Triangles

(i) similar (ii) similar

(iii) equilateral (iv) equal

(v) proportional (vi) equal, proportional

(i) False (ii) True

(iii) False (iv) False

(v) True (vi) True

(i)

we have

DEBC

Therefore by basic proportionally theorem

AD/DB=AE/EC

6/9=8/EC

2/3=8/EC

EC=3x8/2

EC=3x4

EC=12 cm

(ii)

we have

DEBC

Therefore by basic proportionally theorem

AD/DB=AE/EC

Adding 1 both side

AD/DB +1=AE/EC +1

3/4 +1=AE+BC/BC

3+4/4=AC/EC [AE+EC=AC]

7/4= 15/EC

EC=15x4/7

EC=60/7

Now AE+EC=AC

AE+60/7=15

AE=15-60/7

AE=105-60/7

AE=45/7

AE=6.43 cm

(iii)

we have

DEBC

Therefore by basic proportionally theorem

AD/DB=AE/EC

Adding 1 both side

AD/DB +1=AE/EC +1

+1= +1

=

= AC/AE [AE+EC=AC]

5/2=18/AE

AE=

AE=36/5

AE=7.2 cm

(iv)

we have

DEBC

Therefore by basic proportionally theorem

AD/DB=AE/EC

=

4(3x-19)=8(x-4)

12x-76=8x-32

12x-8x=76-32

4x=44

x=44/4

x=11 cm

(v)

AD=8cm,AB=12cm

since BD=AB-AC

BD=12-8

BD=4 cm

DEBC

Therefore by basic proportionally theorem

AD/DB=AE/EC

8/4=12/EC

EC=

EC =6 cm

(vi)

we have

DEBC

Therefore by basic proportionally theorem

AD/DB=AE/EC

4/4.5=8/EC

EC=

EC=9cm

Now AE+EC=AC

AC=8+9

AC=17 cm

(vii)

AD=2cm, AB=6cm

Since BD=AB-AC

BD=6-2

BD=4 cm

DEBC

Therefore by basic proportionally theorem

AD/DB=AE/EC

Taking reciprocal on both side

DB/AD=EC/AE

4/2=EC/AE

Adding 1 both side

AD/DB +1=AE/EC +1

+1= +1

=

= AC/AE [AE+EC=AC]

3=9/AE

AE=

AE=3 cm

(viii) we have

DEBC

Therefore by basic proportionally theorem

AD/DB=AE/EC

4/5=AE/2.5

AE=4x2.5/5

AE=10/5

AE=2 cm

(ix) we have

DEBC

Therefore by basic proportionally theorem

AD/DB=AE/EC

=

x(x-1)=(x+2)(x-2)

x²-x=x²-2²

-x=-4

x=4 cm

(x) we have

DEBC

Therefore by basic proportionally theorem

AD/DB=AE/EC

=

(8x-7)(3x-1)=(4x-3)(5x-3)

8x(3x-1)-7(3x-1)=4x(5x-3)-3(5x-3)

24x²-8x-21x+7=20x²-12x-15x+9

24x²-20x²-29x+27x+7-9=0

4x²-2x-2=0

2[2x²-x-1]=0

2x²-x-1=0

2x²-2x-x-1=0

2x(x-1)+1(x-1)=0

(x-1)(2x+1)=0

x-1=0

x=1

or 2x+1=0

or x=-1/2

-1/2 is not possible.

So x=1

(xi) we have

DEBC

Therefore by basic proportionally theorem

AD/DB=AE/EC

=

(8x-7)(3x-1)=(4x-3)(5x-3)

24x²-8x-21x+7=20x²-12x-15x+9

24x²-20x²-29x+27x+7-9=0

4x²-2x-2=0

2[2x²-x-1]=0

2x²-x-1=0

2x²-2x-x-1=0

2x(x-1)+1(x-1)=0

(x-1)(2x+1)=0

x-1=0

x=1

or 2x+1=0

or x=-1/2

-1/2 is not possible.

So x=1

(xii) we have

DEBC

Therefore by basic proportionally theorem

AD/DB=AE/EC

2.5/3=3.75/EC

EC=3.75x3/2.5

EC=375x3/250

EC=15x3/10

EC=9/2

EC=4.5 cm

Now AC=AE+EC

AC=3.75+4.5

AC=8.25 cm

(i) AB = 12 cm, AD = 8 cm, and AC = 18 cm.

∴ DB=AB-AD

= 12-8

=4 cm

EC=AC-AE

= 18-12

= 6 cm

Now AD/DB=8/4=2

AE/EC=12/6=2

Thus DE divides side AB and AC of ⊿ ABC in same ratio

Then by the converse of basic proportionality theorem.

(ii)

AB = 5.6 cm, AD = 1.4 cm, AE = 1.8 cm and AC = 7.2 cm

∴ DB=AB-AD

DB=5.6-1.4

DB= 4.2 cm

And EC=AC-AE

EC= 7.2-1.8

EC=5.4

Now AD/DB=1.4/4.2=1/3

AE/EC=1.8/5.4=1/3

Thus DE divides side AB and AC of ⊿ ABC in same ratio

Then by the converse of basic proportionality theorem.

(iii)

we have

AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm and AE = 2.8 cm

∴ AD=AB-DB

AD=10.8-4.5

AD= 6.3 cm

And EC=AC-AE

EC= 4.8-2.8

EC=2 cm

Now AD/DB=6.3/4.5=7/5

AE/EC=2.8/2=28/20=7/5

Thus DE divides side AB and AC of ⊿ ABC in same ratio

Then by the converse of basic proportionality theorem.

(iv)

DE∥BC

We have,

AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and EC = 5.5 cm

Now AD/DB=5.7/9.5=57/95 =3/5

AE/EC=3.3/5.5=33/55=3/5

Thus DE divides side AB and AC of ⊿ ABC in same ratio

Then by the converse of basic proportionality theorem.

WE have,

PQ∥BC

We have AP/PB=AQ/QC

2.4/PB=2/3

PB=3x2.4/2

PB=3x1.2

PB=3.6 cm

Now AB=AP+PB

AB=2.4+3.6

AB=6 cm

Now IN ⊿ APQ and ⊿ ABC

∠A=∠A [Common]

∠APQ=∠ABC [PQ∥BC]

⊿ APQ ~⊿ ABC [By AA criteria]

AB/AP=BC/PQ

PQ=6x2.4/6

PQ=2.4 cm

DP/PE=3.9/3=1.3/1=13/10

DQ/QF=3.6/2.4=36/24=3/2

DP/PE≠DQ/QF

So PQ is not parallel to EF

(i) we have PM=4cm, QM=4.5 cm,PN=4 cm and NR=4.5 cm

Hence PM/QM=4/4.5=40/45=8/9

PN/NR=4/4.5=40/45=8/9

PM/QM= PN/NR

by the converse of proportionality theorem

MN∥QR

(ii) we have PQ=1.28cm, PR=2.56 cm,PM=0.16 cm and PN=0.32 cm

Hence PQ/PR=1.28/2.56=128/256=1/2

PM/PN=0.16/0.32=16/32=1/2

PQ/PR = PM/PN

by the converse of proportionality theorem

MN∥QR

Given: In three line segments OA, OB, and OC, points L, M, N respectively are so chosen that and but neither of L, M, N nor of A, B, C are collinear.

To show :

Solution:




We have LM∥AB and MN∥BC

by the basic proportionality theorem

OL/AL=OM/MB ……….(i)

ON/NC=OM/MB ………(ii)

Comparing equ.(i)and(ii)

OL/AL=ON/NC

Thus LN divides side OA and OC of ⊿ OAC in same ratio

Then by the converse of basic proportionality theorem

We have DE∥BC

by the converse of proportionality theorem

AD/DB=AE/EC

AD/DB=AE/DB [BD=CE]

AD=AE

Adding D both sides

AD+BD=AE+DB

AD+BD=AE+EC [BD=CE]

AB=AC

⊿ABC is isosceles

(i) we have

Angle BAD=CAD

Here AD bisects ∠A

BD/DC=AB/AC

2.5/DC=5/4.2

DC=2.5*4.2/5

DC=2.1 cm

(ii) Here AD bisects ∠A

AB/DC=AB/AC

2/3=5/AC

AC=15/2

AC=7.5 cm

(iii) in △ ABC A bisects ∠A

BD/DC=AB/BC

BD/2.8=3.5/4.2

BD=3.5*2.8/4.2

BD=7/3

BD=2.33 cm

(iv) In△ABC, AD bisects ∠A

BD/DC=AB/AC

X/6-x =10/14

14x=60-10x

14x+10x=60

24x=60

x= 60/24

x=5/2

x=2.5

BD=2.5

DC= 6-2.5

DC=3.5

(v) AB/AC=BD/DC

AB/4.2=BC-DC/DC

AB/4.2=10-6/6

AB/4.2=4/6

AB=4*4.2/6

AB=2.8 cm

(vi) BD/DC=AB/AC

BD/6=5.6/6

BD=5.6

BC= BD+DC

BC=5.6+6

BC=11.6 cm

(viii) In△ABC, AD bisects ∠A

AB/AC=BD/DC

5.6/AC=3.2/BC-BD

5.6/AC=3.2/6-3.2

5.6/AC=3.2/2.8

AC*3.2=2.8*5.6

AC=2.8*5.6/3.2

AC=7*0.7

AC=4.9 cm

(ix) let BD=x,then DC=12-X

BD/DC=AB/BC

x/12-x= 10/6

6x=120-10x

6x+10x=120

16x=120

x=120/16

x= 7.5

BD=7.5 cm

DC =12-x

DC=12-7.5

DC=4.5 cm

AE is the bisector of ∠A

We know that external bisector of an angle of a triangle divides the opposite side externally in
the ratio of the sides containing the angles.

