RD Sharma 2020 solution class 9 chapter 13 Quadrilaterals MCQS

MCQS

Page-13.68

Question 1:

PQRS is a quadrilateral, PR and QS intersect each other at O. In which of the following cases, PQRS is a parallelogram?

(a) ∠P = 100°, ∠Q = 80°, ∠R = 95°

(b)  ∠P =85°, ∠Q = 85°, ∠R = 95°

(c) PQ = 7 cm, QR = 7 cm, RS = 8 cm, SP = 8 cm

(d) OP = 6.5 cm, OQ = 6.5 cm, OR = 5.2 cm, OS = 5.2 cm

Answer 1:

Let us analyze each case one by one.
We have a quadrilateral named PQRS, with diagonals PR and QS intersecting at O.
(a) ,,
By angle sum property of a quadrilateral, we get:

Clearly,
And
Thus we have PQRS a quadrilateral with opposite angles are equal.
Therefore,
PQRS is a parallelogram.
(b) ,,
By angle sum property of a quadrilateral, we get:

Clearly,
And
Thus we have PQRS a quadrilateral with opposite angles are not equal.
Therefore,
PQRS is not a parallelogram.
(c) ,,,
Clearly,
And
Thus we have PQRS a quadrilateral with opposite sides are not equal
Therefore,
PQRS is not a parallelogram.
(c) ,,,
We know that the diagonals of a parallelogram bisect each other.
But, here we have

And
Therefore,
PQRS is not a parallelogram. 
Hence, the correct choice is (a).

Question 2:

Mark the correct alternative in each of the following:
The opposite sides of a quadrilateral have

(a) no common point

(b) one common point

(c) two common points

(d) infinitely many common points

Answer 2:

We can look at a quadrilateral as:

The opposite sides of the above quadrilateral AB and CD have no point in common.
Hence the correct choice is (a).

Question 3:

The consecutive sides of a quadrilateral have

(a) no common point

(b) one common point

(c) two common points

(d) infinitely many common points

Answer 3:

We can look at a quadrilateral as:

The consecutive sides of the above quadrilateral AB and BC have one point in common.
Hence the correct choice is (b).

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Question 4:

Which  of the following quadrilateral is not a rhombus?

(a) All four sides are equal

(b) Diagonals bisect each other

(c) Diagonals bisect opposite angles

(d) One angle between the diagonals is 60°

Answer 4:

Let us consider the rhombus ABCD as:

We have the following properties of a rhombus:
All four sides are equal.
Diagonals bisect each other at right angles.
Hence the correct choice is (d).

Question 5:

Diagonals necessarily bisect opposite angles in a

(a) rectangle

(b) parallelogram

(c) isosceles trapezium

(d) square

Answer 5:

From the given choices, only in a square the diagonals bisect the opposite angles.
Let us prove it.
Take the following square ABCD with diagonal AD.

In and:
(Opposite sides of a square are equal.)
(Common)
(Opposite sides of a square are equal.)
Thus,
(By SSS Congruence Rule)
By Corresponding parts of congruent triangles property we have:


Therefore, in a square the diagonals bisect the opposite angles.
Hence the correct choice is (d).

Question 6:

The two diagonals are equal in a

(a) parallelogram

(b) rhombus

(c) rectangle

(d) trapezium

Answer 6:

Two diagonals are equal only in a rectangle.
This can be proved as follows:

The rectangle is given as ABCD, with the two diagonals as AD and BC.
In and:
(Opposite sides of a rectangle are equal.)
(Common)
(Each angle in a rectangle is a right angle)
Thus,
(By SAS Congruence Rule)
By Corresponding parts of congruent triangles property we have:

Therefore, in a rectangle the two diagonals are equal.
Hence the correct choice is (c).

Question 7:

We get a rhombus by joining the mid-points of the sides of a

(a) parallelogram

(b) rhombus

(c) rectangle

(d) triangle

Answer 7:

We get a rhombus by joining the mid-points of the sides of a rectangle.

It is given a rectangle ABCD in which P,Q,R and S are the mid-points AB,BC,CD and DA respectively.
PQ,QR,RS and SP are joined.
In , P and Q are the mid-points AB and BC respectively.
Therefore,
and ……(i)
Similarly, In , R and S are the mid-points CD and AD respectively.
Therefore,
and ……(ii)
From (i) and (ii), we get
and
Therefore, is a parallelogram. …… (iii)
Now is a rectangle.
Therefore,

…… (iv)
In and , we have:
(P is the mid point of AB)
(Each is a right angle)
(From equation (iv))
So, by SAS congruence criteria, we get:

By Corresponding parts of congruent triangles property we have:
…… (v)
From (iii) and (v) we obtain that is a parallelogram such that and
Thus, the two adjacent sides are equal.
Thus, is a rhombus.
Hence the correct choice is (c).

