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Page No 305:

Question 1:

In a quadrilateral ABCD, the measure of angles A, B, C and D are in the ratio 1 : 2 : 3 : 4. Find the measure of all the angles.

Answer:

We have , A:B:C:D=1:2:3:4

Let , A=x,B=2x,C=3x,D=4x

By angle sum property of a quadrilateral, we get
A+B+C+D=360°x+2x+3x+4x=360°10x=360°x=36°=A2x=2×36°=72°=B3x=3×36°=108°=C4x=4×36°=144°=D

Hence, the four angles are 36°, 72°, 108°, 144°.

Page No 305:

Question 2:

Can all the four angles of a quadrilateral be obtuse angles? Give reason for your answer.

Answer:

We know,
Sum of interior angles of a quadrilateral = 360°
So, at least one angle of a quadrilateral must be an acute angle.

Hence, all four angles of a quadrilateral cannot be obtuse angles.

Page No 305:

Question 3:

Can all the four angles of a quadrilateral be acute angles? Give reason for your answer.

Answer:

We know,
Sum of interior angles of a quadrilateral = 360°
So, a maximum of three angles of a quadrilateral can be an acute angle.

Hence, all four angles of a quadrilateral cannot be acute angles.

Page No 305:

Question 4:

Can all the angles of a quadrilateral be right angle? Give reason for your answer.

Answer:

We know,
Sum of angles of a quadrilateral = 360°
So, all the angles of a quadrilateral can be right angles.

Hence, all the angles of a quadrilateral can be right angle.

Page No 305:

Question 5:

In a quadrilateral ABCD, if ∠A : ∠C = 1 : 3 and ∠B : ∠D = 5 : 6 and ∠A + ∠C = 140°; then find all the angles of the quadrilateral.

Answer:

Given that, ∠A : ∠C = 1 : 3 and ∠B : ∠D = 5 : 6
Let the angles A, C, B and D of the quadrilateral be x,3x,5y and 6y respectively.
Also,
A+C=140°x+3x=140°4x=140°x=35°=A3x=3×35°=105°=C

Since, A+B+C+D=360°
A+C+B+D=360°B+D=360°-140°=220°5y+6y=220°11y=220°y=20°5y=5×20°=100°=B6y=6×20°=120°=C

Thus, the angles of the quadrilateral are 35°, 100°, 105° and 120°.



Page No 318:

Question 1:

Two opposite angles of a parallelogram are (3x − 2)° and (50 − x)°. Find the measure of each angle of the parallelogram.

Answer:

It is given that the two opposite angles of a parallelogram are and .

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

Therefore,

…… (i)

Thus, the given angles become

Also, .

Therefore the sum of consecutive interior angles must be supplementary.

That is;

Since opposite angles of a parallelogram are equal.

Therefore,

And

Hence the four angles of the parallelogram are , , and .

Page No 318:

Question 2:

If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.

Answer:

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

Since, opposite angles of a parallelogram are equal.

Therefore, the four angles in sequence are ,,and.

Page No 318:

Question 3:

Find the measure of all the angles of a parallelogram, if one angle is 24° less than twice the smallest angle.

Answer:

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

Hence, the four angles of the parallelogram are , , and .

Page No 318:

Question 4:

The perimeter of a parallelogram is 22 cm. If the longer side measures 6.5 cm what is the measure of the shorter side?

Answer:

Let the shorter side of the parallelogram be cm.

The longer side is given ascm.

Perimeter of the parallelogram is given as 22 cm

Therefore,

Hence, the measure of the shorter side is cm.

Page No 318:

Question 5:

In a parallelogram ABCD, ∠D  = 135°, determine the measures of ∠A and ∠B.

Answer:

It is given that ABCD is a parallelogram with

We know that the opposite angles of the parallelogram are equal.

Therefore,

Also, and are adjacent angles, which must be supplementary.

Therefore,

Hence , and .

Page No 318:

Question 6:

ABCD is a parallelogram in which ∠A = 70°. Compute ∠B, ∠C and ∠D.

Answer:

It is given that ABCD is a parallelogram with

We know that the opposite angles of the parallelogram are equal.

Therefore,

Also, and are adjacent angles, which must be supplementary.

