PROOF – ar.( ABC)= ⅟2 x BC x AM

ar.( PQR)= ⅟2 x QR x PN

ar. ( ABC) ⅟2 x BC x AM ar. ( PQR) ⅟2 x QR x PN

BC AM i

QR PN

In ABM & PQN, ABM = PQN ( ABC PQR) AMB = PNQ (Each angle 90)

By AA criteria of similarity, ABM PQN

AB⁄PQ = AM⁄PN ii

But AB⁄PQ = BC⁄QR = AC⁄PR ( ABC PQR)

iii

From (ii) & (iii), AM⁄PN = BC⁄QR

iv

ar. ( ABC) BC AM ar. ( PQR) QR PN

BC BC from i QR QR BC 2

QR From (iii) ar. of ABC AB 2 BC 2 AC2

ar. of PQR PQ QR PR

Hence Proved

EXAMPLE PROBLEM

In the given fig., PB and QA are perpendiculars to segment AB. If PO=4cm, QO=7cm and area of POB=80cm2, find the area of QOA.

P

GIVEN – PB and QA are s to AB. 4cm

PO=4cm A B

OQ=7cm O

Area of POB=80cm2 7cm

TO FIND – Area of QOA Q

PROOF – In QOA and POB AOQ= BOP (Vertically opposite

angles) OAP = OBP (Each angle is 90) By AA similarity,

QOA POB P

ar.( QOA) (QO)2 4cm

ar.( POB) (PO) 2 A ar.( QOA) (7) 2 O

B 80 (4) 2

ar.( QOA) 49 7cm 80 16 ar.( QOA) 49 x 80 Q

16 245

Area of QOA = 245 sq.cm.

If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

EXAMPLE PROBLEM

ABC is a triangle right-angled at C. If p is the length of perpendicular from C to AB and AB=c, BC=a and CA=b, show that pc=ab.Area of a triangle = ⅟2 x base x heightTaking AC as base, area of ABC = ⅟2x(AC)x(BC)

= ⅟2(b)(a) (i)Taking AB as base, area of ABC = ⅟2x(AB)x(CD)

= ⅟2(c)(p) (ii)From (i) and (ii), B

⅟2(p)(c) = ⅟2(a)(b)

pc=ba a D c

C b A

p

GIVEN – ABC is a triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

TO PROVE – ABC is a right triangle, right angled at ‘B’.

CONSTRUCTION – Draw altitude BD to AC. B

A D C

PROOF – In ADB & ABC, ADB = ABC (Each angle is 90) BAD = BAC (Common angle) By AA criteria of similarity, B

ADB ABC

AB⁄AC = AD⁄AB

AB2=AC x AD i A

CSimilarly, D

In BDC & ABC, BDC = ABC (Each angle is 90) BCD = BCA (Common angle)By AA criteria of similarity, BDC ABC

BC⁄AC = DC⁄BC

BC2=AC x DC ii

From (i) & (ii),

AB2+ BC2= AC x AD + AC x DC = AC x (AD+DC) A = AC x AC = AC2

B C

AB2+BC2 =AC2 D

Hence Proved

EXAMPLE PROBLEM

In the figure, find PR and PQ, when QR=26cm, PO=6cm and OR=8cm.

GIVEN – PQR and POR are two right triangles, right angled at P and O respectively. QR=26cm P PO=6cm OR=8cm

TO FIND – Lengths of PR and PQ Q 26cm

R

6cm

8cmO

6cm

8cmO

PROOF – PR2=OP2+OR2 (Pythagoras theorem)

=(6)2+(8)2

=36+84=100

PR=√100 PR =10cm

In QPR,QR2=PQ2+PR2 (Pythagoras theorem)

(26)2=PQ2+(10) 2 P

676=PQ2+100PQ2=676-100PQ2=576PQ= √576 Q 26cm R PQ=24cm

O

6cm

8cm

STATEMENT – In a triangle, if square of one side is equal to the sum of squares of other two sides, then the angle opposite to the first side is a right triangle.

GIVEN – ABC is a triangle in which AC2=AB2+BC2

TO PROVE – ABC =90

CONSTRUCTION – Construct PQR such that AB=PQ, BC=QR and PQR=

90 A P

B C Q R

90

PROOF – AC2=AB2+BC2 (Given) In PQR, PQR = 90 PR2=PQ2+QR2 (By Pythagoras theorem)

=AB2+BC2 (PQ=AB;QR=BC-construction)=AC2 (Given)

PR2=AC2 A

PR=AC

In ABC & PQR, AB=PQ(Construction) B C BC=QR(Construction) AC=PR (Proved) P

ABC PQR (SSS congruency criteria) By C.P.C.T.

ABC= PQR= 90Hence Proved Q R

90