⇒ 10X=72+6X

⇒ 10X-6X=72

⇒ 4X =72

⇒ x=72/4

⇒ x=18

We have

AB/AC=BD/DC

∴ ∠1=∠2

IN ⊿ABC

∠A+∠B+∠C=180

∠A+70+50=180

∠A+120=180

∠A=180-120

∠A=60

∠1+∠2=60 (∠1+∠2=∠A)

∠1+∠1=60 (∠1=∠2)

2∠1=60

∠1=60/2

∠1=30

∠BAD=30

∠1=∠2 (Given)

Draw a line EC∥AD

AC bisects them

∴∠2=∠3 (by alternate angle) ………………. (i)

∠1=∠4 (corresponding angle) ……………..(ii)

∠1=∠2 (given)

From equ (i) and equ (ii)

∠3=∠4

or AE=AC ……………..(III)

Now ,⊿ BCE

BD/DC=BA/AE ( BY PROPORTIONALITY THEORAM)

BD/DC=AB/AC ( ∵ BA=AB AND AE=AC from equ (iii))

Hence AB/AC=BA/DC Proved

in⊿ ABC

CF bisects ∠A

∴ AF/FB=AE/AC

AF/5-AF=4/8

2AF=5-AF

2AF+AF=5

3AF=5

AF=5/3 cm

⊿ABC, BE bisects ∠B

∴ AE/AC=AB/BC

4-CE/CE=5/8

5CE=32-8CE

5CE+8CE=32

13CE=32

CE=32/13 cm

Similarly

BD/DC=AB/AC

BD/8-BD=5/4

4BD=40-5BD

4BD+5BD=40

9BD=40

BD=40/9 cm

(i) BD/DC=AB/AC

1.5/3.5=5/10

15/35*10/10=1/2

3/7=1/2

Not bisects

(ii) 1.6/2.4=4/6

16/24=2/3

2/3=2/3

bisects

(iii) BD/CD=AB/AC

BD/BC-BD=AB/AC

BD/24-6=8/24

6/18=1/3

1/3=1/3

bisects

(iv) 1.5/2= 6/8

3/4=3/4

bisects

(v) BD/CD=AB/AC

BD/BC-BD=AB/AC

BD/9-2.5=5/12

2.5/6.5=5/12

5/13=5/12

Not bisects

AD bisects ∠A

∴ AB/AC=BD/CD

12/20=5/CD

CD =100/12

CD=8.33 cm

Diagonal of trapezium divide each other proportiona

AO/OC=BO/OD

4/4X-2=x+1/2x+4

4x²-2x+4x-2=8x+16

4x²+2x-2-8x-16=0

4x²-6x-18=0

2(2x²-3x-9)=0

2x²-3x-9=0

2x²-6x+3x-9=0

2x(x-3)+3(x-3)=0

(x-3)(2x+3)=0

x-3=0

x=3

or,2x+3=0

2x=-3

x= -3/2

x=-3/2 is not possible

So x=3

AO/OC=BO/OD

3x-1/5x-3=2x+1/6x-5

(3x-1)(6x-5)= (2x+1) (5x-3)

18x²-15x-6x+5=10x²-6x+5x-3

18x²-21x+5=10x²-x-3

18x²-21x+5-10x²+x+3=0

8x²-20x+8=0

4(2x²-5x+2)=0

2x²-5x+2=0

2x²-4x-x+2=0

2x(x-2)-1(x-2)=0

(x-2)(2x-1)=0

x-2=0

x=2

Or, 2x-1=0

2x=1

x=1/2

But x=1/2 is not possible

So x=2

AO/OC=BO/OD

3X-19/X-3=X-4/4

(x-3)(x-4)=4(3x-19)

X² -4x-3x+12=12x-76

X² -7x+12-12x+76=0

X² -19x+88=0

X² -11x-8x+88=0

X(x-11)-8(x-11)=0

(x-11)(x-8)=0

x-11=0

x=11

or x-8=0

x=8

x=11 or 8

Given ΔACB ~ ΔAPQ

Then, AC/AP = BC/PQ = AB/AQ

Or AC/2.8 = 8/4 = 6.5/AQ

Or AC/2.8 = 8/4 and 8/4 = 6.5/AQ

Or AC = 8/4 x 2.8 and AQ = 6.5 x4/8

Or AC=5.6cm and AQ = 3.25cm

Length of stick = 10cm

Length of shadow stick= 8cm

Length of shadow of tower = hcm

In ΔABC and ΔPQR

<B = <C = 90° And <C = <R (Angular elevation of sum)

Then ΔABC ~ ΔPQR (By AA similarty)

So,

Or =

Or h= x 3000

Or 3750cm

Or 37.5m

We have ΔPAB and ΔPQR

<P = <P (Common)

<PAB = <PQR (Corresponding angles)

Then, ΔPAB ~ ΔPQR (BY AA similarity)

So, (Corresponding parts of similar triangle area proportion)

Or ,

Or PB = x 6

Or PB= 2cm

We have , XY||BC

In Δ AXY and ΔABC

<A = <A (Common)

<AXY = <ABC (Corresponding angles)

Then, Δ AXY ~ΔABC (By AA Similarity)

So, (Corresponding parts of similar triangle area proportion)

Or

Or XY = 6/4

Or XY = 1.5cm

Given: In a right angled triangle with sides a and b and hypotenuse c, the altitude drawn on the hypotenuse is x.

To prove: ab = cx

Proof:

Let in a right-angled triangle ABC at B, a perpendicular from C to AB is drawn such that
BC = a
AC = b
BA = c
BD = x


In ΔABC and ΔCDB

∠B = ∠B (Common)

∠ABC = ∠CDB (Both 90°)

Then, ΔABC ~ ΔCDB (By AA Similarity)

So, (Corresponding parts of similar triangle area proportion)

Or

Or ab = cx

We have, <ABC = 90° and BD perpendicular AC

Now, <ABD + <DBC – 90° ………..(I) (<ABC – 90°)

And <C + <DBC – 90° ………..(II) (By angle sum Prop. in ΔBCD) Compare equation I &II

<ABD = <C ………..(III)

In ΔABD and ΔBCD

<ABD = <C (From equation I)

<ADB = <BDC (Each 90°)

Then, ΔABD ~ΔBCD (By AA similarity)

So, (Corresponding parts of similar triangle area proportion)

Or,

Or CD =

Or CD = 16cm

We have , <ABC = 90° and BD Perpendicular AC

In Δ ABY and ΔBDC

<C = <C (Common)

<ABC = <BDC (Each 90° angles)

Then, Δ ABC ~ΔBDC (By AA Similarity)

So, (Corresponding parts of similar triangle area proportion)

Or

Or BC = 5.7/3.8 x 8.1

Or BC = 12.15cm

We have, DE||BC, AB = 6cm and AE = 1/4 AC

In ΔADE and ΔABC

<A = <A (Common)

<ADE = <ABC (Corresponding angles)

Then, ΔADE ~ ΔABC (By AA similarity)

So, (Corresponding parts of similar triangle area proportion)

Or (AE = 1/4 AC Given)

Or ,

Or, AD = 6/4

Or, AD = 1.5cm

We have, PA ⏊ AC, and RC ⏊ AC

Let AB = a and BC = b

In ΔCQB and ΔCPA

<QCB = <PCA (Common)

<QBC = <PAC (Each 90°)

Then, ΔCQB ~ ΔCPA (By AA similarity)

So, (Corresponding parts of similar triangle area proportion)

Or, -----------(i)

In ΔAQB and ΔARC

<QAB = <RAC (Common)

<ABQ = <ACR (Each 90°)

Then, ΔAQB ~ ΔARC (By AA similarity)

So, (Corresponding parts of similar triangle area proportion)

Or, -----------(ii)

Adding equation i & ii

=

Or, y () =

Or, y () = 1

Or, =

We have, <A = <CED

In ΔCAB and ΔCED

<C = <C (Common)

<A = <CED (Given)

Then, ΔCAB ~ ΔCED (By AA similarity)

So, (Corresponding parts of similar triangle area proportion)

Or,15/9 = 9/x

Or, 15x = 90

Or, x = 90/6

Or, x = 6cm.

Assume ABC and PQR to be 2 triangle.

We, have

ΔABC ~ΔPQR

Perimeter of ΔABC = 25cm

Perimeter of ΔPQR = 15cm

AB = 9cm

PQ = ?