Therefore,

Also, and are opposite angles of a parallelogram.

Therefore,

Hence, the angles of a parallelogram are , , and .

Page No 318:

Question 7:

In the given figure, ABCD is a parallelogram in which ∠DAB = 75° and ∠DBC = 60°. Compute ∠CDB and ∠ADB.

Answer:

The figure is given as follows:

It is given that ABCD is a parallelogram.

Thus

And are alternate interior opposite angles.

Therefore,

…… (i)

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

Also, we have

Therefore,

…… (ii)

In

By angle sum property of a triangle.

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

Hence, the required value for is

And is .

Page No 318:

Question 8:

In the given figure, ABCD is a parallelogram in which ∠A = 60°. If the bisectors of ∠A and ∠B meet at P, prove that AD = DP, PC = BC and DC = 2AD.
 

Answer:

The figure is given as follows:

It is given that ABCD is a parallelogram.

Thus,

Opposite angles of a parallelogram are equal.

Therefore,

Also, we have AP as the bisector of

Therefore,

…… (i)

Similarly,

…… (ii)

We have ,

From (i)

Thus, sides opposite to equal angles are equal.

Similarly,

From (ii)

Thus, sides opposite to equal angles are equal.

Also,

Page No 318:

Question 9:

In the given figure, ABCD is a parallelogram and E is the mid-point of side BC. If DE and AB when produced meet at F, prove that AF = 2AB.
 

Answer:

Figure is given as follows:

It is given that ABCD is a parallelogram.

DE and AB when produced meet at F.

We need to prove that

It is given that

Thus, the alternate interior opposite angles must be equal.

In and , we have

(Proved above)

(Given)

(Vertically opposite angles)

Therefore,

(By ASA Congruency )

By corresponding parts of congruent triangles property, we get

DC = BF …… (i)

It is given that ABCD is a parallelogram. Thus, the opposite sides should be equal. Therefore,

…… (ii)

But,

From (i), we get:

From (ii), we get:

Hence proved.



Page No 330:

Question 1:

In a parallelogram ABCD, determine the sum of angles ∠C and ∠D.

Answer:

The parallelogram can be drawn as:

We have ,thus and are consecutive interior angles.

These must be supplementary.

Therefore,

Page No 330:

Question 2:

In a parallelogram ABCD, if ∠B = 135°, determine the measure of its other angles.

Answer:

Since ABCD is a parallelogram with .

Opposite angles of a parallelogram are equal.

Therefore,

Also, let

Similarly,

We know that the sum of the angles of a quadrilateral is .

Hence the measure of other angles are , and .

Page No 330:

Question 3:

ABCD is a rectangle with ∠ABD = 40°. Determine ∠DBC.

Answer:

The rectangle is given as follows with

We have to find .

An angle of a rectangle is equal to .

Therefore,

Hence, the measure for is .

Page No 330:

Question 4:

In the given figure, ABCD is a parallelogram in which AB is produced to E so that BE = AB. Prove that ED bisects BC.

Answer:

Given that, ABCD is a parallelogram in which AB is produced to E so that BE = AB.
In ∆AED, we have
ADBO          ABCD is parallelogram, BCAD
B is the midpoint of AE in ∆AED.
So, by the midpoint theorem, we have 
O is the midpoint of DE.
DO=OE
Thus, ED bisects BC.

Hence, proved.

Page No 330:

Question 5:

In a parallelogram ABCD, points M and N have been taken on opposite sides AB and CD respectively such that AM = CN. Show that AC and MN bisect each other.

Answer:

Given that, ABCD is a parallelogram such that AM = CN
To prove: AC bisects MN

Now, in parallelogram ABCD
AB=CD     Opposite sides of parallelogram are equalAM=CN     GivenAB-AM=CD-CNBM=DN
Therefore, M and N are midpoints of AB and CD.
In AMP and CPN
PAM=PCN    Alternate interier anglesAM=CN   GivenPMA=PNC    Alternate interier angles PAMPCN    By ASAAP=PC and PM=PN  By CPCT

Hence, proved.

