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Page No 38:

Question 1:

Identify, with reason, which of the following are Pythagorean triplets.
(i) (3, 5, 4)
(ii) (4, 9, 12)
(iii) (5, 12, 13)
(iv) (24, 70, 74)
(v) (10, 24, 27)
(vi) (11, 60, 61)

Answer:

(i) In the triplet (3, 5, 4),
32 = 9, 52 = 25, 42 = 16 and 9 + 16 = 25

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ (3, 5, 4) is a pythagorean triplet.

(ii) In the triplet (4, 9, 12),
42 = 16, 92 = 81, 122 = 144 and 16 + 81 = 97 ≠ 144

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴ (4, 9, 12) is not a pythagorean triplet.

(iii) In the triplet (5, 12, 13),
52 = 25, 122 = 144, 132 = 169 and 25 + 144 = 169

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ (5, 12, 13) is a pythagorean triplet.

(iv) In the triplet (24, 70, 74),
242 = 576, 702 = 4900, 742 = 5476 and 576 + 4900 = 5476

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ (24, 70, 74) is a pythagorean triplet.

(v) In the triplet (10, 24, 27),
102 = 100, 242 = 576, 272 = 729 and 100 + 576 = 676 ≠ 729

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴ (10, 24, 27) is not a pythagorean triplet.

(vi) In the triplet (11, 60, 61),
112 = 121, 602 = 3600, 612 = 3721 and 121 + 3600 = 3721

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ (11, 60, 61) is a pythagorean triplet.

Page No 38:

Question 2:

In the given figure, ∠MNP = 90°, seg NQ ⊥seg MP, MQ = 9, QP = 4, find NQ.

Answer:

We know that,
In a right angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex, is the geometric mean of the segments into which the hypotenuse is divided.

 QN2=MQ×QP             =9×4             =36QN=36      =6

Hence, NQ = 6.

Page No 38:

Question 3:

In the given figure, ∠QPR = 90°, seg PM ⊥ seg QR and Q–M–R, PM = 10, QM = 8, find QR.

Answer:

We know that,
In a right angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex, is the geometric mean of the segments into which the hypotenuse is divided.

Here, seg PM ⊥ seg QR

 PM2=QM×MR102=8×MR100=8×MRMR=1008MR=252Now,QR=QM+MR      =8+252      =16+252      =412      =20.5

Hence, QR = 20.5



Page No 39:

Question 4:

In the given figure. Find RP and PS using the information given in ∆PSR.

Answer:

In ∆PSR,
∠S = 90, ∠P = 30, ∴ ∠R = 60

By 30− 60 − 90 theorem,

RS=12×RP6=12×RP6×2=RPRP=12           ...1PS=32×RP      =32×12      =63           ...2

Hence, RP = 12 and PS = 63.

Page No 39:

Question 5:

For finding AB and BC with the help of information given in the figure, complete following activity.

AB = BC ..........           

BAC=          AB=BC=         ×AC                    =         ×8                    =         ×22                    =         

Answer:

In ∆ABC,
∠B = 90, AC = 8, AB = BC, ∴ ∠A = ∠C = 45

By 45− 45 − 90 theorem,

AB=BC=12×AC               =12×8               =12×22               =2

Hence, AB = 2 and BC = 2.

Hence, the completed activity is

AB = BC ..........    given    

BAC=    45°   AB=BC=   12   ×AC                    =   12   ×8                    =   12   ×22                    =   2   

Page No 39:

Question 6:

Find the side and perimeter of a square whose diagonal is 10 cm.

Answer:


It is given that ABCD is a square.

∴ AB = BC = CD = DA = a (say)

According to Pythagoras theorem, in ∆ABD

AB2+AD2=BD2a2+a2=1022a2=100a2=50a=50a=52 cm

Hence, the side of the square is 52 cm.

Now,
Perimeter of a square = 4×side
                                   = 4×a
                                   = 4×52
                                   = 202 cm

Hence, the perimeter of the square is 202 cm.

Page No 39:

Question 7:

In the given figure, ∠DFE = 90°, FG ⊥ ED, If GD = 8, FG = 12, find (1) EG (2) FD and (3) EF

Answer:

We know that,
In a right angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex, is the geometric mean of the segments into which the hypotenuse is divided.

Here, seg GF ⊥ seg ED

 GF2=EG×GD122=EG×8144=EG×8EG=1448EG=18

Hence, EG = 18.

