Triangles

Given: AD = 3.6 cm, AB = 10 cm and AE = 4.5 cm.

Applying Thale’s Theorem,

⇒

⇒ [∵ DB = AB – AD ⇒ DB = 10 – 3.6 = 6.4]

⇒

⇒ EC = 8

Now, AC = AE + EC

⇒ AC = 4.5 + 8 = 12.5

Hence, EC = 8 cm and AC = 12.5 cm

Given: AB = 13.3 cm, AC = 11.9 cm and EC = 5.1 cm.

Applying Thale’s Theorem,

Since we need to find DB first, we add 1 on both sides

⇒

⇒

⇒

⇒

⇒ DB = 5.7

AD is given by,

AD = AB – DB

⇒ AD = 13.3 – 5.7

⇒ AD = 7.6 cm

Hence, AD is 7.6

Given: AD/DB = 4/7 or AD = 4 cm, DB = 7 cm, and AC = 6.6

Applying Thale’s Theorem,

We have AE at RHS but we need AC, as the value of AC is given. So by adding 1 to both sides of the equation, we can get the desired result

⇒

⇒

⇒

⇒

⇒

⇒ EC = 4.2

AE is given by,

AE = AC – EC

⇒ AE = 6.6 – 4.2

⇒ AE = 2.4

Hence, AE is 2.4 cm.

Given: AD/AB = 8/15 or AD = 8 cm, AB = 15 cm, and EC = 3.5 cm

By applying Thale’s Theorem,

⇒

⇒

⇒ 8×(AE + 3.5) = 15×AE

⇒ 8×AE + 28 = 15×AE

⇒ 15×AE – 8×AE = 28

⇒ 7×AE = 28

⇒ AE = 28/7 = 4

Hence, AE is 4 cm.

Given: AD = x cm,

DB = (x – 2) cm,

AE = (x + 2) cm and,

EC = (x – 1) cm

By applying Thale’s Theorem,

⇒

⇒ x(x – 1) = (x + 2)(x – 2)

⇒ x² – x = x² – 4

⇒ x = 4

Thus, x = 4 cm

Given: AD = 4 cm, DB = (x – 4) cm, AE = 8 cm and EC = (3x – 19) cm

By Thale’s theorem,

⇒

⇒ 4(3x – 19) = 8(x – 4)

⇒ 12x – 76 = 8x – 32

⇒ 12x – 8x = 76 – 32

⇒ 4x = 44

⇒ x = 44/4 = 11

Thus, x = 11 cm

Given: AD = (7x – 4) cm, AE = (5x – 2), DB = (3x + 4) cm and EC = 3x cm

By Thale’s theorem,

⇒

⇒ 3x(7x – 4) = (5x – 2)(3x + 4)

⇒ 21x² – 12x = 15x² + 20x – 6x – 8

⇒ 21x² – 12x = 15x² + 14x – 8

⇒ 21x² – 15x² – 12x – 14x + 8 = 0

⇒ 6x² – 26x + 8 = 0

⇒ 2×(3x² – 13x + 4) = 0 [Simplifying the equation]

⇒ 3x² – 13x + 4 = 0

⇒ 3x² – 12x – x + 4 = 0

⇒ 3x(x – 4) – (x – 4) = 0

⇒ (3x – 1)(x – 4) = 0

⇒ (3x – 1) = 0 or (x – 4) = 0

⇒ x = 1/3 or x = 4

Now since we’ve got two values of x, that is, 1/3 and 4. We shall check for its feasibility.

Substitute x = 1/3 in AD = (7x – 4), we get

AD = 7×(1/3) – 4 = -1.67, which is not possible since side of a triangle cannot be negative.

Hence, x = 4 cm.

Here, by applying converse of Thale’s theorem we can conclude whether or not DE ∥ BC.

By Thale’s theorem,

Solving for ,

…(i)

Solving for ,

…(ii)

As equation (i) is equal to equation (ii),

it satisfies Thale’s theorem.

Hence, we can say DE ∥ BC.

Here, by applying converse of Thale’s theorem we can conclude whether or not DE ∥ BC.

By Thale’s theorem,

Solving for ,

We need to find AD from given AB = 11.7 cm and BD = 6.5 cm.

AD = AB – BD

⇒ AD = 11.7 – 6.5

⇒ AD = 5.2

…(i)

Solving for ,

We need to find EC from given AC = 11.2 cm and AE = 4.2 cm.

EC = AC – AE

⇒ EC = 11.2 – 4.2

⇒ EC = 7

…(ii)

As equation (i) is not equal to equation (ii),

it doesn’t satisfies Thale’s theorem.

Hence, we can say DE not parallel to BC.

Here, by applying converse of Thale’s theorem we can conclude whether or not DE ∥ BC.

By Thale’s theorem,

Solving for ,

We need to find DB from given AB = 10.8 cm and AD = 6.3 cm.

DB = AB – AD

⇒ DB = 10.8 – 6.3

⇒ DB = 4.5

…(i)

Solving for ,

We need to find AE from given AC = 9.6 cm and EC = 4 cm.

AE = AC – EC

⇒ AE = 9.6 – 4

⇒ AE = 5.6

…(ii)

As equation (i) is equal to equation (ii),

it satisfies Thale’s theorem.

Hence, we can say DE ∥ BC.

Here, by applying converse of Thale’s theorem we can conclude whether or not DE ∥ BC.

By Thale’s theorem,

Solving for ,

We need to find DB from given AB = 12 cm and AD = 7.2 cm.

DB = AB – AD

⇒ DB = 12 – 7.2

⇒ DB = 4.8

…(i)

Solving for ,

We need to find EC from given AC = 10 cm and AE = 6.4 cm.

EC = AC – AE

⇒ EC = 10 – 6.4

⇒ EC = 3.6

…(ii)

As equation (i) is not equal to equation (ii),

it doesn’t satisfies Thale’s theorem.

Hence, we can say DE is not parallel to BC.

Given: AB = 6.4 cm, AC = 8 cm and BD = 5.6 cm

Since AD bisects ∠A, we can apply angle-bisector theorem in ∆ABC,

Substituting given values, we get

⇒

⇒ DC = 7

Thus, DC is 7 cm.

Given: AB = 10 cm, AC = 14 cm and BC = 6 cm

Since AD bisects ∠A, we can apply angle-bisector theorem in ∆ABC,

Substituting given values, we get

To find BD and DC,

Let BD = x cm, and it’s given that BC = 6 cm, then DC = (6 – x) cm

Then

⇒ 14x = 10(6 – x)

⇒ 14x = 60 – 10x

⇒ 14x + 10x = 60

⇒ 24x = 60

⇒ x = 60/24 = 2.5

⇒ BD = 2.5 cm

If BD = 2.5 cm and BC = 6 cm, then DC = (6 – x) = (6 – 2.5) = 3.5

Thus, BD is 2.5 cm and DC = 3.5 cm.

Given: AB = 5.6 cm, BC = 6 cm and BD = 3.2 cm

Since AD bisects ∠A, we can apply angle-bisector theorem in ∆ABC,

Substituting given values, we get

Here, DC is given by

DC = BC – BD

⇒ DC = 6 – 3.2 = 2.8

⇒

⇒ AC = 4.9

Thus, AC is 4.9 cm.

Given: AB = 5.6 cm, AC = 4 cm and DC = 3 cm

Since AD bisects ∠A, we can apply angle-bisector theorem in ∆ABC,

Substituting given values, we get

⇒

⇒ BD = 4.2

Now, BC = BD + DC

⇒ BC = 4.2 + 3 = 7.2

Thus, BC is 7.2 cm.

(i). Given: ABCD is a parallelogram.

To Prove:

Proof: In ∆DMC and ∆NMB,

∠DMC = ∠NMB [∵ they are vertically opposite angles]

∠DCM = ∠NBM [∵ they are alternate angles]

∠CDM = ∠MNB [∵ they are alternate angles]

By AAA-similarity, we can say

∆DMC ∼ ∆NMB

So, from similarity of the triangle, we can say

Hence, proved.

(ii). Given: ABCD is a parallelogram.

To Prove:

Proof: As we have already derived

Add 1 on both sides of the equation, we get

⇒

⇒ [∵ ABCD is a parallelogram and a parallelogram’s opposite sides are always equal ⇒ DC = AB]

⇒

Hence, proved.

We can draw the trapezium as

Here, let EF be the line segment joining the oblique sides of the trapezium at midpoints E and F (say) correspondingly.

Construction: Extend AD and BC such that it meets at P.

To Prove: EF ∥ DC and EF ∥ AB

Proof: Given that, ABCD is trapezium which means DC ∥ AB. …(statement (i))

In ∆PAB,

DC ∥ AB (by statement (i))

So, this means we can apply Thale’s theorem in ∆PAB. We get

…(ii)

∵ E and F are midpoints of AD and BC respectively, we can write

DA = DE + EA

Or DA = 2DE …(iii)

CB = CF + FB

Or CB = 2CF …(iv)

Substituting equation (iii) and (iv) in equation (ii), we get

⇒

By applying converse of Thale’s theorem, we can write DC ∥ EF.

Now if DC ∥ EF, and we already know that DC ∥ AB.

⇒ EF is also parallel to AB, that is, EF ∥ AB.

This means, DC ∥ EF ∥ AB.

Hence, proved.

Given: In the adjoining figure, ABCD is a trapezium in which CD || AB and its diagonals intersect at O. If AO = (5x – 7) cm, OC = (2x + 1) cm, DO = (7x – 5) cm and OB = (7x + 1) cm.