Page No 330:

Question 6:

In the given figure, through the vertices A, B and C of ∆ABC, lines QR, PQ and RP have been drawn, respectively parallel to sides BC, CA and AB of ∆ABC. Prove that the perimeter of ∆PQR is double the perimeter of ∆ABC.

Answer:

Given that, ABPR, BCQR and CAPQ.
We have, ABCR and AQBC are parallelogram.
⇒ BC = AR and BC = AQ
⇒ AQ = AR
Therefore, A is the midpoint of QR.

Similarly, B and C are midpoints of PQ and PR.
AB=12PRBC=12QRCA=12PQPR=2AB, QR=2BC and PQ=2CAPQ+QR+PR=2AB+BC+CA
Therefore, Perimeter of PQR = 2(Perimeter of  ABC)

Hence, proved.



Page No 337:

Question 1:

ABCD is a rectangle with ∠ABD = 40°. Determine ∠DBC.

Answer:

Given that, ∠ABD = 40°

DBA+DBC=90°     Each angle of rectangle is 90°40°+DBC=90°  DBC=50°

Thus, ∠DBC is 50°.

Page No 337:

Question 2:

ABCD is a square. A line segment DX meets the side BC at X and the diagonal AC at O such that ∠COD = 105°. Find ∠OXC.

Answer:

Given that, ABCD is a square and ∠COD = 105°


Since
DOC+COX=180°         Linear pair105°+COX=180°COX=75°

Also, the diagonals of a square bisect its angles.

DCO=OCX=45° In COXCOX+OCX+OXC=180°75°+45°+OXC=180°120°+OXC=180°OXC=60° 

Thus, ∠OXC = 60°

Page No 337:

Question 3:

ABCD is a rhombus whose diagonals AC and BD intersect at O. If ∠BAC = 35°, find ∠ABO, ∠BCD and ∠CDA.

Answer:

Given that, ABCD is a rhombus and ∠BAC = 35°.

Since diagonals of a rhombus are perpendicular bisector of each other, i.e., AOB=BOC=COD=DOA=90°.

In AOBAOB+OBA+OAB=180°90°+OBA+35°=180°OBA=55°

We know, the diagonals of a rhombus bisect vertex angles.

ABC=2×ABOABC=2×55°ABC=110°Again,ABC+BCD=180°              Adjacent angles of a parallelogram are supplementary     110°+BCD=180    BCD=70°Since opposite angles of a parallelogram are equal.CDA=ABC=110°

Thus, ∠ABO = 55º, ∠BCD = 70º and ∠CDA = 110º

Page No 337:

Question 4:

Diagonals of rhombus are of lengths 16 cm and 12 cm respectively, find the perimeter of the rhombus.

Answer:

Given: Diagonals of  the rhombus ABCD are AC = 16 cm and BD = 12 cm.

Since diagonals of a rhombus are perpendicular bisectors of each other, i.e., AOB=BOC=COD=DOA=90°.

Also, AO=OC=8 cm and BO=OD=6 cm

In OAD, by Pythagoras TheoremAO2+OD2=AD282+62=AD2AD2=100AD=10 cm

Therefore, the perimeter of the rhombus = 4 × side of the rhombus
                                                                 = 4 × 10
                                                                 = 40 cm   

Thus, the perimeter of the rhombus is 40 cm.

Page No 337:

Question 5:

ABCD is a rhombus such that ∠ACB = 40°. Find ∠ADB.

Answer:



Given that, ABCD is a rhombus such that ∠ACB = 40°

Since diagonals of a rhombus are perpendicular bisectors of each other, i.e., AOB=BOC=COD=DOA=90°.

Now, in BOC
BOC+OBC+OCB=180°90°+OBC+40°=180°130°+OBC=180°OBC=50°

Now, AD || BC and DB is the transversal.
ADB=OBC=50°     Alternate interier angles

Thus, ∠ADB = 50º

Page No 337:

Question 6:

In a rhombus, the altitude from D to the side AB bisects AB. Find the angles of the rhombus.

