PYTHAGORAS THEOREM
PYTHAGORAS THEOREM – “ IN A RIGHT ANGLED TRIANGLE, THE SQUARE ON
HYPOTENUSE IS EQUAL TO SUM OF SQUARES ON OTHER TWO SIDES”
EXERCISE 2.1
*THE SIDES OF A RIGHT ANGLED TRIANGLE CONTAINING THE RIGHT ANGLE
ARE 5 AND 12. FIND HYPOTENUSE?
P
Q
R
PQ = 12 ; QR = 5; PR = ?
ANS - IN TRIANGLE PQR,
( PR )2 = (PQ)2 + (QR)2
= (12)2 + (5)2
= 144 + 25
= 169
= 13.
*FIND THE LENGTH OF A DIAGONAL OF A SQUARE OF SIDE 12 CM.
ANS-
A
B
AB = 12 CM
LENGTH OF A DIAGONAL = x square root of two [ BOUDHAYANA THEOREM ]
= 12 square root of two.
*THE LENGTH OF A DIAGONAL OF A RECTANGULAR PLAYGROUND IS 125
METRES AND THE LENGTH OF ONE SIDE IS 75 METRES. FIND THE OTHER
SIDE.
ANS:
N
M
K
L
KM=125 MTRS
NK=75 MTRS , THEREFORE ML=75 MTRS
IN TRIANGLE KLM, BY PYTHAGORAS THM
(KM)2=(KL)2+ (ML)2
(125)2= (KL)2+ (75)2
15625= (KL)2 + 5625
15625-5625= (KL)2
10000 = (KL)2
SQUARE ROOT OF 10000 = (KL)
100 = KL
*IN RIGHT ANGLED TRIANGLE LAW, ANGLE LAW = 90, ANGLE LNA = 90. LW=26
CM, LN=6 CM, AN = 8 CM. CALCULATE LENGTH OF WA.
ANS:
L
N
N
A
W
LW= 26 CM
NA= 8 CM
LN= 6 CM
IN TRIANGLE LNA , BY PYTHAGORAS THM,
( LA )2= ( LN )2+ ( NA)2
( LA)2 = 36 + 64
( LA) 2 = 100
( LA) 2 = SQUARE ROOT OF 100
( LA)2 = 10
IN TRIANGLE LAW, BY PYTHAGORAS THM,
(LW)2 = (LA )2+( WA)2
(26 )2 = (10 )2+ (WA)2
( 676 ) = 100 + (WA)2
576 =( WA)2
SQUARE ROOT OF 576=( WA)
24 = (WA)
*A DOOR OF WIDTH 6 MTRS HAS AN ARCH ABOVE IT HAVING A HEIGHT OF
2 MTRS. FIND THE RADIUS OF THE ARCH.
ANS : { COPY DIAGRAM FROM TEXTBOOK}
LET “O” BE THE CENTRE OF ARCH. JOIN OR AND OQ. LET OR = x.
OM=OQ=x+2.
IN TRIANGLE ORQ, BY PYTHAGORAS THM
( OQ )2 = ( OR)2 +( RQ)2
( X+2 )2 = ( X )2+ (3)2
X+4X+4 =
4X+4 =
X+ 9
9
4X =
9-4
4X =
5
X =
5∕4
X =
1.25.
THEREFORE RADIUS OF THE ARCH= X +2
= 1.25+2
=3.25 MTRS.
*A PEACOCK ON A PILLAR OF 9 FEET HEIGHT ON SEEING A SNAKE COMING
TOARDS ITS HOLE SITUSTED JUST BELOW THE PILLAR FROM A DISTANCE OF
27 FEET, AWAY FROM A PILLAR WILL FLY TO CATCH IT. IF BOTH POSSESS
THE SAME SPEED, HOW FAR FROM THE PILLAR ARE THEY GOING TO MEET.
ANS:
P
Q
R
S
HEIGHT OF THE PILLAR PQ= 9 FEET
DISTANCE OF THE HOLE QS, FROM PILLAR = 27 FT
THEREFORE, RQ=27- X
IN TRIANGLE PQR, BY PYTHAGORAS THM,
(PR)2 = (PQ)2 + (QR)2
(X)2 = (9)2 + (27 –X)2
X2 = 81 + 729 - 54X + X2
54X= 81 + 729
54X= 810
X = 810 / 54
X = 15 FT
THEREFORE, 27-X = 27-15 =12 FEET.
THEREFORE, THEY ARE GOING TO MEET 12 FEET AWAY FROM PILLAR.