Since, ΔABC ~ΔPQR

Then, ratio of perimeter of triangles = ratio of corresponding sides

So, (Corresponding parts of similar triangle area proportion)

Or

Or PQ = 135/25

Or PQ = 5.4 cm

Since = = =

Then, _{ΔABC ~ ΔDEF (By SS similarity)}

Now, In _{ΔABL ~ ΔDEM}

_{<B = <E (ΔABC ~ΔDEF)}

_{<ALB =<DME (Each 90°)}

_{Then, ΔABL ~ ΔDEM (By SS similarity)}

_So, (Corresponding parts of similar triangle area proportion)

Or

Or,

We have ,

= =

And, = =

Since,

Then , by converse of basic proportionality theorem.

DE||BC

In Δ ADE and Δ ABC

<A= <A (Common)

<ADE = <B (Corresponding angles)

Then, Δ ADE ~ Δ ABC (By AA similarity)

(Corresponding parts of similar triangle are proportion)

BC = 5/2 DE

Given:- In ΔABC, D is the midpoint of BC and E is the midpoint of AD.

To prove:- BE: EX = 3 : 1

Proof:

Const:- Through D, Draw DF||BX

In ΔEAX and Δ ADF

∠EAX = ∠DAF (Common)

∠AXE = ∠DFA (Corresponding angles)

ΔEAX ~ Δ ADF

So, (Corresponding parts of similar triangle are proportion)

⇒

Or, DF = 2EX. ……………(i)

In ΔDCF and ΔBCX


∠DCY = ∠BCX (common)

∠CFD = ∠CXB (Corresponding angles)

By AA similarity,

ΔDCF ~ ΔBCX

SO, (Corresponding parts of similar triangle area proportion)

Or

Or BE + EX = 2DF

From (i)

BE + EX = 4EX

⇒ BE = 4EX – EX

⇒ BE = 4EX – EX

⇒ BE = 3EX

⇒ BE/EX =3/1

Given :- ABCD is a parallelogram

To prove :- BP x DQ = AB x BC

Proof:-








In ΔABP and ΔQDA

<B = <D (Opposite angles of parallelogram)

<BAP = <AQD (Alternative interior angle)

Then, ΔABP ~ ΔQDA

SO, (Corresponding parts of similar triangle area proportion) But, DA = BC (Opposite side of parallelogram)


But DA = BC ( opposite sides of parallelogram)

Then,

Or, AB x BC = QD X BP

Hence proved

_{We have}

_{AL⏊BC and CM⏊AB}

_{IN ΔOMA and ΔOLC}

_{<MOA = <LOC(Vertically opposite angles)}

_{<AMO = <LOC (Each 90°)}

_{Then, ΔOMA ~ΔOLC (BY AA Similarity)}

SO, (Corresponding parts of similar triangle area proportion)

We have AB||CD. If AB = 6cm, CD = xcm, EF = 10 cm, BD = 4cm and DE = ycm

In ΔECD and ΔEAB

<ECD = <EAB (Corresponding angles)

Then, ΔECD ~ ΔEAB ………(i) (By AA similarity)

SO, (Corresponding parts of similar triangle are proportion)

Or ………………(ii)

In ΔACD and ΔAEF

<CAD = <EAF (Common)

<ACD = <AEF (Corresponding angles)

Then, ΔACD ~ ΔAEF (By AA similarity)

SO,

Or, …………………….. (iii)

Adding equation iii & ii

So,

Or,

Or, 1 =

Or, x =

Or, x = 3.75cm

From (i) =

Or, =

Or, 6y = 3.75y +15

Or, 2.25y = 15

Or, y =

Or, y = 6.67cm

Given: ABCD is a quadrilateral in which AD = BC. P, Q, R, S be the mid-points of AB, AC, CD and BD respectively.
To show: PQRS is a rhombus.
Solution:
So, we have, a quadrilateral ABCD where AD = BC

And P, Q, R and S are the mid-point of the sides AB, AC, and BD.

We need to prove that PQRS is a rhombus.

In ΔBAD, P and S are the mid points of the sides AB and BD respectively,
By midpoint theorem which states that the line joining mid-points of a triangle is parallel to third side we get,

PS||AD and PS = 1/2 AD…………(i)

In ΔCAD, Q and R are the mid points of the sides CA and CD respectively,
by midpoint theorem we get,

QR||AD and QR = 1/2 AD …………..(ii)

Compare (i) and (ii)

PS||QR and PS = QR

Since one pair of opposite sides is equal and parallel,

Then, we can say that PQRS is a parallelogram…………(iii)

Now, In ΔABC,P and Q are the mid points of the sides AB and AC respectively,
by midpoint theorem,

PQ||BC and PQ = 1/2 BC…………..(iv)

And AD = BC …………………………..(v) (given)

Compare equations (i) (iv) and (v), we get,

PS = PQ ………………………………….(vi)

From (iii) and (vi), we get,

PS = QR = PQ
Therefore, PQRS is a rhombus.

Given AB⏊BC, DC ⏊ BC and DE ⏊AC

To prove:- ΔCED ~ΔABC

Proof:-

<BAC + <BCA = 90° …………..(i) (By angle sum property)

And, <BCA + <ECD = 90°……(ii) (DC ⏊ BC given)

Compare equation (i) and (ii)

<BAC = <ECD……………..(iii)

In ΔCED and ΔABC

<CED = <ABC (Each 90°)

<ECD = <BAC (From equation iii)

Then, ΔCED ~ΔABC.

Given : In ΔABC , CA – CB and AP x BQ = AC²

To prove :- ΔAPC ~ BCQ

Proof:-

AP X BQ = AC² (Given)

Or, AP x BC = AC x AC

Or, AP x BC = AC x BC (AC = BC given)

Or, AP/BC = AC/PQ ………………(i)

Since, CA = CB (Given)

Then, <CAB = <CBA …………….(ii) (Opposite angle to equal sides)

NOW, <CAB +<CAP = 180° …………(iii) (Linear pair of angle)

And <CBA + <CBQ = 180° …………..(iv) (Linear pair of angle)

Compare equation (ii) (iii) & (iv)

<CAP = <CBQ……………..(v)

In ΔAPC and ΔBCQ

<CAP = < CBQ (From equation v)

AP/BC = AC/PQ (From equation i)

Then , ΔAPC ~ ΔBCQ (By SAS similarity)

We have,

Let BC be a girl, such that CQ is distance she covered and Let AC be her shadow,

Height of girl = AB = 90cm = 0.9m

Height of lamp post = PQ = 3.6m

Speed of girl = 1.2 m/sec

So, Distance moved by the girl(CQ) = Speed x time

= 1.2 x 4 = 4.8m

Let length of shadow (AC) = 'x' cm

In ΔABC and ΔAPQ

∠ACB = ∠AQP (Each 90 °)

∠BAC = ∠PAQ (Common)

Then , ΔABC ~ ΔAPQ (By AA similarity)

So, AC/AQ = BC/ PQ(Corresponding parts of similar triangle are proportional)

Or, x/x +4.8 = 0.9/3.6

Or, x/x +4.8 = 1/4

Or, 4x – x = 4.8

Or, 3x = 4.8

Or x = 4.8/3

Or x = 1.6m

i.e. length of shadow is 1.6 m.

We have,

ABCD is a trapezium with AB || DC

In ΔAOB and ΔCOD <AOB = <COD (Vertically opposite angle)

<OAB = <OCD (Alternate interior angle)

Then, ΔAOB ~ΔCOD (By AA similarity)

So, OA/OC = OB/OD(Corresponding parts of similar triangle are proportional)

We have,

<B = <M = 90°

And, <BAC = <MAP

In ΔABC and ΔAMP

<B = <M (each 90°)

<BAC = <MAP (Given)

Then, ΔABC ~ΔAMP (By AA similarity)

So, CA/PM = BC/MP(Corresponding parts of similar triangle are proportional)

Let AB be a tower

CD be a stick , CD = 6m

Shadow of AB is BE = 28cm

Shadow of CD is DF = 4m

At same time light rays from sun will fail on tower and stick at the same angle

So, <DCF = <BAE

And <DFC = <BEA

< CDF = <ABE (Tower and stick are vertically to ground)

Therefore ΔABE ~ ΔCDF (By AAA similarity)

So, AB/CD = BE/DF

AB/6 = 28/4

AB/6 = 7

AB = 7 x 6

AB = 42 m

So, height of tower will be 42 meter.

