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Page No 27:

Question 1:

In the given figure, ACD is an exterior angle of ABC. B = 40°, A = 70°.

Find the measure of ACD.
 

Answer:

In ABC,
∠ACD = ∠A + ∠B   (Exterior angle property)
= 70∘ + 40∘          
= 110∘
Hence, the measure of  ACD is 110∘​.
 

Page No 27:

Question 2:

In PQR, P = 70°Q = 65 ° then find  R.

Answer:

In PQR,
∠P + ∠Q + ∠R = 180∘   (Angle sum property)
⇒ 70∘ + 65∘ + ∠R = 180∘         
⇒ 135∘ + ∠R = 180∘  
⇒ ∠R = 180∘ − 135∘  
= 45∘
Hence, the measure of  R is 45∘.
 

Page No 27:

Question 3:

The measures of angles of a triangle are x°, ( x-20)°, (x-40)°. Find the measure of each angle.

Answer:

Let us suppose the angles ∠P, ∠Q, ∠Rof a PQR be x°, (- 20)°, (- 40)° respectively.
∠P + ∠Q + ∠R = 180∘   (Angle sum property)
x∘ + (- 20)° + (- 40)° = 180∘         
⇒ 3- 60 = 180  
⇒ 3= 240
​= 80
Therefore,
∠P = 80∘
∠R = (80 - 20)°
= 60∘
∠R = (80 - 40)°
= 40∘
Hence, the measure of each angle is 80∘, 60∘ and 40∘respectively.

Page No 27:

Question 4:

The measure of one of the angles of a triangle is twice the measure of its smallest angle and the measure of the other is thrice the measure of the smallest angle. Find the measures of the three angles.

Answer:

Let us suppose the angles of a PQR such that ∠P < ∠Q < ∠R.
A.T.Q, 
∠Q = 2∠P
∠R = 2∠P
Now, ∠P + ∠Q + ∠R = 180∘   (Angle sum property)
⇒ ∠P + 2∠P + 3∠P = 180∘         
⇒ 6∠P = 180∘  
⇒ ∠P = 30∘
Therefore,
∠P = 30∘
∠R = 60∘
∠R = 90∘
Hence, the measure of each angle is 30∘, 60∘ and 90∘respectively.



Page No 28:

Question 5:

In the given figure, measures of some angles are given. Using the measures find the values of x, y, z. 

Answer:

∠NEM + ∠NET =  180∘      (Linear angle property)
⇒ y + 100∘ = 180∘       
⇒ y = 80∘
Also, ∠NME + ∠EMR =  180∘      (Linear angle property)
⇒ z + 140∘ = 180∘       
⇒ z = 40∘
Now, In â–³NEM
∠N + ∠E + ∠M = 180∘   (Angle sum property)
⇒ x + y + z = 180∘         
+ 80∘ + 40∘ = 180∘  
⇒ + 120∘ = 180∘ 
⇒ x = 60∘  
Hence, the values of xy and z are 60∘, 80∘ and 40∘respectively.

Page No 28:

Question 6:

In the given figure, line AB || line DE. Find the measures of DRE and ARE using given measures of some angles.

Answer:

AB || DE and AD is a transversal line.
∠BAR = ∠RDE = 70∘      (Alternate angles)
In â–³DER
∠D + ∠E + ∠DRE = 180∘   (Angle sum property)
⇒ 70∘ + 40∘ + ∠DRE = 180∘         
⇒ 110∘ + ∠DRE = 180∘ 
⇒ ∠DRE = 70∘
Now, ∠ARE = ∠DRE + ∠RDE   (Exterior angle property)
= 70∘ + 40∘          
= 110∘
Hence, the measures of ∠DRE and ∠ARE are 70∘ and 110∘respectively.

Page No 28:

Question 7:

In ABC, bisectors of A and B intersect at point O. If C = 70° . Find measure of AOB.

