Triangles

The solutions of the fill ups are:

(i) Similar

(ii) Similar

(iii) Equilateral

(iv) (a) Equal

(b) Proportional

Two examples of similar figures are:

(i) Two equilateral triangles with sides 1 cm and 2 cm

(ii) Two squares with sides 1 cm and 2 cm

Now two examples of non-similar figures are:

(i) Trapezium and square

(ii) Triangle and parallelogram

The given quadrilateral PQRS and ABCD are not similar because though their corresponding sides are proportional, i.e. 1:2, but their corresponding angles are not equal.

(i) Let us take EC = x cm

Given: DE || BC

Now, using basic proportionality theorem, we get:

=

x =

x = 2 cm

Hence, EC = 2 cm

(ii) Let us take AD = x cm

Given: DE || BC

Now, using basic proportionality theorem, we get

x =

Hence, AD = 2.4 cm

(i)

Given :
PE = 3.9 cm,

EQ = 3 cm,

PF = 3.6 cm,

FR = 2.4 cm

Now we know,

Triangle Proportionality Theorem: If a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides these two sides proportionally.

So, if the lines EF and QR are to be parallel, then ratio PE:EQ should be proportional to PF:PR

Therefore, EF is not parallel to QR

(ii)

We know that,
Triangle Proportionality Theorem: If a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides these two sides proportionally.

Hence,

Therefore, EF is parallel to QR

(iii)

In this we know that,

Triangle Proportionality Theorem: If a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides these two sides proportionally.

So, if the lines EF and QR are to be parallel, then ratio PE:EQ should be proportional to PF:PR

Hence,

EF is parallel to QR

Given: LM ll CB and LN ll CD
From the given figure:
In ΔALM and ΔABC

LM || CB
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

Now, using basic proportionality theorem that the corresponding sides will have same proportional lengths, we get:

(i)

And in ΔALN and ΔACD corresponding sides will be proportional

Therefore,

(ii)

From (i) and (ii), we obtain

Hence, Proved.

Given:

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 triangle ABC, DE is parallel to AC

Therefore,

......(1)

In triangle BAE, DF is parallel to AE


Therefore,

By Basic proportionality theorem

...... (2)

From (1) and (2), we get

To Prove: EF ll QR

Given: In triangle POQ, DE parallel to OQ

Proof:

In triangle POQ, DE parallel to OQ

Hence,

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.

(Basic proportionality theorem) (i)

Now,

Hence,

(Basic proportionality theorem) (ii)

From (i) and (ii), we get

Therefore,

EF is parallel to QR (Converse of basic proportionality theorem)

To Prove: BC ll QR

Given that in triangle POQ, AB parallel to PQ

Hence,

(Basic proportionality theorem)

Now,

Therefore,

Using Basic proportionality theorem, we get:

From above equations, we get

BC is parallel to QR (By the converse of Basic proportionality theorem)

Consider the given figure in which PQ is a line segment drawn through the mid-point P of line AB, such that PQ is parallel to BC.

To Prove: PQ bisects AC
Given: PQ ll BC and PQ bisects AB
Proof:
According to Theorem 6.1: 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.
Now, using basic proportionality theorem, we get

[As AP = PB coz P is the mid-point of AB]

AQ = QC

Or, Q is the mid-point of AC
Hence proved.

To Prove: PQ ll BC
Given: P and Q are midpoints of AB and AC
Proof:
Let us take the given figure in which PQ is a line segment which joins the mid-points P and Q of line AB and AC respectively

i.e., AP = PB and AQ = QC

We observe that,

And,

Therefore,

Hence, using basic proportionality theorem we get:

PQ parallel to BC
Hence, Proved.

The figure is given below:

Given: ABCD is a trapezium
AB ll CD
Diagonals intersect at O
To Prove =

Construction: Construct a line EF through point O, such that EF is parallel to CD.

Proof:

In ΔADC, EO is parallel to CD
According to basic propotionality theorem, if a side is drawn parallel to any side of the triangle then the corresponding sides formed are propotional

Now, using basic proportionality theorem in ΔABD and ΔADC, we obtain

In ΔABD, OE is parallel to AB

So, using basic proportionality theorem in ΔEOD and ΔABD, we get

From (i) and (ii), we get

Therefore by cross multiplying we get,

Hence, Proved

The quadrilateral ABCD is shown below, BD and AC are the diagonals.

