RD SHARMA SOLUTION CHAPTER -12 Congruent Triangles| CLASS 9TH MATHEMATICS-EDUGROWN

Chapter 12 – Congruent Triangles Exercise Ex. 12.1

Question 1

In fig., the sides BA and CA have been produced such that BA = AD and CA = AE.

Prove that segment DE || BC

Solution 1

Question 2

Solution 2

Question 3

Prove that the medians of an equilateral triangle are equal.Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

The vertical angle of an isosceles triangle is 100^o. Find its base angles.Solution 6

Question 7

 In fig., AB = Ac and ∠ACD = 105°, find ∠BAC. 

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

In fig., AB =AC and DB = DC, find the ratio ∠ABD = ∠ACD. 

Solution 10

Question 11

Determine the measure of each of the equal angles of a right-angled isosceles triangle.

                                                                                   OR

ABC is a right-angled triangle in which A = 90^o and AB = AC. Find B and C.Solution 11

Question 12

Solution 12

Question 13

AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (See fig.). Show that the line PQ is perpendicular bisector of AB.

Solution 13

Chapter 12 – Congruent Triangles Exercise Ex. 12.2

Question 1

Solution 1

Question 2

In fig., it is given RT = TS, ∠1 = 2∠2 and ∠4 = 2∠3 prove that ΔRBT ≅ ΔSAT.

Solution 2

Question 3

Solution 3

Chapter 12 – Congruent Triangles Exercise Ex. 12.3

Question 1

In two right triangles one side and acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.Solution 1

Let ABC and DEF be two right triangles.

Question 2

Solution 2

Question 3

Solution 3

Question 4Show that the angles of an equilateral triangle are 60^o each.Solution 4

 Let us consider that ABC is an equilateral triangle.
So, AB = BC = AC
Now, AB = AC⇒ ∠C = ∠B         (angles opposite to equal sides of a triangle are equal)

We also have
AC = BC    
⇒ ∠B = ∠A             (angles opposite to equal sides of a triangle are equal)

So, we have
∠A = ∠B = ∠C
    Now, in ΔABC
∠A + ∠B + ∠C = 180^o
⇒ ∠A + ∠A + ∠A = 180^o
⇒ 3∠A = 180^o
⇒ ∠A = 60^o
⇒ ∠A = ∠B = ∠C = 60^o
Hence, in an equilateral triangle all interior angles are of 60^o.

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 12 – Congruent Triangles Exercise Ex. 12.4

Question 1

In fig., it is given that Ab = CD and AD = BC. prove that ΔADC ≅ ΔCBA

Solution 1

Question 2

Solution 2

Chapter 12 – Congruent Triangles Exercise Ex. 12.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

In fig., AD ⊥ CD and CB ⊥ CD. If AQ = BP an DP = CQ, prove that ∠DAQ = ∠CBP.

Solution 4

Question 5

Which of the following statements are True (T) and which are False (f):

(i) Sides opposite to equal angles of a triangle may be unequal.

(ii) Angles opposite to equal sides of a triangle are equal.

(iii) The measure of each angle of an equilaterial triangle is 60^o.

(iv) If the altitude from one vertex of a triangle bisects the opposite side, then the triangle may be isoscles.

(v) The bisectors of two equal angles of a traingle are equal.

(vi) If the bisector of the vertical angle of a triangle bisects the base, then the triangle may be isosceles.

(vii) The two altitudes corresponding to two equal sides of a triangle need not be equal.

(viii) If any two sides of a right triangle are respectively equal to two sides of other right triagnle, then the two triangles are congruent.

(ix) Two right triangles are congruent if hypotenuse and a side of one triangle are respectively equal to the hypotenuse and a side of the other triangle.Solution 5

(i) False

(ii) True

(iii) True

(iv) False

(v) True

(vi) False

(vii) False

(viii) False

(ix) TrueQuestion 6

Solution 6

(i) equal

(ii) equal

(iii) equal

(iv) BC

(v) AC

(vi) equal to

(vii) EFDQuestion 7

Solution 7

Chapter 12 – Congruent Triangles Exercise Ex. 12.6

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Is it possible to draw a triangle with sides of length 2cm, 3cm and 7 cm?Solution 4

Here, 2 + 3 < 7

Hence, it is not possible because triangle can be drawn only if the sum of any two sides is greater than third side.Question 5

Solution 5

Question 6

Solution 6

Question 7

In fig., prove that:

i. CD + DA + AB + BC > 2AC

ii. CD + DA + AB > BC

Solution 7

Question 8

Which of the following statements 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 8

(i) False

(ii) True

(iii) True

(iv) False

(v) True

(vi) TrueQuestion 9

Solution 9

(i) largest

(ii) less

(iii) greater

(iv) smaller

(v) less

(vi) greaterQuestion 10

Solution 10

Question 11

Solution 11