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,