RS Aggarwal Solutions Class 9 Ex 9B

Q.1: In the adjoining figure ABCD is a trapezium in which ABDC and E is the midpoint of AD. A line segment EFAB meets BC at F. Show that F is the midpoint of BC.

                

                          https://lh3.googleusercontent.com/0fds2tcwiW1CdXTVugPCOOgwp32u5hYDQLmthUSSAyriOTESG_RayP3Cyz_hyFVxhljHasdWnyOp00WlrgtfL84jS7EXiRhKqfSWKBOe4yjqmSI1QxEcxmOkezWrgQTlJ7-Xr7DGlsMd0ulDWQ

Sol:                            

Join BD to cut EF at M

 https://lh3.googleusercontent.com/0fds2tcwiW1CdXTVugPCOOgwp32u5hYDQLmthUSSAyriOTESG_RayP3Cyz_hyFVxhljHasdWnyOp00WlrgtfL84jS7EXiRhKqfSWKBOe4yjqmSI1QxEcxmOkezWrgQTlJ7-Xr7DGlsMd0ulDWQ

Now, in ΔDAB, E is the midpoint of AD and EM || AB

M is the midpoint of BD. (By converse of midpoint theorem)

Again, in ΔBDC, M is the midpoint of BD and MF || DC

Therefore, F is the midpoint of BC (By converse of midpoint theorem)

 

Q.2: In the adjoining figure ABCD is a parallelogram in which E and F are the midpoints of AB and CD respectively. If GH is a line segment that cuts AD, EF and BC at G, P and H respectively, prove that GP = PH.

                                  C:\Users\user\Desktop\1.PNG

Sol:

In parallelogram ABCD, we have:

AD || BC and AB || DC

AD = BC and AB = DC

AB = AE + BE and DC = DF + FC

AE = BE = DF = FC

Now, DF = AE and OF || AE.

i.e., AEFD is a parallelogram.

AD || EF

Similarly, BEFC is also a parallelogram.

EF || BC and AD || EF || BC

Thus, AD, EF and BC are three parallel lines cut by the transversal line DC at D, F and C, respectively such that DF = FC.

These lines AD, EF and BC are also cut by the transversal AB at A, E and B, respectively such that AE = BE

Similarly, they are also cut by GH

Therefore, GP = PH (By intercept theorem)

 

Q.3: In the adjoining figure ABCD is a trapezium in which ABDC and P and Q are the midpoints of AD and BC respectively.  DQ and AB when produced meet at E. Also AC and PQ intersect at R . Prove that

(i) DQ = QE

(ii) PRAB

(iii) ARRC

                            

                                  C:\Users\user\Desktop\1.PNG                             

Sol:

Given: AB || DC, AP = PD and BQ = CQ

(i) In ΔQCD and ΔQBE, we have:

∠DQC = ∠BQE (Vertically opposite angles)

∠DCQ = ∠EBQ (Alternate angles, as AE || DC)

BQ = CQ (P is the midpoints)

ΔQCD =ΔQBE

Hence, DQ = QE (CPCT)

(ii) Now, in ΔADE, P and Q are the midpoints of AD and DE, respectively.

PQ || AE, PQ AB || DC and AB || PR || DC

(iii) PQ AB and DC are the three lines cut by transversal AD at P such that AP = PD. These lines PQ, AB, DC are also cut by transversal BC at Q such that BQ = QC. Similarly, lines PQ, AB and DC are also cut by AC at R. Therefore, AR = RC (By intercept theorem).

 

Q.4: In the adjoining figure AD is a median of ABC and DEBA.  Show that BE is also a median of triangle ABC

 

                                     C:\Users\user\Desktop\1.PNG

Sol:

AD is a median of ΔABC and BD = DC

We know that the line drawn through the midpoint of one side of a triangle and parallel to another side bisects the third side.

Here, in ΔABC, D is the midpoint of BC and DE || BA (given). Then DE bisects AC.

i.e., AE = EC

E is the midpoint of AC

Therefore, BE is the median of Δ ABC

 

Q.5: In the adjoining figure AD and BE are the medians of ABC and DFBE. Show that CF=14AC.

                        C:\Users\user\Desktop\1.PNG

 

Sol:

In ΔABC, we have:

AC = AE + EC . . . . . . (i)

AE = EC . . . . . . (ii) [BE is the median of ABDC]

AC = 2EC . . . . . . . (iii)

In Δ BEC, DF || BE.

