Quadrilaterals - Test Papers

 CBSE Test Paper 01

CH-8 Quadrilaterals

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1. E Divides AB in the ratio 1 : 3 and also, F divides AC in the ratio 1 : 3. EF = 2.8cm, Find BC
   1. 11.2cm
   2. 11cm
   3. 11.5cm
   4. 12cm

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

   1. ABCD is a Rhombus

   2. Diagonals of ABCD are equal and perpendicular

   3. Diagonals of ABCD are perpendicular

   4. Diagonals of ABCD are equal

3. In fig if DE = 8 cm, DE ∥$\parallel$ BC and D is the mid-Point of AB, then the true statement is
   

   1. E is the mid-Point of AC

   2. AB = BC

   3. DE = BC

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

4. In fig D is mid-point of AB and DE ∥$\parallel$ BC then AE is equal to
   

   1. AD

   2. DB

   3. BC

   4. EC

5. 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?

   1.   20cm

   2.  80cm

   3.  51cm

   4.   52cm

6. Fill in the blanks:

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

7. Fill in the blanks:

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

8. ABCD is a parallelogram. If its diagonals are equal, then find the value of ∠$\mathrm{\angle }$ABC.

9. In a parallelogram PQRS, if ∠$\mathrm{\angle }$P = (3x - 5)o and ∠$\mathrm{\angle }$Q = (2x +15)o. Find the value of x.

10. In figure, ∠$\mathrm{\angle }$B < ∠$\mathrm{\angle }$A and∠$\mathrm{\angle }$C < ∠$\mathrm{\angle }$D. Show that AD < BC.
    

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

12. In △$\mathrm{△}$ABC, if ∠$\mathrm{\angle }$A = 120o and AB = AC. Find ∠$\mathrm{\angle }$B and ∠$\mathrm{\angle }$C.

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

14. 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.

15. 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

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Solution

1. (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
2. (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.
3. (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.
4. (d) EC
   Explanation: By midpoint theorem of a triangle E is the midpoint of AC, hence AE = EC
5. (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 .

6. Square

7. isosceles

8. As diagonals of the parallelogram ABCD are equal, hence it is a rectangle. We know that, each angle of the rectangle is 90° .
   ∴$\therefore$ ∠$\mathrm{\angle }$ABC = 90o 

9. ∠P+∠Q=180∘$\mathrm{\angle }P+\mathrm{\angle }Q={180}^{\circ }$ (Angles on the same side of a transversal are supplementary)
   ⇒ 3x−5+2x+15=180∘$\Rightarrow 3x-5+2x+15={180}^{\circ }$
   5x + 10 = 180o
   ⇒$\Rightarrow$ 5x + 170o ⇒$\Rightarrow$ x = 34o

10. In △$\mathrm{△}$AOB,
    ∠$\mathrm{\angle }$B <∠$\mathrm{\angle }$ A [Given]
    ⇒$\Rightarrow$OA < OB ...(i) [Side opposite to greater angle is longer]
    InΔ$\mathrm{\Delta }$ COD,
    ∠$\mathrm{\angle }$C < ∠$\mathrm{\angle }$D [Given]
    ⇒$\Rightarrow$OD < OC ...(ii) [Side opposite to greater angle is longer]
    Adding eq. (i) and (ii),
    OA + OD < OB + OC
    ⇒$\Rightarrow$AD < BC

11. Let the angles be x, 2x, 3x
    ∴$\therefore$ x + 2x + 3x = 180o (sum of all angles of a triangle)
    ⇒$\Rightarrow$ 6x = 180o
    ⇒$\Rightarrow$ x = 30o
    Since x = 30o
    2x = 2 ×$\times$ 30o = 60o
    3x = 3 ×$\times$ 30o = 90o
    ∴$\therefore$ angles are 30o, 60o, 90o