Now,
According to Pythagoras theorem, in ∆DGF

DG2+GF2=FD282+122=FD264+144=FD2FD2=208FD=413

In ∆EGF

EG2+GF2=EF2182+122=EF2324+144=EF2EF2=468EF=613

Hence, FD = 413 and EF = 613.

Page No 39:

Question 8:

Find the diagonal of a rectangle whose length is 35 cm and breadth is 12 cm.

Answer:



According to Pythagoras theorem, in ∆ABC

AB2+BC2=AC2352+122=AC21225+144=AC2AC2=1369AC=37 cm

Hence, the length of the diagonal is 37 cm.

Page No 39:

Question 9:

In the given figure, M is the midpoint of QR. ∠PRQ = 90°. Prove that, PQ2 = 4PM2 – 3PR2

Answer:

According to Pythagoras theorem,

In ∆PRM

PR2+RM2=PM2RM2=PM2-PR2        ...1

In ∆PRQ

PR2+RQ2=PQ2PQ2=PR2+RM+MQ2PQ2=PR2+RM+RM2PQ2=PR2+2RM2PQ2=PR2+4RM2PQ2=PR2+4PM2-PR2         from 1PQ2=PR2+4PM2-4PR2PQ2=4PM2-3PR2

Hence, PQ2 = 4PM2 – 3PR2.

Page No 39:

Question 10:

Walls of two buildings on either side of a street are parellel to each other. A ladder 5.8 m long is placed on the street such that its top just reaches the window of a building at the height of 4 m. On turning the ladder over to the other side of the street , its top touches the window of the other building at a height 4.2 m. Find the width of the street.

Answer:



Let the length of the ladder be 5.8 m.

According to Pythagoras theorem, in ∆EAB

EA2+AB2=EB24.22+AB2=5.8217.64+AB2=33.64AB2=33.64-17.64AB2=16AB=4 m          ...1

In ∆DCB

DC2+CB2=DB242+CB2=5.8216+CB2=33.64CB2=33.64-16CB2=17.64CB=4.2 m          ...2

From (1) and (2), we get

AB+BC=4+4.2 m              =8.2 m

Hence, the width of the street is 8.2 m.



Page No 43:

Question 1:

In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS =13, find QR.

Answer:

 

In ∆PQR, point S is the midpoint of side QR.

QS=SR=12QR

PQ2+PR2=2PS2+2QS2     by Apollonius theorem112+172=2132+2QS2121+289=2169+2QS2410=338+2QS22QS2=410-3382QS2=72QS2=36QS=6 QR=2×QS           =2×6           =12


Hence, QR = 12.

Page No 43:

Question 2:

In ∆ABC, AB = 10, AC = 7, BC = 9 then find the length of the median drawn from point C to side AB

Answer:

 

In ∆ACB, point D is the midpoint of side AB.

AD=BD=12AB=5

CA2+CB2=2DC2+2AD2     by Apollonius theorem72+92=2DC2+25249+81=2DC2+225130=2DC2+502DC2=130-502DC2=80DC2=40DC=210

Hence, the length of the median drawn from point C to side AB is 210.

Page No 43:

Question 3:

In the given figure seg PS is the median of ∆PQR and PT ⊥ QR. Prove that,

(1) PR2=PS2+QR×ST+QR22

(2) PQ2=PS2-QR×ST+QR22

Answer:

According to Pythagoras theorem, in ∆PTQ

PQ2=PT2+QT2          ...1

In ∆PTS

PS2=PT2+TS2         ...2

In ∆PTR

PR2=PT2+RT2         ...3

In ∆PQR, point S is the midpoint of side QR.

QS=SR=12QR      ...4

PQ2+PR2=2PS2+2QS2     by Apollonius theoremPR2=2PS2+2QS2-PQ2PR2=PS2+PS2+QS2+QS2-PQ2PR2=PS2+PT2+TS2+QS2+QR22-PT2+QT2      From 1, 2 and 4PR2=PS2+QR22+PT2+TS2+QS2-PT2-QT2PR2=PS2+QR22+QS2+TS2-QT2PR2=PS2+QR22+QS2+TS+QTTS-QTPR2=PS2+QR22+QS2+QSTS-QTPR2=PS2+QR22+QS2+QS×TS-QS×QTPR2=PS2+QR22+QS×TS+QS2-QS×QTPR2=PS2+QR22+QS×TS+QSQS-QTPR2=PS2+QR22+QS×TS+QS×TSPR2=PS2+QR22+2QS×TSPR2=PS2+QR22+QR×TS

Hence, PR2=PS2+QR×ST+QR22.