In ΔABC, by Pythagoras theorem

AB² = AC² + BC²

Or, AB² = 5² + 12²

Or, AB² = 25 + 144

Or, AB² = = 169

Or AB = 13 (Square root both side)

In Δ AED and Δ ACB

<A = <A (Common)

<AED = <ACB (Each 90°)

Then, Δ AED ~ Δ ACB(Corresponding parts of similar triangle are proportional)

So, AE/AC = DE/ CB =AD/ AB

Or, AE/5 = DE/12 = 3/13

Or, AE/5 = 3/13 and DE/12 = 3/13

Or, AE = 15/13cm and DE = 36/13cm

(i) We have

ΔABC ~ΔDEF

Area (ΔABC) = 16cm²

Area (ΔDEF) = 25cm²

And BC = 2.3cm

Since, ΔABC ~ΔDEF

Then, Area (ΔABC)/Area (ΔDEF)

= BC²/EF² (By are of similar triangle theorem)

Or, 16/25 = (23)²/ EF²

Or, 4/5 = 2.3/EF (By taking square root)

Or, EF = 11.5/4

Or, EF = 2.875cm

(ii) We have

ΔABC ~ΔDEF

Area (ΔABC) = 9cm²

Area (ΔDEF) = 64cm²

And BC = 5.1cm

Since, ΔABC ~ΔDEF

Then, Area (ΔABC)/Area (ΔDEF)

= AB²/DE² (By are of similar triangle theorem)

Or, 9/64 = AB²/(5.1)²

Or, AB = 3 x 5.1/8 (By taking square root)

Or, AB = 1.9125cm

(iii) We have,

ΔABC ~ ΔDEF

AC = 19cm and DF = 8cm

By area of similar triangle theorem

Then, Area of ΔABC/Area of ΔDEF = AC² /DE²(Br area of similar triangle theorem)

(19)²/(8)² = 364/64

(iv) We have

Area ΔABC = 36cm²

Area ΔDEF = 64 cm²

DE = 6.2 cm

And , ΔABC ~ΔDEF

By area of similar triangle theorem

Area of ΔABC/Area of ΔDEF = AB² /DE²

Or, 36/64 = 6x 6.2/8 (By taking square root)

Or, AB = 4.65cm

(V) We have

ΔABC ~ ΔDEF

AB = 12cm and DF = 1.4 cm

By area of similar triangle theorem

Area of ΔABC/Area of ΔDEF = AB² /DE²

Or, (1.2)²/(1.4)² = 1.44x/1.96

We have,

ΔACB ~ ΔAPQ

Then, AC/AP = CB/PQ = AB/AQ[Corresponding parts of similar Δ are proportional]

Or, AC/2.8 = 10/5 = 6.5/AQ

Or, AC/2.8 = 10/5 and 10/5 = 6.5/AQ

Or, AC = 5.6cm and AQ = 3.25cm

By area of similar triangle theorem

Area of ΔACB/Area of ΔAPQ = BC² /PQ²

= (10)²/(5)²

= 100/25

= 4 cm

Given : ΔABC ~ ΔPQR

Area (ΔABC) = 81 cm²

Area (ΔPQR) = 49 cm²

And AD and PS are the altitudes

By area of similar triangle theorem: The ratio of the areas of two similar triangles equal to the ratio of squares of the corresponding sides of triangles.

We also know that:

So, Ratio of altitudes = 9/7

Hence, ratio of altitudes = Ratio of medians = 9:7

We have,

ΔABC ~ Δ PQR

Area (ΔABC) = 169cm²

Area (PQR) = 121 cm²

And AB = 26 cm

By area of similar triangle theorem

Area of ΔABC/Area of ΔPQR = AB² /PQ²

Or, 169/125 = 26²/ PQ²

Or, 13/11 = 26/PQ (Taking square root)

Or, PQ = 11/13 x 26

Or, PQ = 22cm

Given : - AB = AC, PQ = PR and <A = <P

And AD and PS are altitudes

And, Area (ΔABC)/Area of( ΔPQR) = 36/25………………..(i)

To find: AD/PS

Proof:- Since, AB = AC and PQ = PR

Then, AB/AC = 1 and PQ/PR = 1

So, AB/AC = PQ/PR

Or, AB/PQ = AC/PR……………….(ii)

In ΔABC and ΔPQR

<A = <P (Given)

AB/PQ = AC/PR (From equation ii)

Then, ΔABC ~ ΔPQR (BY AA similarity)

So, Area of ΔABC/Area of ΔPQR = AB² /PQ²…..(iii) (By area of similar triangle)

Compare equation I and II

AB²/PQ² = 36/25

Or, AB/PQ = 6/5

In ΔABD and ΔPQS

<B = <Q (ΔABC ~ ΔPQR)

<ADB = <PSO (Each 90°)

Then , ΔABD ~ ΔPQS (By AA similarity)

So, AB/ PQ = AD/PS

6/5 = AD/ PS (From iv)

We have,

ΔABC ~ Δ PQR

Area (ΔABC) = 25 cm²

Area (PQR) = 36 cm²

And AD = 2.4 cm

And AD and PS are the altitudes

To find: PS

Proof: Since, ΔABC ~ ΔPQR

Then, by area of similar triangle theorem

Area of ΔABC/Area of ΔPQR = AB² /PQ²

25/36 = AB²/PQ²

5/6 = AB/PQ………………..(i)

In ΔABD and Δ PQS

<B = <Q (ΔABC ~ ΔPQR)

<ADB = <PSQ (Each 90°)

Then, ΔABD ~ Δ PQS (By AA similarity)

So, AB/PS = AD/PS…………(ii) (Corresponding parts of similar Δ are proportional )

Compare (i) and (ii)

AD/PS = 5/6

2.4/PS = 5/6

PS = 2.4 x 6/5

PS = 2.88cm

We have,

ΔABC ~ ΔPQR

AD = 6cm

PS = 9cm

By area of similar triangle theorem

Area of ΔABC/Area of ΔPQR = AB² /PQ²…………(i)

In ΔABD and ΔPQS

<B = <Q (ΔABC ~ ΔPQS)

<ADB = <PSQ (Each 90°)

Then, ΔABD ~ ΔPQS (By AA Similarity)

So, AB/PQ = AD/PS (Corresponding parts of similar Δ are proportional)

Or, AB/PQ = 6/9

Or, AB/PQ = 2/3 ……………….(ii)

Compare equation (i) and (ii)

Area of ΔABC/Area of ΔPQR = (2/3)² = 4/9

In Δ ANC and Δ ABC

<C = <C (Common)

<ANC = <BAC (Each 90°)

Then, Δ ANC ~ Δ ABC (By AA similarity)

By area of similarity triangle theorem.

Area of ΔABC/Area of ΔPQR = AC² /BC²

Or, 5²/12²

Or, 25/144

(i) We have , DE||BC, DE = 4cm, BC = 6cm and area (ΔADE) = 16cm²

In ΔADE and ΔABC

<A = <A (Common)

<ADE = <ABC (Corresponding angles)

Then, ΔADE ~ ΔABC (BY AA similarity)

So, By area of similar triangle theorem

Area of ΔADE/Area of ΔABC = DE² /BC²

16/Area of ΔABC = 4²/6²

Or, Area (ΔABC) = 16 x 36/16

= 36cm²

(ii) We have , DE||BC, DE = 4cm, BC = 8cm and area (ΔADE) = 25cm²

In ΔADE and ΔABC

<A = <A (Common)

<ADE = <ABC (Corresponding angles)

Then, ΔADE ~ ΔABC (BY AA similarity)

So, By area of similar triangle theorem

Area of ΔADE/Area of ΔABC = DE² /BC²

25/Area of ΔABC = 4²/8²

Or, Area (ΔABC) = 25 x 64/16

= 100 cm²

(iii) We have DE||BC, And DE/BC = 3/5 ……………(i)

In ΔADE and ΔABC

<A = <A (Common)

<ADE = <ABC (Corresponding angles)

Then, ΔADE ~ ΔABC (BY AA similarity)

So, By area of similar triangle theorem

Area of ΔADE/Area of ΔABC = DE² /BC²

Area of ΔADE/Area of ΔADE + Area of trap. DECB = 3²/5²

Or, 25 area ΔADE = 9 Area of ΔADE +9 Area of trap. DECB

Or 25 area ΔADE - 9 Area of ΔADE = 9 Area of trap. DECB

Or, 16 area ΔADE = 9 Area of trap. DECB

Or, area ΔADE / Area of trap. DECB = 9/16

We have, D and E as the midpoint of AB and AC

So, according to the midpoint therom

DE||BC and DE = 1/2 BC……………..(i)

In ΔADE and ΔABC

<A = <A (Common)

<ADE = <B (Corresponding angles)

Then, ΔADE ~ ΔABC (By AA similarity)

By area of similar triangle theorem

Area ΔADE/ Area ΔABC = DE²/BC²

Or, (1/2BC)²/(BC)²

Or, 1/4

We know that area of a triangle = 1/2 x base x height

Since, ΔABC and ΔDBC are one same base.

Therefore ratio between their areas will be as ratio of their heights.