To find: the value of x.

Solution:




In the trapezium ABCD, AB || DC and its diagonals intersect at O.

Through O draw EO || AB meeting AD at E.

Now In Δ ADC

As EO || AB || DC

By thales theorem which 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.

...... (i)

In Δ DAB,

EO || AB

By thales theorem,



...... (ii)

From (i) and (ii)

⇒

⇒ (5x – 7)(7x + 1) = (7x – 5)(2x + 1)

⇒ 35x² + 5x – 49x – 7 = 14x² – 10x + 7x – 5

⇒ 35x² – 44x – 7 = 14x² – 3x – 5

⇒ 35x² – 14x² – 44x + 3x – 7 + 5 = 0

⇒ 21x² – 41x – 2 = 0

⇒ 21x² – 42x + x – 2 = 0

⇒ 21x(x – 2) + (x – 2) = 0

⇒ (21x + 1)(x – 2) = 0

⇒ (21x + 1) = 0 or (x – 2) = 0

⇒ x = -1/21 or x = 2

But x = -1/21 doesn’t satisfy the length of intersected lines.

So x ≠ -1/21

And thus, x = 2.

We have

To show that, MN ∥ BC.

Given that, ∠B = ∠C and BM = CN.

So, AB = AC [sides opposite to equal angles (∠B = ∠C) are equal]

Subtract BM from both sides, we get

AB – BM = AC – BM

⇒ AB – BM = AC – CN

⇒ AM = AN

⇒ ∠AMN = ∠ANM

[angles opposite to equal sides (AM = AN) are equal] …(i)

We know in ∆ABC,

∠A + ∠B + ∠C = 180° [∵ sum of angles of a triangle is 180°] …(ii)

And in ∆AMN,

∠A + ∠AMN + ∠ANM = 180° [∵ sum of angles of a triangle is 180°] …(iii)

Comparing equations (ii) and (iii), we get

∠A + ∠B + ∠C = ∠A + ∠AMN + ∠ANM

⇒ ∠B + ∠C = ∠AMN + ∠ANM

⇒ 2∠B = 2∠AMN [∵ from equation (i), and also ∠B = ∠C]

⇒ ∠B = ∠AMN

Thus, MN ∥ BC since the corresponding angles, ∠AMN = ∠B.

We can observe two triangles in the figure.

In ∆ABC,

PQ ∥ AB

Applying Thale’s theorem, we get

…(i)

In ∆BDC,

PR ∥ BP

Applying Thale’s theorem, we get

…(ii)

Comparing equations (i) and (ii),

Now, applying converse of Thale’s theorem, we get

QR ∥ AD

Hence, QR is parallel to the AD.

We have the diagram as,

Given: BD = DC & OD = DX

To Prove: and also, EF ∥ BC

Proof: Since, from the diagram we can see that diagonals OX and BC bisect each other in quadrilateral BOCX. Thus, BOCX is a parallelogram.

If BOCX is a parallelogram, BX ∥ OC, and BO ∥ CX.

⇒ BX ∥ FC (as OC extends to FC) and CX ∥ BE (BO extends to BE)

⇒ BX ∥ OF and CX ∥ OE

∵ BX ∥ OF, applying Thale’s theorem in ∆ABX, we get

…(i)

Now since CX ∥ OE, applying Thale’s theorem in ∆ACX, we get

…(ii)

By equations (i) and (ii), we get

By applying converse of Thale’s theorem in the above equation, we can write

EF ∥ BC

Hence, proved.

We have the diagram as

Given: DP = PC &

CQ = (1/4)AC …(i)

To Prove: CR = RB

Proof: Join B to D

As diagonals of a parallelogram bisect each other at S.

…(ii)

Dividing equation (i) by (ii), we get

⇒

⇒ CQ = CS/2

⇒ Q is the midpoint of CS.

According to midpoint theorem in ∆CSD, we have

PQ ∥ DS

Similarly, in ∆CSB, we have

QR ∥ SB

Also, given that CQ = QS

We can conclude that, by the converse of midpoint theorem, CR = RB.

That is, R is the midpoint of CB.

Hence, proved.

Given: AD = AE …(i)

& AB = AC …(ii)

Subtracting AD from both sides of equation (ii), we get

AB – AD = AC – AD

⇒ AB – AD = AC – AE [from equation (i)]

⇒ DB = EC [∵ AB – AD = DB & AC – AE = EC] …(iii)

Now, divide equation (i) by (iii), we get

By converse of Thale’s theorem, we can conclude by this equation that DE ∥ BC.

So, ∠DEC + ∠ECB = 180° [∵ sum of interior angles on the same transversal line is 180°]

Or ∠DEC + ∠DBC = 180° [∵ AB = AC ⇒ ∠C = ∠B]

Hence, we can write DEBC is cyclic and points D, E, B and C are concyclic.

Given: ∠PBR = ∠QBR & PQ ∥ AC.

In ∆BQP,

BR bisects ∠B such that ∠PBR = ∠QBR.

Since angle-bisector theorem says that, if two angles are bisected in a triangle then it equates their relative lengths to the relative lengths of the other two sides of the triangles.

So by applying angle-bisector theorem, we get

⇒ QR × BP = PR × BQ

Hence, proved.

In these triangles ABC and PQR, observe that

∠BAC = ∠PQR = 50°

∠ABC = ∠QPR = 60°

∠ACB = ∠PRQ = 70°

Thus, by angle-angle-angle similarity, i.e., AAA similarity,

∆ABC ∼ ∆PQR

In triangles ABC & EFD,

∠ABC ≠ ∠EDF

So, clearly, since no criteria satisfies, ∆ABC is not similar to ∆EFD.

In triangles ABC & PQR,

∠ACB = ∠PQR

By SAS criteria, we can say

∆ABC ∼ ∆PQR

In triangles DEF & PQR,

By SSS criteria, we can write

∆DEF ∼ ∆PQR

In ∆ABC, we can find ∠ABC.

∠ABC + ∠BCA + ∠CAB = 180° [∵ sum of all the angles of a triangle is 180°]

⇒ ∠ABC + 70° + 80° = 180°

⇒ ∠ABC + 150° = 180°

⇒ ∠ABC = 180° - 150°

⇒ ∠ABC = 30°

We can observe from triangles ABC & MNR,

∠ABC = ∠MNR

∠CAB = ∠RMN

Hence, by AA similarity we can say, ∆ABC ∼ ∆MNR

(i) To find ∠DOC, we can observe the straight line DB.

∠DOC + ∠COB = 180° [∵ sum of all angles in a straight line is 180°]

⇒ ∠DOC + 115° = 180°

⇒ ∠DOC = 180° - 115°

⇒ ∠DOC = 65°

(ii) In ∆DOC,

And given that, ∠CDO = 70°, ∠DOC = 65° (from (i))

∠DOC + ∠DCO + ∠CDO = 180°

⇒ 65° + ∠DCO + 70° = 180°

⇒ ∠DCO + 135° = 180°

⇒ ∠DCO = 180° - 135°

⇒ ∠DCO = 45°

(iii) We have derived ∠DCO from (ii), ∠DCO = 45°

Thus, ∠OAB = 45° [∵ ∠OAB = ∠DCO as ∆ODC ∼ ∆OBA]

(iv) It’s given that, ∠CDO = 70°

Thus, ∠OBA = 70° [∵ ∠OBA = ∠CDO as ∆ODC ∼ ∆OBA]

(i). Given that, AB = 8 cm

BO = 6.4 cm,

OC = 3.5 cm

& CD = 5 cm

∆OAB ∼ ∆OCD

When two triangles are similar, they can be written in the ratio as

Substitute gave values in the above equations,

⇒

⇒ OA = 5.6

Thus, OA = 5.6 cm

(ii). Given that, AB = 8 cm

BO = 6.4 cm,

OC = 3.5 cm

& CD = 5 cm

∆OAB ∼ ∆OCD

When two triangles are similar, they can be written in the ratio as

Substitute gave values in the above equations,

⇒

⇒ DO = 4

Thus, DO = 4 cm

Given is that ∠ADE = ∠B

From the diagram clearly, ∠EAD = ∠BAC [∵ they are common angles]

Now, since two of the angles are correspondingly equal. Then by AA similarity criteria, we can say

∆ADE ∼ ∆ABC

Further, it’s given that

AD = 3.8 cm

AE = 3.6 cm

BE = 2.1 cm

BC = 4.2 cm

DE =?

To find AB, we can express it in the form AB = AE + BE = 3.6 + 2.1

⇒ AB = 5.7

So for the condition that ∆ADE ∼ ∆ABC,

Substituting given values in the above equation,

⇒

⇒

⇒ DE = 2.8

Thus, DE = 2.8 cm

Given that, ∆ABC ∼ ∆PQR

And perimeter of ∆ABC = 32 cm & perimeter of ∆PQR = 24 cm

We can write relationship as,

⇒

⇒

⇒ AB = 16

Thus, AB = 16 cm.

Given that, ∆ABC ∼ ∆DEF

Also, BC = 9.1 cm & EF = 6.5 cm

And perimeter of ∆DEF = 25 cm

We need to find perimeter of ∆ABC = ?

