Question 2: State giving reasons, whether the following pairs of triangles are congruent or Not.

i) and in which

ii) and in which

iii) and in which

iv) and in which

Answer:

i) and in which

Congruent by ASA

ii) and in which

iii) and in which

ASA Theorem applied, Triangles are Congruent.

iv) and in which

Applying Pythagoras theorem,

Hence Triangles are congruent by SAS

Question 3: In the adjoining figure, P in the mid point of AB and . Prove that:

Answer:

, (given)

(given)

(opposite angles)

Hence

Question 4: In the adjoining figure, . Prove that

Answer:

Given

Therefore, by ASA

Question 5: In the adjoining figure,

Answer:

In

and DA = CB

Therefore, by ASA,

Question 6: In the adjoining figure, ABC is a triangle in which .

Answer:

Take

is common and

Therefore, by ASA,

Question 7:

Answer:

By SSS,

Therefore

Question 8: In , it in given that in bisector of , meeting BC at D. Prove: i)

Answer:

(angles opposite equal sides of a triangle)

(angle bisector)

Therefore by ASA:

Hence

Question 9: In the adjoining figure, in such that . Prove that:

Answer:

Given

Therefore,

AB = AC (given)

AO is Common

Because

Therefore, SAS applies, Hence

Since

Question 10: In the adjoining figure,

Answer:

In

(alternate angles)

(alternate angles)

Hence,

Given is common.

Therefore,

Applying ASA,

Question 11: In the adjoining figure, ABCD as a square and CEB is an isosceles triangle in which EC = EB show:

Answer:

Given (Sides of a square)

(Sides of an isosceles triangle)

Hence SAS applies,

Therefore,

Question 12: Find the vales of x and y in each of the following cases:

Answers:

i) Given

(alternate angles)

Therefore, AAS applies Hence,

Therefore

ii) In

Therefore by SSS,

Hence,

iii) In

And PQ in common

Therefore, by SSS,

Hence

Therefore

Since,

Now Calculate,

iv) In

Hence, ASA applies, Therefore,

Hence

And

Question 13: In adjoining figure, the sides BA and CA of have been produced to D and E such that BA = AD and CA = AE Prove ED ∥ BC

Answer:

In

(Opposite Angles)

Hence

Hence

Question 14: Equilateral triangle ABC and ACE have been drawn on the sides AB and AC respectively of , as shown in the adjoining figure, prove: i)

Answer:

In

Hence Therefore,

Question 15: In a regular pentagon ABCDE, prove that is isosceles.

Answer:

Since ABCD in a regular pentagon, all sides are equal and all internal angles are .

In

Therefore

Hence Therefore, isosceles triangle

Question 16: In the adjoining figure, in a square, and are points on the side respectively such that,

Answer:

Given ABCD in a square

(given)

Similarly

Now Consider

(Square)

Question 17: In the adjoining figure,

Answer:

Since

And Similarly,

Given

Therefore, using AAS,

Question 18: In the adjoining figure ABCD is a square, , and R is mid point of EF. Prove:

Answer:

In ,

(Median would bisect the angle)

Common

In

(Side of a square)

(Proved above)

Because

Applying SAS, Proves that

Therefore, . Hence Proven.

Question 19: In the adjoining figure, in the Mid point of PQ Prove:

Answer:

In

(given)

(given)

Since

Question 20: In adjoining figure,

Answer:

Consider

is common

(given)

Hence,

Therefore,

Now Consider

is common

Therefore,