
 = AD
2
 + 
 
 
  
 
 
  AB
2
 = AD
2
 + 
  
 
 
 
  AB
2
 = AD
2
 + 
  
 
 
  ------ [Since, AB = BC = AC – sides of an equilateral triangle] 
  AB
2 
- 
    
 
 
  
 = AD
2 
  
   
 
 
 = AD
2
  
  3AB
2 
 =  4AD
2
 
                                                                                                                        
 
       
Triangles 
 
 
 
 
 
 
 
3) In the given figure, PQR is a right triangle right angled at Q & QS   PR. If PQ = 6cm and PS 
= 4cm, find the QS, RS & QR. 
Solution: 
Given: In PQR,   PQR= 90 . 
PQ = 6cm, PS = 4cm. 
To find: QS , RS & QR? 
 
Since,QS   PR, 
  PS = SR 
And PS =4cm ----- (given ) 
      SR = 4cm. 
PR = PS + SR = 4 + 4 =8 
 PQR is right angled triangle. 
  by Pythagoras theorem, 
PR
2
  =  PQ
2
 + QR
2
 
 QR
2 
= PR
2
 - PQ
2 
  QR
2
 = 8
2 
 - 6
2 
  QR
2
  = 64 -36 = 28 
  QR   =  2v  
 
QS
2 
 = PS   SR  -------- [ Since QS is perpendicular to PR] 
  QS
2 
 = 4   4  = 16  
  QS = 4cm. 
 
 
 
 
 
 
 
 
 
 
 
R 
Q 
P 
S 
6 
4 
Page 3


                                                                                                                        
 
A 
C B 
D 
   
  
Triangles 
 
 
 
 
1)  Let  ABC  DEF and their areas be, respectively, 64cm
2 
and 121 cm
2. 
If EF=15.4 cm, find 
BC. 
Solution:           
Given:  ABC    DEF and A( ABC)=64cm
2 
& A( DEF)= 121cm
2 
and EF = 15.4cm. 
To find: BC = ? 
We know that, the ratio of the areas of two similar triangles is equal to the square of the ratio of  
their sides. 
  
       
       
  
    
 
    
 
  
i.e. 
  
   
  
    
 
      
 
 
   BC
2 
 = 
         
   
 = 
        
   
  = 125.44 
  BC = 11.2 cm 
 
2) In an equilateral triangle ABC, AD   BC. Prove that 3AB
2 
= 4AD
2
. 
Solution: 
Given:  ABC is an equilateral triangle and  AD   BC. 
To prove that:  3AB
2 
= 4AD
2 
Proof:  
Since, AD   BC therefore, BD = DC. 
Now, in right triangle  ABD, we have 
AB
2
 = AD
2
 + BD
2 
---
 
[Pythagoras theorem] 
Since BD = DC therefore, BD = 
 
 
BC 
  AB
2
 = AD
2
 +  
 
 
   
 
 
  AB
2
 = AD
2
 + 
 
 
  
 
 
  AB
2
 = AD
2
 + 
  
 
 
 
  AB
2
 = AD
2
 + 
  
 
 
  ------ [Since, AB = BC = AC – sides of an equilateral triangle] 
  AB
2 
- 
    
 
 
  
 = AD
2 
  
   
 
 
 = AD
2
  
  3AB
2 
 =  4AD
2
 
                                                                                                                        
 
       
Triangles 
 
 
 
 
 
 
 
3) In the given figure, PQR is a right triangle right angled at Q & QS   PR. If PQ = 6cm and PS 
= 4cm, find the QS, RS & QR. 
Solution: 
Given: In PQR,   PQR= 90 . 
PQ = 6cm, PS = 4cm. 
To find: QS , RS & QR? 
 
Since,QS   PR, 
  PS = SR 
And PS =4cm ----- (given ) 
      SR = 4cm. 
PR = PS + SR = 4 + 4 =8 
 PQR is right angled triangle. 
  by Pythagoras theorem, 
PR
2
  =  PQ
2
 + QR
2
 
 QR
2 
= PR
2
 - PQ
2 
  QR
2
 = 8
2 
 - 6
2 
  QR
2
  = 64 -36 = 28 
  QR   =  2v  
 
QS
2 
 = PS   SR  -------- [ Since QS is perpendicular to PR] 
  QS
2 
 = 4   4  = 16  
  QS = 4cm. 
 