Answer:


Given that, ABCD is a rhombus DOAB.
Also, the altitude from D to the side AB bisects AB, i.e., AO = OB.
Since ABCD is a rhombus, i.e., AB = BC = CD = DA

In ADO and BDO
AO=OB                   GivenAOD=BO        Each  90°DO=OD                   CommonADOBDO    By SAS congruence ruleAD=BD                    By CPCT                                  .....2      From 1 and 2, we getAD=BD=DB  

Therefore ADB is an equilateral triangle.
DAB=60°Since adjacent angles of a parallelogram are supplementary.DAB+ABC=180°60°+ABC=180°ABC=120°

Since the opposite angles of a parallelogram are equal. 
A=C=60° and B=D=120°

Thus, the angles of the rhombus are A=C=60° and B=D=120°

Page No 337:

Question 7:

ABCD is a square E, F, G and H are points on AB, BC, CD and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is a square.

Answer:

Given that, ABCD is a square such that AE = BF = CG = DH.

Since, AE = BF = CG = DH.
BE=CF=DG=AH               .....(1)
In AEH and BEF
AE=BF                 GivenA=B              Each 90°AH=BE                 From 1AEHBEF   By SAS congruence rule1=2  and 3=4        By CPCT             .....(2) 

In AEH, by Angle Sum Property of triangleA+1+3=180° 1+3=90°                   .....3In BEF, by Angle Sum Property of triangleB+2+4=180° 2+4=90°                    .....4

Adding 3 and 41+2+3+4=180°                

21+4=180°            From2 and 41+4=90°

1+4 +HEF=180°      Staright lineHEF=90°

Since, HE = EF = FG = GH and HEF=90°, EFGH is a square.



Page No 353:

Question 1:

In a ΔABC, D, E and F are, respectively, the mid-points of BC, CA and AB. If the lengths of side AB, BC and CA are 7 cm, 8 cm and 9 cm, respectively, find the perimeter of ΔDEF.

Answer:

is given with D,E and F as the mid-points of BC , CA and AB respectively as shown below:

Also, , and .

We need to find the perimeter of

In , E and F are the mid-points of CA and AB respectively.

Theorem states, the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

Therefore, we get:

Similarly, we get

And

Perimeter of

Hence, the perimeter of is .

Page No 353:

Question 2:

In a triangle ∠ABC, ∠A = 50°, ∠B = 60° and C =∠70°. Find the measures of the angles of the triangle formed by joining the mid-points of the sides of this triangle.

Answer:

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 measure of angles are , and.

Page No 353:

Question 3:

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

Answer:

It is given that P, Q and R are the mid-points of BC, CA and AB respectively.

Also, we have , and

We need to find the perimeter of quadrilateral ARPQ

In , P and R are the mid-points of CB and AB respectively.

Theorem states, the line segment joining the mid-points of any two sides of a traingle is parallel to the third side and equal to half of it.

Therefore, we get:

Similarly, we get

We have Q and R as the mid points of AC and AB respectively.

Therefore,

And

Perimeter of

Hence, the perimeter of quadrilateral ARPQ is.

Page No 353:

Question 4:

In the given figure, M, N and P are the mid-points of AB, AC and BC respectively. If MN = 3 cm, NP = 3.5 cm and MP = 2.5 cm, calculate BC, AB and AC.
 

Answer:

We have as follows:

M, N and P are the mid-points of sides AB ,AC and BC respectively.

Also, , and

We need to calculate BC, AB and AC.

In , M and N are the mid-points of AB and AC respectively.

Theorem states, the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

Therefore,

Similarly,

And

Hence, the measure for BC, AB and AC is , and respectively.

Page No 353:

Question 5:

In a ABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC intersects FE at Q. Prove that AQ = QP.

Answer:

is given with E and F as the mid points of sides AB and AC.

Also, intersecting EF at Q.

We need to prove that

In , E and F are the mid-points of AB and AC respectively.

Theorem states, the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

Therefore, we get:

Since, Q lies on EF.

Therefore,

This means,

Q is the mid-point of AP.

Thus, (Because, F is the mid point of AC and)

Hence proved.

Page No 353:

Question 6:

ABC is a triangle, D is a point on AB such that AD = 14 AB and E is a point on AC such that AE = 14 AC. Prove that DE = 14 BC.

Answer:

is given with D a point on AB such that .

Also, E is point on AC such that.

We need to prove that

Let P and Q be the mid points of AB and AC respectively.