Now,

PQ2+PR2=2PS2+2QS2     by Apollonius theoremPQ2=2PS2+2QS2-PR2PQ2=PS2+PS2+QS2+QS2-PR2PQ2=PS2+PT2+TS2+QS2+QR22-PT2+RT2      From 2, 3 and 4PQ2=PS2+QR22+PT2+TS2+QS2-PT2-RT2PQ2=PS2+QR22+QS2-RT2-TS2PQ2=PS2+QR22+QS2-TS+RTRT-TSPQ2=PS2+QR22+QS2-TS+RTRSPQ2=PS2+QR22+QS2-TS+RTQSPQ2=PS2+QR22+QS2-QS×TS-QS×RTPQ2=PS2+QR22-QS×TS+QS2-QS×RTPQ2=PS2+QR22-QS×TS-QSRT-QSPQ2=PS2+QR22-QS×TS-QSRT-SRPQ2=PS2+QR22-QS×TS-QS×TSPQ2=PS2+QR22-2QS×TSPQ2=PS2+QR22-QR×TS

Hence, PQ2=PS2-QR×ST+QR22.

Page No 43:

Question 4:

In ∆ABC, point M is the midpoint of side BC.
If, AB2 + AC2 = 290 cm2, AM = 8 cm, find BC.

Answer:

In ∆ABC, point M is the midpoint of side BC.

BM=MC=12BC

AB2+AC2=2AM2+2BM2     by Apollonius theorem290=282+2BM2290=264+2BM2290=128+2BM22BM2=290-1282BM2=162BM2=81BM=9 BC=2×BM           =2×9           =18 cm

Hence, BC = 18 cm.

Page No 43:

Question 5:

In the given figure, point T is in the interior of rectangle PQRS, Prove that, TS2 + TQ2 = TP2 + TR2 (As shown in the figure, draw seg AB || side SR and A-T-B)

Answer:

According to Pythagoras theorem, in ∆PAT

PT2=PA2+AT2          ...1

In ∆ATS

TS2=AT2+AS2         ...2

In ∆QBT

QT2=QB2+BT2         ...3

In ∆BRT

TR2=BT2+BR2         ...4

Now,
TS2+TQ2=AS2+AT2+QB2+BT2         From 2 and 3TS2+TQ2=BR2+AT2+PA2+BT2          AS=BR and PA=QBTS2+TQ2=BR2+BT2+PA2+AT2TS2+TQ2=TR2+PT2         From 1 and 4TS2+TQ2=TR2+PT2

Hence, TS2 + TQ2 = TP2 + TR2.

Page No 43:

Question 1:

Some questions and their alternative answers are given. Select the correct alternative.
(1) Out of the following which is the Pythagorean triplet?

(A) (1, 5, 10) (B) (3, 4, 5) (C) (2, 2, 2) (D) (5, 5, 2)

(2) In a right angled triangle, if sum of the squares of the sides making right angle is 169 then what is the length of the hypotenuse?
(A) 15 (B) 13 (C) 5 (D) 12

(3) Out of the dates given below which date constitutes a Pythagorean triplet ?
(A) 15/08/17 (B) 16/08/16 (C) 3/5/17 (D) 4/9/15

(4) If a, b, c are sides of a triangle and a2 + b2 = c2, name the type of triangle.
(A) Obtuse angled triangle (B) Acute angled triangle (C) Right angled triangle (D) Equilateral triangle

(5) Find perimeter of a square if its diagonal is 102 cm.
(A)10 cm (B) 402 cm (C) 20 cm (D) 40 cm

(6) Altitude on the hypotenuse of a right angled triangle divides it in two parts of lengths 4 cm and 9 cm. Find the length of the altitude.
(A) 9 cm (B) 4 cm (C) 6 cm (D) 26 cm

(7) Height and base of a right angled triangle are 24 cm and 18 cm find the length of its hypotenuse
(A) 24 cm (B) 30 cm (C) 15 cm (D) 18 cm

(8) In ∆ABC, AB = 63 cm, AC = 12 cm, BC = 6 cm. Find measure of ∠A.
(A) 30° (B) 60° (C) 90° (D) 45°

Answer:

(1) (A) In the triplet (1, 5, 10),
12 = 1, 52 = 25, 102 = 100 and 1 + 25 = 26 ≠ 100

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴ (1, 5, 10) is not a pythagorean triplet.