 
 
 
 
 
 
 
 
 
 
R 
Q 
P 
S 
6 
4 
                                                                                                                        
 
 
Triangles 
 
 
 
 
 
4) In the given figure,       and        If 
1
,
4
BX BC ? find the ratio of areas of  ?ABC 
and  ?YCX. 
Solution: 
 Since                   
  
            Similarly       ACX CXY ? ? ? 
 ?  ?ABC   ?YCX …….(AA – similarity criterion) 
                         
                  
2
2
2
) (
) (
?
?
?
?
?
?
? ?
?
?
XC
BC
XC
BC
YCX ar
ABC ar
 
     Since, 
1
4
BX BC ? ….. (Given)                
? ?
1
4
BC XC BX BC ? ? ? 
  ? 4BC – 4XC = BC      
            ? 3BC = 4XC        
                        ? 
4
3
BC
XC
?              
 Putting values in eq.( i ), we get 9 : 16
9
16
3
4
) (
) (
2
? ? ?
?
?
?
?
?
?
?
?
YCX ar
ABC ar
    
 
 
 
 
5) PQR is an isosceles triangle right angled at B. Two equilateral triangles are constructed on 
side QR and PR as shown in the given figure. Prove that area of ?QRS = 
1
2
area of ?PRT. 
Solution: 
 In ?PQR, 90 Q ?? and consider PQ = QR = x 
 ? PR
2
 = PQ
2
 + QR
2 
--- (Pythagoras theorem) 
 ? PR
2
 = 2 x
2
           ? PR = 2 x 
  
 
 
A 
B 
C 
X 
 
 
 
P Q 
R 
S 
T 
Page 4


                                                                                                                        
 
A 
C B 
D 
   
  
Triangles 
 
 
 
 
1)  Let  ABC  DEF and their areas be, respectively, 64cm
2 
and 121 cm
2. 
If EF=15.4 cm, find 
BC. 
Solution:           
Given:  ABC    DEF and A( ABC)=64cm
2 
& A( DEF)= 121cm
2 
and EF = 15.4cm. 
To find: BC = ? 
We know that, the ratio of the areas of two similar triangles is equal to the square of the ratio of  
their sides. 
  
       
       
  
    
 
    
 
  
i.e. 
  
   
  
    
 
      
 
 
   BC
2 
 = 
         
   
 = 
        
   
  = 125.44 
  BC = 11.2 cm 
 
2) In an equilateral triangle ABC, AD   BC. Prove that 3AB
2 
= 4AD
2
. 
Solution: 
Given:  ABC is an equilateral triangle and  AD   BC. 
To prove that:  3AB
2 
= 4AD
2 
Proof:  
Since, AD   BC therefore, BD = DC. 
Now, in right triangle  ABD, we have 
AB
2
 = AD
2
 + BD
2 
---
 
[Pythagoras theorem] 
Since BD = DC therefore, BD = 
 
 
BC 
  AB
2
 = AD
2
 +  
 
 
   
 
 
  AB
2
 = AD
2
 + 
 
 
  
 
 
  AB
2
 = AD
2
 + 
  
 
 
 
  AB
2
 = AD
2
 + 
  
 
 
  ------ [Since, AB = BC = AC – sides of an equilateral triangle] 
  AB
2 
- 
    
 
 
  
 = AD
2 
  
   
 
 
 = AD
2
  
  3AB
2 
 =  4AD
2
 
                                                                                                                        
 
       
Triangles 
 
 
 
 
 
 
 
3) In the given figure, PQR is a right triangle right angled at Q & QS   PR. If PQ = 6cm and PS 
= 4cm, find the QS, RS & QR. 
Solution: 
Given: In PQR,   PQR= 90 . 
PQ = 6cm, PS = 4cm. 
To find: QS , RS & QR? 
 
Since,QS   PR, 
  PS = SR 
And PS =4cm ----- (given ) 
      SR = 4cm. 
PR = PS + SR = 4 + 4 =8 
 PQR is right angled triangle. 
  by Pythagoras theorem, 
PR
2
  =  PQ
2
 + QR
2
 
 QR
2 
= PR
2
 - PQ
2 
  QR
2
 = 8
2 
 - 6
2 
  QR
2
  = 64 -36 = 28 
  QR   =  2v  
 
QS
2 
 = PS   SR  -------- [ Since QS is perpendicular to PR] 
  QS
2 
 = 4   4  = 16  
  QS = 4cm. 
 
