Question 1: In a parallelogram , determine the sum of angles
and
.
Answer:
Since is a parallelogram, therefore
Since, and transversal
intersects them at
and
respectively.
Since the sum of the interior angles on the same side of the transversal is
Therefore
Question 2: In a parallelogram , if
, determine the measures of its other angles.
Answer:
Given
Also is a parallelogram.
Therefore and
and
Question 3: is a square.
and
intersect at
. State the measure of
.
Answer:
Consider and
is common
(side of a square)
(diagonals of a square bisect each other)
Therefore
Hence
Since
Question 4: is a rectangle with
. Determine
.
Answer:
Given is a rectangle with
Question 5: The sides and,
of a parallelogram
are bisected at
and
. Prove that
is a parallelogram.
Answer:
Given: is a parallelogram,
bisects
and
bisects
Therefore and
also
and
Therefore is a parallelogram.
Question 6: and
are the points of trisection of the diagonal
of parallelogram
. Prove
is parallel to
. Prove also that
bisects
.
Answer:
Given:
To prove: i) ii)
bisects
Since diagonals of a parallelogram bisects each other
and
… … … … … i)
Given … … … … … ii)
From i) and ii) we get
Now consider parallelogram
We know that diagonals of a parallelogram bisects each other
Hence is a parallelogram
Therefore
Question 7: is a square
and
are points on
and
respectively, such that
. Prove that
is a square.
Answer:
Given:
Since is a square
Now consider and
Therefore
Therefore and
Similarly, we can prove that
Therefore is a square
is a parallelogram
Question 8: is a rhombus,
is a straight line such that
. Prove that
and
when produced meet at right angles.
Answer:
To prove:
We know that the diagonals of a rhombus bisect each other and are perpendicular to each other.
Given: and
(sides of a rhombus)
In , since
is midpoint of
and
is midpoint of
The line joining the midpoints i.e
Similarly
Therefore in parallelogram and
Therefore
Hence proved.
Question 9: is a parallelogram,
is produced to
so that
and
produced meets
produced in
. Prove that
.
Answer:
To prove:
Given:
Consider and
In
Since
… … … … … i)
Also since
… … … … … i)
From i) and ii)
Therefore or
. Hence proved.