Question 8:

The bisectors of any two adjacent  angles of a parallelogram intersect at

(a) 30°

(b) 45°

(c) 60°

(d) 90°

Answer 8:

Let the figure be as follows:

ABCD is a parallelogram.
We have to find
AN and AD is the bisectors of and .
Therefore,
…… (i)
And
…… (ii)
We know that .
Therefore, the sum of consecutive interior angles must be supplementary.

From (i) and (ii), we get
…… (iii)
By angle sum property of a triangle:

Hence the correct choice is (d).

Question 9:

The bisectors of the angle of a parallelogram enclose a

(a) parallelogram

(b) rhombus

(c) rectangle

(d) square

Answer 9:

We have ABCD, a parallelogram given below:

Therefore, we have
Now, and transversal AB intersects them at A and B respectively. Therefore,
Sum of consecutive interior angle is supplementary. That is;

We have AR and BR as bisectors of and respectively.
…… (i)
Now, in , by angle sum property of a triangle, we get:

From equation (i), we get:

Similarly, we can prove that.
Now, and transversal ADintersects them at A and D respectively. Therefore,
Sum of consecutive interior angle is supplementary. That is;

We have AR and DP as bisectors of and respectively.
…… (ii)
Now, in , by angle sum property of a triangle, we get:

From equation (i), we get:

We know that and are vertically opposite angles, thus,

Similarly, we can prove that.
Therefore, PQRS is a rectangle.
Hence, the correct choice is (c).

Question 10:

The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a

(a) parallelogram

(b) rectangle

(c) square

(d) rhombus

Answer 10:

Figure can be drawn as:

Let ABCD be a quadrilateral such that P,Q,R and S are the mid-points of side AB,BC,CD and DA respectively.
In,P and Q are the mid-points of AB and BC respectively.
Therefore,
and
Similarly, we have
and
Thus,
and
Therefore, PQRSis a parallelogram.
The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a
parallelogram.
Hence the correct choice is (a).

Question 11:

The figure formed by joining the mid-points of the adjacent sides of a rectangle is a

(a) square

(b) rhombus

(c) trapezium

(d) none of these

Answer 11:

We get a rhombus by joining the mid-points of the sides of a rectangle.

It is given that rectangle ABCD in which P,Q,R and S are the mid-points AB,BC,CD and DA respectively.
PQ,QR,RS and SP are joined.
In , P and Q are the mid-points AB and BC respectively.
Therefore,
and ……(i)
Similarly, In , R and S are the mid-points CD and AD respectively.
Therefore,
and ……(ii)
From (i) and (ii), we get
and
Therefore, is a parallelogram. …… (iii)
Now is a rectangle.
Therefore,

…… (iv)
In and , we have:
(P is the mid point of AB)
(Each is a right angle)
(From equation (iv))
So, by SAS congruence criteria, we get:

By Corresponding parts of congruent triangles property we have:
…… (v)
From (iii) and (v) we obtain that is a parallelogram such that .
Thus, the two adjacent sides are equal.
Thus, is a rhombus .
Hence the correct choice is (b).

Question 12:

The figure formed by joining the mid-points of the adjacent sides of a rhombus is a

(a) square

(b) rectangle

(c) trapezium

(d) none of these

Answer 12:

Figure is given as :

A rhombus ABCD is given in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively.
In , P and Q are the mid-points AB and BC respectively.
Therefore,
and ……(i)
Similarly, In , R and S are the mid-points CD and AD respectively.
Therefore,
and ……(ii)
From (i) and (ii), we get
and
Therefore, is a parallelogram. …… (iii)
Now, we shall find one of the angles of a parallelogram.
Since ABCD is a rhombus
Therefore,
(Sides of rhombus are equal)

(P and Q are the mid-points AB and BC respectively)
In , we have

(Angle opposite to equal sides are equal)
Therefore, ABCD is a rhombus


…… (iii)
Also,


…… (iv)
Now, in and , we have
[From (iii)]
[From (iv)]
And ( is a parallelogram)
So by SSS criteria of congruence, we have

By Corresponding parts of congruent triangles property we have:
…… (v)
Now,
And
Therefore,

From (ii), we get
From (v), we get
Therefore, …… (vi)
Now, transversal PQ cuts parallel lines SP and RQ at P and Q respectively.