Let us draw two perpendiculars AP and DM on line BC

In ΔALO and ΔDMO,

∠ALO =∠DMO (Each is 90°)

∠AOL = ∠DOM (Vertically opposite angle)

∠OAL = ∠ODM (remaining angle)

Therefore ΔALO ~ ΔDMO (By AAA rule)

Therefore AL/DM = AO/DO

Therefore,

We have,

AB||DC

In ΔAOB and ΔCOD


∠AOB =∠COD (Vertically opposite angles)

∠OAB = ∠OCD (Alternate interior angle)

Then , ΔAOB ~ ΔCOD (By AA similarity)

(a) By area of similar triangle theorem.

b) Draw DP ⏊ AC

We know

PQ∥BC

1= AP

2 PB

In

∠A=∠A [Common]

∠APQ=∠B [Corresponding angle]

ABCAPQ

Area() =AP²

Area () AB²

ar ()___________ = 1^2/3²

ar()+ar()

9ar()= ar()+ar()

9ar()- ar()=ar()

8ar()=ar()

ar() =

ar()

We have,

ABCPQR

Area () =100cm²

Area () =49 cm²

AD= 5cm

AD and PS are the altitudes

by area of similar triangle theorem

Area() =AB²

Area () PQ²

AB² = 100/49

PQ²

AB/PQ= 10/7 ………..(i)

In ABD and PQS

∠B=∠Q [ABCPQR]

∠ADB=∠PQS=90°

ABD ~ PQS [By AA similarity]

AB/PQ=AD/PS …….(ii)

Compare equ. (i)and(ii)

AD/PS=10/7

5/PS=10/7

PS=35/10

PS=3.5 cm

We have,

ABCPQR

Area () =121cm²

Area () =64cm²

AD= 12.1cm

AD and PS are the medians

By area of similar triangle theorem

Area() =AB²

Area () PQ²

AB² =121

PQ² 64

AB =11 ………… (i)

PQ 8

ABCPQR

AB/PQ=BC/QR [Corresponding parts of similar triangles are proportional] AB/PQ=2BD/2QS [AD and BD are medians]

AB/PQ=BD/QS ………… (ii)

In ABD and PQS

∠B=∠Q [ABCPQS]

AB/PQ=BD/QS [from (ii)]

ABD ~ PQS [By AA similarity]

AB/PQ=AD/PS Compare equ. (i)and(ii)

AD/PS=11/8

12.1/PS=11/8

PS=12.1x8/8

PS= 8.8 cm

We have

ABCDEF

Where AB= 5cm

Area () =20cm²

Area () =45cm²

By area of similar triangle theorem

Area () =AB²

Area () DE²

5²/DE²=20/25

25/DE²=4/9

5/DE=2/3

DE=3x5/2

DE=7.5 cm

We know

PQ∥BC

Area () =Area ()

Area () =Area ()- Area ()

2Area ()= Area () ………(i)

In

∠A=∠A [Common]

∠APQ=∠B [Corresponding angle]

ABCAPQ

Area() =AP²

Area () AB²

Area() =AP²

Area () AB² [By using (I)]

1= AP²

2 AB²

⁼AP/AB

⁼

^=-

⁼1-BP/AB

BP/AB=1-

BP/AB=

We have,

ABCPQR

Area() =BC²

Area () QR²

(4.5)²/QR²=9/16

4.5/QR=3/4

QR=4x4.5/3

QR=6cm

AP=1 cm, PB=3 cm,AQ=1.5cm,and QC=4.5 m

In APQ and ABC

∠A=∠A [Common]

AP/AB=AQ/AC [Each equal to 1/4]

APQABC [By SAS]

By area of similar triangle theorem

Area () = 16 x ar()

We have

AD/DB=3/2

In BDE and BAC

∠B=∠B [Common]

∠BDE=∠A [Corresponding]

BDEBAC

Area() =AB²

Area () BD²

=5²/2² [AD/DB=3/2]

=25/4

Area()

Area () =25:4

ABC and BDE is an equilateral triangles

ABCDEF [By SAS]

By area of similar triangle theorem

Area() =AB² [D is the midpoint of BC]

Area () BD²

=4BD²/BD²

=4/1

Area() = 4:1 Area ()

Construct the figure according to the conditions given.

We have,

ABC is an equilateral triangle

⇒ AB=BC=AC= 2X

∵ AD⊥BC
Since perpendicular bisects the given side into two equal parts,
then BD=DC=x

Now, In ADB

AD²=AB² - BD²
AD² = (2x)²-(x)²
AD² =3x²

AD= cm

ABC and ADE both are equilateral triangles

∴ABCADE [By AA similarity]

By the theorem which states that the areas of two similar triangles are in the ratio
of the squares of the any two corresponding sides.

We have,

AB=3cm, BC=4cm, AC=6cm

∴ AB^{2 =} 3²=9

BC^{2 =} 4² =16

AC²=6²=36

Since AB²+BC²≠ AC²

SO Triangle is not a right angle.

(i) a= 7, b= 24, c=25

Here a²=49, b²=576, c²=625

=a²+b²

=49+576

=625=c²

∴ So given triangle is a right angle.

(ii) a=9, b=16, c= 18

Here a²=81, b²= 256, c²= 324

=a²+b²

=81+256

=337 ≠ c²

So given Triangle is not a right angle.

(iii) a=1.6, b=3.8, c= 4

Here a²=2.56, b²= 14.44, c²= 16

=a²+b²

=2.56+14.44

=17 ≠ c²

So given Triangle is not a right angle.

(iv) a=8, b=10, c= 6

Here a²=64, b²= 100, c²= 36

=a²+c²

=64+36

=100 = b²

So given Triangle is a right angle.

Let the man starts walk from point A and finished at

Point C.

∴ In ⊿ ABC

SO AC²=AB²+BC²

AC²=8²+15²

AC²=64+225

AC²=289

AC=

AC=17 m

The man is 17 m far from the starting point.

In ⊿ ABC

AC²=AB²+BC²

17²=15²+BC²

289=225+BC²

BC²=289 – 225

BC²=64

BC=

BC=8 m

Distance of the foot of ladder is 8 m from the building.

Let AB and CD be the poles.

AB=PD= 6m, CD=11m

BD=AP=12m

CP=CD-PD

CP=11-6

CP=5

In ⊿ APC

AC²=CP²+AP²

AC²=12²+5²

AC²=144+25

AC²=169

AC=

AC= 13m

We have,

AB=AC=25cm

BC=14cm

In ⊿ ACD and ⊿ ABD

∠ADB=∠ADB=90

AB=AC=25cm

AD=AD (Common)

⊿ ABD≅∠ACD

∴BD=CD=7cm (By c.p.c.t)

In ⊿ACD

AB²=AD²+BD²

25²=AD²+72

625=AD²+49

AD²=625-49

AD²=576

AD=

AD=24 cm

Let length of ladder be AD=BE=1m

InACD

AD²=AC²+CD²

t²= 8²+6² ………………..(i)

In BCE

BE²=BC²+EC²

t²= BC²+8² …………….. (II)

From (i) and (ii)

BC²+8²=8²+6²

BC²=6²

BC=6m

We have,

AC=14m, DC=12m, ED=BC=9m

Draw EB ⊥ AC

∴ AB=AC-BC

AB= 14-9=5m

EB=DC=12m

In ABE

AE²=AB²+BE²

AE²=5²+12²

AE²=25+144

AE²=169

AE=

AE=13m

In ABC

BC²=AB²+AC²

BC²=c²+b²

BC= ………………(i)

In ABC and In CBA

∠B= ∠B (Common)

∠ADB=∠BAC=90°

∴ ABD ~ CBA

∴ AB/CB=AD/CA

c/ =AD/b

AD=bc/

Here AB=5cm,BC=12cm, AC=13cm.

AC2=AB²+BC²

ABC is a right angled triangle at ∠B.

Area ABC=1/2(BCxBA)

=1/2(12x5)

=1/2x60

=30cm²

Also Area of ABC=1/2xACxBD

=1/2(13xBD)

30=1/2(13xBD)

13XBD=30x2

BD=60/13

BD=4.6 cm

According to the question, the figure is :

∵ ABCD is a square. Hence, AB = BC = CD = DA

∵ F is the midpoint of AB.

∴ Length of BF = AB/2 = BC/2 (∵ AB = BC)

Given that, BE = BC/3

In ΔFBE, ∠B = 90° and Area of ΔFBE = 108 cm²

⇒ BC² = 108 × 12

⇒ BC² = 36 × 36

⇒ BC = 36 cm²

AC is the diagonal of the ABCD.

⇒ AC = 36√2 = 50.904 cm

Given: isosceles triangle ABC, where AB = AC = 13 cm and the altitude from A on BC is 5 cm.

To find: The value of BC.

Solution:

In ADB

AD²+BD²=AB²

5²+BD²=13²

25+BD²=169

BD²=169-25

BD²=144

BD=

BD=12cm

In ADB and ADC

∠ADB=∠ADC =90

AB=AC=13cm

AD=AD (Common)

ADB≅ADC (By RHS condition)

BD=CD=12cm (c.p.c.t)

As BC=BD+DC

BC=12+12

BC = 24cm

(i) In ABD and ACD

ADB=ADC=90

AB=AC (given)

AD=AD (common)

ADBACD

BD=CD=a (By c.p.c.t)

In ADB

AD²+BD²=AB²

AD²+a²=(2a)²

AD²=4a²-a²

AD²=3a²

AD=a

(ii) Area of ABC=1/2xBCxAD

= 1/2x2axa

=3a²

We have,

ABCD is a rhombus

AC and BD are the diagonals with length 10cm and 24 cm respectively.

