RD Chapter 14- Quadrilaterals Ex-14.1 |
RD Chapter 14- Quadrilaterals Ex-14.2 |
RD Chapter 14- Quadrilaterals Ex-14.4 |
RD Chapter 14- Quadrilaterals Ex-VSAQS |

**Answer
1** :

The parallelogram can be drawn as:

We have ,thus andare consecutive interior angles.

These must be supplementary.

Therefore,

**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 .

Hence the measure of other angles are , and .

**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 .

**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.

**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.

**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.

**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,

Hence proved.

**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.

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