(B) In the triplet (3, 4, 5),
32 = 9, 42 = 16, 52 = 25 and 9 + 16 = 25

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ (3, 4, 5) is a pythagorean triplet.

(C) In the triplet (2, 2, 2),
22 = 4, 22 = 4, 22 = 4 and 4 + 4 = 8 ≠ 4

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴ (2, 2, 2) is not a pythagorean triplet.

(D) In the triplet (5, 5, 2),
22 = 4, 52 = 25, 52 = 25 and 4 + 25 = 29 ≠ 25

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴ (5, 5, 2) is not a pythagorean triplet.

Hence, the correct option is (B).


(2) According to the pythagoras theorem,
Sum of the squares of the sides making right angle is equal to the square of the third side.

∴ 169 = square of the hypotenuse
⇒ Length of the hypotenuse = 169
                                              = 13

Hence, the correct option is (B).


(3) (A) In the triplet 15/08/17,
152 = 225, 82 = 64, 172 = 289 and 225 + 64 = 289

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ 15/08/17 is a pythagorean triplet.

(B) In the triplet 16/08/16,
162 = 256, 82 = 64, 162 = 256 and 256 + 64 = 320 ≠ 256

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴ 16/08/16 is not a pythagorean triplet.

(C) In the triplet 3/5/17,
32 = 9, 52 = 25, 172 = 289 and 9 + 25 = 34 ≠ 289

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴ 3/5/17 is not a pythagorean triplet.

(D) In the triplet 4/9/15,
42 = 16, 92 = 81, 152 = 225 and 16 + 81 = 97 ≠ 225

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴ 4/9/15 is not a pythagorean triplet.

Hence, the correct option is (A).


(4) In a triangle, if the square of one side is equal to the sum of the squares of the remaining two sides, then the triangle is a right angled triangle.

Hence, the correct option is (C).


(5)



It is given that ABCD is a square.

∴ AB = BC = CD = DA = x (say)

According to Pythagoras theorem, in ∆ABD

AB2+AD2=BD2x2+x2=10222x2=200x2=100x=100x=10 cm

Hence, the side of the square is 10 cm.

Now,
Perimeter of a square = 4×side
                                   = 4×x
                                   = 4×10
                                   = 40 cm

Hence, the correct option is (D).


(6)


We know that,
In a right angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex, is the geometric mean of the segments into which the hypotenuse is divided.

 AD2=CD×DB             =4×9             =36AD=6 cm

Hence, the  correct option is (C).


(7) According to Pythagoras theorem,

Hypotenuse2=Base2+Height2                         =182+242                         =324+576                         =900 Hypotenuse=30

Hence, the correct option is (B).


(8) In ∆ABC, AB = 63 cm, AC = 12 cm, BC = 6 cm.

AB2 = (63)2 = 108
AC2 = (12)2 = 144
BC2 = (6)2 = 36

108 + 36 = 144

In a triangle, if the square of one side is equal to the sum of the squares of the remaining two sides, then the triangle is a right angled triangle.

In a right angled triangle, if one side is half of the hypotenuse then the angle opposite to that side is 30°.
Here, BC is half of AC.

Thus, measure of ∠A is 30°

Hence, the correct option is (A).



Page No 44:

Question 2:

Solve the following examples.
(1) Find the height of an equilateral triangle having side 2a.
(2) Do sides 7 cm, 24 cm, 25 cm form a right angled triangle ? Give reason.
(3) Find the length a diagonal of a rectangle having sides 11 cm and 60 cm.
(4) Find the length of the hypotenuse of a right angled triangle if remaining sides are 9 cm and 12 cm.
(5) A side of an isosceles right angled triangle is x. Find its hypotenuse.
(6) In ∆PQR; PQ = 8, QR = 5, PR = 3. Is ∆PQR a right angled triangle ? If yes, which angle is of 90°?

Answer:

(1)
 

Since, ABC is an equilateral triangle, AD is the perpendicular bisector of BC.

Now, According to Pythagoras theorem,
In ∆ABD

AB2=AD2+BD22a2=AD2+a24a2-a2=AD2AD2=3a2AD=3a

Hence, the height of an equilateral triangle is 3a.