It is given that

and

But, we have taken P and Q as the mid points of AB and AC respectively.

Therefore, D and E are the mid-points of AP and AQ respectively.

In , P and Q are the mid-points of AB and AC respectively.

Theorem states, the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

Therefore, we get and …… (i)

In , D and E are the mid-points of AP and AQ respectively.

Therefore, we get and …… (ii)

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

Hence proved.

Page No 353:

Question 7:

In the given figure, BEAC. AD is any line from A to BC intersecting BE in H. P, Q and R are respectively the mid-points of AH, AB and BC. Prove that ∠PQR = 90°

Answer:

is given with

AD is any line from A to BC intersecting BE in H.

P,Q and R respectively are the mid-points of AH,AB and BC.

We need to prove that

Let us extend QP to meet AC at M.

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

Theorem states, the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

Therefore, we get:

…… (i)

Similarly, in,

…… (ii)

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

and

We get, is a parallelogram.

Also,

Therefore, is a rectangle.

Thus,

Or,

Hence proved.

Page No 353:

Question 8:

In the given figure, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = 14 AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.

Answer:

Figure is given as follows:

ABCD is a parallelogram, where P is the mid-point of DC and Q is a point on AC such that

.

PQ produced meets BC at R.

We need to prove that R is a mid-point of BC.

Let us join BD to meet AC at O.

It is given that ABCD is a parallelogram.

Therefore, (Because diagonals of a parallelogram bisect each other)

Also,

Therefore,

In , P and Q are the mid-points of CD and OC respectively.

Theorem states, the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

Therefore, we get:

Also, in , Q is the mid-point of OC and

Therefore, R is a mid-point of BC.

Hence proved.

Page No 353:

Question 9:

In the given figure, ABCD and PQRC are rectangles and Q is the mid-point of AC. Prove that

(i) DP = PC

(ii) PR = 12 AC

Answer:

Rectangles ABCD and PQRC are given as follows:

Q is the mid-point of AC.

In , Q is the mid-point of AC such that

Using the converse of mid-point theorem, we get:

P is the mid-point of DC

That is;

Similarly, R is the mid-point of BC.

Now, in , P and R are the mid-points of DC and BC respectively.

Then, by mid-point theorem, we get:

Now, diagonals of a rectangle are equal.

Therefore putting ,we get:

Hence Proved.

Page No 353:

Question 10:

ABCD is a parallelogram, E and F are the mid-points of AB and CD respectively. GH is any line intersecting AD, EF and BC at G, P and H respectively. Prove that GPPH.

Answer:

ABCD is a parallelogram with E and F as the mid-points of AB and CD respectively.

We need to prove that

Since E and F are the mid-points of AB and CD respectively.

Therefore,

,

And

,

Also, ABCD is a parallelogram. Therefore, the opposite sides should be equal.

Thus,

Also, (Because )

Therefore, BEFC is a parallelogram

Then, and …… (i)

Now,

Thus, (Because as ABCD is a parallelogram)

We get,

AEFD is a parallelogram

Then, we get:

…… (ii)

But, E is the mid-point of AB.

Therefore,

Using (i) and (ii), we get:

Hence proved.



Page No 354:

Question 11:

In a parallelogram PQRS, PQ = 12 cm and PS = 9 cm. The bisector of ∠P meets SR in M. PM and QR both when produced meet at T. Find the length of RT.

Answer:


Given that, PM is bisector of P
QPM=SPM        .....(1)
Also,
QPM=PMS        Alternate interier angles    .....2

From (1) and (2), we have
SPM=∠PMS                 .....(3)

In SPM, SPM =PMS    

As in a triangle if two angles of a triangle are equal, their opposite sides are equal.

MS=PS=9 cmRM=SR-MSRM=12-9RM=3 cm            .....4RMT and PMS are vertically opposite anglesRMT =PMS    .....5

Since PSQT and PT is a transversal
RTM=SPMRTM=RMT         From 5InRTM,RTM=RMT As in a triangle if two angles of a triangle are equal, their opposite sides are equal.RM=TR=3 cm          From 4

Thus, the length of RT is 3 cm.

Page No 354:

Question 12:

ABCD is a trapezium in which AB || DC and AD = BC. If P, Q, R, S be respectively the mid-points of BA, BD, CD, CA, then show that PQRS is a rhombus.