EF = CF (By midpoint theorem, as D is the midpoint of BC)

But EC = EF + CF

EC = 2 (CF) . . . . . . (iv)

From (iii) and (iv), we get:

AC = 2 (2CF)

Therefore, CF = 14 AC

 

Q.6: In the adjoining figure ABCD is a parallelogram, E is the midpoint of DC and through D, a line segment is drawn parallel to EB to meet CB produced at G and it cuts AB at F. Prove that:

(i). qAB=12GC

(ii). DG = 2 EB

                              C:\Users\user\Desktop\2.PNG

Sol:

(i) In Δ DCG, we have:

DG || EB and DE = EC (E is the midpoint of DC)

Also, GB = BC (By midpoint theorem)

B is the midpoint of GC

Now, GC = GB + BC

GC = 2BC = 2 (AD)

AD = 12 GC

In ADCG, DG || EB and E is the midpoint of DC and B is the midpoint of GC.

By midpoint theorem, the line segment joining the mid points of any two sides of a triangle is parallel to the third side and is half of it.

i.e., EB = 12DG

DG = 2 (EB)

 

Q.7: Prove that the line segment joining the midpoints of the sides of a triangle divided it into 4 congruent triangles.

https://lh3.googleusercontent.com/YLi89szON5pGP5N83iH29z_2izz1z39hVZ90FHeN6mDVbJTs30lkssfIYJNOdarZPA9YZYYtcTsjK9Jui1icfpd1fcqzK9bkLm4jIkEa7yxAxBr12uitpVF88-oRhclHPIXJr8-gNkyM1boTCQ

Sol:

Δ ABC is shown in the figure

D, E and F are the midpoints of sides AB, BC and CA, respectively

As, D and E are the mid points of sides AB, and BC of Δ ABC

DE || AC (By midpoint theorem)

Similarly, DF || BC and EF || AB

Therefore, ADEF, BDFE, and DFCE are all parallelograms.

Now, DE is the diagonal of the parallelogram BDFE.

Δ BDE = Δ FED

Similarly, DF is the diagonal of the parallelogram ADEF.

Δ DAF = Δ FED

And, EF is the diagonal of the parallelogram DFCE.

Δ EFC = Δ FED.

So, all the four triangles are congruent.

 

Q.8: In the adjoining figure D,E and F are the midpoints of the sides BC CA and AB respectively of ABC. Show that EDF=A,DEF=BandDFE=C.

 

                    C:\Users\user\Desktop\3.PNG

Sol:

Δ ABC is shown below. D, E and F are the midpoints of sides BC, CA and AB, respectively.

As F and E are the mid points of sides AB and AC of Δ

ABC.

FE || BC (By mid point theorem)

Similarly, DE || FB and FD || AC

Therefore, AFDE, BDEF, and DCEF are all parallelograms.

In parallelogram AFDE, we have:

∠A = ∠EDF (Opposite angles are equal)

In parallelogram BDEF, we have:

∠B = ∠DEF (Opposite angles are equal)

In parallelogram DCEF, we have:

∠C = ∠DFE (Opposite angles are equal)

 

Q.9: Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rectangle is a rhombus.

Sol:

Let, ABCD be the rectangle and P, Q, R and S be the midpoints of AB, BC, CD, and DA, respectively.

Join AC, a diagonal of the rectangle.
In Δ ABC, we have:

PQ || AC and PQ = 12AC [By midpoint theorem]

Again, in Δ DAC, the points S and R are the mid points of AD and DC, respectively.

SR || AC and SR = 12AC [By midpoint theorem]

Now, PQ || AC and SR || AC

PQ || SR

Also, PQ = SR [Each equal to 12 AC] . . . . . . . (i)

So, PQRS is a parallelogram.

Now, in ΔSAP and ΔQBP, we have:

AS = BQ, ∠A = ∠B = 90° and AP = BP

i.e., Δ SAP = Δ QBP

PS = PQ . . . . . . . . . (ii)

Similarly, ASDR = ACICR

SR = RQ . . . . . . . . (iii)

From (i), (ii) and (iii), we have:

PQ = PQ = SR = RQ

Hence, PQRS is a rhombus.

 

Q.10: Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rhombus is a rectangle.

Sol:

https://lh6.googleusercontent.com/XE0A-lsTJOd6zQkvEbIzt9y0RqG1jNq9UKravwaiPJwJcL25_uL_r_VfrQExAGzSWF2vnVxqNrp_fUk6MTiiMZOz8pDz2MpC1tLnjabuMUiO55wOYPANzp0pwMyorfCHP2t5yNp7QYfyDfrr3A

Let, ABCD be a rhombus and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively. Join the diagonals, AC and BD.

In Δ ABC, we have:

PQ || AC and PQ = 12AC [By midpoint theorem]

Again, in ADAC, the points S and R are the midpoints of AD and DC, respectively.

SR || AC and SR = 12AC [By midpoint theorem]

Now, PQ || AC and SR || AC   => PQ || SR

Also, PQ = SR [Each equal to 12 AC] . . . . . . . . . (i)

So, PQRS is a parallelogram.

We know that the diagonals of a rhombus bisect each other at right angles ∠EOF = 90°.

