### Quadrilaterals - Test Papers

CBSE Test Paper 01

CH-8 Quadrilaterals

- E Divides AB in the ratio 1 : 3 and also, F divides AC in the ratio 1 : 3. EF = 2.8cm, Find BC
- 11.2cm
- 11cm
- 11.5cm
- 12cm

The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if

ABCD is a Rhombus

Diagonals of ABCD are equal and perpendicular

Diagonals of ABCD are perpendicular

Diagonals of ABCD are equal

In fig if DE = 8 cm, DE BC and D is the mid-Point of AB, then the true statement is

E is the mid-Point of AC

AB = BC

DE = BC

DE and BC meet at some point if we extend both of them indefinitely.

In fig D is mid-point of AB and DE BC then AE is equal to

AD

DB

BC

EC

In a triangle ABC, P, Q and R are the mid-points of the sides BC, CA and AB respectively. If AC = 21cm, BC = 29cm and AB = 30cm, find the perimeter of the quadrilateral ARPQ?

20cm

80cm

51cm

52cm

- Fill in the blanks:
If in a parallelogram its diagonals bisect each other at right angles and are equal, then it is a ________.

- Fill in the blanks:
If the two non-parallel sides of a trapezium are equal, then it is called an ________ trapezium.

ABCD is a parallelogram. If its diagonals are equal, then find the value of ABC.

In a parallelogram PQRS, if P = (3x - 5)o and Q = (2x +15)o. Find the value of x.

In figure, B < A andC < D. Show that AD < BC.

If the angles of a triangle are in the ratio 1 : 2 : 3, determine three angles.

In ABC, if A = 120o and AB = AC. Find B and C.

Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

PQ and RS are two 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 line segments MN and PQ are equal and parallel to each other.

Show that the line segments joining the mid-points of opposite sides of a quadrilateral bisect each other.

CBSE Test Paper 01

CH-8 Quadrilaterals

Solution

- (a) 11.2cm

Explanation:

Let AE = x and EB = 3x , AF = y and FC = 3y.

EF = 2.8 cm

AE + AF = 2.8 implies x + y = 2.8

BC = CF + FA + AE + EB

= 3y + y + x + 3x

= 4 ( x + y ) = 4 ( 2.8 ) = 11.2 cm - (b) Diagonals of ABCD are equal and perpendicular

Explanation: A quadrilateral formed by joining the mid points of a square is a square. So, ABCD is a square. In Square, diagonals are equal and perpendicular. - (a) E is the mid-Point of AC

Explanation: By the converse of Mid Point Theorem, which states that," If a line segment is drawn passing through the midpoint of any one side of a triangle and parallel to another side, then the line segment bisects the remaining third side. - (d) EC

Explanation: By midpoint theorem of a triangle E is the midpoint of AC, hence AE = EC - (c) 51cm

Explanation:

Given:

A B C is a triangle .P Q R are the mid points of sides BC , CA nad AB .

AC = 21 cm

BC = 29 cm

AB = 30 cm

To find :

perimeter of quadrilateral ARQP ?

Q is the mid point of AC

P is the mid point of BC

QP is parallel to AB

QP = half of AB ( according to mid point theorem )

AB = 30 cm , QP = 15 CM ( QP is half of BA ) ( proved above )

R is the mid point of side AB

QP is also parallel to AR ( half of side AB )

PR is parallel to AC

PR = half of AC (according to mid point theorem )

AC = 21 cm , PR = 10.5 cm ( PR is half of AC ) ( proved above )

PR is parallel to AQ ( AQ is half of AC )

Since , in quadrilateral ARQP both the opposite sides are parallel it is a parallelogram.

Therefore , ARQP is a parallelogram .

WE know that

In parallelogram , opp sides are equal .

Therefore ,

PR = AQ = 10.5 cm

QP = AR = 15 cm

10.5 cm + 10.5 cm + 15 cm + 15 cm = 51 cm .

Therefore the perimeter of quadrilateral ARQP = 51 cm .

Square

isosceles

As diagonals of the parallelogram ABCD are equal, hence it is a rectangle. We know that, each angle of the rectangle is 90° .

ABC = 90o(Angles on the same side of a transversal are supplementary)

5x + 10 = 180o

5x + 170o x = 34oIn AOB,

B < A [Given]

OA < OB ...(i) [Side opposite to greater angle is longer]

In COD,

C < D [Given]

OD < OC ...(ii) [Side opposite to greater angle is longer]

Adding eq. (i) and (ii),

OA + OD < OB + OC

AD < BCLet the angles be x, 2x, 3x

x + 2x + 3x = 180o (sum of all angles of a triangle)

6x = 180o

x = 30o

Since x = 30o

2x = 2 30o = 60o

3x = 3 30o = 90o

angles are 30o, 60o, 90o

In ABC

AB = AC

= x [Angle opposite to equal sides are equal]

Now in ABC,

120+ x + x = 180

2x = 180 - 120

x = 30ABCD is a quadrilateral P, Q, R and S are the mid-points of the sides DC, CB, BA and AD respectively.

To prove : PR and QS bisect each other.

Construction : Join PQ, QR, RS, SP, AC and BD.

Proof : In ABC,

As R and Q are the mid-points of AB and BC respectively.

RQ || AC and RQ = AC

Similarly, we can show that

PS || AC and PS = AC

RQ || PS and RQ = PS.

Thus a pair of opposite sides of a quadrilateral PQRS are parallel and equal.

PQRS is a parallelogram.

Since the diagonals of a parallelogram bisect each other.

PR and QS bisect each other.__Given__: PQ = RS, PQ RS, PN QM, RN MS__To prove:__MN = PQ, MN PQ__Proof:__Since PQ = RS and PQ RS, therefore PQSR is a parallelogram.

PR = QS, PR QS

Since PN QM and MN is the transversal, we have

1 = 3 (Corresponding angles) ......................................(i)

Similarly, RN MS, we have

2 = 4 ...........................................(ii)

Adding (i) and (ii), we obtain

1 + 2 = 3 + 4 i.e., PNR - QMS

Again, PRS = QSX ........(Corresponding angles as PR QS)

and 6 = 5 .......................(Corresponding angles as RN SM)

Subtracting the above two equations, we get

PRS - 6 = QSX - 5 i.e., PRN = QSM

Now, in PNR and QMS,

PR - QS (Opp. sides of gm)

PNR = QMS (Proved above)

PRN = QSM

Therefore, By AAS congruence criterion, we havePNR = QMS

PN = QM (CPCT)

Also, PN QM (Given)

Therefore, PNMQ is a parallelogram.

PQ MN and PQ = MN.Given: A quadrilateral ABCD in which EG and FH are the line-segments joining the mid-points of opposite sides of a quadrilateral.

To prove: EG and FH bisect each other.

Construction: Join AC, EF, FG, GH and HE.

Proof: In ABC, E and F are the mid-points of respective sides AB and BC.

EF || AC and EF = AC ……….(i)

Similarly, in ADC,

G and H are the mid-points of respective sides CD and AD.

HG || AC and HG = AC ……….(ii)

From eq. (i) and (ii), we get,

EF || HG and EF = HG

EFGH is a parallelogram.

Since the diagonals of a parallelogram bisect each other, therefore line segments (i.e. diagonals) EG and FH (of parallelogram EFGH) bisect each other.

Hence, Proved.