[Using (vi)]

Thus, is a parallelogram such that .
Therefore, is a rectangle.
Hence the correct choice is (b).

Question 13:

The figure formed by joining the mid-points of the adjacent sides of a square is a

(a) rhombus

(b) square

(c) rectangle

(d) parallelogram

Answer 13:

We get a square by joining the mid-points of the sides of a square.
It is given a square ABCD in which P, Q, R and S are the mid-points AB, BC, CD and DA respectively.
PQ, QR, RS and SP are joined.
Join AC and BD.

In , P and Q are the mid-points AB and BC respectively.
Therefore,
and ……(i)
Similarly, In ΔADC, R and S are the mid-points CD and AD respectively.
Therefore,
and ……(ii)
From (i) and (ii), we get
and
Therefore, is a parallelogram. …… (iii)
Now is a square.
Therefore,

…… (iv)
Similarly,
…… (v)
In and , we have:
(From equation (iv))
(Each is a right angle)
(From equation (v))
So, by SAS congruence criteria, we get:

By Corresponding parts of congruent triangles property we have:
…… (vi)
From (iii) and (vi) we obtain that is a parallelogram such that .
But, is a parallelogram.

So,
and …… (vii)
Now, (From equation (i))
Therefore,
…… (viii)
Since P and S are the mid-points AB and AD respectively

Therefore,
…… (ix)
Thus, in quadrilateral PMON, we have:

and (From equation (viii) and (ix))
Therefore, quadrilateral PMON is a parallelogram.
Also,

(Because )
(Because diagonals of a square are perpendicular)

Therefore, is a quadrilateral such that ,
and .Also, .
Hence, is a square.
Hence the correct choice is (b).

Question 14:

The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a

(a) rectangle

(b) parallelogram

(c) rhombus

(d) square

Answer 14:

It is given a parallelogram ABCD in which P,Q,R and S are the mid-points AB, BC, CD and DA respectively.
PQ, QR, RS and SP are joined.

In , P and Q are the mid-points AB and BC respectively.
Therefore,
and ……(i)
Similarly, In , R and S are the mid-points CD and AD respectively.
Therefore,
and ……(ii)
From (i) and (ii), we get
and
Therefore, is a parallelogram.
Hence the correct choice is (b).

Question 15:

If one angle of a parallelogram is 24° less than twice the smallest angle, then the measure of the largest angle of the parallelogram is

(a) 176°

(b) 68°

(c) 112°

(d) 102°

Answer 15:

Let the smallest angle of the parallelogram be
Therefore, according to the given statement other angle becomes.
Also, the opposite angles of a parallelogram are equal.
Therefore, the four angles become ,,and.
According to the angle sum property of a quadrilateral:


Thus, the other angle becomes

Thus, the largest angle of the parallelogram are is .
Hence the correct choice is (c).

Question 16:

In a parallelogram ABCD, if ∠DAB = 75° and ∠DBC = 60°, then ∠BDC =

(a) 75°

(b) 60°

(c) 45°

(d) 55°

Answer 16:

Parallelogram can be drawn as :

We know that the opposite angles of a parallelogram are equal.
Therefore,

By angle sum property of a triangle:


Thus, measures.
Hence the correct choice is (c).

Question 17:

ABCD is a parallelogram and E and F are the centroids of triangles ABD and BCD respectively, then EF =

(a) AE

(b) BE

(c) CE

(d) DE

Answer 17:

Parallelogram ABCD is given with E and F are the centroids of and.

We have to find EF.
We know that the diagonals of a parallelogram of bisect each other.
Thus, AC and BD bisect each other at point O.
Also, median is the line segment joining the vertex to the mid-point of the opposite side of the triangle. Therefore, the centroids E and F lie on AC.
Now, the centroid divides each median into two segments whose lengths are in the ratio 2:1, with the longest one nearest the vertex.
Then, in ΔABD, we get:

Or,
and …… (I)
Similarly, in ,we get:
and …… (II)
Also,


From (I) and (II), we get:

And …… (III)
Also, from (II) and (III), we get :
…… (IV)
Now, from (I),

From (IV), we get:

From(III):


Hence the correct choice is (a).

