Triangles

In ∆ADB and ∆ADC,

It is given that,

∠D = ∠D = 90° (∵ AD ⊥ BC)

∠DBA = ∠DAC [each angle equals to 90°- <C]

Now by AAA similarity criteria,

∆ADB ∼ ∆ADC

→ BD/AD = AD/CD

→ BD.CD = AD �

We know that,

A rhombus is a simple quadrilateral whose four sides are of same length and diagonals are perpendicular bisector of each other.

The figure is:

Given: AC = 16 cm and BD = 12 cm

∠AOB = 90°

∵ AC and BD bisects each other

→ AO = � AC and BO = � BD

→ AO = 8 cm and BO = 6 cm

Now, in right angled ∆AOB,

As per by Pythagoras theorem,

→ AB � = AO � + OB �

→ AB � = 8 � + 6 � = 64 + 36 = 100

∴ AB = √100 = 10 cm

As we know that a rhombus has all four sides equal.

Hence side of rhombus = 10 cm.

We know that, sides of one triangle are proportional to the side of the other triangle, and then their corresponding angles are also equal, so by SSS similarity, triangles are similar.

∆ABC ∼ ∆EDF

AB/ED = BC/DF = AC/EF

(By similarity property)

Taking first two terms, we get

AB/ED = BC/DF

⇒ AB.DF = ED.BC

So, option (d) is true

Taking last two terms, we get

BC/DF = AC/EF

⇒ BC.EF = AC.DF

So, option (A) is true

Taking first and last terms, we get,

AB/ED = AC/EF

⇒ AB.EF = ED.AC

Hence, option (b) is true.

In ∆ABC and ∆PQR.

AB/QR = BC/PR = CA/PQ

We know that if sides of one triangle are proportional to the side of the other triangle, and then their corresponding angles are also equal, so by SSS similarity, both the triangles are also similar.

So, by SSS Similarity

We have, ∆CAB = ∆PQR

In ∆APB and ∆CPD,

∠APB = ∠CPD = 50° (vertically opposite angles)

AP/PD = 6/5 …(i)

Also, BP/CP = 3/2.5

Or BP/CP = 6/5 …(ii)

From equations (i) and (ii)

AP/PD = BP/CP

∴ ∆APB ∼ ∆DPC [by SAS similarity criterion]

∴ ∠A = ∠D = 30° [corresponding angles of similar triangles]

In ∆APB,

∠A + ∠B + ∠APB = 180°

[Sum of angles of a triangle = 180°]

⇒ 30° + ∠B + 50° = 180°

∴ ∠B = 180° - (50° + 30°)

∠B = 180 – 80° = 100°

∠PBA = 100°

Given: ∆DEF and ∆PQR,

∠D = ∠Q,

∠R = ∠E

As we know that if two corresponding angles of two triangles are congruent then both the triangles are similar because if two angle pairs are the equal then the third angle must also be equal.

So, by AAA similarity criteria,

∴ ∆DEF ∼ ∆QRP

⇒ ∠F = ∠P [corresponding angles of similar triangles]

Hence, except (B) all are true.

In ∆ABC and ∆DEF,

∠B = ∠E,

∠F = ∠C and AB = 3DE

We know that,

If in two triangles corresponding two angles are same, then they are similar by AA similarity criteria.

Also, we know that two triangles are congruent if they have the same shape & size and satisfy the rule of congruency but ∆ABC and ∆DEF do not satisfy any rule of congruency, which are SAS, ASA, AAA and SSS, so both are not congruent.

By Similar Triangle’s Area property, the ratio of the areas of two similar triangles is equal to square of the ratio of their corresponding sides.

Thus, the area of ΔPRQ = 9 times ΔBCA

Given: ∆ABC ∼ ∆DFE,

As ∆ABC ∼ ∆DFE so, by the property of Corresponding angles of similar triangles,

We have,

∠A = ∠D = 30° (Corresponding angles)

∠C = ∠E = 50° (Corresponding angles)

We know that sum of angles of a triangle = 180°

So,

∴ ∠B = ∠F = 180°- (30° + 50°) = 100°

Also Given,

AB = 5 cm,

AC = 8 cm and

DF = 7.5 cm

Hence,

DE = 12 cm, and ∠F = 100°

Given,

∆ABC and ∆EDF,

→ AB/DE = BC/FD

By converse of basic proportionality theorem, that if a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

As we have,

∆ABC ∼ ∆EDF

Then,

∠B = ∠D,

∠A = ∠E and,

∠C = ∠F

Given,

∆ABC ∼ ∆QRP

AB = 18 cm and

BC = 15 cm

By Similar triangles area property, the ratio of area of two similar triangles is equal to the ratio of square of their corresponding sides.