Given: In ΔABD, OE is parallel to AB

Now, using basic proportionality theorem in ΔDOE and ΔABD, we obtain

...(i)

It is given that,

...(ii)

From (i) and (ii), we get

...(iii)
Now for ABCD to be a trapezium AB has to be parallel of CD

EO || DC (By the converse of basic proportionality theorem)

⇒ AB || OE || DC

⇒ AB || CD

Hence,

ABCD is a trapezium

(i) From the figure:

∠A = ∠P = 60°

∠B = ∠Q = 80°

∠C = ∠R = 40°

Therefore, ΔABC ΔPQR [By AAA similarity]

Now corresponding sides of triangles will be propotional,

(ii)
From the triangle,



Hence the corresponding sides are propotional

Thus the corresponding angles will be equal

The triangles ABC and QRP are similar to each other by SSS similarity

(iii) The given triangles are not similar because the corresponding sides are not proportional

(iv) In triangle MNL and QPR, we have

∠M = ∠Q = 70^o

(v) In triangle ABC and DEF, we have

AB = 2.5, BC = 3

∠A = 80^o

EF = 6

DF = 5

∠F = 80^o

And,

∠B ≠ ∠F

Hence, triangle ABC and DEF are not similar

(vi) In triangle DEF, we have

∠D + ∠E + ∠F = 180^o (Sum of angles of triangle)

70^o + 80^o + ∠F = 180^o

∠F = 30^o

In PQR, we have

∠P + ∠Q + ∠R = 180^o

∠P + 80^o + 30^o = 180^o

∠P = 70^o

In triangle DEF and PQR, we have

∠D = ∠P = 70^o

∠F = ∠Q = 80^o

∠F = ∠R = 30^o

Hence, ΔDEF ~ ΔPQR (AAA similarity)

From the figure,

We see, DOB is a straight line

∠ DOC + ∠ COB = 180° (angles on a straight line form a supplementary pair)

∠ DOC = 180° - 125°

∠ DOC = 55°

Now, In ΔDOC,

∠ DCO + ∠ CDO + ∠ DOC = 180°

(Sum of the measures of the angles of a triangle is 180°)

∠ DCO + 70° + 55° = 180°

∠ DCO = 55°

It is given that ΔODC ΔOBA

∠ OAB = ∠ OCD (Corresponding angles are equal in similar triangles)

Thus, ∠ OAB = 55°

In ΔDOC and ΔBOA,

∠CDO = ∠ABO (Alternate interior angles as AB || CD)

∠DCO = ∠BAO (Alternate interior angles as AB || CD)

∠DOC = ∠BOA (Vertically opposite angles)

Therefore,

ΔDOC ∼ ΔBOA [ BY AAA similarity]

Now in similar triangles, the ratio of corresponding sides are proportional to each other. Therefore,

......... (Corresponding sides are proportional)

or,

Hence, Proved.

To Prove: Δ PQS ∼ Δ TQR

Given: In ΔPQR,

∠PQR = ∠PRQ

As∠PQR = ∠PRQ

PQ = PR [sides opposite to equal angles are equal] (i)

Given,

by (1)

In Δ PQS and Δ TQR, we get

∠Q = ∠Q

Therefore,

Δ PQS ∼ Δ TQR

Hence, Proved.

In ΔRPQ and ΔRST,

∠ RTS = ∠ QPS (Given)

∠ R = ∠ R (Common to both the triangles)
If two angles of two triangles are equal, third angle will also be equal. As the sum of interior angles of triangle is constant and is 180°

∴ ΔRPQ ΔRTS (By AAA similarity)

To Prove: Δ ADE ∼ Δ ABC

Given: ΔABE ≅ ΔACD

Proof:

ΔABE ≅ ΔACD

∴ AB = AC (By CPCT) (i)

And,

AD = AE (By CPCT) (ii)

In ΔADE and ΔABC,

Dividing equation (ii) by (i)

∠A = ∠A (Common)

SAS Similarity: Triangles are similar if two sides in one triangle are in the same proportion to the corresponding sides in the other, and the included angle are equal.

Therefore,

ΔADEΔABC (By SAS similarity)

Hence, Proved.