Question 18:

ABCD is a parallelogram, M is the mid-point of BD and BM bisects ∠B. Then ∠AMB =

(a) 45°

(b) 60°

(c) 90°

(d) 75°

Answer 18:

Figure is given as follows:

ABCD is a parallelogram.
It is given that :
BM bisects
Therefore,

But,
(Alternate interior opposite angles as)
Therefore,

In,

Sides opposite to equal angles are equal.
Thus,

Also,
(Opposite sides of a parallelogram are equal)
Thus,

ABCD is a parallelogram with
Therefore,
ABCD is a rhombus.
And we know that diagonals of the rhombus bisect each other at right angle.
Thus,
Hence the correct choice is (c).

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Question 19:

If an angle of a parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is

(a) 108°

(b) 54°

(c) 72°

(d) 81°

Answer 19:

Let one of the angle of the parallelogram as
Then the adjacent angle becomes
We know that the sum of adjacent angles of the parallelogram is supplementary.
Therefore,

Thus, the angle adjacent to

Therefore, the smallest angle of the parallelogram as
Hence, the correct choice is (c).

Question 20:

If the degree measures of the angles of quadrilateral are 4x, 7x, 9x and 10x, what is the sum of the measures of the smallest angle and largest angle?

(a) 140°

(b) 150°

(c) 168°

(d) 180°

Answer 20:

We have,
,
,

and
By angle sum property of a quadrilateral, we get:


Smallest angle:

Also, Largest angle:

Thus, the sum of the smallest and the largest becomes:

Hence, the correct choice is (c).

Question 21:

If the diagonals of a rhombus are 18 cm and 24 cm respectively, then its side is equal to

(a) 16 cm

(b) 15 cm

(c) 20 cm

(d) 17 cm

Answer 21:

Let ABCD be rhombus with diagonals AC and BD 18cm and 24cm respectively.

We know that diagonals of the rhombus bisect each other at right angles.
Therefore,

Similarly,

Also, is a right angled triangle.
By Pythagoras theorem, we get:


Hence the correct choice is (b).

Question 22:

ABCD is a parallelogram in which diagonal AC bisects ∠BAD. If ∠BAC = 35°, then ∠ABC =

(a) 70°

(b) 110°

(c) 90°

(d) 120°

Answer 22:

ABCD is a parallelogram in which AC bisects.

It is given that
Therefore,

Since,
Therefore,

Hence, the correct choice is (b).

Question 23:

In a rhombus ABCD, if ∠ACB = 40°, then ∠ADB =

(a) 70°

(b) 45°

(c) 50°

(d) 60°

Answer 23:

Rhombus ABCD is given as follows:

It is given that.
Therefore, (Because O lies on AC)
We know that the diagonals of a rhombus intersect at right angle.
Therefore,
By angle sum property of a triangle, we get:

Since, O lies on BD

Also ,
Therefore,

Hence, the correct choice is (c).

Question 24:

In ΔABC, ∠A = 30°, ∠B = 40° and ∠C = 110°. The angles of the triangle formed by joining the mid-points of the sides of this triangle are

(a) 70°, 70°, 40°

(b) 60°, 40°, 80°

(c) 30°, 40°, 110°

(d) 60°, 70°, 50°

Answer 24:

It is given that D, E and F be the mid-points of BC , CA and AB respectively.

Then,
, and .
Now, and transversal CB and CA intersect them at D and E respectively.
Therefore,

[(Given)]
and
[(Given)]
Similarly,
Therefore,

[(Given)]
and
[(Given)]
Similarly,
Therefore,

[ (Given)]
and
[ (Given)]
Now BC is a straight line.

Similarly,
and
Hence the correct choice is (c).

Question 25:

The diagonals of a parallelogram ABCD intersect at O. If ∠BOC = 90° and ∠BDC = 50°, then ∠OAB =

(a) 40°

(b) 50°

(c) 10°

(d) 90°

Answer 25:

ABCD is a parallelogram with diagonals AC and BD intersect at O.

It is given that and.
We need to find
Now,
(Linear pair)

Since, O lies on BD.
Therefore,

By angle sum property of a triangle, we get:

Since, O lies on AC.
Therefore,

Also,
Therefore,

Hence the correct choice is (a).