So,

We have,

(By similar triangles area property).

But given

∴ RP = √100 = 10 cm

Given,

In ∆PQR

PS = QS + RS ……(i)

In ∆PSR

PS = RS ….. [from Equation (i)]

⇒ ∠1 = ∠2 Equation ….(ii)

Similarly,

In ∆RSQ,

⇒ ∠3 = ∠4 Equation……(iii)

[Corresponding angles of equal sides are equal]

[By using Equations (ii) and (iii)]

Now in,

∆PQR, sum of angles = 180°

⇒ ∠P + ∠Q + ∠R = 180°

⇒ ∠2 + ∠4 + ∠1 + ∠3 = 180°

⇒ ∠1 + ∠3 + ∠1 + ∠3 = 180°

⇒∠2 (1 + ∠3) = 180°

⇒ ∠1 + ∠3 = (180°)/2 = 90°

∴ ∠R = 90°

In ∆PQR, by Pythagoras theorem,

PR² + QR² = PQ²

(False)

As per question,

Let suppose,

a = 25 cm,

b = 5 cm and

c = 24 cm

Now,

b² + c² = (5)² + (24)²

b² + c² = 25 + 576

→ b² + c² = 601 ≤ (25)²

According to the Pythagoras theorem, in a right-angle triangle the sum of square of two sides must be equal to square of third side. But Given sides do not make a right triangle because it does not satisfy the property of Pythagoras theorem so the triangle with sides 25cm, 5cm and 24cm is not a right triangle.

(False)

We know that, in similar triangles corresponding angles are also equal.

So,

∠D = ∠R,

∠E = ∠P and

∠F = Q

(True)

Given,

PQ = 12.5 cm,

PA = 5 cm,

BR = 6 cm and

PB = 4 cm

Then,

QA = QP – PA = 12.5 – 5 = 7.5 cm

Form Equations (i) and (ii).

PA/AQ = PB/BR

According to the converse of basic proportionality theorem, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Hence,

AB || QR.

(True)

In ∆PBC and ∆PDE,

∠BPC = ∠EPD [vertically opposite angles]

Since, ∠BPC of ∆PBC is equal to ∠EPD of ∆PDE and the sides including these.

So,

∆PBC ∼ ∆PDE by SAS similarity criteria.

We know that,

The sum of three angles of a triangle is 180°.

In ∆PQR,

∠P + ∠Q + ∠R = 180°

⇒ 55° + 25° + ∠R = 180°

⇒ ∠R = 180° - (55° + 25°) = 180°- 80° = 100°

Similarly, in ∆TSM,

∠T + ∠S + ∠M = 180°

⇒ ∠T + ∠25° + 100° = 180°

⇒ ∠T = 180°- (∠25° + 100°)

⇒ ∠T = 180° - 125° = 55°

In ∆PQR and ∆TSM,

∠P = ∠T,

∠Q = ∠S

And

∠R = ∠M

So, ∆PQR ∼ ∆TSM

Since, all corresponding angles are equal

Hence,

∆QPR is similar to ∆TSM,

(False)

Just one corresponding angle on either quadrilateral isn't enough to say that they would be similar. To be similar, with the corresponding angles of each they should also have proportional corresponding sides.

(True)

If the corresponding two sides and the perimeters of two triangles are proportional, and third sides of both triangles are also in proportion then, the two triangles are similar.

(True)

Let two right angled triangles be,

∆LMO and ∆RST

In which

∠L = ∠R = 90°

And

∠M = ∠S [Acute angle]

By angle sum property of triangle that, sum of interior angles of a triangle is 180°. Angle O of first triangle must be equal to angle T of second triangle.

So,

∠O = ∠T

Hence, by AAA similarity criteria,

We have, ∆LMO ∼ ∆RST

(False)

Given: The ratio of altitudes of similar triangles is 3/5

By the property of area of two similar triangles,

We have,

So, given statement is not correct because it does not satisfy the criteria.

(False)

In ∆PQD and ∆RPD,

Given,

PD = PD [common side]

∠PDQ = ∠PDR [each 90°]

Here, no other sides or angles are equal, so we can say that ∆PQD is not similar to ∆RPD.

