Given below are the **Class 9 Maths** Extra Questions for Quadrilaterals

a. True and False

b. Hard problems

c. Multiple choice questions

d. Long answer questions

e. Fill in the blanks

a. True and False

b. Hard problems

c. Multiple choice questions

d. Long answer questions

e. Fill in the blanks

- The diagonals of a parallelogram bisect each other
- In a parallelogram, opposite sides and angle are equal.
- A diagonal of a parallelogram divides it into two congruent triangles
- The bisectors of the angles of parallelogram create a rectangle
- Sum of all the internal angles is 360
^{0} - Sum of all the exterior angles is 180
^{0} - Square, rectangle and rhombus are all parallelogram
- Consecutive angles are supplementary

- True. It is by definition
- True. It is by definition
- True. This can be proved easily using SSS congruence
- True
- True. This can easily proved by drawing one diagonal and summing all the angles based on triangle angle sum
- False
- True
- True

- All the angles of the quadrilateral are obtuse
- Diagonal of the rhombus are equal and perpendicular to each other
- Diagonal of the square are equal and bisect each other at right angle
- Out of four points A,B,C,D in place, there are collinear. A quadrilateral can be formed from these points
- Trapezium, in which the sides that are not parallel are equal in length and angles formed by parallel sides are equal, such trapezium is called isosceles trapezium
- In a parallelogram, diagonal bisect the angles

- False
- false
- True
- False
- True
- True

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD

Which of the following is true based on given information

a. $AP=CQ$

b. $QD=PB$

c. $DP=QB$

d. $\Delta PAD \cong \Delta QCB$

All are correct

In Triangle $ \Delta PAD \; and \; \Delta QCB$

AD=CB

$\angle P= \angle Q=90^0$

$\angle CBQ=\angle ADP$ ( Alternate interior angles of AB||CD)

So AAS congruence

Also as they are congruent, we get AP=CQ and DP=QB

Now Lets see the triangles $\Delta APB \; and \; \Delta CQD$

AB=CD

$\angle P= \angle Q=90^0$ (Alternate interior angles of AB||CD)

So QD=PB

The angles of the quadrilateral are in the ratio 2:5:4:1?

Which of the following is true?

a. Largest angle in the quadrilateral is $150^0$

b. Smallest angle is $30^0$

c.The second largest angle in the quadrilateral is $80^0$

d.None of these

(a) and (b)

Angles are $2x, 5x,4x,x$

Now

$2x+5x+4x+x=360$

Or $x=30$

Angles are $30^0, 60^0,120^0,150^0$

Two adjacent angles in a parallelogram are in the ratio 2:4. Find the values?

a. 80,100

b. 40,140

c. 60,120

d. None of the above

(c)

Adjacent angles

$2x+4x=180$

$x=30$

60,120 are adjacent angles

ABCD is a trapezium with AB =10cm, AD=5 cm, BC=4 cm and DC =7 cm?

Find the area of the ABCD

a. 34 cm

b. 28cm

c. 20 cm

d. None of these

(a)

BC is the altitude between the two parallel sides AB and DC

So Area of trapezium will be given by

$A=\frac {1}{2}BC(AB+DC)=34cm^2$

PQRS is a quadrilateral whose diagonal bisect each other at right angles

a. PQRS is a Square

b. PQRS is a rectangle

c. PQRS is a rhombus

d. None of these

(c)

ABCD is a trapezium where AB||DC. BD is the diagonal and E is the mid point of AD. A line is draw from point E parallel to AB intersecting BC at F. Which of these is true?

a. BF=FC

b. EA=FB

c. CF=DE

d. None of these

(a)

Let’s call the point of intersection at diagonal as G

Then in triangle DAB

EG||AB and E is the mid point of DA,So by converse of mid point theorem,

G is the mid point of BD

Now in triangle DBC

GF||CD

G is the mid point of DB

So by converse of mid point theorem

F is the mid point of BC

ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively

a. PQRS is a rectangle

b. PQRS is a parallelogram

c. PQRS is a rhombus

d. None of these

(c)

In a parallelogram PQRS, The bisector of angles P and Q meet at point O as shown in below figure. What is the angle O?

a. 80

b. 90

c. 45

d. None of the above

(b)

Rhombus |
Is a quadrilateral with only one pair of parallel sides. |

Rectangles |
Is a quadrilateral whose Two pairs of adjacent sides of a kite are equal, and one pair of opposite angles is equal. Diagonals intersect at right angles. One diagonal is bisected by the other. |

Kite |
Is a quadrilateral whose all the sides are equal and opposite sides are parallel. Opposite angles are equal. |

Right-angled trapezoid |
Is a quadrilateral whose opposite sides of a rectangle are parallel and equal. All angles are 90º. |

Isosceles trapezoid |
A trapezoid having two right angles |

Trapezoid |
Is a trapezoid whose non parallel sides are equal |

Show that the quadrilateral formed by joining the mid- points of adjacent sides of rectangle is a rhombus.

