PYTHAGORAS THEOREM PYTHAGORAS THEOREM IN A RIGHT ANGLED TRIANGLE, THE SQUARE ON HYPOTENUSE IS EQUAL TO SUM OF SQUARES ON OTHER TWO SIDES

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1 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 = = 169 = 13. *FIND THE LENGTH OF A DIAGONAL OF A SQUARE OF SIDE 12 CM. ANS- A B

2 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) = (KL) = (KL) = (KL) 2 SQUARE ROOT OF = (KL)

3 100 = 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 = ( LA) 2 = 100 ( LA) 2 = SQUARE ROOT OF 100 ( LA) 2 = 10 IN TRIANGLE LAW, BY PYTHAGORAS THM, (LW) 2 = (LA ) 2 +( WA) 2

4 (26 ) 2 = (10 ) 2 + (WA) 2 ( 676 ) = (WA) =( 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 = X+ 9 4X+4 = 9 4X = 9-4 4X = 5 X = 5 4 X = THEREFORE RADIUS OF THE ARCH= X +2 =

5 =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 X 2 = X + X 2 54X= X= 810 X = 810 / 54 X = 15 FT THEREFORE, 27-X = =12 FEET.

6 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, A 2 b 2 = (MP + C) - (MP +D) A 2 B 2 =MP 2 - MP 2 + C 2 D 2 A 2 -B 2 =C 2 D 2

7 (A+B) (A-B)= (C+D) (C-D) *IN TRIANGLE ABC, BD PERPENDICULAR AC, IF AB=C, BC=A, AC=B. PROVE THAT 1/a 2 + 1/c 2 =1/p 2. ANS: A D B C AB=c, BC=a, AC=b, BD=p. PROOF: BD 2 =AD DC P 2 =AD DC BC 2 =AC DC A 2 = b DC AB 2 = AC AD C 2 = b AD NOW, 1 a c 2 = 1 b DC + 1 b AD = 1 b [ 1 DC + 1 AD] = 1 b [ AD+DC DC.AD] = 1 b[ AC p 2 ]

8 =1 b[ b p 2 ] = 1 p 2. *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 4a 2. 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

9 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

10 (144) 2 = (144) 2. *m 2 - n 2, 2mn, m 2 +n 2. Ans: (m 2 -n 2 ) 2 = (2mn) 2 + (m 2 +n 2 ) 2 m 4 + 2m 2 n 2 +n 4 = 4m 2 n 2 +m 4-2m 2 n 2 +n 4 m 4 +2m 2 n2+n 4 = m 4 + 2m 2 n 2 +n 4 *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

11 a+b+c = 30. a+b+c = c = 30 c = c = 12 a+b+c = 30 a + 25 = 30 a = a = 5 a+b+c = 30 b+17 = 30 b = b = 13 NOW, (b) 2 = (a) 2 + (c) 2 (13 2 ) = (5) 2 + (12) 2 (169) = (25) + (144) (169) = (169).

12 *IN TRIANGLE ABC, CA=2AD, BD=3AD. PROVE THAT ANGLE BCA=90 DEGREE. ANS: C 2M B 3M D M A 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 4m 2 = CD 2 + m 2 4m 2 -m 2 = CD 2 3m 2 = CD 2 CD = 3m. IN TRIANGLE CBD, BY PYTHAGORAS THEOREM, (BC) 2 = (BD) 2 + (CD) 2 (BC) 2 = (3) (m) + ( 3m)

13 BC 2 = 9m 2 + 3m 2 BC 2 = 12m 2 BC = 12m IN TRIANGLE BCA, BY PYTHAGORAS THEOREM, (BA) 2 = (BC) 2 + (AC) 2 (4m) 2 = ( 12m)+(2m) 2 16m 2 = 12m 2 +4m 2 16m 2 = 16m 2 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 Q P R AP=12

14 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= = 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 6 A B

15 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 AC 2 = 36+9 AC= 45 CM IN TRIANGLE ABC, BY PYTHAGORAS THEOREM, (AB) 2 = (BC) 2 + (AC) 2 (9) 2 = ( 45 ) 2 + (6) 2 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.

16 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, AC 2 = PA 2 + PC 2 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

17 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) = DE = DE 2 256= DE 16 = DE. IN TRIANGLE ADB, BY PYTHAGORAS THEOREM, (AB) 2 = (BD) 2 + (DA) 2 (AB) 2 = (9) 2 + (12) 2 AB 2 = AB 2 = 225 AB 2 = 15

18 IN TRIANGLE ABE, BY PYTHAGORAS THEOREM, (BE) 2 = (AB) 2 + (AE) 2 (25) 2 = (15) 2 + (20) 2 (625) = = 625. THEREFORE ANGLE BAE = 90 DEGREES.

Class Notes on Pythagoras Theorem (Chapter 12) Chapter 12 English Version

Class Notes on Pythagoras Theorem (Chapter 12) Chapter 12 English Version Yakub S.,GHS Nada,Belthangady Tq, ph:9008983286,email:yhokkila@gmail.com Page 0 Pythagoras Theorem In a right angled triangle,