We can write relationship as,

⇒

⇒

⇒ perimeter of ∆ABC = 35

Thus, perimeter of ∆ABC = 35 cm

Given that, ∠CAB = 90°

AC = 75 cm

AB = 1 m

BC = 1.25 m

To show that, ∆BDA ∼ ∆BAC

In the diagram, we can see

∠BDA = ∠BAC = 90°

∠DBA = ∠CBA [They are common angles]

So by AA-similarity theorem,

∆BDA ∼ ∆BAC

Thus, now since ∆BDA ∼ ∆BAC, we can write as

⇒ [∵ AC = 75 cm, AB = 1 m = 100 cm & BC = 1.25 m = 125 cm]

⇒

⇒ AD = 60 cm

Hence, AD = 60 cm or 0.6 m

Given that, ∠ABC = 90°

AB = 5.7 cm

BD = 3.8 cm

CD = 5.4 cm

In order to find BC, we need to prove that ∆BDC and ∆ABC are similar.

∠BDC = ∠ABC = 90°

∠ACB = ∠DCB [They are common angles]

By this we have proved ∆BDC ∼ ∆ABC, by AA-similarity criteria.

So we can write,

⇒

⇒

⇒ BC = 8.1

Hence, BC = 8.1 cm.

Given that, ∠ABC = 90°

AD = 4 cm

BD = 8 cm

In order to find CD, we need to prove that ∆BDC and ∆ABC are similar.

∠BDC = ∠ADB = 90°

∠DBA = ∠DCB

We have proved ∆DBA ∼ ∆DCB, by AA-similarity criteria.

So we can write,

⇒

⇒

⇒ CD = 16

Hence, CD = 16 cm.

There are two triangles here, ∆APQ and ∆ABC. We shall prove these triangles to be similar.

&

⇒

Also, ∠A = ∠A [common angle]

So by AA-similarity criteria,

∆APQ ∼ ∆ABC

Thus,

And we know

⇒ BC = 3×PQ

Hence, proved.

Given that, AB ∥ DC & AD ∥ BC

To Prove: AF × FB = EF × FD

Proof: In ∆DAF & ∆BEF

∠DAF = ∠BEF [∵ they are alternate angles]

∠AFD = ∠EFB [∵ they are vertically opposite angles]

This implies that ∆DAF ∼ ∆BEF by AA-similarity criteria.

⇒

Now cross-multiply them,

AF × FB = FD × EF

Hence, proved.

Observe in ∆BED & ∆ACB, we have

∠BED = ∠ACB = 90°

Now according to what’s given, DB ┴ BC and AC ┴ BC we can write,

∠B + ∠C = 180°

This clearly means BD ∥ CA

⇒ ∠EBD = ∠CAB [They are alternate angles]

Thus, by AA-similarity theorem, ∆BED ∼ ∆ACB
Now, by property of similarity of triangles,

So,

⇒

Hence, proved.

We have

Let the two triangles be ∆ABC and ∆PQR.

Given that, AB = 7.5 cm

BC = 5 m = 500 cm

QR = 24 m = 2400 cm

We have to find PQ = x (say).

We need to prove ∆ABC is similar to ∆PQR.

We can observe that,

∠ABC = ∠PQR = 90°

∠ACB = ∠PRQ [∵ the sum castes same angle at all places at the same time]

Thus, by AA-similarity criteria, we can say

∆ABC ∼ ∆PQR

So,

Substitute the given values in this equation,

⇒

⇒ x = 36 cm

Thus, height of the tower is 36 cm.

To prove: ∆ACP ∼ ∆BCQ

Proof:

Given that, ∆ABC is an isosceles triangle. ⇒ AC = BC

Also, if ∆ABC is an isosceles triangle,

then ∠CAB = ∠CBA …(i)

Subtracting it by 180° from both sides, we get

180° - ∠CAB = 180° - ∠CBA

⇒ ∠CAP = ∠CBQ …(ii)

Also, given that AP × BQ = AC × AC

Or

Or [∵ AC = BC] …(iii)

Recollecting equations (i), (ii) and (iii),

By SAS-similarity criteria, we get

∆ACP ∼ ∆BCQ

Hence, proved.

To Prove: ∆ACB ∼ ∆DCE

Proof:

Given that, ∠1 = ∠2

⇒ ∠DBC = ∠DCE

Also in ∆ABC & ∆DCE, we get

∠DCE = ∠ACB [they are common angles to both triangles]

And

Or

Or [∵ BD = DC as ∠1 = ∠2]

Thus by SAS-similarity criteria, we get

∆ACB ∼ ∆DCE

Hence, proved.

Given: AD = BC

P, Q, R and S are the midpoints of AB, AC, CD and BD respectively.

So in ∆ABC, if P and Q are midpoints of AB and A respectively ⇒ PQ ∥ BC

And PQ = (1/2)BC …(i)

Similarly in ∆ADC,

QR = (1/2)AD …(ii)

In ∆BCD,

SR = (1/2)BC …(iii)

In ∆ABD,

PS = (1/2)AD = (1/2)BC [∵ AD = BC]

Using equations (i), (ii), (iii) & (iv), we get

PQ = QR = SR = PS

All these sides are equal.

⇒ PQRS is a rhombus.

Hence, shown that PQRS is a rhombus.

Given: AB and CD are chords of the circle, intersecting at point P.

(a). To Prove: ∆PAC ∼ ∆PDB

Proof: In ∆PAC and ∆PDB,

∠APC = ∠DPB [∵ they are vertically opposite angles]

∠CAP = ∠PDB [∵ angles in the same segment are equal]

Thus, by AA-similarity criteria, we can say that,

∆PAC ∼ ∆PDB

Hence, proved.

(b). To Prove: PA × PB = PC × PD

Proof: As already proved that ∆PAC ∼ ∆PDB

We can write as,

By cross-multiplying, we get

PA × PB = PC × PD

Hence, proved.

Given: AB and CD are chords of a circle intersecting at point P outside the circle.

(a). To Prove: ∆PAC ∼ ∆PDB

Proof: We know

∠ABD + ∠ACD = 180° [∵ opposite angles of cyclic quadrilateral are supplementary] …(i)

∠PCA + ∠ACD = 180° [∵ they are linear pair angle] …(ii)

Comparing equations (i) & (ii), we get

∠ABD + ∠ACD = ∠PCA + ∠ACD

⇒ ∠ABD = ∠PCA

Also, ∠APC = ∠BPD [∵ they are common angles]

Thus, by AA-similarity criteria, ∆PAC ∼ ∆PDB

Hence, proved.

(b). To Prove: PA × PB = PC × PD

Proof: We have already proved that, ∆PAC ∼ ∆PDB

Thus the ratios can be written as,

By cross-multiplication, we get

PA × PB = PC × PD

Hence, proved.

By the property that says, if a perpendicular is drawn from the vertex of a right triangle to the hypotenuse then the triangles on both the sides of the perpendicular are similar to the whole triangle and also to each other.

We can conclude by the property in ∆BDC,

∆CQD ∼ ∆DQB

(a). To Prove: DQ² = DP × QC

Proof: As already proved, ∆CQD ∼ ∆DQB

We can write the ratios as,

By cross-multiplication, we get

DQ² = QB × QC …(i)

Now since, quadrilateral PDQB forms a rectangle as all angles are 90° in PDQB.

⇒ DP = QB & PB = DQ

And thus replacing QB by DP in equation (i), we get

DQ² = DP × QC

Hence, proved.

(b). To Prove: DP² = DQ × AP

Prof: Similarly using same property, we get

∆APD ∼ ∆DPB

We can write the ratios as,

By cross-multiplication, we get

DP² = PB × AP

⇒ DP² = DQ × AP [∵ PB = DQ]

Hence, proved.

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

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

Given that

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

Let the two triangles be ABC and PQR and their longest sides are BC and QR.

We know that if two triangles are similar then the ratio of their areas is equal to the ratio of the squares of their longest sides.

Let the two triangles ABC and DEF have their altitudes as AS and DT.

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

Let the two triangles ABC and DEF have their altitudes as AS and DT.

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

Let the two triangles ABC and DEF have their altitudes as AS and DT.

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

Let the two triangles ABC and DEF have their medians as AS and DT.

We know that if two triangles are similar then the ratio of their areas is equal to the ratio of the squares of their corresponding medians.

We have

Also ∠ A = ∠ A

So, by SAS similarity criterion ΔAPQ ~ ΔABC

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

Hence, proved.

It is given that DE || BC

∴∠ ADE = ∠ ABC (Corresponding angles)

∠ AED = ∠ ACB (Corresponding angles)

So, by AA similarity criterion ΔADE ~ ΔABC

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

Hence, proved.

In ΔABC and ΔADC

∴∠ BAC = ∠ ADC (90° angle)

∠ ACB = ∠ ACD (Common)

So, by AA similarity criterion ΔADC ~ ΔABC

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

It is given that DE || BC

∴∠ ADE = ∠ ABC (Corresponding angles)

∠ AED = ∠ ACB (Corresponding angles)

So, by AA similarity criterion ΔADE ~ ΔABC

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

Hence, proved.

In ΔABC and ΔADE

It is given that AD = DB and AE = EC

Also ∠ A = ∠ A

So, by SAS similarity criterion ΔADE ~ ΔABC

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

In a right angled triangle

(Hypotenuse) ² = (Base)² + (Height)²

where hypotenuse is the longest side.

(i) L.H.S. = (Hypotenuse)² = (18)² = 324

R.H.S. = (Base)² + (Height)² = (9)² + (16)² = 81 + 256 = 337

⇒L.H.S. ≠ R.H.S.

∴It is not a right triangle.

(ii) L.H.S. = (Hypotenuse)² = (27)² = 729

R.H.S. = (Base)² + (Height)² = (7)² + (25) ² = 49 + 625 = 674

⇒ L.H.S. ≠ R.H.S.