 
 
 
 
 
 
 
 
 
 
R 
Q 
P 
S 
6 
4 
                                                                                                                        
 
 
Triangles 
 
 
 
 
 
4) In the given figure,       and        If 
1
,
4
BX BC ? find the ratio of areas of  ?ABC 
and  ?YCX. 
Solution: 
 Since                   
  
            Similarly       ACX CXY ? ? ? 
 ?  ?ABC   ?YCX …….(AA – similarity criterion) 
                         
                  
2
2
2
) (
) (
?
?
?
?
?
?
? ?
?
?
XC
BC
XC
BC
YCX ar
ABC ar
 
     Since, 
1
4
BX BC ? ….. (Given)                
? ?
1
4
BC XC BX BC ? ? ? 
  ? 4BC – 4XC = BC      
            ? 3BC = 4XC        
                        ? 
4
3
BC
XC
?              
 Putting values in eq.( i ), we get 9 : 16
9
16
3
4
) (
) (
2
? ? ?
?
?
?
?
?
?
?
?
YCX ar
ABC ar
    
 
 
 
 
5) PQR is an isosceles triangle right angled at B. Two equilateral triangles are constructed on 
side QR and PR as shown in the given figure. Prove that area of ?QRS = 
1
2
area of ?PRT. 
Solution: 
 In ?PQR, 90 Q ?? and consider PQ = QR = x 
 ? PR
2
 = PQ
2
 + QR
2 
--- (Pythagoras theorem) 
 ? PR
2
 = 2 x
2
           ? PR = 2 x 
  
 
 
A 
B 
C 
X 
 
 
 
P Q 
R 
S 
T 
                                                                                                                        
 
 
Triangles 
 
 
 
 
 
Area of equilateral triangle ?QRS = 
2
3
4
x 
Area of equilateral triangle ?PRT = 
? ?
? ?
2
2 33
2
44
PR x ? 
Therefore area of ?PRT = 
2
3
2
x
   
  
      =
2
2
2 3 2 3
2 2 4
x
x ??
                                                     
 
      
                               = 2 area of ?QRS 
 
6) The areas of two similar triangles are144 cm
2
 and 81 cm
2 
 respectively. If the median of the 
first triangle is 14.4 cm find the corresponding median of the other. 
Solution: 
 Let two triangles be LMN and XYZ, such that ?LMN   ?XYZ 
 
2
2
)] ( [
)] ( [
) (
) (
XYZ Median
LMN Median
XYZ ar
LMN ar
?
?
?
?
?
 
 
?
 
2
) (
4 . 14
81
144
?
?
?
?
?
?
?
?
?
?
XYZ Median
  
 
? (Median of ?XYZ)
2 
 = 
14.4 14.4 81
144
??
 
          Median = 10.8 cm. 
 
 
 
 
 
 
 
 
 
 
L 
N 
R 
M 
K P 
T S 
Page 5


                                                                                                                        
 
A 
C B 
D 
   
  
Triangles 
 
 
 
 
1)  Let  ABC  DEF and their areas be, respectively, 64cm
2 
and 121 cm
2. 
If EF=15.4 cm, find 
BC. 
Solution:           
Given:  ABC    DEF and A( ABC)=64cm
2 
& A( DEF)= 121cm
2 
and EF = 15.4cm. 
To find: BC = ? 
We know that, the ratio of the areas of two similar triangles is equal to the square of the ratio of  
their sides. 
  
       
       
  
    
 
    
 
  
i.e. 
  
   
  
    
 
      
 
 
   BC
2 
 = 
         
   
 = 
        
   
  = 125.44 
  BC = 11.2 cm 
 
2) In an equilateral triangle ABC, AD   BC. Prove that 3AB
2 
= 4AD
2
. 
Solution: 
Given:  ABC is an equilateral triangle and  AD   BC. 
To prove that:  3AB
2 
= 4AD
2 
Proof:  
Since, AD   BC therefore, BD = DC. 
Now, in right triangle  ABD, we have 
AB
2
 = AD
2
 + BD
2 
---
 
[Pythagoras theorem] 
Since BD = DC therefore, BD = 
 
 
BC 
  AB
2
 = AD
2
 +  
 
 
   
 
 
  AB
2
 = AD
2
 + 
 
 
  
 
 
  AB
2
 = AD
2
 + 
  
 
 
 
  AB
2
 = AD
2
 + 
  
 
 
  ------ [Since, AB = BC = AC – sides of an equilateral triangle] 
  AB
2 
- 
    
 
 
  
 = AD
2 
  
   
 
 
 = AD
2
  
  3AB
2 
 =  4AD
2
 
                                                                                                                        
 
       
Triangles 
 
 
 
 
 
 
 
3) In the given figure, PQR is a right triangle right angled at Q & QS   PR. If PQ = 6cm and PS 
= 4cm, find the QS, RS & QR. 
Solution: 
Given: In PQR,   PQR= 90 . 
PQ = 6cm, PS = 4cm. 
To find: QS , RS & QR? 
 