Answer:

If the bisectors of ∠X and ∠Y of a â–³XYZ intersect at point O, then XOY=90°+12XZY
AOB=90°+12ACB=90°+1270°
= 90∘ + 35∘ 
= 125∘  
Hence, the measure of ∠AOB is 125∘.

Page No 28:

Question 8:

In the given Figure, line AB || line CD and line PQ is the transversal. Ray PT and ray QT are bisectors of BPQ and PQD respectively.
Prove that m PTQ = 90 °.

Answer:

AB || CD and PQ is a transversal line.
∠BPQ + ∠DQP = 180∘      (Angles on the same side of a transversal line are supplementary angles)
12BPQ+12DQP=180°2
⇒ ∠QPT + ∠PQT = 90∘
In â–³PQT
∠QPT + ∠PQT + ∠PTQ = 180∘   (Angle sum property)
⇒ 90∘ + ∠PTQ = 180∘         
⇒ ∠PTQ = 90∘ 
Hence proved.

Page No 28:

Question 9:

Using the information shown in figure, find the measures of a, b and c.

Answer:

c + 100∘ = 180∘                (Linear angle property)      
⇒ c = 80∘
Now, b = 70∘     (Vertically opposite angles)      
a + b + c = 180∘       (Angle sum property)   
⇒ + 70∘ + 80∘ = 180∘  
⇒ + 150∘ = 180∘ 
⇒ = 30∘  
Hence, the values of ab and c are 30∘, 70∘ and 80∘respectively.

Page No 28:

Question 10:

In the given figure, line DE ||   line GF ray EG and ray FG are bisectors of DEF and DFM respectively. Prove that,
(i)  DEG = 12 EDF    (ii)  EF = FG.   

Answer:

(i) Given: DE || GF
Now, DEF = GFM              (Corresponding angles as DM is a transversal line)
⇒ 2DEG = DFG         (Ray EG and ray FG are bisectors of DEF and DFM)
⇒ 2DEG = EDF          (∵ EDF = DFG, alternate angles as DF is a transversal line)
⇒ DEG = 12 EDF

(ii) Given: DE || GF
 DEG = EGF              (Alternate angles as EG is a transversal line)
∴ GEF = EGF            (∵ DEG = GEF)
∴ EF = FG                         (Sides opposite to equal angles)


 



Page No 31:

Question 1:

In each of the examples given below, a pair of triangles is shown. Equal parts of triangles in each pair are marked with the same signs. Observe the figures and state the test by which the triangles in each pair are congruent.

(i)

 By . . . . . . . . . . test
ABC  PQR

(ii)

 By . . . . . . . . . . test
XYZ  LMN

(iii)
 
 By . . . . . . . . . . test
LMN  PTR

(iv) 

By . . . . . . . . . . test
 LMN   PTR

Answer:

(i) SSC Test
(ii) SAS Test
(iii) ASA Test
(iv) Hypotenuse Side Test. 



Page No 32:

Question 2:

Observe the information shown in pairs of triangles given below. State the test by which the two triangles are congruent. Write the remaining congruent parts of the triangles.

(i)

 
From the information shown in the figure,
in ABC and  PQR
ABC  PQR
seg BC seg QR
ACBPRQ
 ABC  PQR ....... 3243432 test
 BAC  3243432 .......corresponding angles of congruent triangles.

seg AB 1234and  1234  seg PR...corresponding sides of congruent triangles


(ii)  

 
From the information shown in the figure,
in PTQ and  STR
seg PT  seg ST 
PTQ STR...... vertically opposite angles
 PTQ  STR ....... 3243432 test
 
 TPQ 1234and  1234  TRS...corresponding angles of congruent triangles

seg PQ  3243432  corresponding sides of congruent triangles
 

Answer:

(i)
From the information shown in the figure,
in ABC and  PQR
ABC  PQR
seg BC  seg QR 
ACB PRQ
 ABC  PQR ....... ASA test
 BAC  QPR .......corresponding angles of congruent triangles.

seg AB seg PQ123Aand  seg AC  seg PR...corresponding sides of congruent triangles


(ii)  
From the information shown in the figure,
in PTQ and  STR
seg PT  seg ST 
PTQ STR...... vertically opposite angles
 PTQ  STR ....... SAS test
 
 TPQ TSRand  TQP  TRS...corresponding angles of congruent triangles

seg PQ  seg SR  corresponding sides of congruent triangles

Page No 32:

Question 3:

From the information shown in the figure, state the test assuring the congruence of  ABC and  PQR. Write the remaining congruent parts of the triangles.