∴It is not a right triangle.

(iii) L.H.S. = (Hypotenuse)² = (5)² = 25

R.H.S. = (Base)² + (Height)² = (1.4)² + (4.8) ² = 1.96 + 23.04 = 25

⇒ L.H.S. = R.H.S.

∴It is a right triangle.

(iv) L.H.S. = (Hypotenuse)² = (4)² = 16

R.H.S. = (Base)² + (Height)² = (1.6)² + (3.8)² = 2.56 + 14.44 = 17

⇒ L.H.S. ≠ R.H.S.

∴It is not a right triangle.

(v) L.H.S. = (Hypotenuse)² = (a + 1)²

R.H.S. = (Base)² + (Height)² = (a-1)² + (2√a)² = a² + 1-2a + 4a = a² + 1 + 2a = (a + 1)²

⇒ L.H.S. = R.H.S.

∴It is a right triangle.

The starting point of the man is A and the last point is B so we need to find AB. From the figure, ΔABC is a right triangle.

In a right angled triangle

(Hypotenuse) ² = (Base)² + (Height)²

where hypotenuse is the longest side.

(AB)² = (AC)² + (BC)²

⇒ AB² = (80) ² + (150) ² = 6400 + 22500 = 28900

⇒ AB = 170 m

The starting point of the man is B and the last point is A so we need to find AB. From the figure, ΔABC is a right triangle.

In a right angled triangle

(Hypotenuse)² = (Base)² + (Height)²

where hypotenuse is the longest side.

(AB)² = (AC)² + (BC)²

⇒ AB² = (24) ² + (10) ² = 576 + 100 = 676

⇒ AB = 26 m

Ladder AB = 13 m and distance from the window BC = 12 m.

AC is the distance of the ladder from the building.

From the figure, ΔABC is a right triangle.

In a right angled triangle

(Hypotenuse) ² = (Base)² + (Height)²

where hypotenuse is the longest side.

(AB)² = (AC)² + (BC)²

⇒13² = (AC) ² + (12) ²

⇒ AC² = 169- 144 = 25

⇒ AC = 5 m

Ladder AB and distance from the window BC = 20 m.

AC is the distance of the ladder from the building = 15 m.

From the figure, ΔABC is a right triangle.

In a right angled triangle

(Hypotenuse) ² = (Base)² + (Height)²

where hypotenuse is the longest side.

(AB)² = (AC)² + (BC)²

⇒AB² = (20) ² + (15) ²

⇒ AB² = 400 + 225 = 625

⇒AB = 25 m

AE(height of the first building) = 14 m , CD(height of the second building) = 9 m , ED(distance between their feet) = BC = 12 m

AE – AB = 14 m – 9 m = 5 m

From the figure, ΔABC is a right triangle.

In a right angled triangle

(Hypotenuse) ² = (Base)² + (Height)²

where hypotenuse is the longest side.

(AC)² = (AB)² + (BC)²

⇒ AC² = (5) ² + (12) ²

⇒ AB² = 25 + 144 = 169

⇒ AB = 13 m

Pole AB = 18 m and distance from the window BC.

AC is the length of the wire = 24 m.

From the figure, ΔABC is a right triangle.

In a right angled triangle

(Hypotenuse) ² = (Base)² + (Height)²

where hypotenuse is the longest side.

(AC)² = (AB)² + (BC)²

⇒ 24² = (18) ² + (BC) ²

⇒ BC² = 576 - 324 = 252

⇒ BC = 6√7 m

ΔPOR is a right triangle because ∠O = 90°.

In a right angled triangle

(Hypotenuse)² = (Base)² + (Height)²

where hypotenuse is the longest side.

(PR)² = (OP)² + (OR)²

⇒ PR² = (6) ² + (8) ²

⇒ PR² = 36 + 64 = 100

⇒ PR = 10 m

Now, PR² + PQ² = 10² + 24² = 100 + 576 = 676

Also, QR² = 26² = 676

⇒ PR² + PQ² = QR²

which satisfies Pythagoras theorem.

Hence, ∆PQR is right angled triangle.

Δ ABC is an isosceles triangle.

Also, AB = AC = 13 cm

Suppose the altitude from A on BC meets BC at D. Therefore, D is the midpoint of BC.

AD = 5 cm

ΔADB and ΔADC are right-angled triangles.

Applying Pythagoras theorem,

AB² = BD² + AD²

⇒BD² = 13² - 5²

⇒ BD² = 169 – 25 = 144

⇒ BD = 12 cm

So, BC = 2× 12 = 24 cm

Δ ABC is an isosceles triangle.

Also, AB = AC = 2a

The AD is the altitude. Therefore, D is the midpoint of BC.

ΔADB and ΔADC are right-angled triangles.

Applying Pythagoras theorem,

AB² = BD² + AD²

Δ ABC is an equilateral triangle.

Also, BC = AB = AC = 2a

The AD, CE, and BF are the altitude at BC, AB and AC respectively. Therefore, D, E, and F are the midpoint of BC, AB and AC respectively.

Now, ΔADB and ΔADC are right-angled triangles.

Applying Pythagoras theorem,

AB² = BD² + AD²

⇒ (2a) ² = a² + AD²

⇒ AD² = 3a²

⇒ AD = a√3 units

Similarly ΔACE and ΔBEC are right-angled triangles.

Applying Pythagoras theorem,

CE = a√3 units

And ΔABF and ΔBFC are right-angled triangles.

Applying Pythagoras theorem,

BF = a√3 units

Δ ABC is an equilateral triangle.

Also, BC = AB = AC = 12 cm

The AD is the altitude at BC. Therefore, D is the midpoint of BC.

Now, ΔADB and ΔADC are right-angled triangles.

Applying Pythagoras theorem,

AB² = BD² + AD²

⇒ (12) ² = 6² + AD²

⇒ AD² = 144 – 36 = 108

⇒ AD = 6√3 cm

Given that AB = 30cm and AD = 16 cm

∵ ∠ A = 90°

∴ ΔADB is a right-angled triangle.

Applying Pythagoras theorem,

BD² = BA² + AD²

⇒ BD ² = 30² + 16²

⇒ BD² = 900 + 256 = 1156

⇒ BD = 34 cm

∵ Diagonals of a rectangle are equal

∴ AC = 34 cm

ABCD is a rhombus where AC = 24 cm and BD = 10 cm.

We know that diagonals of a rhombus bisect each other at 90°.

⇒ ∠AOB = 90°, OA = 12 cm and OB = 5 cm

∴ ΔAOB is a right-angled triangle.

Applying Pythagoras theorem,

BA² = BO² + AO²

⇒ BA ² = 5² + 12²

⇒ BA² = 25 + 144 = 169

⇒ BA = AD = CD = BC = 13 cm

∵Sides of a rhombus are equal.

In right-angled triangle AED, applying Pythagoras theorem,

AB² = AE² + BE²

⇒ AE² = AB² – BE² ….(i)

In right-angled triangle AED, applying Pythagoras theorem,

AD² = AE² + ED²

⇒ AE² = AD² – ED² ….(ii)

Therefore,

AB² – BE² = AD² – ED²

In ΔACB and ΔCDB,

∠ABC = ∠CBD (Common)

∠ACB = ∠CDB (90°)

So, by AA similarity criterion ΔACB ~ ΔCDB

Similarly, In ΔACB and ΔADC,

∠ABC = ∠ADC (Common)

∠ACB = ∠ADC (90°)

So, by AA similarity criterion ΔACB ~ ΔADC

We know that if two triangles are similar then the ratio of their corresponding sides is equal.

⇒ BC² = AB×BD ….(i)

And AC² = AB×AD …..(ii)

Dividing (i) and (ii), we get

Hence, proved.

(i) ΔAEC and ΔAED are right triangles.

Applying Pythagoras theorem we get,

AC² = EC² + AE²

And AD² = ED² + AE²

….(i)

And p² = h² + x² ….(ii)

Putting (ii) in (i),

…..(iii)

Hence, proved.

(ii) ΔAEB is a right triangle.

Applying Pythagoras theorem we get,

AB² = EB² + AE²

….(iv)

Putting (ii) in (iv ),

…..(v)

Hence, proved.

(iii) Adding (iii) and (v),

Hence, proved.

(iv) Subtracting (iii) and (v),

Hence, proved.

Draw AE⊥BC. Applying Pythagoras theorem in right-angled triangle AED,

Since, ABC is an isosceles triangle and AE is the altitude and we know that the altitude is also the median of the isosceles triangle.

So, BE = CE

And DE + CE = DE + BE = BD

AD² = AE² + ED²

⇒AE² = AD² - ED² …(i)

In ΔACE,

AC² = AE² + EC²

⇒ AE² = AC² –EC² …(ii)

Using (i) and (ii),

⇒ AD² - ED² = AC² –EC²

⇒ AD² - AC² = ED²–EC²

⇒ AD² - AC² = (DE + CE) (DE – CE)

⇒ AD² - AC² = (DE + BE) CD

⇒ AD² - AC² = BD.CD

ΔABC is right triangle.

Applying Pythagoras theorem we get,

AC² = AB² + BC² {∵ AB = BC}

⇒ AC² = 2AB²

Given that the two triangles ΔACD and ΔABE are similar.