Since,QS   PR, 
  PS = SR 
And PS =4cm ----- (given ) 
      SR = 4cm. 
PR = PS + SR = 4 + 4 =8 
 PQR is right angled triangle. 
  by Pythagoras theorem, 
PR
2
  =  PQ
2
 + QR
2
 
 QR
2 
= PR
2
 - PQ
2 
  QR
2
 = 8
2 
 - 6
2 
  QR
2
  = 64 -36 = 28 
  QR   =  2v  
 
QS
2 
 = PS   SR  -------- [ Since QS is perpendicular to PR] 
  QS
2 
 = 4   4  = 16  
  QS = 4cm. 
 
 
 
 
 
 
 
 
 
 
 
R 
Q 
P 
S 
6 
4 
                                                                                                                        
 
 
Triangles 
 
 
 
 
 
4) In the given figure,       and        If 
1
,
4
BX BC ? find the ratio of areas of  ?ABC 
and  ?YCX. 
Solution: 
 Since                   
  
            Similarly       ACX CXY ? ? ? 
 ?  ?ABC   ?YCX …….(AA – similarity criterion) 
                         
                  
2
2
2
) (
) (
?
?
?
?
?
?
? ?
?
?
XC
BC
XC
BC
YCX ar
ABC ar
 
     Since, 
1
4
BX BC ? ….. (Given)                
? ?
1
4
BC XC BX BC ? ? ? 
  ? 4BC – 4XC = BC      
            ? 3BC = 4XC        
                        ? 
4
3
BC
XC
?              
 Putting values in eq.( i ), we get 9 : 16
9
16
3
4
) (
) (
2
? ? ?
?
?
?
?
?
?
?
?
YCX ar
ABC ar
    
 
 
 
 
5) PQR is an isosceles triangle right angled at B. Two equilateral triangles are constructed on 
side QR and PR as shown in the given figure. Prove that area of ?QRS = 
1
2
area of ?PRT. 
Solution: 
 In ?PQR, 90 Q ?? and consider PQ = QR = x 
 ? PR
2
 = PQ
2
 + QR
2 
--- (Pythagoras theorem) 
 ? PR
2
 = 2 x
2
           ? PR = 2 x 
  
 
 
A 
B 
C 
X 
 
 
 
P Q 
R 
S 
T 
                                                                                                                        
 
 
Triangles 
 
 
 
 
 
Area of equilateral triangle ?QRS = 
2
3
4
x 
Area of equilateral triangle ?PRT = 
? ?
? ?
2
2 33
2
44
PR x ? 
Therefore area of ?PRT = 
2
3
2
x
   
  
      =
2
2
2 3 2 3
2 2 4
x
x ??
                                                     
 
      
                               = 2 area of ?QRS 
 
6) The areas of two similar triangles are144 cm
2
 and 81 cm
2 
 respectively. If the median of the 
first triangle is 14.4 cm find the corresponding median of the other. 
Solution: 
 Let two triangles be LMN and XYZ, such that ?LMN   ?XYZ 
 
2
2
)] ( [
)] ( [
) (
) (
XYZ Median
LMN Median
XYZ ar
LMN ar
?
?
?
?
?
 
 
?
 
2
) (
4 . 14
81
144
?
?
?
?
?
?
?
?
?
?
XYZ Median
  
 
? (Median of ?XYZ)
2 
 = 
14.4 14.4 81
144
??
 
          Median = 10.8 cm. 
 
 
 
 
 
 
 
 
 
 
L 
N 
R 
M 
K P 
T S 
                                                                                                                        
 
 
Triangles 
 
 
 
 
7) The areas of two similar triangles are 144 cm
2
 and 81 cm
2
 respectively. If the altitude of 
bigger triangle is 4 cm, find the corresponding altitude of other triangle. 
Solution:           
Let two triangles be LMN and RST,     
2
2
) (
) (
RP
LK
RST ar
LMN ar
?
?
?
 
  
  
 
 ? 
2
2
81
144 4
LK
? 
? 
2
81 16
144
LK
?
?      
 ? LK = 3 cm.      
 
8) If the diagonals of a quadrilateral PQRS divide each other proportionally, then prove that it is 
a trapezium. 
Solution: 
In quad. PQRS, diagonals PR and QS intersect at M such that 
MP MQ
MR MS
? 
Draw MT   SR 
Proof:  In ?PSR, MT   SR 
PT PM
TS MR
?  ( i ) [ B.P. Theorem ]  
   
But 
PM QM
MR MS
?  ( ii ) 
Comparing eq. ( i ) and ( ii ), we get              
PT QM
TS MS
?  ?
 
MT   PQ  
Also MT   SR   PQ
 
  SR
     
 
Hence PQRS is a trapezium. 
 
 
 
 
 
P 
Q 
S R 
T 
M 
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