Answer:


Given that, in ABC 
P and S are the midpoints of AB and AC respectively then by Mid Point Theorem.
PSBC and PS=12BC    .....1

Similarly, ADC
R and S are the midpoints of DC and AC respectively then by Mid Point Theorem.
RS||AD and RS=12ADRS=12AD=12BC        .....2

In BCD
R and Q are the midpoints of DC and DB respectively then by Mid Point Theorem.
RQ||BC and RQ=12BC        .....3

In ABD
P and Q are the midpoints of AB and DB respectively then by Mid Point Theorem.
QP||AD and QP=12ADQP=12AD =12BC       .....4

From (1), (2), (3) and (4), we get
PQ = QR = RS = SP
Hence, PQRS is a rhombus.
 

Page No 354:

Question 13:

ABCD is a rhombus, EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced meet at right angles.

Answer:

Since diagonals of a rhombus are perpendicular bisectors of each other.
OA=OC, OB=OD, And, AOD=COD=AOB=COB=90°



In ∆BDE , A and O are midpoints of BE and BD then by Mid Point Theorem.
 OADEOCDG               .....1


In ∆CFA , B and O are midpoints of AF and AC then by Mid Point Theorem.
OBCFODGC               .....2

Thus, in quadrilateral DOCG,
OCDG and ODGC                From 1 and 2

Since  a quadrilateral  whose both pairs of opposite sides are parallel is a parallelogram.
Thus, DOCG is a parallelogram.

 DOC=DGC   Opposite angles of a parallelogram are equalDGC=90°        DOC=90°

Hence proved.

 

Page No 354:

Question 14:

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.

Answer:


Given that, ABCD is a parallelogram with DE = DC

In DCE,DE=DCAs in a triangle if two sides of a triangle are equal, their opposite angles are equal.DCE=DEC      .....1Also, ABCDAFCDAFCD and EF is a transversalDCE=BFC                Corresponding angles     .....2From 1 and 2, we getDEC=BFCIn AFE,AFE=AEFAs in a triangle if two sides of a triangle are equal, their opposite angles are equal.AE=AF AD+DE=AB+BFBC+DC=AB+BF       GivenBC+AB=AB+BF        DC=ABBC=BF  

Hence proved.

Page No 354:

Question 15:

In the given figure, ABCD is a trapezium in which AB || DC and P, Q are the mid-points 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) PR || AB
(iii) AR = RC

Answer:


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

(i) In ∆QCD and ∆BQE, 

DQC=BQE     Vertically Opposite anglesDCQ=EBQ     Alternate anglesBQ=CQ                   GivenQCDQBE     By ASA congruence ruleDQ=QE   By CPCT

(ii) In ∆ADE, P and Q are the midpoints AD and DE then by Mid Point Theorem.
PQAE and PQ=12AEPQABPRAB

(iii) 
PRABPRDC        DC||AB
In ∆ADC, P is the midpoint of AD and PR || DC then by Converse of Mid Point Theorem.
R is the midpoint of the side AC.
⇒ AR = RC



 



Page No 356:

Question 1:

In a parallelogram ABCD, write the sum of angles A and B.

Answer:

In Parallelogram ABCD, and are adjacent angles.

Thus, .

Then, we have and as consecutive interior angles which must be supplementary.

Hence, the sum of and is .

Page No 356:

Question 2:

In a parallelogram ABCD, if ∠D = 115°, then write the measure of ∠A.

Answer:

In Parallelogram ABCD , and are Adjacent angles.

We know that in a parallelogram, adjacent angles are supplementary.

Now, A+D=180°A+115°=180°A=180°-115°A=65°So, measure of A is 65°.

 

Page No 356:

Question 3:

If PQRS is a square, then write the measure of ∠SRP.

Answer:

The square PQRS is given as:

Since PQRS is a square.

Therefore,

and

Now, in , we have

That is, (Angles opposite to equal sides are equal)

By angle sum property of a triangle.

()

Hence, the measure of is.

Page No 356:

Question 4:

The perimeter of a parallelogram is 22 cm. If the longer side measures 6.5 cm, what is the measure of shorter side?