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

Let A be the first aeroplane flied due north at a speed of 1000 km/hr and B be the second aeroplane flied due west at a speed of 1200 km/hr

Distance covered by plane A in 1.5 hrs = 1000 × 32 = 1500km

Distance covered by plane B in 1.5 hrs = 1200 × 32 = 1800km

Now, in right triangle ABC

By using Pythagoras theorem, we have

AB² = BC² + AC²

⇒ AB² = (1800)² + (1500)²

⇒ AB² = 3240000 + 2250000

⇒ AB² = 5490000

⇒ AB = 300√61 km

(a) In right triangle ALC

Using Pythagoras theorem, we have

AC² = AL² + LC²

⇒ AC² = AD² − DL² + (DL + DC) ²

….(1)

(b) In right triangle ALD

Using Pythagoras theorem, we have

AL² = AD² - LD²

Again, in ΔABL

Using Pythagoras theorem, we have

AB² = AL² + LB²

⇒ AB² = AD² –DL² + LB²

⇒ AB² = AD²–DL² + (BD – DL)²

……(2)

(c) Adding (1) and (2)

Naman pulls in the string at the rate of 5 cm per second.

Hence, after 12 seconds the length of the string he will pull is given by

12 × 5 = 60 cm or 0.6 m

Now, in ΔBMC

By using Pythagoras theorem, we have

BC² = CM² + MB²

⇒ BC² = (2.4)² + (1.8)² = 9

∴ BC = 3 m

Now, BC’ = BC – 0.6 = 3 – 0.6 = 2.4 m

Now, in ΔBC’M

By using Pythagoras theorem, we have

𝐶′𝑀² = 𝐵𝐶′²−𝑀𝐵²

⇒ C’M² = (2.4)²− (1.8)² = 2.52

∴ C’M = 1.6 m

The horizontal distance of the fly from him after 12 seconds is given by

C’A = C’M + MA = 1.6 + 1.2 = 2.8 m

Two triangles are similar, if

(i) their corresponding angles are equal and

(ii) their corresponding sides are in the same ratio (or proportion).

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

Converse of Thales’ Theorem: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Midpoint Theorem: The line joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar. This criterion is referred to as the AAA (Angle–Angle–Angle) criterion of similarity of two triangles.

If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar. This is referred to as the AA-similarity criterion for two triangles.

If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar. This criterion is referred to as the SSS (Side–Side–Side)-similarity criterion for two triangles.

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. This criterion is referred to as the SAS (Side–Angle–Side) similarity criterion for two triangles.

In a right angled trianglPythagoras’ Theorem: In a right 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 the first side is a right angle.

The figure is shown below:

We know that the midpoint theorem states that the line joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

Since D, E and F are respectively the midpoints of sides AB, BC and CA of ΔABC,

DE = AB/2; EF = BC/2; DF = AC/2

⇒ DE/AB = 1/2; EF/BC = 1/2; DF/AC = 1/2

⇒ DE/AB = EF/BC = DF/AC = 1/2

We know that if in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar (SSS criteria).

So Δ ABC ~ Δ DEF.

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

∴ ar(Δ ABC)/ar(Δ DEF) = (AB/DE)²

⇒ ar(ΔABC)/ar(ΔDEF) = (2DE/DE)²

⇒ ar(ΔABC)/ar(ΔDEF) = (2/1)²

⇒ ar(ΔABC)/ar(ΔDEF) = (4/1)

But we need to find the ratio of the areas of ΔDEF and ΔABC.

∴ ar(ΔDEF)/ar(ΔABC) = (1/4)

∴ ar(ΔABC):ar(ΔDEF) = 1:4

1: 4

We know 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 (SAS criteria).

Here in the given triangles, ∠A = ∠P = 70° .

And AB/PQ = AC/PR

i.e. 6/4.5 = 6/9

⇒ 2/3 = 2/3

Hence ΔABC ~ ΔPQR.

SAS-similarity

Given: ABC ~DEF such that 2AB = DE and BC = 6 cm.

From SSS-similarity criterion,

We get

AB/DE = BC/EF

Substituting the given values,

AB/2AB = 6cm/EF

1/2 = 6cm/EF

EF = 2 × 6cm

EF = 12cm

12cm

We know that the 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, the other two sides are divided in the same ratio.”

So if DE || BC,

Then AD/DB = AE/EC

By substituting the given values,

⇒ x cm/(3x + 4)cm = (x + 3)cm/(3x + 19)cm

Cross multiplying, we get

⇒ 3x² + 19x = 3x²+ 9x + 4x + 12

⇒ 3x² + 19x - 3x²- 9x - 4x = 12

⇒ 6x = 12

⇒ x = 2

x = 2

Let AB be the ladder and CA be the wall with the window at A.

Let the distance of foot of ladder from base of wall BC be x.

Also, AB = 10m and CA = 8m

From Pythagoras Theorem,

we have: AB² = BC² + CA²

⇒ (10)² = x² + 8²

⇒ x² = 100 – 64

⇒ x² = 36

⇒ x = 6m

So, BC = 6m.

Length of the ladder is 6m.

Let ABC be the equilateral triangle whose side is 2a cm.

Let us draw altitude AD such that AD ⊥ BC.

We know that altitude bisects the opposite side.

So, BD = DC = a cm.

In ADC, ∠ADC = 90°.

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

So, by applying Pythagoras Theorem,

AC² = AD² + DC²

(2a cm)² = AD² + (a cm)²

4a² cm² = AD² + a² cm²

AD² = 3a² cm²

AD = √3 a cm

The length of altitude is √3 a cm.

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

i.e. ar(ABC)/ar(DEF) = (BC/EF)²

Substituting the given values, we get

⇒ 64cm²/169cm² = (4cm/EF cm)²

⇒ 64/169 = 16/EF²

⇒ EF² = 42.25

⇒ EF = 6.5cm

6.5 cm

Let us consider AOB and COD.

∠AOB = ∠COD (∵ vertically opposite angles)

∠OBA = ∠ODC (∵ alternate interior angles)

∠OAB = ∠OCD (∵ alternate interior angles)

We know that if in two triangles, corresponding angles are equal,

then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar (AAA criteria).

So, ΔAOB ≅ ΔCOD.

Given, AB = 2CD and ar(ΔAOB) = 84 cm²

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

∴ ar(AOB)/ar(COD) = (AB/CD)²

⇒ 84cm²/ar(ΔCOD) = (2CD/CD)²

⇒ 84cm²/ar(ΔCOD) = 4

⇒ ar(ΔCOD) = 84cm²/4

⇒ ar(ΔCOD) = 21cm²

ar(ΔCOD) = 21cm²

Let the smaller triangle be ABC and larger triangle beDEF.

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

i.e. ar(ABC)/ar(DEF) = (AB/DE)²

Substituting the given values, we get

⇒ 48cm²/ ar(DEF) = (2/3)²

⇒ 48cm²/ ar(DEF) = 4/9

⇒ ar(DEF)= (48 × 9)/4 cm²

⇒ ar(DEF) = 108cm²

108cm²

Let ABC be the equilateral triangle whose side is a cm.

Let us draw altitude AD(h) such that AD ⊥ BC.

We know that altitude bisects the opposite side.

So, BD = DC = a cm.

In ADC, ∠ADC = 90°.

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

So, by applying Pythagoras Theorem,

AC² = AD² + DC²

(a cm)² = AD² + (a/2 cm)²

a² cm² = AD² + a²/4 cm²

AD² = 3a²/4 cm²

AD = √3 a/2 cm = h

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

Ar(ΔABC) = 1/2 × a cm × √3 a/2 cm

⇒ ar(ΔABC) = √3 a²/4 cm²

Hence proved.

ar(ΔABC) = √3 a²/4 cm²

The diagonals of a rhombus bisect each other at right angles.

Let the intersecting point be O.

So, we get right angled triangles.

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

Let us consider ΔAOB.

By Pythagoras Theorem,

AB² = AO² + OB²

AB² = 12² + 5²

AB² = 144 + 25

AB² = 169

AB = 13cm

The length of side of the rhombus is 13cm.

Given that ΔDEF ≅ ΔGHK.

We know that if in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar (AAA criteria).

∴ ∠D = 48° = ∠G

∠H = 57° = ∠E

∠F = ∠K = x°

We know that the sum of angles in a triangle = 180°.

So, in ΔDEF,

⇒ 48° + 57° + x° = 180°

⇒ 105° + x° = 180°

⇒ x° = 180° - 105°

⇒ x° = 75° = ∠F

Ans.∠F = 75^o

We have MN || BC,

So, ∠AMN = ∠B and ∠ANM = ∠C (Corresponding angles)

We know that if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar (AA criteria).

∴ ΔAMN ~ ΔABC.

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

i.e. ar(ΔAMN)/ ar(ΔABC) = (AM/AB)²

Given that AM: MB = 1: 2.

Since AB = AM + MB,

AB = 1 + 2 = 3.

⇒ ar(ΔAMN)/ ar(ΔABC) = (1/3)²

⇒ ar(ΔAMN)/ ar(ΔABC) = 1/9

area(ΔAMN)/ area(ΔABC) = 1/9

Given: PB = 5 cm,

MP = 6 cm,

BM = 9 cm and,

NR = 9 cm

Now, it is also given that: ΔBMP ~ ΔCNR

When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.

⇒ …(i)

⇒

⇒

⇒

⇒ CN = 54/6 = 13.5 cm.

Similarly,

⇒

⇒

⇒

⇒ CR = 7.5 cm

∴ Perimeter of ΔCNR = CN + NR+ CR = 13.5+9+7.5=30 cm

Let ABC be the isosceles triangle whose sides are AB = AC = 25cm, BC = 14cm.