Now, RQ || DB => RE || FO

Also, SR || AC => FR || OE

Therefore, OERF is a parallelogram.

So, ∠FRE = ∠EOF = 90 (Opposite angles are equal)

Thus, PQRS is a parallelogram with ∠R = 90°.

Therefore, PQRS is a rectangle.

 

Q.11: Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a square is a square

Sol:

https://lh4.googleusercontent.com/ddKge7ecS3b9IjJKHkZElllt4tAk5FGtXlt2CQz2_H4Ev8wwGJK5r0XsKr7iO5KjHyBEpnNUhuVtsw1MsVCEAoV8wb7xEsx2pK57nS2F8oQ6X0pWvBKCOtie3E5zQFWYhVuWRMVDLZfw1V3KEg

Let, ABCD be a square and P, Q, R and S are the midpoints of AB, BC, CD and DA, respectively.

Join the diagonals AC and BD. Let BD cut SR at F and AC cut RQ at E. Let 0 be the intersection point of AC and BD.

In Δ ABC, we have:

PQ || AC and PQ = 12AC [By midpoint theorem]

Again, in ADAC, the points S and R are the midpoints of AD and DC, respectively. SR || AC and SR = 12AC [By midpoint theorem]

Now, PQ || AC and SR || AC => PQ || SR

Also, PQ = SR     [Each equal to 12 AC]

So, PQRS is a parallelogram.

Now, in Δ SAP and Δ QBP, we have:

AS = BQ

AP = BP

i.e., Δ SAP = ΔQBP

PS = PQ . . . . . . . . (ii)

Similarly, ΔSDR = ΔRCQ

SR = RQ . . . . . . . . . . (iii)

From (i), (ii) and (iii), we have:

PQ = PS = SR = RQ . . . . . . . . (iv)

We know that the diagonals of a square bisect each other at right angles.

∠EOF = 90°

Now, RQ || DB

=> RE || FO

Also, SR || AC

=>FR || OE

Therefore, OERF is a parallelogram.

So, ∠FRE = ∠EOF = 90° (Opposite angles are equal)

Thus, PQRS is a parallelogram with ∠R = 90° and PQ = PS = SR = RQ.

Therefore, PQRS is a square.

 

Q.12: Prove that the line segment joining the midpoints of opposite sides of a quadrilateral bisect each other.

Sol:

Let ABCD be the quadrilateral in which P, Q, R, and S are the midpoints of the sides AB, BC, CD and DA respectively.

Join PQ, QR, RS, SP and BD. BD is a diagonal of ABCD.

https://lh5.googleusercontent.com/YxrdnjKtJmXaJ0Oq05vDNlu-e2FJCgG_f14IsfPOMzr4nwHbgzHjbVaiPL-kxIuQzjknDLR4HoAiI2NYsiHl6YdqWL-TNOOX-Ef1fZ2LkDIeZjK3CsVrrXZ7TWPi5Te6HgfUhPe4Qpr1DLSVgQ

In ΔABD, S and P are the midpoints of AD and AB, respectively. SP || BD and SP = 12 BD … (i) (By midpoint theorem)

Similarly in ΔBCD. we have:

QR || BD and QR =12  BD … (ii) (By midpoint theorem)

From equations (i) and (ii), we get:

SP || BD || QR

SP || QR and SP = OR   [Each equal to 12  BD]

In quadrilateral SPQR, one pair of the opposite sides is equal and parallel to each other.

Therefore, SPQR is a parallelogram.

We know that the diagonals of a parallelogram bisect each other.

Therefore, PR and OS bisect each other.

Q.13: In the given figure, ABCD is a quadrilateral whose diagonals intersect at right angles. Show that the quadrilateral formed by joining the midpoints of the pair of adjacent it is a rectangle.

 

                                C:\Users\user\Desktop\4.PNG

Sol:

ABCD is a quadrilateral and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.

AC and BD are the diagonals which intersect each other at 0. RQ intersects AC at E and SR intersects BD at F.

In ΔABC, we have:

PC || AC and PO = 12AC [By midpoint theorem]

Again, in ADAC, the points S and R are the midpoints of AD and DC, respectively.  SR || AC and SR = 12AC [By midpoint theorem]

Now, PC || AC and SR || AC

PQ || SR

Also, PQ = SR [Each equal to 12AC]

So, PARS is a parallelogram.

We know that the diagonals of the given quadrilateral bisect each other at right angles.

∠EOF = 90°

Now, RQ || DB

i.e. RE || FO

Also, SR || AC

i.e. FR || OE

OERF is a parallelogram.

So, ∠FRE = ∠EOF = 90° (Opposite angles are equal)

Thus, PQRS is a parallelogram with ∠R = 90°.

PQRS is a rectangle.

 

 


Practise This Question

During a race, you are accelerating on a leveled track and acquire a velocity 3 times your initial velocity. What was the change in your potential energy?