Answer:

​In â–³ABC and â–³QPR
AB = PQ                        (Given)
BC = PR                        (Given)
∠A = ∠Q = 90∘             (Given)
By RHS test of congruency
â–³ABC ≅ â–³QPR
∠B = ∠P                (corresponding angles of congruent triangles)
∠C = ∠R               (corresponding angles of congruent triangles)
​AC = QR         (corresponding sides of congruent triangles)

Page No 32:

Question 4:

As shown in the following figure, in  LMN and  PNM, LM = PN, LN = PM. Write the test which assures the congruence of the two triangles. Write their remaining congruent parts.

Answer:

​In â–³LMNand â–³PNM
LM = PN                         (Given)
LN = PM                        (Given)
MN = NM                      (Common)
By SSS test of congruency
â–³LMN ≅ â–³PNM
∠L = ∠P                (corresponding angles of congruent triangles)
∠LMN = ∠PNM   (corresponding angles of congruent triangles)
∠MNL = ∠NMP   (corresponding angles of congruent triangles)

Page No 32:

Question 5:

In the given figure, seg AB  seg CB and seg ADseg CD.

Prove that ABD CBD
 

Answer:

In â–³ABD and â–³CBD
AB = CB                         (Given)
AD = CD                        (Given)
BD = BD                      (Common)
By SSS test of congruency
â–³ABD ≅ â–³CBD



Page No 33:

Question 6:

In the given figure, P R seg PQ seg RQ

Prove that,  PQT RQS

Answer:

In â–³PQT and â–³RQS
∠P = ∠R                         (Given)
PQ = RQ                        (Given)
∠Q = ∠Q                      (Common)
By ASA test of congruency
â–³PQT ≅ â–³RQS



Page No 38:

Question 1:

Find the values of x and y using the information shown in figure .

Find the measure of ABD and mACD.
 

Answer:

In â–³ABC, AB = AC
∴ ∠ABC = ∠ACB            (Angles opposite to equal sides are equal) 
⇒ x = 50∘
In â–³BDC, DB = DC
∴ ∠DBC= ∠DCB            (Angles opposite to equal sides are equal) 
⇒ y = 60∘
Now, ∠ABD = ∠ABC + ∠DBC
= 50∘ + 60∘
= 110∘
Also, ∠ACD = ∠ACB + ∠DCB
= 50∘ + 60∘
= 110∘
Hence, the values of x and y are 50∘ and 60∘respectively.

Page No 38:

Question 2:

The length of hypotenuse of a right angled triangle is 15. Find the length of median of its hypotenuse.

Answer:

In a right angled triangle, the length of the median on its hypotenuse is half the length of the hypotenuse. 
Therefore, the length of median of its hypotenuse is 7.5 units.

Page No 38:

Question 3:

In PQR, Q = 90° , PQ = 12, QR = 5 and QS is a median. Find l(QS).

Answer:

In â–³PQR,
PR2 = PQ2 + QR2           (By Pythagoras theorem)
⇒ PR= 122 + 52
⇒ PR= 144 + 25
⇒ PR= 132
⇒ PR = 13
In a right angled triangle, the length of the median on its hypotenuse is half the length of the hypotenuse. 
lQS=12PR=1213
= 6.5 units.
Hence, the length of QR is 6.5 units.

Page No 38:

Question 4:

In the given figure, P point G is the point of concurrence of the medians of PQR . If GT = 2.5, find the lengths of PG and PT.