Let us draw altitude AD such that AD ⊥ BC.

We know that altitude bisects the opposite side.

So, BD = DC = 7cm.

In ADC, ∠ADC = 90°.

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

So, by applying Pythagoras Theorem,

AC² = AD² + DC²

(25 cm)² = AD² + (7 cm)²

625 cm² = AD² + 49 cm²

AD² = 576 cm²

AD = 24 cm

The length of altitude is 24 cm.

From ΔABC, we note that

A is the starting point.

AB = 12m, BC = 35m

CA = distance from starting point = x m

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

By Pythagoras Theorem,

CA² = AB² + BC²

CA² = 12² + 35²

CA² = 144 + 1225

CA² = 1369

CA = 37m

The man is 37 m far from the starting point.

Given that ΔABC is the triangle whose sides are AB = c, AC = b, BC = a

And AD is the bisector of ∠A.

We know that altitude bisects the opposite side.

So, let BD = DC = x.

Since AD bisects ∠A,

AC/AB = CD/DB

Substituting the given values,

b/c = CD/(a-CD)

Cross multiplying,

⇒ b( a – CD) = c (CD)

⇒ ba – b(CD) = c (CD)

⇒ ba = CD (b + c)

⇒ CD = ba/ (b + c)

Since CD = BD,

BD = ba/ (b + c)

BD = ba/(b + c) and DC = ba/(b + c)

In ΔAMN and ΔABC

∠AMN = ∠ABC = 76° (Given)

∠A = ∠A (common)

By AA Similarity criterion, ΔAMN ~ Δ ABC

If two triangles are similar, then the ratio or the their corresponding sides are proportional

The diagonals of a rhombus bisect each other at right angles.

Let the intersecting point be O.

So, we get right angled triangles.

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

Let us consider ΔAOB.

By Pythagoras Theorem,

AB² = AO² + OB²

AB² = 21² + 20²

AB² = 441 + 400

AB² = 841

AB = 29cm

The length of each side of the rhombus is 29cm.

(i) T

Two similar figures have the same shape but not necessarily the same size. Therefore, all circles are similar.

(ii) F

Two polygons of the same number of sides are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion).

Consider an example,

Let a rectangle have sides 2cm and 3cm and another rectangle have sides 2cm and 5cm.

Here, the corresponding angles are equal but the corresponding sides are not in the same ratio.

(iii) F

Two triangles are similar, if

(i) their corresponding angles are equal and

(ii) their corresponding sides are in the same ratio (or proportion).

(iv) T

Midpoint Theorem states that the line joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

(v) F

Two triangles are similar, if

(i) their corresponding angles are equal and

(ii) their corresponding sides are in the same ratio (or proportion).

But here, the corresponding sides are

AB/DE = 6/12 = 1/2 and AC/DF = 8/9

AB/DE ≠ AC/DF

(vi) F

The polygon formed by joining the midpoints of sides of any quadrilateral is a parallelogram.

(vii) T

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

(viii) T

The perimeters of the two triangles are in the same ratio as the sides. The corresponding medians also will be in this same ratio.

(ix) T

Let us construct perpendiculars OP, OQ, OR and OS from point O.

Let us take LHS = OA² + OC²

From Pythagoras theorem,

= (AS² + OS²) + (OQ² + QC²)

As also AS = BQ, QC = DS and OQ = BP = OS,

= (BQ² + OQ²) + (OS² + DC²)

Again by Pythagoras theorem,

= OB² + OD² = RHS

As LHS = RHS, hence proved.

(x) T

In rhombus ABCD, AB = BC = CD = DA.
We know that diagonals of a rhombus bisect each other perpendicularly.
i.e. AC ⊥ BD, ∠AOB = ∠BOC = ∠COD = ∠AOD = 90° and
OA = OC = AC/2, OB = OD = BD/2
Let us consider right angled triangle AOB.

By Pythagoras theorem,
AB² = OA² + OB²

⇒ AB² = (AC/2)² + (BD/2)²

⇒ AB² = AC²/4 + BD²/4
⇒ 4AB² = AC²+ BD²
⇒ AB² + AB² + AB² + AB² = AC²+ BD²
∴ AB² + BC² + CD² + DA² = AC²+ BD²

Since the man goes C to A = 24 m west and then A to B = 10 m north, he is forming a right angle triangle with respect to starting point C.

His distance from the starting point can be calculated by using Pythagoras theorem

(AC)² = (AB) ² + (BC)²

⇒ (AC)² = (24)² + (10)²

⇒ (AC)² = 576 + 100

⇒ (AC)² = 676

⇒ AC = 26

Let AB and CE be the two poles of the height 13 cm and 7 cm each which are perpendicular to the ground. The distance between them is 8 cm.

Now since CE and AB are ⊥ ground AE

BD ⊥ to CE and BD = 8 cm

Top of pole AB is B and top of pole CE is C

Now Δ BDC is right angled at D and BC, the hypotenuse is the distance between the top of the poles and CD = 13 – 7 = 6

(BC)² = (BD) ² + (CD)²

⇒ (BC)² = 64 + 36

⇒ (BC)² = 100

⇒ (BC)= 10 cm

The distance between the top of the poles is 10 cm

Let DF be the stick of 1.8 m height and AB be the pole of 6 m height.

AC and FE are the shadows of the pole and stick respectively.

FE = 45cm = .45 m

Since the shadows are formed at the same time, the two Δs are similar by AA similarity criterion

So

⇒

⇒ x = 1.5 m

Let DE be the pole of 6 m length casting shadow of 3.6 m . Let AB be the tower x meter height casting shadow of 18mat the same time.

Since pole and tower stands vertical to the ground, they form right angled triangle with ground.

Δ ABC and Δ EDF are similar by AA similarity criterion

∴ x/6 = 18 /3.6

⇒ x = 30

The height of the tower is 30 m

SINCE BOTH the tree and the stick are forming shadows at the same time the sides of the triangles so formed, would be in same ration ∵ of AA similarity criterian

12.5 /5 = x / 2

⇒ x = 5

Shadow of the tree would be 5 m long.

Let BC be the ladder placed against the wall AB. The distance of the ladder from the wall is the base of the right angled triangle as building stands vertically straight to the ground.

By Pythagoras theorem

(BC)² = (AB)² + (AC)²

(25)² = (24)² + (x)²

x = 7

the distance of ladder from the wall is 7m

The ΔMOP is right angled at O so MP is hypotenuse

(MP)² = (OM)² + (OP)²

(MP)² = (16) ² + (12)²

(MP)² = 400

MP = 20 cm

Δ NMP is right angled at M so NP is the hypotenuse so

(NP)² = (21)² + (20)²

NP = 29

Given (BC) = 25 cm

By Pythagoras theorem

(BC)² = (AB)² + (AC)²

(25) ² = (x + 5)² + x²

625 = x² + 25 + 10x + x²∵ (a + b) ² = a² + b² + 2ab

x² + 5x – 300 = 0

x (x + 15) – 10(x + 15) = 0

Since x = -15 is not possible so side of the triangle is 15 cm and 20 cm

Since Δ ABC is an equilateral triangle so the altitude (Height = h) from the C is the median for AB dividing AB into two equal halves of 6 cm each

Now there are two right angled Δs

h² = a² - 1/2 (AB) ²

h² = (12)² - 6²

h = 6 √3

The given triangle is isosceles so the altitude from the one of the vertex is median for the side opposite to it.

AB = AC = 13 cm

h = 5 cm (altitude)

Δ ABX is a right angled triangle, right angled at X

(AB)² = h² + (BX)² (BX = 1/2 BC)

169 = 25 + (BX) ²

BX = 12

⇒ BC = 24

By internal angle bisector theorem, the bisector of vertical angle of a triangle divides the base in the ratio of the other two sides.

Hence in ΔABC, we have

Here AB = 6 cm, AC = 8 cm

So

By internal angle bisector theorem, the bisector of vertical angle of a triangle divides the base in the ratio of the other two sides”
Hence in ΔABC, we have

⇒ x = 7.5cm

By internal angle bisector theorem, the bisector of vertical angle of a triangle divides the base in the ratio of the other two sides”
Hence in Δ ABC

⇒ 10x -84 + 14x = 0

⇒ x = CD = 3.5 cm

In Δ ABC, AD bisects ∠ A and meets BC in D such that BD = DC

Extend AD to E and join C to E such that CE is ∥ to AB

∠ BAD = ∠ CAD

Now AB ∥ CE and AE is transversal

∠ BAD = ∠ CED (alternate interior ∠s)

But ∠ BAD = ∠ CED = ∠CAD

In Δ AEC

∠ CEA = ∠CAE

∴ AC = CE………………. 1

In Δ ABD and Δ DCE

∠BAD = ∠CED (alternate interior ∠s)

∠ADB = ∠ CDE (vertically opposite ∠s)

BD = BC (given)

Δ ABD ≅ Δ DCE

AB = EC (CPCT)

AC = EC (from 1)

⇒AB = AC

⇒ ABC is an isosceles Δ with AB = AC

Δ ABC is an equilateral triangle

By Pythagoras theorem in triangle ABD

AB² = AD² + BD²

but BD = 1/2 BC (∵ In a triangle, the perpendicular from the vertex to the base bisects the base)

thus AB² = AD² + {1/2 BC}²

AB² = AD² + 1/4 BC²

4 AB² = 4AD² + BC²

4 AB² - BC² = 4 AD²

(as AB = BC we can subtract them )