We know that rhombus of diagonal bisects each other at 90

∴ AO=OC=5cm and BO=OD=12cm

In AOB

AB²=AO²+BO²

AB²=5²+12²

AB²=25+144

AB²=169

AB=

AB=13 cm

We have,

ABCD is a rhombus with side 10 cm and diagonal BD=16 CM

We know that rhombus of diagonal bisects each other at 90

BO=OD=8cm

In AOB

AB²=AO²+BO²

10²= AO² +8²

100= AO²+64

AO²=100-64

AO²=36

AO=

AO=6 cm

∴ AC=AO+OC

AC=6+6

AC=12 cm

_{We have}

In ABC, AD is median

AE⊥BC

In AEB

AB²=AE²+BE²

AB²=AD²-DE²+(BD-DE)²

AB²=AD²-DE²+BD²-2xBDxDE+DE²

AB²=AD²+BD²-2xBDxDE

AB²=AD²+BC²/4-BCxDE …………. (I) [GIVEN BC=2BD]

In AEC

AC²=AE²+EC²

AC²=AD²-DE²+ (DE+CD)²

AC²=AD²-DE²+2CDxDE

AC²=AD²+BC²/4+BCxDE ……….(II) [BC=2CD]

By adding equ. (i) and (ii) we get

AB²+AC²=2AD²+BC²/2

2AB²+2AC²=4AD²+BC² [MULTIPLY BY 2]

4AD²=2AB²+2AC²-BC²

AD²=2AB²+2AC²-BC²

ABC is an equilateral triangle with side 12cm

AE⊥BC

In ABD andACD

∠ADB=∠ADC=90

AB=AC=12cm

AD=AD (COMMON)

ABDACD

AD²+BD²=AB²

AD²+6²=12²

AD²+36=144

AD²=144-36

AD²=108

AD=

AD=10.39 cm

Given: In right-angled triangle ABC in which, if D is the mid-point of BC.

To prove:


Solution:





We have

∠C=90 and D is the midpoint of BC

In ABC

AB²=AC²+BC²

⇒ AB²=AC²+ (2CD)²

⇒ AB²=AC²+4CD²

⇒ AB²=AC²+4(AD²-AC²)

AB²=AC²+4AD²-4AC²

AB²=4AD²-3AC²

We have

D is the midpoint of BC

(i) In AEC

AC²=AE²+EC²

b²=AE²+(ED+DC)²

b²=AD²+DC²+2xEDxDC (Given BC=2CD)

b²=p²+(a/2)²+2(a/2)x

b²=p²+a²/4+ax

b²=p² +ax+a²/4 ………….. (i)

(ii) In AEB

AB²=AE²+BE²

c²=AD²-ED²+(BD-ED)²

c²=p²-ED²+BD²+ED²-2BDxED

c²=P²+(a/2)²-2(a/2)²x

c²=p²-ax+a²/4 ……………….(ii)

(iii) Adding equ. (i)and(ii) we get

b²+c²=2p²+a²/2

In ADC

AC²=AD²+DC²

b²=h²+(a-x)²

b²=h²+a²-2ax+x² ………… (i)

b²=h²+x²-2ax

b²=a²+ (h²+x2)-2ax (from equ.i)

b²=a²+c²-2ax [h²+x²=c²]

Draw the diagram according to given questions.







(I)

In ⊿APB and ⊿AQC

∠A = ∠A (common)

∠P = ∠Q ( both 90^° )

∴ ⊿APB~⊿AQC [By AA similarity]

{Corresponding part of similar triangle are proportional}

AP x AC = AQ x AB ………….(1)

In ⊿BPC

By pythagoras theoram,

BC² = BP² + PC²

Also in ⊿BPA

⇒ BC² = AB² - AP² + (AP + AC)²

⇒ BC² = AB² - AP² + AP² + AC² + 2AP x AC

BC² = AB² + AC² + 2AP x AC ……..(ii)

In ⊿BQC

BC² = CQ² + BQ²

BC² = AC² - AQ² + (AB + AQ)²

BC² = AC² - AQ² + AB² + 2AB x AQ

BC² = AC² + AB² + AQ² + 2AB x AQ ………….(iii)

Adding equ. (ii)and(iii)

⇒ 2BC² = 2AC² + 2AB² + 2AP x AC + 2AB x AQ

⇒ 2BC² = 2AC[AC + AP] + AB[AB + AQ]

⇒ 2BC² = 2AC x PC + 2AB x BQ

⇒ BC² = AC x PC + AB x BQ

Hence proved.

We have

∠C=90 and D is the midpoint of BC

LHS=BC²

=(2CD)²

=4CD²

=4(AD²-AC²) =RHS

We have

∠B = 90 and

AD² = AB² + BC² + CD² (Given)

But AB² + BC² = AC²

AD² = AC² + CD²

By converse of by Pythagoras

∠ACD = 90

We have _{⊿ABC is an equilateral triangle and AD⊥BC}

_{In⊿ADB⊿ADC}

_{∠ADB=∠ADC=90°}
AB=AC (Given)

_{AD=AD (Common)}

_{⊿ADB≅⊿ADC (By RHS condition)}

_{∴BD=CD=BC/2 ……. (i)}

_In⊿ABD

_BC²=AD²+BD²

_BC²=AD²+BD² [Given AB=BC]

_(2BD)²= AD²+BD² [From (i)]

_4BD²-BD²=AD²

_AD²=3BD²

_{(i) In⊿ABD and In⊿CAB}

_{∠DAB=∠ACB=90°}

_{∠ABD=∠CBA [Common]}

_{∠ADB=∠CAB [remaining angle]}

_{So,⊿ADB≅⊿CAB [By AAA Similarity]}

_{∴AB/CB=BD/AB}

_AB²=BCxBD

_(ii)

Let <CAB= x

InΔCBA=180-90°-x

<CBA=90°-x

Similarly in ΔCAD

<CAD=90°-<CAD=90°-x

<CDA=90°-<CAB

=90°-x

<CDA=180°-90°-(90°-x)

<CDA=x

Now in ΔCBA and ΔCAD we may observe that

<CBA=<CAD

<CAB=<CDA

<ACB=<DCA=90°

Therefore ΔCBA~ΔCAD ( by AAA rule)

Therefore AC/DC=BC/AC

AC²=DCxBC

(iii) In DCA and ΔDAB

<DCA=<DAB (both angles are equal to 90°)

<CDA=. <ADB (common)

<DAC=<DBA

ΔDCA= ΔDAB (AAA condition)

Therefore DC/DA=DA/DB

AD²=BDxCD

(iv) From part (I) AB²=CBxBD

From part (II) AC²=DCxBC

Hence AB²/AC²=CBxBD/DCxBC

AB²/AC²=BD/DC

Hence proved

Let OB be the pole and AB be the wire.

AB² =OB²+OA²

24²=18²+OA²

OA²=576-324

OA²=252

AO=

AO=6√7 m.

Distance from base=6√7 m

Distance traveled by the plane flying towards north in 11/2 hrs

=1000x1=1500km

Similarly Distance traveled by the plane flying towards west in 11/2hrs

=1200x1 =1800 km

Let this distance is represented by OA and OB

Distance between these place after 1hrs AB=

= =

===300

=300x7.8102

= 2343.07 km

So, distance between these places will be 2343 km (Approx) km, after 1 1/2 hrs

Let ABC be the triangle

Where AB=(a-1)² cm

BC=2√a cm

CA=(a+1) cm

AB²=(a-1)²=a²+1-2a

BC²=(2√a)²=4a²

CA²=(a+1)²=a²+1+2a

Hence AB²+BC²=AC²

SO ΔABC is a right angles triangle at B

Given A vertical stick 20 m long casts a shadow 10 m long on the ground and a tower casts a shadow 50 m long on the ground.

We know that AAA similarity criterion states that in two triangles, if corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.

In ΔABC and ΔDEF,

∠A = ∠D = 90°, ∠C = ∠F

∴ ΔABC ~ ΔDEF

We know that if two triangles are similar then their sides are proportional.

∴ DE = 100 m

Basic Proportionality Theorem: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

Converse of Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

From the given figure ΔABC, DE || BC.

Let EC = x cm.

We know that basic proportionality theorem states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

Then

⇒ x = 12 cm = EC

Here, AC = AE + EC

⇒ AC = 8 + 12 = 20 cm

∴ AC = 20 cm

Given sides of two similar triangles are in the ratio 4: 9.

We know that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

∴ ar (Δ1): ar (Δ2) = 16: 81

Given AD is the bisector of ∠A in ΔABC. Let AC be x cm.

We know that the angle bisector theorem states that the internal bisector of an angle of a triangle divides the opposite side in two segments that are proportional to the other two sides of the triangle.

⇒ x = 4 cm

∴ AC = 4 cm

Given that area of two similar triangles are 9 cm² and 16 cm².

We know that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

∴ Ratio of their corresponding sides is 3: 4.

Given that area of two similar triangles ΔABC and ΔDEF are 144 cm² and 81 cm². Also the longest side of larger ΔABC is 36 cm.