 

Answer:

The point of concurrence of medians of a triangle divides each median in the ratio 2 : 1.
Let us suppose the length of PG and GT be 2x and x.
Therefore, the value of x is 2.5
Now, PG = 2x
= 2(2.5)
= 5 units
Now, PT = PG + GT
= 5 + 2.5
= 7.5 units
Hence, the lenght of PG and GT is 5 and 7.5units respectively.



Page No 43:

Question 1:

In the given figure, point A is on the bisector of XYZ. If AX = 2 cm then find AZ.

Answer:

By Part I of the angle bisector theorem, every point on the bisector of an angle is equidistant from the sides of the angle.
∴ AZ = AX
= 2 cm
Hence, the length of AZ is 2 cm.

Page No 43:

Question 2:



In the given figure, RST = 56° , seg PT ray ST, seg PR ray SR and seg PR seg PT Find the measure of RSP. State the reason for your answer.

Answer:

By Part II of the angle bisector theorem, any point equidistant from sides of an angle is on the bisector of the angle
∠TSP = ∠RSP
Now, ∠TSP + ∠RSP = ∠TSR
⇒ ∠RSP + ∠RSP = 56∘
⇒ 2∠RSP = 56∘
⇒ ∠RSP = 28∘
Hence, the measure of ∠RSP is 28∘.

Page No 43:

Question 3:

In PQR, PQ = 10 cm, QR = 12 cm, PR = 8 cm. Find out the greatest and the smallest angle of the triangle.

Answer:

In â–³PQR, QR and PR are the longest and smallest side.
The angle opposite to the longest side is the largest angle and the angle opposite to the smallest side is the smallest angle .
Hence, the greatest and the smallest angle of the triangle are ∠QPR and ∠PQR respectively.

Page No 43:

Question 4:

In FAN, F = 80° A = 40° . Find out the greatest and the smallest side of the triangle. State the reason.

Answer:

In âˆ†FAN,
∠F + ∠A + ∠N = 180∘   (Angle sum property)
⇒ 80∘ + 40∘ + ∠N = 180∘   
⇒ 120∘ + ∠N = 180∘
⇒ ∠N = 180∘ − 120∘
= 60∘
Now, in â–³FAN,  ∠F and ∠A are the largest and smallest angle.
The side opposite to the largest angle is the greatest side and the side opposite to the smallest angle is the smallest side .
Hence, the greatest and the smallest side of the triangle are AN and ​FN respectively.

Page No 43:

Question 5:

Prove that an equilateral triangle is equiangular.

Answer:

Consider an equilateral triangle ABC.

In â–³ABC, AB = BC
∴ ∠C = ∠A               ...(1)          (Angles opposite to equal sides)
In â–³ABC, AB = CA
∴ ∠C = ∠B               ...(2)          (Angles opposite to equal sides)
From (1) and (2), we get
 ∠A = ∠B =  ∠C

Hence, an equilateral triangle is equiangular.



Page No 44:

Question 6:

Prove that, if the bisector of BAC of ABC is perpendicular to side BC, then ABC is an isosceles triangle.
 

Answer:


In â–³ADB and â–³ADC
∠BAD = ∠CAD                         (AD is the bisector of  ∠BAC)
∠BDA = ∠CDA = 90∘                (Given)
AD = DA                                     (Common)
By ASA test of congruency
â–³ADB ≅ â–³ADC
∴ ∠B = ∠C        (corresponding angles of congruent triangles)
If two angles of a traingle are equal, then the traingle is said to be a isosceles triangle.

Page No 44:

Question 7:

In the given figure, if seg PRseg PQ, show that seg PS > seg PQ.