12. 
    In △$\mathrm{△}$ABC
    ∵$\because$ AB = AC
    ⇒∠B=∠C$\Rightarrow \mathrm{\angle }B=\mathrm{\angle }C$ = x [Angle opposite to equal sides are equal]
    Now in △$\mathrm{△}$ABC,
    ∠A+∠B+∠C=180∘$\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }$
    ⇒$\Rightarrow$120∘$\circ$+ x + x = 180∘$\circ$
    ⇒$\Rightarrow$ 2x = 180∘$\circ$ - 120∘$\circ$
    ⇒2x=60∘$\Rightarrow 2x={60}^{\circ }$
    ⇒$\Rightarrow$ x = 30∘$\circ$
    ⇒∠B=∠C=30∘$\Rightarrow \mathrm{\angle }B=\mathrm{\angle }C={30}^{\circ }$

13. ABCD 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 △$\mathrm{△}$ABC,
    As R and Q are the mid-points of AB and BC respectively.
    ∴$\therefore$ RQ || AC and RQ = 12$\frac{1}{2}$ AC
    Similarly, we can show that
    PS || AC and PS = 12$\frac{1}{2}$ AC
    ∴$\therefore$ RQ || PS and RQ = PS.
    Thus a pair of opposite sides of a quadrilateral PQRS are parallel and equal.
    ∴$\therefore$ PQRS is a parallelogram.
    Since the diagonals of a parallelogram bisect each other.
    ∴$\therefore$ PR and QS bisect each other.

14. Given: PQ = RS, PQ ∥$\parallel$ RS, PN ∥$\parallel$ QM, RN ∥$\parallel$ MS
    To prove: MN = PQ, MN ∥$\parallel$ PQ
    
    Proof: Since PQ = RS and PQ ∥$\parallel$ RS, therefore PQSR is a parallelogram.
    ⇒$\Rightarrow$ PR = QS, PR ∥$\parallel$ QS
    Since PN ∥$\parallel$ QM and MN is the transversal, we have
    ∠$\mathrm{\angle }$1 = ∠$\mathrm{\angle }$3   (Corresponding angles) ......................................(i)
    Similarly, RN ∥$\parallel$ MS, we have
    ∠$\mathrm{\angle }$2 = ∠$\mathrm{\angle }$4                         ...........................................(ii)
    Adding (i) and (ii), we obtain
    ∠$\mathrm{\angle }$1 + ∠$\mathrm{\angle }$2 = ∠$\mathrm{\angle }$3 + ∠$\mathrm{\angle }$4 i.e., ∠$\mathrm{\angle }$PNR - ∠$\mathrm{\angle }$QMS
    Again, ∠$\mathrm{\angle }$PRS = ∠$\mathrm{\angle }$QSX ........(Corresponding angles as PR ∥$\parallel$ QS)
    and ∠$\mathrm{\angle }$6 = ∠$\mathrm{\angle }$5 .......................(Corresponding angles as RN ∥$\parallel$ SM)
    Subtracting the above two equations, we get
    ∠$\mathrm{\angle }$PRS - ∠$\mathrm{\angle }$6 = ∠$\mathrm{\angle }$QSX - ∠$\mathrm{\angle }$5 i.e., ∠$\mathrm{\angle }$PRN = ∠$\mathrm{\angle }$QSM
    Now, in ∠$\mathrm{\angle }$PNR and ∠$\mathrm{\angle }$QMS,
    PR - QS (Opp. sides of ∥$\parallel$gm)
    ∠$\mathrm{\angle }$PNR = ∠$\mathrm{\angle }$QMS (Proved above)
    ∠$\mathrm{\angle }$PRN = ∠$\mathrm{\angle }$QSM
    Therefore, By AAS congruence criterion, we have 

    △$\mathrm{△}$PNR = △$\mathrm{△}$QMS 
    ⇒$\Rightarrow$ PN = QM (CPCT)
    Also, PN ∥$\parallel$ QM (Given)
    Therefore, PNMQ is a parallelogram.
    ⇒$\Rightarrow$ PQ ∥$\parallel$ MN and PQ = MN.

15. 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.
    ∴$\therefore$ EF || AC and EF = 12$\frac{1}{2}$ AC ……….(i)
    Similarly, in ADC,
    G and H are the mid-points of respective sides CD and AD.
    ∴$\therefore$HG || AC and HG = 12$\frac{1}{2}$ AC ……….(ii)
    From eq. (i) and (ii), we get,
    EF || HG and EF = HG
    ∴$\therefore$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.