Thus 3AB² = 4AD²

Since the diagonals of the rhombus bisects each other at 90°

∴ DO = OB = 6 cm

∠ AOD = ∠ DOC = ∠ COB = ∠BOA = 90°

Δ AOD is right angled Δ with

AD = 10 cm (given)

OD = 6 cm

∠ AOD = 90°

So DA = 10 = hypotenuse

(DA)² = (DO)² + (AO)²

100 – 36 = (AO)²

8 = AO

∴ AC = 16

In a rhombus the diagonals bisect each other at 90°

AC = 24cm (given)

∴ AD = 12cm

BD = 10cm (given)

∴ BO = 5cm

In right angled Δ AOB

AB² = AO² + OB²

AB² = (12)² + 5²

AB² = 144 + 25

AB² = 169

AB = 13cm

Hence the length of the sides of the rhombus is 13 cm

Given that ABCD is a quadrilateral and diagonals AC and BD intersect at O such that

IN Δ AOD and ΔBOC

∠ AOD = ∠COB

Thus Δ AOC ∼ ΔBOC (SAS similarity criterion)

⇒ ∠OAD = ∠ OCB ……………..1

Now transversal AC intersect AD and BC such the ∠CAD = ∠ACB

(from …..1 ) (alternate opposite angles)

So AD ∥ BC

Hence ABCD is a trapezium

ABCD is a Trapezium with AC and BD as diagonals and AB ∥ DC

In Δ AOB and Δ DOC

∠AOB = ∠DOC (vertically opposite angles)

∠CDO = ∠ OBA (alternate interior angles) (AB ∥ DC and BD is transversal)

Δ AOB ∼ Δ DOC (AA similarity criterion)

=

18 x² – 21x + 5 = 10 x² – x – 3

8 x² – 20 x + 8 = 0

2x² – 5 x + 2 = 0

2x² – 4 x – x + 2 = 0

2x (x- 2) – 1(x – 2) = 0

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

x = 1/2, 2

x = 1/2 is not possible so x = 2 cm

In the given quadrilateral ABCD

P, Q, R, S are the midpoints of the sides AB, BC, CD and AD respectively.

Construction: - Join AC

In Δ ABC and Δ ADC

P and Q are midpoints of AB and CB

S and R are midpoints of AD and DC

So by Mid Point Theorem

PQ ∥ AC and PQ = 1/2 AC………….1

And SR ∥ AC and SR = 1/2 AC………………2

From 1 and 2

PQ ∥ SR and PQ = SR

Since a pair of opposite side is equal (= ) and parallel (∥)

PQRS is a parallelogram

Given in Δ ABC, AD bisects the ∠A meeting BC at D

BD = DC and ∠BAD = ∠ CAD…………. 1

Construction:- Extend BA to E and join C to E such CE ∥ AD……… 4

∠BAD = ∠AEC (corresponding ∠s)……………… 2

∠CAD = ∠ ACE (alternate interior ∠s)……………….. 3

From 1 , 2 and 3

∠ ACE = ∠AEC

In Δ AEC

∠ ACE = ∠AEC

∴ AC = AE (sides opposite to equal angles are equal)……….. 5

In Δ BEC

AD ∥ CE (From ….4)

And D is midpoint of BC (given)

By converse of midpoint theorem

A line drawn from the midpoint of a side, parallel to the opposite side of the triangle meets the third side in its middle and is half of it

∴ A is midpoint of BE

BA = AE……… 6

From 5 and 6

AB = BC

⇒ ΔABC is an isosceles triangle

It is given that in Δ ABC,

∠B = 70° and ∠C = 50°

∠A = 180° – (70° + 50°) (∠ sum property of triangle)

= 180° - 120°

= 60°

Since,

∴ AD is the bisector of ∠A

Hence, ∠BAD = 60°/2 = 30°

By Basic Proportionality Theorem

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

In Δ ABC, DE ∥ BC

DB = 3.6 cm

AB = AD + DB

AB = 6 cm

BY Basic Proportionality Theorem:

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

Or

Adding 1 to both sides

=

=

=

AE = 4 cm

By Basic Proportionality Theorem:

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

=

3x (7x — 4) = (5x — 2) (3x + 4)

21 x² – 12x = 15x² + 14x – 8

6 x² – 26x + 8 = 0

3 x² – 13x + 4 = 0

3 x² – 12x – x + 4 = 0

3x(x - 4) – 1(x - 4) = 0

X = 1/3, 4

Since x cannot be 1/3 so x = 4

BY Basic Proportionality Theorem:

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

=

=

3 ( = 5 AE

16.8 = 8AE

AE = 2.1 cm

Since the ABC ∼ DEF

Their sides will be same ratios. Let the ratio be K

= = = K………….1

AB + BC + AC = K (DE + EF + DF)

= K

1.67 = k

From…… 1

= 1.67

EF = 5.4 cm

Since the ABC ∼ DEF

So the sides of the triangles are in the same ratio be k

= = = K………….1

= = K

K = 1.4

= K

= 1.4

= 1.4 × 25

= 35 cm

Perimeter of Δ ABC = AB + BC + CA

= 9 + 6 + 7.5

= 22.5 cm

Since the Δ ABC ∼ DEF

= (ratio of perimeter of triangles is equal to the ratio of the sides of the triangle )

=

=

= 30 cm

Δ ABC and Δ BDE are two equilateral triangles

Let a be the side of Δ ABC

Since D is midpoint of BC

So the side of equilateral ΔBDE =

Area of equilateral Δ = (side)2

Area of Δ ABC = ………….1

Area of Δ BDE =

=

Putting value of × …………from 1

Area of Δ BDE = Area of Δ ABC

⁼

In ΔABC,
∠ A + ∠B + ∠ C = 180°

30° + ∠B + 50° = 180°

∠ B = 100°
Given that Δ ABCΔDEF
∠D = ∠A = 30°

∠ E = ∠ B = 100°
∠ F = ∠ C = 50°
Also,

DE = 4.6875
And as neither BC nor EF is given we can not find either of them. So, none of the given options is correct.
Now,
Δ ABCΔDFE,
Then,
∠D = ∠ A = 30°
∠ F = ∠ B = 100°
∠ E = ∠ C = 50°
Also,

DE = 12cm
Therefore, the correct option is (b).

∠ABD + ∠ BAD = 90°

∠ ABD + (90-∠CAD) = 90°
∠ ABD = ∠ DAC
In ΔBDA and Δ ADC,
∠ ABD = ∠ CAD
∠ BDA = ∠ ADC = 90°
Therefore, ΔBDA and Δ ADC are similar by AAA.
BD.CD = AD²
Therefore the correct option is (c).

AB = 6√3cm.
In ΔABC,
AB² + BC² = AC²

Since the square of the longest side is equal to the sum of the squares of the remaining two sides of Δ ABC. Therefore ABC is right angled at B.

∠ B = 90°

With the ratio given, we can observe that Δ ABC Δ EDF,
∠ A = ∠E,
∠ B = ∠D,
∠ C = ∠ F

Given ∠ D = ∠ Q and ∠E = ∠R
By AA similarity, ΔDEFΔQRP
We have to find the option which is not true.

ΔABCΔEDF,
Therefore,

We have to find the option which is not true.
∴ The correct option is (c) .

Δ ABCΔ DEF
By AA similarity, the triangles are similar

For triangles to be congruent, AB = DE, but given that AB = 3DE.

Given
Therefore ΔABCΔQRP
or ΔPQRΔCAB.

In ΔAPB and Δ DPC,
∠APB = ∠DPC = 50°

By SAS property, Δ APBDPC
∠PBA = ∠DPC
In Δ DPC,
∠D + ∠ P + ∠C = 180°

∠ C = 100°
∴ ∠PBA = ∠DPC = 100°

If two triangles are similar, the ratio of the area of triangle is equal to the square of the ratio of the sides.
Ratio of Area = (Ratio of Side)² = (² = 16:81
∴ The correct option is (d).

If two triangles are similar, the ratio of the area of triangle is equal to the square of the ratio of the sides.

Given that D and E are of AB and AC respectively,
Therefore,

Δ ABCΔ ADE
If two triangles are similar, the ratio of the area of triangle is equal to the square of the ratio of the sides.
∴ The correct option is (b).

Therefore Δ ABCΔ DEF
If two triangles are similar, the ratio of the area of triangle is equal to the square of the ratio of the sides.

It is given that the corresponding angles are equal, that implies that the triangles are similar.

In this figure,
As given that the inner triangle is formed by joining the midpoints of the sides.
Therefore the outer three triangles are similar to bigger triangle.
By Basic Proportionality Theorem,
The inner triangle is also similar to the bigger triangle.

ΔABCΔQRP

PR =
PR = = 10cm

In Δ DOB and Δ AOC,
∠DOB = ∠AOC = 45° (vertically opposite angle)
∠OAC = ∠ODB (angles in the same segment)

∠OCA = ∠OBD (angles in the same segment)
Therefore, Δ DOB Δ AOC by AA similarity,

Therefore, OC = OA.

AB² = 2AC²

AB² = AC² + AC²

AB² = AC² + BC²

Therefore, it is an isosceles triangle right angled at C. ∠ C = 90°

AB² + BC² = 16² + 12² = 256 + 144 = 400 = 20² = AC²

Therefore, ABC is a right angled triangle.

If two triangles ABC and PQR are similar,

That is their corresponding sides are proportional.

The ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

A)

Given that DE||BC,
by B.P.T.,

Let AE = x

Then, from the figure, EC = 5.6-x

5x = 3(5.6-x)

5x = 16.8-3x

8x = 16.8

x = 2.1cm

Therefore, (A)-(s)

B)
As Δ ABCΔDEF,

3EF = 2BC

3EF = 2 × 6

EF = 4cm

Therefore,(B)-(q)

C)

QR =
QR = = 6cm

Therefore (C)-(p)

D)

(BPT)

⇒ 3(2x + 4) = (2x-1)(9x-21)

⇒ 6x + 12 = 18x²-42x-9x + 21

⇒ 18x²-57x + 9 = 0

⇒18x2-54x-3x + 9 = 0

⇒ 18x(x-3)-3(x-3) = 0

⇒ (18x-3)(x-3) = 0

So, x = 3 or x =

But for x = , 2x-1<0 which is not possible.
Therefore, (D)-(r)

A)

The man starts from A, goes east 10m to B. From B, he goes 20m to C.

AC² = AB² + BC²

AC² = 10² + 20²

AC² = 100 + 400 = 500

AC = √ 500 = 10√5

Therefore, (A)-(R)

B)

In ΔABD,

AB² = AD² + BD²

10² = AD² + 5²

AD² = 100-25 = 75

AD = √ 75 = 5√3

Therefore, (B)-(Q)

C)

Area of an equilateral triangle = cm

Therefore,(C)-(P)

D)

In ΔABC,

AC² = AB² + BC²

AC² = 6² + 8²

AD² = 36 + 64 = 100

AD = √ 100 = 10

Therefore, (D)-(S)

Given: ∆ABC ∼ ∆DEF

Perimeter of ∆ABC = 32 cm

Perimeter of ∆DEF = 24 cm

AB = 10 cm

To find: DE

∵ ∆ABC ∼ ∆DEF

∴ The ratio of the corresponding sides of ∆ ABC and ∆ DEF are equal to the ratio of the perimeter of the corresponding triangles.

Given: DE ∥ BC

DE = 5 cm

BC = 8 cm

AD = 3.5 cm

To find: AB

∵ DE ∥ BC

∴ By Basic proportionality theorem, we have

………..(i)

Now, in ∆ ADE and ∆ ABC, we have

[By (i)]

∠DAE = ∠BAC [Common angle]

∴ By SAS criterion,

∆ ADE ∼ ∆ ABC

Given: Height of pole 1 = 6 m

Height of pole 2 = 11 m

Distance between the feet of pole 1 and pole 2 = 12 m

To find: Distance between the tops of both the poles

Clearly, In ∆ ADE,

DE = 5 m

AD = 12 m

Also, ∠ADE = 90° [∵ Both the poles stand vertically upright]

∴ By applying Pythagoras theorem, we have

AE² = AD² + DE²

⇒ AE² = (12)² + (5)² = 144 + 25 = 169

⇒ AE = √169 = 13 m

Given: Area of triangle 1 = 25 cm²

Area of triangle 2 = 36 cm²

Altitude of triangle 1 = 3.5 cm

To find: Altitude of triangle 2

Let the altitude of triangle 2 be x.

∵ The areas of two similar triangles are in the ratio of the squares of the corresponding altitudes.

∴ We have,

⇒ x = √17.64 = 42 cm

Given: ∆ABC ∼ ∆DEF

2AB = DE ……….(i)

BC = 6 cm

To find: EF

∵ ∆ABC ∼ ∆DEF

∴ Ratio of all the corresponding sides of ∆ ABC and ∆ DEF are equal.

……….(ii)

Also, from (i), we have

2AB = DE

……….(iii)

[By (ii) and (iii)]

Given: DE ∥ BC

AD = x cm

DB = (3x + 4) cm

AE = (x + 3) cm

EC = (3x + 19) cm

To find: x

∵ DE ∥ BC

∴ By Basic Proportionality theorem, we have

⇒ x (3x + 19) = (x + 3) (3x + 4)

⇒ 3x² + 19x = 3x² + 13x + 12

⇒ 19x – 13x = 3x² + 12 – 3x²

⇒ 6x = 12 or x = 2

Given: Height of the window from the ground = 8 m

Length of the ladder = 10 m

To find: Distance of the foot of the ladder from the base of the wall.

Consider the following diagram corresponding to the question.

Here, AB = Height of the window form the ground = 8 m

AC = Length of the ladder = 10 m

BC = Distance of the foot of the ladder from the base of the wall

Now, in ∆ ABC,

By Pythagoras theorem, we have

AC² = AB² + BC²

⇒ BC² = AC² – AB²

⇒ BC² = (10)² – (8)² = 100 – 64 = 36

⇒ BC = √36 = 6 m

Given: Side of equilateral triangle = 2a cm

To find: Length of altitude

Let ∆ ABC be an equilateral triangle with side 2a cm.

Let AD be the altitude of ∆ ABC.

Here, BD = DC = a

In ∆ ABD,

Using Pythagoras theorem, we have

AB² = AD² + BD²⇒ AD² = AB² – BD²⇒ AD² = (2a)² – (a)² = 4a² – a² = 3a²

⇒ AD² = 3a²

⇒ AD = √3a² = √3 a cm

Given: ∆ ABC ∼ ∆ DEF

ar (∆ ABC) = 64 cm², ar (∆ DEF) = 169 cm²

BC = 4 cm

To find: EF

∵ The ratios of the areas of two similar triangles are equal to the ratio of squares of any two corresponding sides.

∴ We have

Given: AB ∥ CD

AB = 2CD ……….(i)

ar (∆ AOB) = 84 cm²

To find: ar (∆ COD)

In ∆ AOB and ∆ COD,

∠ AOB = ∠ COD [Vertically Opposite angles]

∠ OAB = ∠ OCD [Alternate interior angles (AB ∥ CD)]

∠ OBA = ∠ ODC [Alternate interior angles (AB ∥ CD)]

⇒ ∆ AOB ∼ ∆ COD [By AAA criterion]

Now,

∵ The ratios of the areas of two similar triangles are equal to the ratio of squares of any two corresponding sides.

∴ We have

Also, from (i), we have

Given: Let the smaller triangle be ∆ ABC and the larger triangle be ∆ DEF.

The ratio of AB and DE = 2 : 3

……….(i)

ar (∆ ABC) = 48 cm²

To find: ar (∆ DEF)

∵ The ratios of the areas of two similar triangles are equal to the ratio of squares of any two corresponding sides.

∴ We have

Given: LM ∥ CB and LN ∥ CD

To prove:

In ∆ AML, LM ∥ CB

∴ By Basic proportionality theorem, we have

……….(i)

In ∆ ALN, LN ∥ CD

∴ By Basic proportionality theorem, we have

……….(ii)

By (i) and (ii), we have

Given: ∆ ABC with the internal bisector AD of ∠A which intersects BC at D.

To prove:

First, we construct a line EC ∥ AD which meets BA produced in E.

Now, we have

CE ∥ DA ⇒ ∠2 = ∠3 [Alternate interior angles are equal (transversal AC)]

Also, ∠1 = ∠4 [Corresponding angles are equal (transversal AE)]

We know that AD bisects ∠A ⇒ ∠1 = ∠2

⇒ ∠4 = ∠1 = ∠2 = ∠3

⇒ ∠3 = ∠4

⇒ AE = AC [Sides opposite to equal angles are equal] ……….(i)

Now, consider ∆ BCE,

AD ∥ EC

[By Basic Proportionality theorem]

[∵ BA = AB and AE = AC (From (i))]

Let ∆ ABC be an equilateral triangle with side a.

To prove: Area of ∆ ABC =

In ∆ ABC, AD bisects BC

Now, in ∆ ACD

Using Pythagoras theorem,

AC² = AD² + DC²

⇒ AD² = AC² – DC²

Now, in ∆ ABC

Area of ∆ ABC =

Given: Length of one of the diagonals = 24 cm

Length of the other diagonal = 10 cm

To find: Length of the side of the rhombus

∵ The length of all sides of rhombus is equal.

∴ Let side of rhombus ABCD be x cm.

Also, we know that the diagonals of a rhombus are perpendicular bisectors of each other.

⇒ AO = OC = 12 cm and BO = OD = 5 cm

Also, ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°

Now, consider ∆ AOD

AO = 12 cm and OD = 5 cm

∠AOD = 90°

So, using Pythagoras theorem, we have

AD² = AO² + OD² = 12² + 5² = 144 + 25 = 169

⇒ AD = √169 = 13 cm

Let ∆ ABC and ∆ DEF be two similar triangles, i.e., ∆ ABC ∼ ∆ DEF.

⇒ Ratio of all the corresponding sides of ∆ ABC and ∆ DEF are equal.

Let these ratios be equal to some number α.

⇒ AB = α DE, BC = α EF, AC = α DF ……….(i)

Now, perimeter of ∆ ABC = AB + BC + AC

= α DE + α EF + α DF [ From (i)]

= α (DE + EF + DF)

= α (perimeter of ∆ DEF)

Given: ∆ ABC and ∆ DBC have the same base BC.

AD and BC intersect at O.

To show:

First, we construct the altitudes, AE and DF, of ∆ ABC and ∆ DBC, respectively.

Consider, ∆ AOE and ∆ DOF,

∠DFO = ∠AEO [Right angles]

∠DOF = ∠AOE [Vertically Opposite angles]

So, by AA criterion,

∆AOE ∼ ∆DOF

⇒ Ratio of all the corresponding sides of ∆ AOE and ∆ DOF are equal.

……….(i)

Now, we know that

……….(ii)

Similarly, ……….(iii)

Dividing (ii) by (iii),