Answer:

In â–³PRS, ∠PRS is an obtuse angle.
Since, a traingle can have maximum one obtuse angle.
Hence, ∠PRS is the greatest angle.
Therefore, the side opposite to ∠PRS i.e., PS is the longest side.
Now, seg PS > seg PR > seg RS
∴ seg PS > seg PQ                  [∵seg PRseg PQ]

 

Page No 44:

Question 8:

In the given figure, in ABC, seg AD and seg BE are altitudes and AE = BD. Prove that seg AD seg BE

Answer:

In â–³ADB and â–³BEA
∠ADB = ∠BEA = 90∘                (Given)
BD = AE                                     (Given)
AB = BA                                     (Common)
By RHS test of congruency
â–³ADB ≅ â–³BEA
∴ AD = BE        (corresponding angles of congruent triangles)
Hence proved.



Page No 47:

Question 1:

If XYZ ~   LMN, write the corresponding angles of the two triangles and also write the ratios of corresponding sides.

Answer:

Consider,  XYZ ~   LMN
XYLM=YZMN=XZLNXL, YM, and ZN

Page No 47:

Question 2:

In XYZ, XY = 4 cm, YZ = 6 cm, XZ = 5 cm, If   XYZ ~ â€‹ PQR  and PQ = 8 cm then find the lengths of remaining sides of PQR.
 

Answer:

Consider,  XYZ ~   PQR
XYPQ=YZQR=XZPR48=6QR=5PR48=6QR and 48=5PR
QR=12 cm and PR=10 cm
Hence, the lengths of remaining sides of PQR are 12 and 10 cm.

Page No 47:

Question 3:

Draw a sketch of a pair of similar triangles. Label them. Show their corresponding angles by the same signs. Show the lengths of corresponding sides by numbers in proportion.

Answer:

The pair of similar triangles is given below:



Page No 49:

Question 1:

Choose the correct alternative answer for the following questions.

(i) If two sides of a triangle are 5 cm and 1.5 cm, the lenght of its third side cannot be . . . . . . . .
(A) 3.7 cm (B) 4.1 cm (C) 3.8 cm (D) 3.4 cm

(ii)   In PQR, If R >  Q then . . . . . . . . 

(A) QR > PR
(B) PQ > PR
(C) PQ < PR
(D) QR < PR

(iii) In TPQ, T = 65° , P = 95° which of the following is a true statement ?
(A) PQ < TP
(B) PQ < TQ
(C) TQ < TP < PQ
(D) PQ < TP < TQ

Answer:

(i) 
According to the triangle inequality "the sum of the lengths of any two sides of a triangle must be greater than or equal to the length of the third side".
Therefore, the lenght of the third side cannot be greater than 3.4 cm.
Hence, the correct option is (D).

(ii)  In PQR, If  R >  Q
∴ PQ > PR                       (The sides opposite to the greater angle is greater than side opposite to the smaller angle)
Hence, the correct option is (B).

(iii) In TPQ,  T = 65° ,  P = 95°
A triangle can have maximum one obtuse angle.
T <  P
PQ < TQ                         (The sides opposite to the greater angle is greater than side opposite to the smaller angle)
Hence, the correct option is (B).
 

Page No 49:

Question 2:

ABC is isosceles in which AB = AC. Seg BD and seg CE are medians. Show that BD = CE.

Answer:


​In △ABC, AB = AC
∴ ∠B = ∠C                    (Angles opposite to equal sides)
Dividing both sides by 2, we get
12B=12C
⇒ ∠EBC = ∠DCB               ...(2)       
In â–³BEC and â–³CDB
∠EBC = ∠DCB                         [From (2)]
BE = CD                                   (E and D are the mid points of AB & AC respectively and AB = AC)
BC = CB                                     (Common)
By SAS test of congruency
â–³BEC ≅ â–³CDB
∴ BD = CE        (corresponding sides of congruent triangles)

Page No 49:

Question 3:

InPQR, If PQ > PR and bisectors of Q and R intersect at S. Show that SQ > SR.
 