(2) In the triplet (7, 24, 25),
72 = 49, 242 = 576, 252 = 625 and 49 + 576 = 625

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ Sides 7 cm, 24 cm, 25 cm form a right angled triangle.


(3)


According to Pythagoras theorem,

In ∆ABC

AB2+BC2=AC2602+112=AC23600+121=AC2AC2=3721AC=61 cm

Hence, the length of a diagonal of the rectangle is 61 cm.


(4) In a right angled triangle,

According to Pythagoras theorem.

Hypotenuse2=Base2+Height2                         =92+122                         =81+144                         =225 Hypotenuse=15 cm

Hence, the length of the hypotenuse is 15 cm.


(5) It is given that, a side of an isosceles right angled triangle is x.

Then, the other side of the triangle is also x.

According to Pythagoras theorem.

Hypotenuse2=x2+x2                         =2x2 Hypotenuse=2x

Hence, its hypotenuse is 2x.


(6) In ∆PQR, PQ = 8, QR = 5, PR = 3.
 (8)2 = 8,  (5)2 = 5,  (3)2 = 3 and 3 + 5 = 8

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ ∆PQR form a right angled triangle, where angle R is of 90°.

Page No 44:

Question 3:

In ∆RST, ∠S = 90°, ∠T = 30°, RT = 12 cm then find RS and ST.

Answer:

In ∆RST,
∠S = 90, ∠T = 30, ∴ ∠R = 60

By 30− 60 − 90 theorem,

RS=12×RTRS=12×12RS=6 cm          ...1ST=32×RTST=32×12ST=63 cm           ...2

Hence, RS = 6 cm and ST = 63 cm.

Page No 44:

Question 4:

Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq.cm.

Answer:



It is given that, area of rectangle is 192 sq.cm.

Area=Length×Breadth192=16×BCBC=19216BC=12 cm           ...1

According to Pythagoras theorem,

In ∆ABC

AB2+BC2=AC2162+122=AC2256+144=AC2AC2=400AC=20 cm

Hence, the length of a diagonal of the rectangle is 20 cm.

Page No 44:

Question 5:

Find the length of the side and perimeter of an equilateral triangle whose height is 3 cm.

Answer:



Since, ABC is an equilateral triangle, CD is the perpendicular bisector of AB.

Now, According to Pythagoras theorem,
In ∆ACD

AC2=AD2+CD22a2=a2+324a2-a2=33a2=3a2=1a=1 cm

Hence, the length of the side of an equilateral triangle is 2 cm.

Now,
Perimeter of the triangle = (2 + 2 + 2) cm
                                        = 6 cm

Hence, perimeter of an equilateral triangle is 6 cm.

Page No 44:

Question 6:

In ∆ABC seg AP is a median. If BC = 18, AB2 + AC2 = 260 Find AP.

Answer:

In ∆ABC, point P is the midpoint of side BC.

BP=PC=12BC=9

AB2+AC2=2AP2+2BP2     by Apollonius theorem260=2AP2+292260=2AP2+281260=2AP2+1622AP2=260-1622AP2=98AP2=49AP=7

Hence, AP = 7.



Page No 45:

Question 7:

∆ABC is an equilateral triangle. Point P is on base BC such that PC = 13 BC, if AB = 6 cm find AP.

Answer:



∆ABC is an equilateral triangle.

It is given that,
PC=13BCPC=13×6PC=2 cmBP=4 cm

Since, ABC is an equilateral triangle, OA is the perpendicular bisector of BC.
∴ OC = 3 cm
⇒ OP = OC − PC
           = 3 − 2
           = 1              ...(1)

Now, According to Pythagoras theorem,
In ∆AOB,

AB2=AO2+OB262=AO2+3236-9=AO2AO2=27AO=33 cm          ...2

In ∆AOP,

AP2=AO2+OP2AP2=332+12            From 1 and 2AP2=27+1AP2=28AP=27 cm

Hence, AP = 27 cm.