(True)

In ∆ADE and ∆ACB,

Given,

∠D = ∠C

∠A = ∠A [common angle]

As we know sum of all the angles of a triangle is equals to 180∘.

So, by angle sum property of triangle the third angle of both triangles must be equal.

∠E = ∠B

∴ ∆ADE ∼ ∆ACB [by AAA similarity criterion]

(False)

The given statement is not correct.

As here, one angle and two sides of two triangles are equal but these sides not including equal angle.

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

Given,

∆PQR,

PR² = QR² and QM⊥PR

Since,

By Pythagoras theorem,

PR² – PQ² = QR²

⇒ PR² = PQ² + QR²

So, ∆PQR is right angled triangle at Q.

In ∆QMR and ∆PMQ,

∠M = ∠M [each]

∠MQR = ∠QPM [each equal to 90°-∠R]

So, by AAA similarity criteria,

∴ ∆QMR ∼ ∆PMQ

Now, using property of area of similar triangles, we get,

[∵ Area of triangles = × base × height]

⇒ QM² = PM × RM

Hence proved.

Given,

DE || AB

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

So, the line drawn is equals to the third side of the triangle.

⇒ (x + 3) (3x + 4) = x (3x + 19)

⇒ 3x² + 4x + 9x + 12 = 3x² + 19x

⇒ 19x - 13x = 12

⇒ 6x = 12

∴ x = 12/6 = 2

[By basic proportionality theorem]

Hence, the required value of x is 2.

Given: ∆ NSQ ≅ ∆MTR

∠1 = ∠2

Since,

∆NSQ = ∆MTR

So,

SQ = TR ….(i)

Also,

∠1 = ∠2 ⇒ PT = PS….(ii)

[Since, sides opposite to equal angles are also equal]

From Equation (i) and (ii).

PS/SQ = PT/TR

⇒ ST || QR

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

∴ ∠1 = PQR

And

∠2 = ∠PRQ

In ∆PTS and ∆PRQ.

∠P = ∠P [Common angles]

∠1 = ∠PQR (proved)

∠2 = ∠PRQ (proved)

∴ ∆PTS - ∆PRQ

[By AAA similarity criteria]

Hence proved.

Given,

PQRS is a trapezium in which PQ‖RS and PQ = 3RS

In ∆POQ and ∆ROS,

∠SOR = ∠QOP [vertically opposite angles]

∠SRP = ∠RPQ [alternate angles]

∴ ∆POQ ∼ ∆ROS [by AAA similarity criterion]

By property of area of similar triangle,

[from Equation (i)]

Hence, the required ratio is 9:1.

Given,

AC and PQ intersect each other at the point O and AB‖DC.

in ∆AOP and ∆COQ,

∠AOP = ∠COQ [vertically opposite angles]

∠APO = ∠CQO [since, AB‖DC and PQ is transversal, so alternate angles]

∴ ∆AOP ∼ ∆COQ [by AAA similarity criterion]

Then, OA/OC = AP/CQ

[since, corresponding sides are proportional]

⇒ OA × CQ = OC × AP

Hence Proved.

Let LMN be an equilateral triangle of side 8 cm

LM = MN = NL = 8 cm. (all sides of an equilateral triangle is equal)

Draw altitude LD which is perpendicular to MN.

Then, D is the mid-point of MN.

∴ MD = ND = 1/2

MN = 8/2 = 4 cm

Now, by Pythagoras theorem

LM² = LD² + MD²

⇒ (8)² = LD² + (4)²

⇒ 64 = LD² + 16

⇒ LD = √48 = 4√3 cm.

Hence, altitude of an equilateral triangle is 4√3 cm.

Given,

AB = 4 cm,

DE = 6 cm

EF = 9 cm and

FD = 12 cm

Also,

∆ABC ∼ ∆DEF

We have,

By taking first two terms, we have

And by taking last two terms, we have,

Now,

Perimeter of ∆ABC = AB + BC + AC

= 4 + 6 + 8 = 18 cm

Thus, the perimeter of the triangle is 18 cm.