Answer:

Let the shorter side of the parallelogram be cm.

The longer side is given ascm.

Perimeter of the parallelogram is given as 22 cm

Therefore,

Hence, the measure of the shorter side is cm.

Page No 356:

Question 5:

In a parallelogram ABCD, if ∠A = (3x − 20)°, ∠B = (y + 15)°, ∠C = (x + 40)°, then find the values of x and y.

Answer:

In parallelogram ABCD, and are opposite angles.

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

Therefore,

We have and

Therefore,

Therefore,

Similarly,

Also,

Therefore,

By angle sum property of a quadrilateral, we have:

Hence the required values for x and y are and respectively.

Page No 356:

Question 6:

If measures opposite angles of a parallelogram are (60 − x)° and (3x − 4)°, then find the measures of angles of the parallelogram.

Answer:

Let ABCD be a parallelogram, with and .

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

Therefore,

Thus, the given angles become

Similarly,

Also, adjacent angles in a parallelogram form the consecutive interior angles of parallel lines, which must be supplementary.

Therefore,

Similarly,

Thus, the angles of a parallelogram are ,, and .

Page No 356:

Question 7:

In a parallelogram ABCD, the bisector of ∠A also bisects BC at X. Find AB : AD.

Answer:

Parallelogram ABCD is given as follows:

We have AX bisects bisecting BC at X.

That is,

We need to find

Since, AX is the bisector

That is,

…… (i)

Also, ABCD is a parallelogram

Therefore, and AB intersects them

…… (ii)

In by angle sum property of a triangle:

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

…… (iii)

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

Sides opposite to equal angles are equal. Therefore,

As X is the mid point of BC. Therefore,

Also, ABCD is a parallelogram, then,

Thus,

Hence the ratio of is .

Page No 356:

Question 8:

In the given figure, PQRS is an isosceles trapezium. Find x and y.

Answer:

Trapezium is given as follows:

We know that PQRS is a trapezium with

Therefore,

Hence, the required value for x is .

Page No 356:

Question 9:

In the given figure, ABCD is a trapezium. Find the values of x and y.

Answer:

The figure is given as follows:

We know that ABCD is a trapezium with

Therefore,

It is given that and .

Similarly,

Hence, the required values for x and y is and respectively.

Page No 356:

Question 10:

In the given figure, ABCD and AEFG are two parallelograms. If ∠C = 58°, find ∠F.

Answer:

ABCD and AEFG are two parallelograms as shown below:

Since ABCD is a parallelogram, with

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

Therefore,

Similarly, AEFG is a parallelogram, with

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

Therefore,

Hence, the required measure for is .

Page No 356:

Question 11:

Complete each of the following statements by means of one of those given in brackets against each:
(i) If one pair of opposite sides are equal and parallel, then the figure is ........................
(parallelogram, rectangle, trapezium)

(ii) If in a quadrilateral only one pair of opposite sides are parallel, the quadrilateral is ................ (square, rectangle, trapezium)

(iii) A line drawn from the mid-point of one side of a triangle .............. another side intersects the third side at its mid-point. (perpendicular to parallel to, to meet)

(iv) If one angle of a parallelogram is a right angle, then it is necessarily a .................
(rectangle, square, rhombus)

(v) Consecutive angles of a parallelogram are ...................
(supplementary, complementary)

(vi) If both pairs of opposite sides of a quadrilateral are equal, then it is necessarily a ...............
(rectangle, parallelogram, rhombus)

(vii) If opposite angles of a quadrilateral are equal, then it is necessarily a ....................
(parallelogram, rhombus, rectangle)

(viii) If consecutive sides of a parallelogram are equal, then it is necessarily a ..................
(kite, rhombus, square)

Answer:

(i) If one pair of opposite sides are equal and parallel, then the figure is parallelogram.

Reason:

In and,

(Given)

(Common)

(Because , Alternate interior angles are equal)

So, by SAS Congruence rule, we have

Also,

(Corresponding parts of congruent triangles are equal)

But, these are alternate interior angles, which are equal.

Thus, and

Hence, quadrilateral ABCD is parallelogram

(ii) If in a quadrilateral only one pair of opposite sides are parallel, the quadrilateral is trapezium.