Page No 45:

Question 8:

From the information given in the figure, prove that PM = PN = 3 × a

Answer:

Since, ∆PQR is an equilateral triangle, PS is the perpendicular bisector of QR.
∴ QS = SR = a2       ...(1)

Now, According to Pythagoras theorem,
In ∆PQS,

PQ2=QS2+PS2a2=a22+PS2PS2=a2-a24PS2=4a2-a24PS2=3a24PS=3a2          ...2


In ∆PMS,

PM2=MS2+PS2PM2=a+a22+32a2PM2=3a22+32a2PM2=9a24+3a24PM2=12a24PM2=3a2PM=3a          ...3


In ∆PNS,

PN2=NS2+PS2PN2=a+a22+32a2PN2=3a22+32a2PN2=9a24+3a24PN2=12a24PN2=3a2PN=3a          ...4

From (3) and (4), we get
PM = PN = 3 × a

Hence, PM = PN = 3 × a.

Page No 45:

Question 9:

Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Answer:



Diagonals of a parallelogram bisect each other.
i.e. O is the mid point of AC and BD.

In ∆ABD, point O is the midpoint of side BD.

BO=OD=12BD

AB2+AD2=2AO2+2BO2     by Apollonius theorem       ...1

In ∆CBD, point O is the midpoint of side BD.

BO=OD=12BD

CB2+CD2=2CO2+2BO2     by Apollonius theorem        ...2

Adding (1) and (2), we get

AB2+AD2+CB2+CD2=2AO2+2BO2+2CO2+2BO2AB2+AD2+CB2+CD2=2AO2+4BO2+2AO2       OC=OAAB2+AD2+CB2+CD2=4AO2+4BO2AB2+AD2+CB2+CD2=2AO2+2BO2AB2+AD2+CB2+CD2=AC2+BD2AB2+AD2+CB2+CD2=AC2+BD2

Hence, the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Page No 45:

Question 10:

Pranali and Prasad started walking to the East and to the North respectively, from the same point and at the same speed. After 2 hours distance between them was 152 km. Find their speed per hour.

Answer:



It is given that, Pranali and Prasad have same speed.
Thus, they cover same distance in 2 hours.
i.e. OA = OB

Let the speed be x km per hour.

According to Pythagoras theorem,
In ∆AOB

AB2=AO2+OB21522=AO2+OA2450=2AO2AO2=4502AO2=225AO=15 kmBO=15 km

Speed=DistanceTime            =152            =7.5 km per hour

Hence, their speed is 7.5 km per hour.

Page No 45:

Question 11:

In ∆ABC, ∠BAC = 90°, seg BL and seg CM are medians of ∆ABC. Then prove that:
4(BL2 + CM2) = 5 BC2

Answer:


According to Pythagoras theorem,
In ∆ABC

AB2+AC2=CB2        ...1

In ∆ABC, point M is the midpoint of side BD.

BM=MA=12AB

AC2+BC2=2CM2+2AM2     by Apollonius theorem       ...2

In ∆CBA, point L is the midpoint of side AC.

CL=LA=12AC

CB2+AB2=2BL2+2AL2     by Apollonius theorem        ...3

Adding (2) and (3), we get

AC2+BC2+CB2+AB2=2CM2+2AM2+2BL2+2AL22CM2+2BL2=AC2+BC2+CB2+AB2-2AM2-2AL22CM2+BL2=AC2+2BC2+AB2-2AM2-2AL22CM2+BL2=AC2+AB2+2BC2-212AB2-212AC22CM2+BL2=BC2+2BC2-12AB2-12AC2            From 12CM2+BL2=3BC2-12AB2+AC22CM2+BL2=3BC2-12BC24CM2+BL2=23BC2-12BC24CM2+BL2=6BC2-BC24CM2+BL2=5BC2

Hence, 4(BL2 + CM2) = 5 BC2.

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Question 12:

Sum of the squares of adjacent sides of a parallelogram is 130 sq.cm and length of one of its diagonals is 14 cm. Find the length of the other diagonal.

Answer:



It is given that,
AB2 + AD2 = 130 sq. cm
BD = 14 cm

Diagonals of a parallelogram bisect each other.
i.e. O is the mid point of AC and BD.

In ∆ABD, point O is the midpoint of side BD.

BO=OD=12BD=7 cm

AB2+AD2=2AO2+2BO2     by Apollonius theorem130=2AO2+272130=2AO2+2×49130=2AO2+982AO2=130-982AO2=32AO2=16AO=4 cm

Sinec, point O is the midpoint of side AC.

 AC=2AO=8 cm

Hence, the length of the other diagonal is 8 cm.