The given figure:

To find: area (ADE) : area (DECB)
Given,

DE || BC,

DE = 6 cm and BC = 12 cm

In ∆ABC and ∆ADE,

∠ABC = ∠ADE [corresponding angle]

∠ACB = ∠AED [corresponding angle]

And

∠A = ∠A [common side]

∴ ∆ABC ∼ ∆AED [by AAA similarity criterion]

Then,

By property of area of similar triangle,

Let area (∆ADE) = k,

Then area (∆ABC) = 4k

Now,

Area (DECB) = area (ABC) – area (ADE) = 4k – k = 3k

∴ Required ratio = area (ADE) : area (DECB) = k : 3k = 1:3
Therefore, area (ADE) : area (DECB) = 1:3

Given,

A trapezium, ABCD in which AB‖DC, P and Q are Points on AD and BC respectively,

Such that PQ || DC.

Thus,

AB||PQ||DC.

In ∆ABD,

PO || AB [∵ PQ || AB]

By basic proportionality theorem,

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

By basic proportionality theorem,

From equation (i) and (ii)

→ AP = 42

∴ AD = AP + DP

AD = 42 + 18 = 60

So, AD = 60 cm

Given.

Ratio of corresponding sides of two similar triangles = 2:3 or

Area of smaller triangle = 48 cm²

By the property of area of two similar triangle,

Ratio of area of both triangles = (Ratio of their corresponding sides)²

⇒ area (larger triangle) =

⇒ area (larger triangle) = 12 × 9 = 108 cm²

So, the area of the larger triangle = 108 cm²

Given,

In ∆PQR,

N is a point on PR, such that QN ⊥ PR

And PN.NR = QN²

We have,

PN.NR = QN²

⇒ PN.NR = QN.QN

And

∠PNQ = ∠RNQ (common angle)

∴ ∆QNP ∼ ∆RNQ

Then, ∆QNP and ∆RNQ are equiangular.

∠PQN + ∠RQN = ∠QRN + ∠QPN

⇒ ∠PQR = ∠QRN + ∠QPN …..(ii)

We know that, sum of angles of a triangle = 180°

In ∆PQR,

∠PQR + ∠QPR + ∠QRP = 180°

⇒∠PQR + ∠QPN + ∠QRN = 180°

[∵ ∠QPR = ∠QPN and ∠QRP = ∠QRN]

⇒∠PQR + ∠PQR = 180°

⇒ 2∠PQR = 180°

[using Equation (ii)]

⇒ ∠PQR = 180°/2 = 90°

∴ ∠PQR = 90°

[each angle is equal to 90° by SAS similarity criteria]

Hence Proved.

Note:

To remember the process first, show that ∆QNP ∼ ∆RNQ, by SAS similarity criterion and then use the property that sum of all angles of a triangle is 180°.

Given,

Area of smaller triangle = 36 cm²

Area of larger triangle = 100 cm²

Also, length of a side of the larger triangle = 20 cm

Let length of the corresponding side of the smaller triangle = x cm

By property of area of similar triangle.

∴ x = √144 = 12 cm

Hence, the length of corresponding side of the smaller triangle is 12 cm.

Given,

AC = 8 cm,

AD = 3 cm and ∠ACB = ∠CDA

From figure,

∠CDA = 90°

∴ ∠ACB = ∠CDA = 90°

In right angled ∆ADC,

AC² = AD² + CD²

⇒ (8)² = (3)² + (CD)²

CD² = 64 – 9 = 55

⇒ CD = √55 cm

In ∆CDB and ADC.

∠BDC = ∠AD [each 90°]

∠DBC = ∠DCA [each equal to 90°-∠A]

∴ ∠CDB ∼ ∆ADC

Then,

⇒ CD² = AD × BD

Given,

Let,

QR = 15 m (height of tower)

PQ = 24 m (shadow of tower)

At that time ∠RPQ = θ

Again, let YZ = h be a telephone pole and its shadow XY = 16 m.

The same time ∠YXZ = θ

Here, ABC and ∆DEF both are right angles triangles.

In ∆PQR and ∆XYZ,

∠RPQ = ∠YXZ = θ

∠Q = ∠Y [each 90°]

∴ ∆PQR ∼ ∆XYZ [by AAA similarity criterion]

Then,

Hence, the height of the telephone pole is 10 m.

Given,

Let PQ be a vertical wall and PR = 10m is a ladder.

The top of the ladder reaches to P and distance of ladder from the base of the wall QR is 6m.

In right angled ∆PQR,

PR² = PQ² + QR² [by Pythagoras theorem]

⇒ (10)² = PQ² + (6)²

⇒ 100 = PQ² + 36

⇒ PQ² = 100 – 36 = 64

∴ PQ = √64 = 8 cm

Hence, the height of the point on the wall where the top of the ladder reaches is 8cm.