(iii) A line drawn from the mid-point of one side of a triangle parallel to another side intersects the third side at its mid-point.

Reason:

This is a theorem.

(iv) If one angle of a parallelogram is a right angle, then it is necessarily a rectangle.

Reason:

Let ABCD be the given parallelogram.

We have,

In a parallelogram, opposite angles are equal.

Therefore,

Similarly,

Also,

Thus, a parallelogram with all the angles being right angle and opposite sides being equal is a rectangle.

(v) Consecutive angles of a parallelogram are supplementary.

Reason:

Let ABCD be the given parallelogram.

Thus, .

Therefore,

Consecutive angles and are supplementary.

(vi) If both pairs of opposite sides of a quadrilateral are equal, then it is necessarily a parallelogram.

Reason:

ABCD is a quadrilateral in which and .

We need to show that ABCD is a parallelogram.

In and , we have

(Common)

(Given)

(Given)

So, by SSS criterion of congruence, we have

By corresponding parts of congruent triangles property.

…… (i)

And

Now lines AC intersects AB and DC at A and C,such that

(From (i))

That is, alternate interior angles are equal.

Therefore, .

Similarly, .

Therefore, ABCD is a parallelogram.

(vii) If opposite angles of a quadrilateral are equal, then it is necessarily a parallelogram.

Reason:

ABCD is a quadrilateral in which and .

We need to show that ABCD is a parallelogram.

In quadrilateral ABCD, we have

Therefore,

…… (i)

Since sum of angles of a quadrilateral is

From equation (i), we get:

Similarly,

Now, line AB intersects AD and BC at A and B respectively

Such that

That is, sum of consecutive interior angles is supplementary.

Therefore, .

Similarly, we get .

Therefore, ABCD is a parallelogram.

(viii) If consecutive sides of a parallelogram are equal, then it is necessarily a rhombus.

We have ABCD, a parallelogram with .

Since ABCD is a parallelogram.

Thus,

And

But,

Therefore,all four sides of the parallelogram are equal, then it is a rhombus.



Page No 357:

Question 12:

In a quadrilateral ABCD, bisectors of angles A and B intersect at O such that ∠AOB = 75°, then write the value of ∠C + ∠D.

Answer:

The quadrilateral can be drawn as follows:

We have AO and BO as the bisectors of angles and respectively.

In ,We have,

…… (1)

By angle sum property of a quadrilateral, we have:

Putting in equation (1):

……(2)

It is given that in equation (2), we get:

Hence, the sum of and is .

Page No 357:

Question 13:

If the bisectors of two adjacent angles A and B of a quadrilateral ABCD intersect at a point O such that ∠C + ∠D = kAOB, then find the value of k.

Answer:

The quadrilateral can be drawn as follows:

We have AO and BO as the bisectors of angles and respectively.

In ,We have,

…… (I)

By angle sum property of a quadrilateral, we have:

Putting in equation (I):

…… (II)

On comparing equation (II) with

We get .

Hence, the value for k is .

Page No 357:

Question 14:

In the given figure, PQRS is a rhombus in which the diagonal PR is produced to T. If ∠SRT = 152°, find x, y and z.

Answer:

Rhombus PQRS is given.

Diagonal PR is produced to T.

Also, .

We know that in a rhombus, the diagonals bisect each other at right angle.

Therefore,

Now,

In , by angle sum property of a triangle, we get:

Or, (Because O lies on SQ)

We have, .Thus the alternate interior opposite angles must be equal.

Therefore,

In ,we have

Since opposite sides of a rhombus are equal.

Therefore,

Also,

Angles opposite to equal sides are equal.

Thus,

But

Thus,

Hence the required values for x,y and z are , and respectively.

Page No 357:

Question 15:

In the given figure, ABCD is a rectangle in which diagonal AC is produced to E. If ∠ECD = 146°, find ∠AOB.

Answer:

ABCD is a rectangle

With diagonal AC produced to point E.

We have

(Linear pair)

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

Thus

Also, angles opposite to equal sides are equal.

Therefore,

By angle sum property of a traingle

Also, and are vertically opposite angles.

Therefore,

Hence, the required measure for is .



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