We have to find the longest side of the smaller triangle ΔDEF. Let it be x.

We know that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

⇒ x = 27 cm

∴ Longest side of ΔDEF is 27 cm.

AAA similarity criterion: In two triangles, if corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.

Given ΔABC and ΔBDE are two equilateral triangles such that D is the midpoint of BC.

Since the given triangles are equilateral, they are similar triangles.

And also since D is the mid-point of BC, BD = DC.

We know that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

∴ ar (ΔABC): ar (ΔBDE) = 4: 1

SSS similarity criterion: If in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.

SAS similarity criterion: If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

Given two isosceles triangles have equal angles and their areas are in the ratio 16 : 25.

We know that SAS similarity criterion states that if one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

In ΔABC and ΔDEF,

if and ∠A = ∠D, then ΔABC ~ ΔDEF

We know that the ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding altitudes.

∴ AG: DH = 4: 5

Given DE || BC, AD = 1 cm and DB = 2 cm.

So, AB = 3 cm.

In ΔABC and ΔADE,

∠ABC = ∠ADE [corresponding angles]

∠ACB = ∠AED [corresponding angles]

∠A = ∠A [common angle]

We know that AAA similarity criterion states that in two triangles, if corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.

∴ ΔABC ~ ΔADE

We know that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

∴ ar (ΔABC): ar (ΔADE) = 9: 1

Given ΔABC and ΔDEF are similar triangles such that 2AB = DE and BC = 8 cm

We know that if two triangles are similar then their sides are proportional.

For ΔABC and ΔDEF,

∴ EF = 16 cm

Given DE || BC, AD = 2.4 cm, DB = 3.6 cm and AC = 5 cm.

We have to find AE.

We know that basic proportionality theorem states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

⇒ 2.4 (5 – AE) = 3.6 AE

⇒ 12 – 2.4 AE = 3.6 AE

⇒ 12 = 3.6 AE + 2.4 AE

⇒ 12 = 6 AE

⇒ AE = 12/6

∴ AE = 2 cm

Given ΔABC and ΔDEF are two triangles such that

We know that if two triangles are similar then their sides are proportional.

Since , ΔABC and ΔDEF are similar.

We know that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

∴ ar (ΔABC): ar (ΔDEF) = 4: 25

Given ΔABC ~ ΔPQR, ar (ΔABC): ar (ΔPQR) = 9: 16 and BC = 4.5 cm

We know that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

⇒ QR² = 2.25 (16)

⇒ QR² = 36

⇒ QR = 6

∴ The length of QR is 6 cm.

Given that ΔABC and ΔDEF are similar triangles such that AB = 3 cm, BC = 2 cm, CA = 2.5 cm and EF = 4 cm.

We know that two triangles are similar if their corresponding sides are proportional.

First consider,

⇒ DE = 6 cm … (1)

Now,

⇒ DF = 5 cm … (2)

Then, perimeter of ΔDEF = DE + EF + DF = 6 + 4 + 5

∴ Perimeter of ΔDEF = 15 cm

Given ΔABC ~ ΔPQR, ar (ΔABC): ar (ΔPQR) = 169: 121 and BC = 26 cm

We know that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

⇒ QR² = 4 (121)

⇒ QR² = 484

⇒ QR = 22

∴ The length of QR is 22 cm.

Given XY is drawn parallel to the base BC of a ΔABC cutting AB at X and AC at Y. AB = 4BX and YC = 2 cm.

In ΔAXY and ΔABC,

∠AXY = ∠ABC [corresponding angles]

∠AYX = ∠ACB [corresponding angles]

∠A = ∠A [common angle]

We know that AAA similarity criterion states that in two triangles, if corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.

∴ ΔAXY ~ ΔABC

Let BX = x, so AB = 4x and AX = 3x.

We know that two triangles are similar if their corresponding sides are proportional.

∴ AY = 6 cm

Given ABC and DEF are two similar triangles, ∠A = 57° and ∠E = 73°

We know that SAS similarity criterion states that if one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

In ΔABC and ΔDEF,

if and ∠A = ∠D, then ΔABC ~ ΔDEF

So, ∠A = ∠D

⇒ ∠D = 57° … (1)

Similarly, ∠B = ∠E

⇒ ∠B = 73° … (2)

We know that the sum of all angles of a triangle is equal to 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ 57° + 73° + ∠C = 180°

⇒ 130° + ∠C = 180°

⇒ ∠C = 180° - 130° = 50°

∴ ∠C = 50°

Given two poles of heights 6 m and 11 m stand vertically upright on a plane ground. Distance between their foot is 12 m.

Let CD be the pole with height 6 m. AB is the pole with height 11m and DB = 12 m

Let us assume a point E on the pole AB which is 6m from the base of AB.

Hence AE = AB – 6 = 11 – 6 = 5m

We know that the Pythagoras theorem state that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Now, in right triangle AEC,

⇒ AC² = AE² + EC²

Since CDEB forms a rectangle and opposite sides of rectangle are equal,

⇒ AC² = 5² + 12²

= 25 + 144

= 169

⇒ AC = 13

∴ The distance between their tops is 13 m.

Given altitudes of two similar triangles are in ratio 2: 3.

Let first triangle be ΔABC and second triangle be ΔPQR.

We know that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

∴ ar (ΔABC): ar (ΔPQR) = 4: 9

Given in ΔABC, XY || BC and BY is a bisector of ∠XYC.

Since XY || BC,

∠YBC = ∠BYC [alternate angles]

Now, in Δ BYC, two angles are equal.

Hence, two corresponding sides will be equal.

∴ BC = CY

Given that ΔABC and ΔDEF are two triangles such that

Here, the corresponding sides are given proportional.

We know that two triangles are similar if their corresponding sides are proportional.

And we know that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

∴ Area (ΔABC): Area (ΔDEF) = 9: 16

From the given figure ΔABC, DE || BC.

Let AC = x cm.

We know that basic proportionality theorem states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

Then

⇒ x = 4.4 cm

∴ AC = 4.4 cm

Given that ΔABC and ΔDEF are similar triangles such that AB = 3 cm, BC = 2 cm, CA = 2.5 cm and EF = 4 cm.

We know that two triangles are similar if their corresponding sides are proportional.

First consider,

⇒ DE = 6 cm … (1)

Now,

⇒ DF = 5 cm … (2)

Then, perimeter of ΔDEF = DE + EF + DF = 6 + 4 + 5

∴ Perimeter of ΔDEF = 15 cm

Given in triangles ABC and DEF, ∠A = ∠E = 40°, AB: ED = AC: EF and ∠F = 65°.

We know that SAS similarity criterion states that if one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

In ΔABC and ΔDEF,

∠A = ∠E and AB: ED = AC: EF then ΔABC ~ ΔDEF

So, ∠A = ∠E = 40°

⇒ ∠C = ∠F = 65°

Similarly, ∠B = ∠D

We know that the sum of all angles of a triangle is equal to 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ 40° + ∠B + 65° = 180°

⇒ ∠B + 115° = 180°

⇒ ∠B = 180° - 115° = 75°

∴ ∠B = 75°

Given ABC and DEF are two similar triangles, ∠A = 47° and ∠E = 83°

We know that SAS similarity criterion states that if one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

In ΔABC and ΔDEF,

if and ∠A = ∠D, then ΔABC ~ ΔDEF

So, ∠A = ∠D

⇒ ∠D = 47° … (1)

Similarly, ∠B = ∠E

⇒ ∠B = 83° … (2)

We know that the sum of all angles of a triangle is equal to 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ 47° + 83° + ∠C = 180°

⇒ 130° + ∠C = 180°

⇒ ∠C = 180° - 130° = 50°

∴ ∠C = 50°

Pythagoras Theorem: In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

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

Given the lengths of the diagonals of a rhombus are 30 cm and 40 cm.

Let the diagonals AC and BD of the rhombus ABCD meet at point O.

We know that the diagonals of the rhombus bisect each other perpendicularly.

Also we know that Pythagoras theorem states that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Consider right triangle AOD,

⇒ AD² = AO² + OD²

= 15² + 20²

= 225 + 400

= 625

⇒ AD = 25 cm

∴ The side of the rhombus is 25 cm.

Given D, E and F are the mid-points of sides BC, CA and AB respectively of ΔABC.

Then DE || AB, DE || FA … (1)

And DF || CA, DF || AE … (2)

From (1) and (2), we get AFDE is a parallelogram.

Similarly, BDEF is a parallelogram.

In ΔADE and ΔABC,

⇒ ∠FDE = ∠A [Opposite angles of ||gm AFDE]

⇒ ∠DEF = ∠B [Opposite angles of ||gm BDEF]

∴ By AA similarity criterion, ΔABC ~ ΔDEF.

We know that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

∴ ar (ΔDEF): ar (ΔABC) = 1: 4

Given in ΔABC, ∠A = 90°, AB = 5 cm, AC = 12 cm and AD ⊥ BC

In ΔACB and ΔADC,

∠CAB = ∠ADC [90°]

∠ABC = ∠CAD [corresponding angles]

∠C = ∠C [common angle]

We know that AAA similarity criterion states that in two triangles, if corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.