Given,

∠A = ∠C,

AB = 6 cm, BP = 15 cm,

AP = 12 cm

CP = 4 cm

In ∆APB and ∆CPD,

∠A = ∠C [given]

∠APB = ∠CPD [vertically opposite angles]

∴ By AAA similarity criteria,

∆APD ∼ ∆CPD

By taking first two terms,

By taking first and last terms,

Hence,

length of PD = 5 cm

length of CD = 2 cm

Given,

∆ABC ∼ ∆EDF,

So the corresponding sides of ∆ABC and ∆EDF are in the same ratio. (Property of similar triangle)

AB/ED = AC/EF = BC/DF ….(i)

Also,

AB = 5cm, AC = 7cm (given)

DF = 15cm and DE = 12cm (given)

On putting these values in Equation (i), we get,

On taking first and second terms, we get,

On taking first and third terms, we get,

Hence, lengths of the remaining sides of the triangles are EF = 16.8 cm and BC = 6.25 cm.

Let a ∆ABC in which a line DE parallel to BC intersects AB at D and AC at E.

To prove DE divides the two sides in the same ratio.

Construction join BE, CD and draw EF ⊥ AB and DG ⊥ AC.

Proof Here,

[ area of triangle = × base × height]

Similarly,

Now,

Since,

∆BDE and ∆DEC lie between the same parallel DE and BC and on the same base DE.

So, area (∆BDE) = area(∆DEC) …..(iii)

From Equation (i), (ii) and (iii),

Hence proved.

Given,

PQRS is a parallelogram,

So, PQ || SR and PS || QR.

Also, AB || PS.

To prove OC || SR

In ∆OPS and OAB,

PS | | AB

∠POS = ∠AOB [common angle]

∠OSP = ∠OBA [corresponding angles]

∴ ∆OPS ∼ ∆OAB [by AAA similarity criteria]

Then,

PS/AB = OS/OB …(i) [by basic proportionality theorem]

In ∆CQR and ∆CAB,

QR || PS || AB

∠QCR = ∠ACB [common angle]

∠CRQ = ∠CBA [corresponding angles]

∴ ∆CQR ∼ ∆CAB

Then, by basic proportionality theorem

[PS ≅ QR Since, PQRS is a parallelogram,]

From Equation (i) and (ii),

On subtracting from both sides, we get,

By converse of basic proportionality theorem, SR || OC

Hence proved.

Let AC be the ladder of length 5 m.

BC = 4 m be the height of the wall, which ladder is placed.

In right angled ∆EBD,

ED² = EB² + BD² [by Pythagoras theorem]

⇒ (5)² = (EB)² + (14)² [ BD = 1.4]

⇒ 25 = (EB)² + 1.96

⇒ (EB)² = 25 –1.96 = 23.04

⇒ EB = √23.04 = 4.8

Now, EC = EB – BC = 4.8 – 4 = 0.8

Hence, the top of the ladder would slide upwards on the wall at distance 0.8 m.

Given,

AC⊥CB,

AC = 2x km,

CB = 2 (x + 7) km and AB = 26 km

On drawing the figure, we get the right angled ∆ ACB right angled at C.

Now, in ∆ACB,

AB² = AC² + BC² by Pythagoras theorem,

⇒ (26)² = (2x)² + {2(x + 7)}²

⇒ 676 = 4x² + 4(x² + 196 + 14x)

⇒ 676 = 4x² + 4x² + 196 + 56x

⇒ 676 = 8² + 56x + 196

⇒ 8x² + 56x – 480 = 0

On dividing by 8, we get

X² + 7x – 60 = 0

⇒ x² + 12x - 5x - 60 = 0

⇒ x(x + 12)-5(x + 12) = 0

⇒ (x + 12)(x - 5) = 0

∴ x = -12, x = 5

As the distance can’t be negative.

∴ x = 5 [∵ x ≠ -12]

Now, AC = 2x = 10km

And BC = 2(x + 7) = 2(5 + 7) = 24 km

The distance covered to each city B from city A via city C

= AC + BC

= 10 + 24

= 34 km

Distance covered to each city B from city A after the construction of the highway = BA = 26 km

Hence, the required saved distance is 34 – 26 = 8 km.