The figure is shown as below

To Prove: quadrilateral PQRS is a rhombus

Proof:

In Δ ABC

P and Q are mid points of sides AB and BC

By Mid point theorem

PQ= 1/2 AC and PQ || AC -(X)

In Δ ACD

S and R are mid points of sides AD and DC

By Mid point theorem

SR= 1/2 AC and SR || AC --(Y)

From (X) and (Y), we have

PQ=SR

PQ ||SR

Hence PQSR is a parallelogram

Now in Δ BCD

Q and R are mid points of sides BC and DC

By Mid point theorem

QR= 1/2 BD

Nowe AC = BC

Hence

PQ=SR=QR

Now a parallelogram whose adjacent sides are equal is a rhombus.

Hence proved

P, Q, R and S are respectively the mid- point of sides AB, BC, CD and DA of a quadrilateral ABCD such that AC=BD. Prove that PQRS is a rhombus.

Given AC=BD

Proof:

In Δ ABC

P and Q are mid points of sides AB and BC

By Mid point theorem

PQ= 1/2 AC and PQ || AC -(X)

In Δ ACD

S and R are mid points of sides AD and DC

By Mid point theorem

SR= 1/2 AC and SR || AC --(Y)

From (X) and (Y), we have

PQ=SR= 1/2 AC ----(1)

Similarly in Δ BCD

Q and R are mid point of BC and CD

By Mid point theorem

QR= 1/2 BD

Similarly in Δ ADB

S and P are mid point of AD and AB

By Mid point theorem

SP= 1/2 BD

Therefore

SP=QR=1/2 BD ---(2)

AC=BD

So from (1) and (2)

PQ=SR=SP=QR

Hence PQRS is a rhombus

Prove that the quadrilateral formed (if possible) by the internal angular bisectors of any quadrilateral is cyclic.

PQ and RS are to equal and parallel line segments. Any point M not lying on PQ or RS is joined to Q and S and lines through P parallel to QM and through R parallel to SM meet at N. Prove that the line segments MN and PQ are equal and parallel to each other.

l, m and n are three parallel lines intersected by transversal's p and q such that l, m and n cut off equal intercepts AB and BC on p. Show that l, m and n cut off equal intercepts DE and EF on q also.

Given

AB=BC

To Prove:

DE=EF

Proof:

Let us join A to F intersecting m at G

The trapezium ACFD is divided into two triangles namely Δ ACF and &Delta AFD

In Δ ACF, it is given that B is the mid-point of AC (AB = BC)

and BG || CF (since m || n).

So, G is the mid-point of AF (by Mid Point Theorem)

Now, in Δ AFD, we can apply the same argument as G is the mid-point of AF,

GE || AD and so by by Mid Point Theorem, E is the mid-point of DF,

i.e., DE = EF.

In other words, l, m and n cut off equal intercepts on q also.

In a quadrilateral ABCD, CO and DO are bisectors of $\angle C$ and $\angle D$ respectively. Prove that $\angle COD = \frac {1}{2} (\angle A + \angle B)$

ABC is a triangle. D is a point on AB such that AD = ¼ AB and E is the point on A such that AE = ¼ AC. Prove that DE = ¼ BC.

If the diagonals of a parallelogram are equal, then show that it is a rectangle.

In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angles.

Show that the line segment joining the mid- point of any two sides of a triangle is parallel to third side and is equal to half of it.

A diagonal of a rectangle is inclined to one side of a rectangle at $25^0$. Find the acute angles between the rectangle diagonals.

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.

ABCD is a parallelogram P is a point on AD such that AP = 1/3 AD and Q is a point on BC such that CQ = 1/3 BC. Prove that AQCP is a parallelogram.

P is the mid- point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R. Prove that

- AR = 2BC
- BR = 2BQ.

ABCD is a kite having AB = AD and BC = CD. Prove that the figure formed by joining the mid- points of the side, in order, is a rectangle.

This Quadrilaterals class 9 extra questions with answers is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.

**Notes****Assignments & NCERT Solutions**

Class 9 Maths Class 9 Science