Answer:

In â–³PQR, PQ > PR
Now, the angle opposite to the greater side is greater than angle opposite to the smaller side.
∴ ∠R > ∠Q
Dividing both sides by 2, we get
12R>12Q
 ⇒ ∠SRQ > ∠SQR
⇒ SQ > SR       (The sides opposite to the greater angle is greater than side opposite to the smaller angle)
Hence proved.

 

Page No 49:

Question 4:

In the figure, point D and E are on side BC of ABC, such that BD = CE and AD = AE.
Show that ABDACE.

Answer:

​In â–³ADE, AD = AE 
∴ ∠ADE = ∠AED                    (Angles opposite to equal sides)
Now, 
∠ADE + ∠ADB = 180∘            ...(1)
∠AED + ∠AEC = 180∘            ...(2)
Subtracting (2) from (1), we get
∠ADB − ∠AEC = 0
⇒ ∠ADB = ∠AEC                  ...(3)
In â–³ABD and â–³ACE
∠ADE = ∠AEC                         [From (3)]
BD = CE                                   (Given)
AD = AE                                   (Given)
By SAS test of congruency
â–³ABD ≅ â–³ACE

Page No 49:

Question 5:

In the given figure, point S is any point on side QR of PQR Prove that : PQ + QR + RP > 2PS

Answer:

In â–³PQS
PQ + QS > PS                     ...(1)     
                                                      (Sum of two sides of a traingle is greater than the third side)
In â–³PRS
RP + RS > PS                     ...(2)     
​                                                      (Sum of two sides of a traingle is greater than the third side)
Adding (1) and (2), we get
PQ + QS + RP + RS  > PS + PS
⇒ PQ + QR + RP > 2PS



Page No 50:

Question 6:

In the given figure, bisector of  BAC intersects side BC at point D. Prove that AB > BD

Answer:

Given: AD is the bisector of ∠A 
∴ ∠DAB = ∠DAC 
Now, ∠BDA is the exterior angle of the ∆DAC 
∴ ∠BDA > ∠DAC 
⇒ ∠BDA > ∠DAB        (∵ ∠DAC = ∠DAB)
⇒ AB > BD                   (The sides opposite to the greater angle is greater than side opposite to the smaller angle)
Hence proved.

Page No 50:

Question 7:

In the given figure, seg PT is the bisector of QPR. A line through R intersects ray QP at point S. Prove that PS = PR

Answer:

According to the figure, we have
PT || SR
If PT || SR and PS is a transversal line, then 
QPT = PSR   (Corresponding angles)           ....(1)
If PT || SR and PR is a transversal line, then 
TPR = PRS  (Alternate angles)              ....(2)
We are given that seg PT is the bisector of QPR.
∴ QPT = TPR         ...(3)
From (1), (2) and (3)
 PSR = PRS
The sides opposite to equal angles are also equal in a triangle. 
∴ PS = PR

Page No 50:

Question 8:


In the given figure, seg AD seg BC. seg AE is the bisector of CAB and C - E - D. Prove that
DAE = 12C - B 

 
 

Answer:

In ∆ABC, since AE bisects ∠CAB, then
∠BAE = ∠CAE                         .......(1)
In ∆ADB,   
∠ADB + ∠DAB + ∠B = 180°   [Angle sum property]
⇒ 90° + ∠DAB + ∠B = 180°
⇒∠B = 90° − ∠DAB                    .....(2)
In ∆ADC, 
∠ADC+∠DAC+∠C = 180°   [Angle sum property]
⇒ 90° + ∠DAC + ∠C = 180°
⇒∠C = 90° − ∠DAC                         .....(3)
Subtracting (2) from (3), we get   
∠C − ∠B = ∠DAC − ∠DAB
⇒ ∠C − ∠B = (∠AEC + ∠DAE) − (∠BAE − ∠DAE)
⇒ ∠C − ∠B = ∠AEC + ∠DAE − ∠BAE + ​∠DAE
​⇒ ∠C − ∠B = 2∠DAE        [From (1)]
​⇒ DAE = 12C - B 



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