Let P be the position of the street bulb fixed on a pole PQ = 6m.

CD = 1.5 m be the height of a woman

ED = 3 m. (Shadow of women)

Let distance between pole and woman be x meter.

Here, woman and pole both are standing vertically.

So,

CD || PQ

In ∆CDE and ∆PQE,

∠E [common angle]

∠PQE = ∠CDE [each equal to 90°]

∴ ∆CDE ∼ ∆PQE [by AAA similarity criterion]

Then,

Hence,

She is at the distance of 9 m from the base of the pole.

Let MN = 18 m be the flag pole and its shadow be LM = 9.6 m.

The distance of the top of the pole, N from the far end, L of the shadow is LN.

In right angled ∆LMN,

LN² = LM² + MN² [by Pythagoras theorem]

⇒ LN² = (9.6)² + (18)²

⇒ LN² = 9.216 + 324

⇒ LN² = 416.16

∴ LN = √416.16 = 20.4 m

Hence, the required distance is 20.4 m

Given,

∆ABC in which ∠B = 90° and BD ⊥ AC

Also, AD = 4 cm and CD = 5 cm

In ∆ADB and ∆CDB,

∠ADB = ∠CDB [each equal 90°]

∠BAD = ∠DBC [each equal to 90°-∠C]

∴ ∆DBA ∼ ∆DCB [by AAA similarity criteria]

Then,

In right angled ∆BDC,

BC² = BD² + CD² [by Pythagoras theorem]

= BC² = (2√5)² + (5)²

= BC² = 20 + 25 = 45

= BC = √45 = 3√5

Again,

Hence, BD = 2√5cm and AB = 6cm.

Given,

∆PQR in which ∠Q = 90°,

QS ⊥ PR and PQ = 6cm.

PS = 4cm

In ∆SQP and ∆SRQ,

∠PSQ = ∠RSQ [each equal to 90°]

∠SPQ = ∠SQR [each equal to 90°-∠R]

∴ ∆SQP ∼ ∆SRQ

Then,

= SQ/PS = SR/SQ

⇒ SQ² = PSXSR …..(i)

In right angled ∆PSQ,

PQ² = PS² + QS² [by Pythagoras theorem]

= (6)² = (4)² + QS²

= 36 = 16 + QS²

= QS² = 36-16 = 20

∴ QS = √20 = 2√5 cm

On putting the value of QS in Eq. (i),

we get,

In right angled ∆QSR,

QR² = QS² + SR²

= QR² = (2√5)² + (5)²

= QR² = 20 + 25

∴ QR = √45 = 3√5cm

Hence,

QS = 2√5 cm, RS = 5 cm and QR = 3√5 cm.

Given,

In ∆PQR,

PD⊥QR,

PQ = a, PR = b, QD = c and DR = d

in right angled ∆PDQ,

PQ² = PD² + QD² [by Pythagoras theorem]

= a² = PD² + c²

= PD² = a² – c² …..(i)

In right angled ∆PDR,

PR² = PD² + DR² [by Pythagoras theorem]

= b² = PD² + d²

= PD² = b² - d² ……(i)

From Equation (i) and (ii),

A² - c² = b² - d²

A² - b² = c² - d²

= (a – b) (a + b) = (c – d) (c + d)

Hence proved.

Given,

Quadrilateral ABCD,

In which ∠A + ∠D = 90°

Construct produce AB and CD to meet at E.

Also, join AC and BD.

in ∆AED,

∠A + ∠D = 90° [given]

∴∠E = 180°- (∠A + ∠D) = 90° [∵ sum of angles of a triangle = 180°]

Then,

By Pythagoras theorem,

AD² = AE² + DE²

In ∆BEC,

by Pythagoras theorem,

BC² = BE² + EF²

On adding both equations, we get,

AD² + BC² = AE² + DE² + BE² + CE²

In ∆AEC,

by Pythagoras theorem,

AC² = AE² + CE²

And in ∆BED,

by Pythagoras theorem,

BD² = BE² + DE²

On adding both equations, we get,

AC² + BD² = AE² + CE² + BE² + DE² ….(i)

From Equation (i) and (ii),

AC² + BD² = AD² + BC²

Hence proved.

Given,

⎩||m and line segment AB, CD and EF are concurrent at point P.