RIDERS:
*IN TRIANGLE MGN, IF MG=A, MN=B, MP PERPENDICULAR GN, GP=C AND PN
= D. PROVE THAT ( A+B)(A-B)= (C+D)(C-D).
ANS:
M
G
N
P
MG=A, GP=C, PN= D, MN=B. MP perpendicular TO GN.
IN TRIANGLE MPG, BY PYTHAGORAS THM,
(MG) = (MP)+(GP)
(A)
= (MP) + (C).
IN TRIANGLE MPN, BY PYTHAGORAS THM,
(MN) = (MP)+ (PN)
(B) = (MP) + (D)
SUBTRACT 1 FROM 2,
A2 – b2= (MP + C) - (MP +D)
A2 – B2=MP2 - MP2+ C2 – D2
A2-B2 =C2 – D2
(A+B) (A-B)= (C+D) (C-D)
*IN TRIANGLE ABC, BD PERPENDICULAR AC, IF AB=C, BC=A, AC=B. PROVE
THAT 1/a2+ 1/c2=1/p2.
ANS:
A
D
B
AB=c, BC=a, AC=b, BD=p.
PROOF: BD2=AD ×DC
P2 =AD×DC
BC2 =AC×DC
A2 = b ×DC
AB2 = AC×AD
C2 = b ×AD
NOW,
1∕a2 + 1∕c2 = 1∕b×DC + 1∕b×AD
= 1∕b [ 1∕DC + 1∕AD]
= 1∕b [ AD+DC∕ DC.AD]
= 1∕b[ AC∕p2]
C
=1∕b[ b∕p2]
= 1∕p2.
*DERIVE THE FORMULA FOR HEIGHT AND AREA OF EQUILATERAL TRIANGLE
OR IF SIDE OF EQUILATERAL TRIANGLE IS a, THEN P.T THE HEIGHT OF
TRIANGLE = √3∕2a AND AREA OF TRIANGLE =√3 ∕ 4𝑎2.
ANS:
A
B
C
P
AB=AC=a, BP=PC=a∕2, AP=h, AP perpendicular to BC.
PROOF: IN TRIANGLE APC, BY PYTHAGORAS THEOREM,
AC= AP+AC
a = h + a∕2
a = h + a∕4
a - a∕4 = h
4a - a∕4 = h
3a∕4 = h
h = √3𝑎⁄4
h = a∕2√3
NOW AREA OF TRIANGLE = 1∕2 × b × h
= 1∕2 ×a×a∕2√3
= √3a∕4.
EXERCISE 12.2
*VERIFY
WHETHER THE FOLLOWING ARE TRIPLETS:
* 1, 2, √3.
ANS: (2)2 = (1)2 + (√3)2
( 4)2 = (1)2 + (3)2
(4)2 = (4)2.
THEREFORE THEY FORM A TRIANGLE.
*√2, √3, √5.
ANS: (√5)2 = (√2)2 +(√3)2
(5)2 = (2)2 + (3)2
(5)2 = (5)2.
*6√3, 12, 6
ANS:
(12)2 = (6√3)2 + (6)2
(144)2= (108)2 + (36)2
(144)2 = (144)2.
*m2- n2 , 2mn, m2+n2.
Ans: (m2-n2)2 = (2mn)2 + (m2+n2)2
m4+ 2m2n2+n4= 4m2n2 +m4-2m2n2 +n4
m4+2m2n2+n4= m4 + 2m2n2+n4
*IN TRIANGLE ABC, a+b=18, b+c=25,c+a=17.WHAT TYPE OF TRIANGLE IS ABC.
GIVE REASON.
A
a
b
B
C
c
ANS: a+b=18
b+c=25
c+a=17
ADD 1, 2, 3.
2A+2B+2C = 60
2( a+b+c ) = 60
a+b+c = 60/2
a+b+c =
a+b+c
30.
=
30
18+c =
30
c
=
30-18
c
=
12
a+b+c
= 30
a + 25
= 30
a
= 30-25
a
=
5
a+b+c
= 30
b+17
= 30
b
=
30-17
b
=
13
NOW, (b)2 = (a)2 + (c)2
(132) = (5)2 + (12)2
(169) = (25) + (144)
(169) = (169) .
*IN TRIANGLE ABC, CA=2AD, BD=3AD. PROVE THAT ANGLE BCA=90 DEGREE.
ANS:
C
2M
B
3M
D
M
T.P.T = ANGLE BCA= 90 DEGREES.