∴ ΔACB ~ ΔADC

∴ AD = 60/13 cm

Given in the given figure PQ || BC and AP: PB = 1: 2

We know that basic proportionality theorem states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

Since Δ APQ and ΔABC are similar,

Given

⇒ PB = 2AP

So,

we know that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

∴ Area (ΔAPB): Area (ΔABC) = 1: 9

Given in equilateral ΔABC, AD ⊥ BC.

We know that the Pythagoras theorem state that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In ΔABD,

⇒ AB² = AD² + BD²

⇒ AB² = AD² + ( 1/2BC)² [∵ BD = 1/2BC]

⇒ AB² = AD² + ( 1/2AB)² [∵ AB = BC]

⇒ AB² = AD² + 1/4AB²

∴ 3AB² = 4AD²

Given ∠M = ∠N = 46°

It forms a pair of corresponding angles, hence LM || PN.

In ΔLMK and ΔPNK,

∠LMK = ∠PNK [corresponding angles]

∠MLK = ∠NPK [corresponding angles]

∠K = ∠K [common angle]

We know that AAA similarity criterion states that in two triangles, if corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.

∴ ΔLMK ~ ΔPNK

We know that two triangles are similar if their corresponding sides are proportional.

Given ST || QR, TR = 4 cm and PT = 2 cm.

So, PR = 6 cm.

In ΔPST and ΔPQR,

∠PST = ∠PQR [corresponding angles]

∠PTS = ∠PRQ [corresponding angles]

∠P = ∠P [common angle]

We know that AAA similarity criterion states that in two triangles, if corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.

∴ ΔPST ~ ΔPQR

We know that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

∴ ar (ΔPST): ar (ΔPQR) = 1: 9

Given in an equilateral ΔABC, AD ⊥ BC

Since AD ⊥ BC, BD = CD = BC/2

We know that the Pythagoras theorem state that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Now, in right triangle ADC,

⇒ AC² = AD² + DC²

⇒ BC² = AD² + DC²

⇒ (2DC)² = AD² + DC²

⇒ 4DC² = AD² + DC²

⇒ 3DC² = AD²

∴ 3CD² = AD²

Given ΔAHK ~ ΔABC, AK = 10 cm, BC = 3.5 cm and HK = 7 cm.

We know that two triangles are similar if their corresponding sides are proportional.

∴ AC = 5 cm

Given in ΔABC, AD ⊥ BC, BD = 8 cm, DC = 2 cm and AD = 4 cm.

We know that the Pythagoras theorem state that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Now, in right triangle ADC,

⇒ AC² = AD² + DC²

⇒ AC² = (4)² + (2)²

= 16 + 4

∴ AC² = 20 … (1)

In ΔADB,

⇒ AB² = AD² + BD² = 4² + 8² = 16 + 64

∴ AB² = 80 … (2)

Now, in ΔABC,

⇒ BC² = (CD + DB)² = (2 + 8)² = 10² = 100

And AB² + CA² = 80 + 20 = 100

∴ AB² + CA² = BC²

Hence, ΔABC is right angled at A.

Given DE || BC, BC = 8 cm, AB = 6 cm and DA = 1.5 cm.

So, PR = 6 cm.

In ΔABC and ΔADE,

∠ABC = ∠ADE [corresponding angles]

∠ACB = ∠AED [corresponding angles]

∠A = ∠A [common angle]

We know that AAA similarity criterion states that in two triangles, if corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.

∴ ΔABC ~ ΔADE

We know that two triangles are similar if their corresponding sides are proportional.

∴ DE = 2 cm

Given in ΔABC, point D is on side AB and point E is on side AC, such that BCED is a trapezium and DE: BC = 3: 5.

In ΔABC and ΔADE,

∠ABC = ∠ADE [corresponding angles]

∠ACB = ∠AED [corresponding angles]

∠A = ∠A [common angle]

We know that AAA similarity criterion states that in two triangles, if corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.

∴ ΔABC ~ ΔADE

We know that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Let ar (ΔADE) = 9x sq. units and ar (ΔABC) = 25x sq. units

⇒ ar (trap BCED) = ar (ΔABC) – ar (ΔADE)

= 25x – 9x

= 16x sq. units

Now,

∴ ar (ΔADE): ar (trap BCED) = 9: 16

Given DE || BC, AD = 1/2 BD and BC = 4.5 cm

In ΔABC and ΔADE,

∠ABC = ∠ADE [corresponding angles]

∠ACB = ∠AED [corresponding angles]

∠A = ∠A [common angle]

We know that AAA similarity criterion states that in two triangles, if corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.

∴ ΔABC ~ ΔADE

We know that two triangles are similar if their corresponding sides are proportional.

∴ DE = 1.5 cm

Given AD is the bisector of ∠BAC. AB = 6 cm, AC = 5 cm and BD = 3 cm.

We know that the internal bisector of angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

∴ DC = 2.5 cm

Given AD is the bisector of ∠BAC. AB = 8 cm, DC = 3 cm and BD = 6 cm.

We know that the internal bisector of angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

∴ AC = 4 cm

Given ABCD is a trapezium in which BC || AD and AD = 4 cm.

Also, the diagonals AC and BD intersect at O such that

In ΔAOD and ΔCOB,

∠OAD = ∠OCB [alternate angles]

∠ODA = ∠OBC [alternate angles]

∠AOD = ∠BOC [vertically opposite angles]

We know that AAA similarity criterion states that in two triangles, if corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.

∴ ΔAOD ~ ΔCOB

We know that two triangles are similar if their corresponding sides are proportional.

∴ BC = 8 cm

Given ABC is an isosceles triangles and AD ⊥ BC.

We know that in an isosceles triangle, the perpendicular from the vertex bisects the base.

∴ BD = DC

We know that the Pythagoras theorem state that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Now, in right triangle ABD,

⇒ AB² = AD² + BD²

⇒ AB² – AD² = BD²

⇒ AB² – AD² = BD (BD)

Since BD = DC,

∴ AB² – AD² = BD (DC)

Given ΔABC is a right triangle right-angled at A and AD ⊥ BC.

⇒ ∠CAD + ∠BAD = 90° … (1)

⇒ ∠BAD + ∠ABD = 90° … (2)

From (1) and (2),

∠CAD = ∠ABD

By AA similarity,

In ΔADB and ΔADC,

⇒ ∠ADB = ∠ADC [90° each]

⇒ ∠ABD = ∠CAD

∴ ΔADB ~ ΔADC

We know that if two triangles are similar, their corresponding angles are equal and corresponding sides are proportional.

Given ABC is a right triangle right-angled at B and M, N are mid-points of AB and BC respectively.

M is the mid-point of AB.

And N is the mid-point of BC.

Now,

⇒ AN² + CM² = (AB² + ( �BC)²) + (( �AB)² + BC²)

= AB² + �BC² + 1/4 AB² + BC²

= 5/4 (AB² + BC²)

∴ 4 (AN² + CM²) = 5AC²

Hence proved.

Given in equilateral ΔABC, BE ⊥ AC.

We know that in an equilateral triangle, the perpendicular from the vertex bisects the base.

∴ CE = AE = AC/2

In ΔABE,

⇒ AB² = BE² + AE²

Since AB = BC = AC,

⇒ AB² = BC² = AC² = BE² + AE²

⇒ AB² + BC² + AC² = 3BE² + 3AE²

Since BE is an altitude,

BE = √3 AE

⇒ AB² + BC² + AC²

= 3BE² + BE²

∴ AB² + BC² + AC² = 4BE²

Given in right triangle ABC right-angled at B, P and Q are points on the sides AB and BC respectively.

Applying Pythagoras Theorem,

In ΔAQB,

⇒ AQ² = AB² + BQ² … (1)

In ΔPBC,

⇒ CP² = PB² + BC² … (2)

Adding (1) and (2),

⇒ AQ² + CP² = AB² + BQ² + PB² + BC² … (3)

In ΔABC,

⇒ AC² = AB² + BC² … (4)

In ΔPBQ,

⇒ QP² = PB² + BQ² … (5)

From (3), (4) and (5),

∴ AQ² + CP² = AC² + PQ²

Given in ΔABC and ΔDEF,

We know that if in two triangles, one pair of corresponding sides are proportional and included angles are equal, then the two triangles are similar.

Hence, ΔABC is similar to ΔDEF, we should have ∠B = ∠D.

Given that ΔABC and ΔDEF are two triangles such that

We know that AAA similarity criterion states that in two triangles, if corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.

⇒ ∠A = ∠D, ∠B = ∠E, ∠C = ∠F

∴ ΔCAB ~ ΔFDE

Hence proved.

Given Ar (ΔABC) = 9 cm², ar (ΔDEF) = 16 cm² and BC = 2.1 cm

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

∴ EF = 2.8 cm

Given that one side of isosceles right triangle is 4√2 cm.

We know that in isosceles triangle two sides are equal.

In isosceles triangle ABC, let AB and AC be two equal sides of measure 4√2 cm.