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Question 13:

In ∆ABC, seg AD ⊥ seg BC DB = 3CD. Prove that :
2AB2 = 2AC2 + BC2

Answer:

It is given that,
DB = 3 CD

∴ BC = 4 CD        ....(1)


According to Pythagoras theorem,
In ∆ABD

AB2=AD2+DB2AD2=AB2-DB2        ...2

In ∆ACD

AC2=AD2+CD2AC2=AB2-DB2+CD2       From 2 AC2=AB2-3CD2+CD2       GivenAC2=AB2-9CD2+CD2AC2=AB2-8CD2AB2=AC2+8CD2AB2=AC2+8BC42          From 1AB2=AC2+BC222AB2=2AC2+BC2

Hence, 2AB2 = 2AC2 + BC2.

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Question 14:

In an isosceles triangle, length of the congruent sides is 13 cm and its base is 10 cm. Find the distance between the vertex opposite the base and the centroid.

Answer:

Area of the triangle = ss-as-bs-cs=a+b+c2   =13+13+102   =362   =18 cmArea of the triangle = 1818-1318-1318-10                                  = 2×3×3×5×5×2×2×2                                  = 60 sq. cmAlso, Area of the triangle = 12×base×height60=12×10×heightheight=605height=12 cm

The centroid is located two third of the distance from any vertex of the triangle.

 Distance between the vertex and the centroid=23×12=8 cm

Hence, the distance between the vertex opposite the base and the centroid is 8 cm.



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Question 15:

In a trapezium ABCD, seg AB || seg DC seg BD ⊥ seg AD, seg AC ⊥ seg BC, If AD = 15, BC = 15 and AB = 25. Find A(▢ABCD)

Answer:


 
According to Pythagoras theorem,
In ∆ABD

AB2=AD2+DB2252=152+BD2625=225+BD2BD2=625-225BD2=400BD=20

Now,
Area of the triangle ABD= ss-as-bs-cs=a+b+c2   =20+25+152   =602   =30Area of the triangle = 3030-2530-2030-15                                  = 30×5×10×15                                  = 150 sq. unitsAlso, Area of the triangle = 12×base×height150=12×25×DPDP=30025DP=12

Therefore, height of the trapezium = 12.

Now,
According to Pythagoras theorem,
In ∆ADP

AD2=AP2+DP2152=122+AP2225=144+AP2AP2=225-144AP2=81AP=9

∴ AP = QB = 9

∴ CD = PQ = 25 − (9 + 9) = 7
Area of Trapezium=12×Sum of parallel sides×Height                                =12×25+7×12                                =12×32×12                                =32×6                                =192 sq. units

Hence, A(▢ABCD) = 192 sq. units.

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Question 16:

In the given figure, ∆PQR is an equilateral triangle. Point S is on seg QR such that
QS = 13 QR.
Prove that : 9 PS2 = 7 PQ2

Answer:

Let the side of equilateral triangle ∆PQR be x.
PT be the altitude of the ∆PQR.

We know that, in equilateral triangle, altitude divides the base in two equal parts.
∴ QT = TR = 12QR=x2

Given: QS = 13 QR = x3

 ST=QT-QS=x2-x3=x6

According to Pythagoras theorem,
In ∆PQT

PQ2=QT2+PT2x2=x22+PT2x2=x24+PT2PT2=x2-x24PT2=3x24PT=3x2

In ∆PST

PS2=ST2+PT2PS2=x62+3x22PS2=x236+3x24PS2=x2+27x236PS2=28x236PS2=7x299PS2=7PQ2

Hence, 9 PS2 = 7 PQ2.

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Question 17:

Seg PM is a median of ∆PQR. If PQ = 40, PR = 42 and PM = 29, find QR.

Answer:



In ∆PQR, point M is the midpoint of side QR.

QM=MR=12QR

PQ2+PR2=2PM2+2QM2     by Apollonius theorem402+422=2292+2QM21600+1764=1682+2QM23364-1682=2QM21682=2QM2QM2=841QM=29QR=2×29QR=58

Hence, QR = 58.

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Question 18:

Seg AM is a median of ∆ABC. If AB = 22, AC = 34, BC = 24, find AM

Answer:



In ∆ABC, point M is the midpoint of side BC.

BM=MC=12BC=12

AB2+AC2=2AM2+2BM2     by Apollonius theorem222+342=2AM2+2122484+1156=2AM2+2881640-288=2AM21352=2AM2AM2=676AM=26

Hence, AM = 26.



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