Chapter 10 Congruent Triangles RD Sharma Solutions Exercise 10.6 Class 9 Maths

Chapter 10 Congruent Triangles RD Sharma Solutions Exercise 10.6 Class 9 Maths

Chapter Name

RD Sharma Chapter 10 Congruent Triangles Exercise 10.6

Book Name

RD Sharma Mathematics for Class 10

Other Exercises

  • Exercise 10.1
  • Exercise 10.2
  • Exercise 10.3
  • Exercise 10.4
  • Exercise 10.5

Related Study

NCERT Solutions for Class 10 Maths

Exercise 10.6 Solutions

1. In ΔABC, if ∠A = 40° and ∠B = 60°. Determine the longest and shortest sides of the triangle. 

Solution

Given that in ΔABC, ∠A = 40° and ∠B = 60°
We have to find longest and shortest side 
We know that, 

Sum of angles in a triangle 180° 

2. In a ΔABC, if ∠B = ∠C = 45° , which is the longest side ?

Solution

Given that in ΔABC, 
∠B = ∠C = 45°
We have to find longest side 
We know that, 
Sum of angles in a triangle = 180°


3. In ΔABC, side AB is produced to D so that BD = BC. If ∠B = 60°, ∠A = 70° , prove that :
(i) AD > CD
(ii) AD > AC
Solution
Given that in ΔABC, side AB is produced to D So that BD = BC and ∠B = 60°, ∠A = 70°
We have to prove that 
(i) AD > CD 
(ii) AD > AC
First join C and D
Now, in ΔABC 

4. Is it possible to draw a triangle with sides of length 2cm, 3cm and 7cm ? 
Solution
Given length of sides are 2cm, 3cm and 7cm we have to check whether it is possible to draw a triangle with ten the given lengths of sides 
We know that, 
A triangle can be drawn only when the sum of any two sides is greater than the third side. 
So, let's check the rule. 
2 + 3 ⊁ 7 or 2 + 3 < 7
2 + 7 > 3
and 3 + 7 > 2
Here,  2 + 3 ⊁ 7 So, the triangle does not exit. 

5. O is any point in the interior of ΔABC. Prove that 
(i) AB + AC > OB + OC
(ii) AB + BC + CA > OA + QB + OC
(iii) OA + OB + OC > 1/2(AB + BC + CA)
Solution
Given that O is any point in the interior of ΔABC 
We have to prove 
(i) AB + AC > OB + OC
(ii) AB + BC + CA > OA + QB + OC
(iii) OA + OB + OC > 1/2(AB + BC + CA)
We know that, in a triangle the sum of any two sides is greater than the third side
So, we have 
In ΔABC 
AB + BC > AC
BC + AC > AB
AC + AB > BC
In ΔOBC 
OB + OC > BC ...(1)
In ΔOAC 
OA + OC > AC ...(2)
In ΔOAB
OA + OB > AB  ...(3)
Now, extend (or) produce BO to meet AC in D. 
Now, in ΔABD, we have 
AB + AD > BD 
⇒ AB + AD > BO + OD  ...(4) [∵ BD = BO + OD]
Similarly in ΔODC, we have 
OD + DC > OC ...(5)

6. Prove that the perimeter of a triangle is greater than the sum of its altitudes. 
Solution
Given A ΔABC in which AD ⊥ BC, BE ⊥ AC and CF ⊥ AB.
To prove : 
AD + BE + CF < AB + BC + AC
Figure : 
Proof: 
We know that of all the segments that can be drawn to a given line, from a point not lying on it, the perpendicular distance i.e., the perpendicular line segment is the shortest. 
Therefore,
AD ⊥ BC 

7. In Fig. 10.131, prove that : 
(i) CD + DA + AB + BC >2AC 
(ii) CD + DA + AB > BC 

Solution
Given to prove 
(i) CD + DA + AB + BC > 2AC 
(ii) CD + DA + AB > BC 
From the given figure, 
We know that, in a triangle sum of any two sides is greater than the third side.

(i) So, 
In ΔABC, we have 
AB + BC > AC    .....(1) 


8. Which of the following statement are true (T) and which are false (F)?
(i) Sum of the three sides of a triangle is less than the sum of its three altitudes.
(ii) Sum of any two sides of a triangle is greater than twice the median drawn to the third side.
(iii) Sum of any two sides of a triangle is greater than the third side.
(iv) Difference of any two sides of a triangle is equal to the third side.
(v) If two angles of a triangle are unequal, then the greater angle has the larger side opposite to it.
(vi) Of all the line segments that can be drawn from a point to a line not containing it, the perpendicular line segment is the shortest one. 

Solution

(i) False (F)
Reason: Sum of these sides of a triangle is greater than sum of its three altitudes

(ii) True (F)

(iii) True (T)

(iv) False (F)
Reason: The difference of any two sides of a triangle is less than third side.

(v) True (T)
Reason: The side opposite to greater angle is longer and smaller angle is shorter in a triangle

(vi) True (T)
Reason: The perpendicular distance is the shortest distance from a point to a line not containing it. 


9. Fill in the blanks to make the following statements true. 
(i) In a right triangle the hypotenuse is the ...... side.
(ii) The sum of three altitudes of a triangle is ..... than its perimeter.
(iii) The sum of any two sides of a triangle is .... than the third side.
(iv) If two angles of a triangle are unequal, then the smaller angle has the side opposite to it.
(v) Difference of any two sides of a triangle is than the third side.
(vi) If two sides of a triangle are unequal, then the larger side has .... angle opposite to it.

Solution

(i) In a right triangle the hypotenuse is the largest side.
Reason : Since a triangle can have only one right angle, other two angles must be less than 90°.
⇒ The right angle is the largest angle.
⇒ Hypotenuse is the largest side. 
(ii) The sum of three altitudes of a triangle is less than its perimeter.
(iii) The sum of any two sides of a triangle is greater than the third side.
(iv) If two angles of a triangle are unequal, then the smaller angle has the smaller side opposite to it.
(v) Difference of any two sides of a triangle is less than the third side.
(vi) If two sides of a triangle are unequal, then the larger side has greater angle opposite to it.

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