Question 26:

ABCD is a trapezium in which AB || DC. M and N are the mid-points of AD and the respectively. If AB = 12 cm, MN = 14 cm, then CD =

(a) 10 cm

(b) 12 cm

(c) 14 cm

(d) 16 cm

Answer 26:

The given trapezium ABCD can be drawn as follows:

Here, .
M and N are the mid-points of AD and BC respectively.
We have,.
We need to find CD.
Join MN to intersect AC at G.
We have and a line MN formed by joining the mid-points of sides AD and BC.
Thus, we can say that
In M is the mid-point of AD and
Therefore, G is the mid point of AC
By using the converse of mid-point theorem, we get:
…… (i)
In , N is the mid point of BC and
By using the converse of mid-point theorem, we get:
…… (ii)
Adding (i) and (ii),we get:

…… (iii)
On substituting and in (iii),we get:

Hence the correct choice is (d).

Question 27:

Diagonals of a quadrilateral ABCD bisect each other. If ∠A= 45°, then ∠B =

(i) 115°

(ii) 120°

(iii) 125°

(iv) 135°

Answer 27:

We know that the diagonals of a parallelogram bisect each other.
Thus, the given quadrilateral ABCD is a parallelogram.
and are consecutive interior angles, which must be supplementary.
Therefore, we have

Hence the correct choice is (d).

Question 28:

P is the mid-point of side BC of a parallelogram ABCD such that ∠BAP = ∠DAP. If AD = 10 cm, then CD =

(a) 5 cm

(b) 6 cm

(c) 8 cm

(d) 10 cm

Answer 28:

Parallelogram ABCD is given such that

We haveand
We need to find the measure of side CD.
Since and AP as transversal

But it is given that

Therefore, we get:

Also, sides opposite to equal angles are equal.
Then,
…… (I)
Also,
Substituting in (I), We get:

It is given that , this means opposite side ,as ABCD is a parallelogram. Therefore,

Or,
Hence the correct choice is (a).

Question 29:

In ΔABC, E is the mid-point of median AD such that BE produced meets AC at F. IF AC = 10.5 cm, then AF =

(a) 3 cm

(b) 3.5 cm

(c) 2.5 cm

(d) 5 cm

Answer 29:

is given with E as the mid-point of median AD.

Also, BE produced meets AC at F.
We have , then we need to find AF.
Through D draw .
In , E is the mid-point of AD and .
Using the converse of mid-point theorem, we get:
…… (i)
In , D is the mid-point of BC and .
…… (ii)
From (i) and (ii),we have:
, …… (iii)
Now,

From (iii) equation, we get:

…… (iv)
We have .Putting this in equation (iv), we get:

Hence the correct choice is (b).

Question 30:

ABCD is a parallelogram and E is the mid-point of BC. DE and AB when produced meet at F. Then, AF =

(a) $\frac{3}{2}AB$

(b) 2 AB

(c) 3 AB

(d) $\frac{5}{4}AB$

Answer 30:

Parallelogram ABCD is given with E as the mid-point of BC.

DE and AB when produced meet at F
We need to find AF.
Since ABCD is a parallelogram, then
Therefore,
Then, the alternate interior angles should be equal.
Thus, …… (I)
In and :
(From(I))
(E is the mid-point of BC)
(Vertically opposite angles)
(by ASA Congruence property)
We know that the corresponding angles of congruent triangles should be equal.
Therefore,

But,
(Opposite sides of a parallelogram are equal)
Therefore,
…… (II)
Now,

From (II),we get:

Hence the correct choice is (b).

Question 31:

In a quadrilateral ABCD, ∠A + ∠C is 2 times ∠B + ∠D. If ∠A = 140° and ∠D = 60°, then ∠B =

(a) 60°

(b) 80°

(c) 120°

(d) None of these

Answer 31:

ABCD is a quadrilateral, with.
By angle sum property of a quadrilateral we get:

But,we have

Then,

The two equations so formed cannot give us the value for with a given value of .
Hence the correct choice is (d).

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Question 32:

The diagonals  AC and BD of a rectangle ABCD intersect each other at P. If ∠ABD = 50°, then ∠DPC =

(a) 70°

(b) 90°

(c) 80°

(d) 100°

Answer 32:

Figure can be drawn as follows:

The diagonals AC and BD of rectangle ABCD intersect at P.
Also, it is given that
We need to find
It is given that
Therefore, (Because P lies on BD)
Also, diagonals of rectangle are equal and they bisect each other.
Therefore,

Thus, (Angles opposite to equal sides are equal)
[(Given)]
By angle sum property of a triangle:

But and are vertically opposite angles.
Therefore, we get:

[(Already proved)]
Hence the correct choice is (c).