PROOF = LET AD=M
CA=2AD=2M
BD=3AD=3M
IN TRIANGLE CDA, BY PYTHAGORAS THEOREM,
(CA)2 = (CD)2 + (DA)2
(2m)2 = (CD)2 + (m)2
4m2 = CD2 + m2
4m2-m2 = CD2
3m2 = CD2
CD = √3m.
IN TRIANGLE CBD, BY PYTHAGORAS THEOREM,
(BC)2 = (BD)2 + (CD)2
(BC)2 = (3)×(m) + (√3𝑚)
A
BC2 = 9m2 + 3m2
BC2 = 12m2
BC = √12𝑚
IN TRIANGLE BCA, BY PYTHAGORAS THEOREM,
(BA)2 = (BC)2+ (AC)2
(4m)2 = (√12𝑚)+(2m)2
16m2= 12m2+4m2
16m2 = 16m2
LHS= RHS
ANGLE BCA= 90 DEGREES.
*THE SHORTEST DISTANCE AP FROM A POINT A TO QR IS 12CM. Q
AND R ARE RESPECTIVELY FROM A, AT 15 AND 20 CM. PROVE THAT ANGLE
QAR = 90 DEGREES.
ANS:
A
15
Q
AP=12
20
P
R
AQ=15
AR=20.
IN TRIANGLE QAR,
(QR)2 = (QA)2 + (AR)2
= (15)2 + (20)2
= (225)+(400)
= 625.
THEREFORE, QR=25.
IN TRIANGLE QAR, BY PYTHAGORAS THEOREM
(QR)2 = (QA)2 + (RA)2
(25)2 = (15)2 + (20)2
625= 225+400
625= 625.
*IN A QUADRILATARAL, ABCD, ANGLE ADC=90 DEGREES, AB=9 CM,
AD=BC=6CM AND DC=3 CM. PROVE THAT ANGLE ACB=90 DEGREES.
ANS:
3
C
D
6
A
6
B
AD=BC=6 CM
DC=3 CM
AB=9 CM.
T.P.T= ANGLE ACB=90 DEGREES.
PROOF = IN TRIANGLE ADC, PYTHGORAS THEOREM,
(AC)2 = (AD)2+ (DC)2
(AC)2 = (6)2 + (3)2
AC2 = 36+9
AC= √45 CM
IN TRIANGLE ABC, BY PYTHAGORAS THEOREM,
(AB)2 = (BC)2 + (AC)2
(9)2
= (√45 )2+ (6)2
81 =
45 + 36
81 =
81.
THEREFORE, ANGLE ACB=90 DEGREES.
*ABCD IS A RECTANGLE, P IS ANY POINT OUTSIDE IT SUCH THAT (PA)2 +
(PC)2 = (BA)2 + (AD)2. PROVE THAT ANGLE APC= 90 DEGREES.
ANS:
P
A
D
B
C
T.P.T= ANGLE APC=90 DEGREES.
CONSTRUTION- JOIN AC.
PROOF: IN TRIANGLE ABC, (AC )2= (AB)2+ (BC)2
(AC)2 = (AB)2+ (AD)2 ( BC=AD)
GIVEN: (PA)2 + (PC)2= (BA) 2 + (AD)2
FROM 1 AND 2,
AC2 = PA2 + PC2
THEREFORE, ANGLE APC= 90 DEGREES.
*IN AN ISOSCELES TRIANGLE ABC, AB=AC, BC=18 CM, AD PERPENDICULAR
BC, AD=12 CM, BC IS PRODUCED TO E, AE=20 CM. PROVE THAT ANGLE
BAE=90 DEGREES.
ANS:
A
B
E
HERE, AD PERPENDICULAR TO BE = 12 CM.
BC= 18 CM.
AE=20 CM
JOIN AC
T.P.T= ANGLE BAE=90 DEGREES.
IN TRIANGLE ADE, BY PYTHAGORAS THEOREM,
(AE)2 = (AD)2 + (DE)2
(20)2 = (12)2 + (DE)2
(400)2= (144)2 + (DE)2
400-144 = DE2
256 = DE2
√256= DE
16 = DE.
IN TRIANGLE ADB, BY PYTHAGORAS THEOREM,
(AB)2 = (BD)2 + (DA)2
(AB)2 = (9)2 + (12)2
AB2 = 81 + 144
AB2 =
225
AB2 =
15
IN TRIANGLE ABE, BY PYTHAGORAS THEOREM,
(BE)2 = (AB)2+ (AE)2
(25)2 = (15)2 + (20)2
(625) = 225 + 400
625 = 625.
THEREFORE ANGLE BAE = 90 DEGREES.