To prove

in ∆APC and ∆BPD,

∠APC = ∠BPD [vertically opposite angles]

∠PAC = ∠PBD [alternate angles]

∴ ∆APS-BPD [by AAA similarity criterion]

Then,

In ∆APE and ∆BPF,

∠APE = ∠BPF [vertically opposite angles]

∠PAE = ∠PBF [alternate angles]

∴ ∆APE ∼ ∆BPF [by AAA similarity criterion]

Then,

In ∆PEC and ∆PED,

∠EPC = ∠FPD [Vertically opposite angles]

∠PCE = ∠PDF [alternate angles]

∴ ∆PEC ∼ ∆PFD [by AAA similarity criterion]

Then,

From Eqs. (i), (ii) and (iii),

Hence proved.

Given,

AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm

Also,

PA, QB, RC and SD are all perpendiculars to line ⎩.

∴ PA:QB:RS = AB:BC:CD

= 6:9:12

Let PQ = 6x, QR = 9x and RS = 12x

Since, length of PS = 36 km

∴ PQ + QR + RS = 36

⇒ 6x + 9x + 12x = 36

⇒ 27x = 36

Given:

ABCD is a trapezium,

Diagonals AC and BD are intersect at 0.

To prove: PQ || AB || DC.

PO = QO

Concepts Used:

AAA Similarity Criterion: If all three angles of a triangle equals to angles of another triangle, then both the triangles are similar.

Basic Proportionality theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion

in ∆ABD and ∆POD,

PO || AB [∵ PQ || AB]

∠D = ∠D [common angle]

∠ABD = ∠POD [corresponding angles]

∴ ∆ABD ∼ ∆POD [by AAA similarity criterion]

Then,

OP/AB = PD/AD …(i) [by basic proportionality theorem]

In ∆ABC and ∆OQC,

OQ || AB [∵ OQ || AB]

∠C = ∠C [common angle]

∠BAC = ∠QOC [corresponding angle]

∴ ∆ABC ∼ ∆OQC [by AAA similarity criterion]

Then,

OQ/AB = QC/BC …(ii) [by basic proportionality theorem]

Now, in ∆ADC,

OP || DC

∴ AP/PD = 0A/0C [by basic proportionality theorem] …(iii)

In ∆ABC, OQ || AB

∴ BQ/QC = OA/OC [by basic proportionality theorem] …(iv)

From Equation (iii) and (iv),

AP/PD = BQ/QC

Adding 1 on both sides, we get,

= AP/PD + 1 = BQ/QC + 1

= ((AP + PD))/PD = (BQ + QC)/QC

= AD/PD = BC/QC

= PD/AD = QC/BC

= OP/AB = OQ/BC [from Equation (i) and (ii)]

⇒ OP/AB = OQ/AB [from Equation (iii)]

⇒ OP = OQ

Hence proved.

Given,

in ∆ABC, E is the mid-point of CA and ÐAEF = ÐAFE

To prove BD/CD = BF/CE

Construction take a point G on AB such that CG‖EF.

since, E is the mid-point of CA.

∴ CE = AE……(i)

In ∆ACG,

CG‖EF and E is mid-point of CA.

So, CE = GF…(ii) [by mid-point theorem]

Now, in ∆BCG and ∆BDF,

CG || EF

Hence proved.

Let RST be a right triangle at S and RS = y, ST = x.

Three semi-circles are draw on the sides RS, ST and RT, respectively A₁, A₂ and A₃.

To prove A₃ = A₁ + A₂

In ∆RST,

by Pythagoras theorem,

RT² = RS² + ST²

= RT² = y² + x²

We know that,

Area of a semi-circle with radius,

∴ Area of semi-circle drawn on RT,

Now, area of semi-circle drawn on RS,

And area of semi-circle drawn on ST,

On adding Equation (ii) and (iii), we get

⇒ A₁ + A₂ = A₃

Hence proved.

Let a right triangle QPR in which ÐP is right angle and PR = y, PQ = x.

Three equilateral triangles ∆PER, ∆PFR and ∆RQD are drawn on the three sides of ∆PQR.

Again, let area of triangles made on PR, PQ are A1, A2 and A3, respectively.

To prove A₃ = A₁ + A₂

In ∆RPQ,

By Pythagoras theorem,

QR² = PR² + PQ²

⇒ QR² = y² + x²

We know that,

Area of an equilateral triangle =

∴ Area of equilateral ∆PER,

And area of equilateral ∆PFQ,

A₁ + A₂

[from Equation (i) and (ii)]

Hence proved.