RD Chapter 14 Quadrilaterals Ex 14.3 Solutions
Question - 1 : - In a parallelogram ABCD, determine the sum of angles ∠C and ∠D.
Answer - 1 : -
The parallelogram can be drawn as:
These must be supplementary.
Therefore,
Question - 2 : - In a parallelogram ABCD, if ∠B = 135°, determine the measure of its other angles.
Answer - 2 : -
Since ABCD is a parallelogram with
.Opposite angles of a parallelogram are equal.
Therefore,
Also, let
Similarly,
We know that the sum of the angles of a quadrilateral is
.Answer - 3 : -
The figure can be drawn as follows:
In
and 
, 
(Sides of a square are equal) 
(Diagonals of a parallelogram bisect each other) 
(Common) So, by SSS Congruence rule, we have
Also,
(Corresponding parts of congruent triangles are equal)
But,
(Linear pairs)We have,
Hence, the required measure of
is 
.
Question - 4 : - ABCD is a rectangle with ∠ABD = 40°. Determine ∠DBC.
Answer - 4 : -
The rectangle is given as follows with
We have to find
.An angle of a rectangle is equal to
.Therefore,
Hence, the measure for
is 
.Answer - 5 : -
Figure is given as follows:
It is given that ABCD is a parallelogram.
E is the mid point of AB
Thus,
,
...... (i) Similarly,
……(ii)
From (i) and (ii)
Also,
Thus,
Therefore, EBFD is a parallelogram.
Question - 6 : - P and Q are the points of trisection of the diagonal BD of a parallelogram ABCD. Prove that CQ is parallel to AP. Prove also that AC bisects PQ.
Answer - 6 : -
Figure can be drawn as follows:
We have P and Q as the points of trisection of the diagonal BD of parallelogram ABCD.
We need to prove that AC bisects PQ. That is,
.Since diagonals of a parallelogram bisect each other.
Therefore, we get:
and P and Q as the points of trisection of the diagonal BD.
Therefore,
and 
Now,
and 
Thus,
AC bisects PQ.
Hence proved.
Question - 7 : - ABCD is a square E, F, G and H are points on AB, BC, CD and DA respectively. such that AE = BF = CG = DH. Prove that EFGH is a square.
Answer - 7 : -
Square ABCD is given:
E, F, G and H are the points on AB, BC, CD and DA respectively, such that :
We need to prove that EFGH is a square.
Say,
As sides of a square are equal. Then, we can also say that:
In
and
,we have: 
(Given) 
(Each equal to 90°) 
(Each equal to y ) By SAS Congruence criteria, we have:
Therefore, EH = EF
Similarly, EF= FG, FG= HG and HG= HE
Thus, HE=EF=FG=HG
Also,
and 
But,
and 
Therefore,
i.e.,
Similarly,
Thus, EFGH is a square.
Hence proved.
Question - 8 : - ABCD is a rhombus, EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced meet at right angles.
Answer - 8 : -
Rhombus ABCD is given:
We have
We need to prove that
We know that the diagonals of a rhombus bisect each other at right angle.
Therefore,
In
A and O are the mid-points of BE and BD respectively.By using mid-point theorem, we get:
Therefore,
In
A and O are the mid-points of BE and BD respectively.By using mid-point theorem, we get:
Therefore,
Thus, in quadrilateral DOCG,we have:
and 
Therefore, DOCG is a parallelogram.
Thus, opposite angles of a parallelogram should be equal.
Also, it is given that
Therefore,
Or,
Question - 9 : - ABCD is a parallelogram, AD is produced to E so that DE = DC and EC produced meets AB produced in F. Prove that BF = BC
Answer - 9 : -
ABCD is a parallelogram, AD produced to E such that .
Also , AB produced to F.
We need to prove that
In
, D and O are the mid-points of AE and AC respectively.By using Mid-point Theorem, we get:
Since, BD is a straight line and O lies on AC.
And, C lies on EF
Therefore,
…… (i)
Also,
is a parallelogram with 
.Thus,
In
and 
,we have:So, by ASA Congruence criterion, we have:
By corresponding parts of congruence triangles property, we get:
From (i) equation, we get:
Hence proved.