(i) In ΔAEP and ΔCDP,

∠AEP = ∠CDP (Each 90°)

∠APE = ∠CPD (Vertically opposite angles)

Hence, by using AA similarity,

ΔAEP ΔCDP

(ii) In ΔABD and ΔCBE,

∠ADB = ∠CEB (Each 90°)

∠ABD = ∠CBE (Common)

Hence, by using AA similarity,

ΔABD ΔCBE

(iii) In ΔAEP and ΔADB,

∠AEP = ∠ADB (Each 90°)

∠PAE = ∠DAB (Common)

Hence, by using AA similarity,

ΔAEP ΔADB

(iv) In ΔPDC and ΔBEC,

∠PDC = ∠BEC (Each 90°)

∠PCD = ∠BCE (Common angle)

Hence, by using AA similarity,

ΔPDC ΔBEC

To Prove: Δ ABE ∼ Δ CFB
Given: E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. As shown in the figure.
Proof:
In ΔABE and ΔCFB,

∠A = ∠C (Opposite angles of a parallelogram are equal)

∠AEB = ∠CBF (Alternate interior angles are equal because AE || BC)

Therefore,

ΔABE ΔCFB (By AA similarity)

(i)

To Prove: Δ ABC ∼ Δ AMP

Given: In ΔABC and ΔAMP,

∠ABC = ∠AMP (Each 90°)

∠A = ∠A (Common)

∴ΔABC ΔAMP (By AA similarity)

Hence, Proved.

(ii) Δ ABC ∼ Δ AMP

Now we get that,

Similarity Theorem - If the lengths of the corresponding sides of two triangles are proportional, then the triangles must be similar. And the converse is also true, so we have

Given, Δ ABC ∼ Δ FEG …..eq(1)

⇒ corresponding angles of similar triangles

⇒ ∠ BAC = ∠ EFG ….eq(2)

And ∠ ABC = ∠ FEG …….eq(3)

⇒ ∠ ACB = ∠ FGE

⇒

⇒ ∠ ACD = ∠ FGH and ∠ BCD = ∠ EGH ……eq(4)

Consider Δ ACD and Δ FGH

⇒ From eq(2) we have

⇒ ∠ DAC = ∠ HFG

⇒ From eq(4) we have

⇒ ∠ ACD = ∠ EGH

Also, ∠ ADC = ∠ FGH

⇒ If the 2 angle of triangle are equal to the 2 angle of another triangle, then by angle sum property of triangle 3^rd angle will also be equal.

⇒ by AAA similarity we have in two triangles if the angles are equal, then sides opposite to the equal angles are in the same ratio (or proportional) and hence the triangles are similar.

∴ Δ ADC ∼ Δ FHG

⇒ By Converse proportionality theorem

⇒

Consider Δ DCB and Δ HGE

From eq(3) we have

⇒ ∠ DBC = ∠ HEG

⇒ From eq(4) we have

⇒ ∠ BCD = ∠ FGH

Also, ∠ BDC = ∠ EHG

∴ Δ DCB ∼ ΔHGE

Hence proved.

To Prove: Δ ABD ∼ Δ ECF

Given: ABC is an isosceles triangle, AD is perpendicular to BC

BC is produced to E and EF is perpendicular to AC

Proof:

Given that ABC is an isosceles triangle

AB = AC

⇒ ∠ABD = ∠ECF

In ΔABD and ΔECF,

∠ADB = ∠EFC (Each 90°)

∠ABD = ∠ECF (Proved above)

ΔABD ΔECF (By using AA similarity criterion)

AA Criterion: If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.

Hence, Proved.

To Prove: Δ ABC ∼ Δ PQR

Given:

Proof:
Median divides the opposite side

BD = and,

QM =

Now,

=

In Δ ABD and Δ PQM,

ΔABD ΔPQM (By SSS similarity)

∠ABD = ∠PQM (Corresponding angles of similar triangles)

In ΔABC and ΔPQR,

∠ABD = ∠PQM (Proved above)

ΔABC ΔPQR (By SAS similarity)

In ΔADC and ΔBAC,

∠ACD = ∠BCA (Common angle)

According to AA similarity, if two corresponding angles of two triangles are equal then the triangles are similar

ΔADC ΔBAC (By AA similarity)

We know that corresponding sides of similar triangles are in proportion

Hence in ΔADC and ΔBAC,

Hence Proved

To Prove: Δ ABC ∼ Δ PQR
Given:

Proof:

Let us extend AD and PM up to point E and L respectively, such that AD = DE and PM = ML.
Then, join B to E, C to

E, Q to L, and R to L

We know that medians divide opposite sides.

Hence, BD = DC and QM = MR

Also, AD = DE (By construction)

And, PM = ML (By construction)

In quadrilateral ABEC,

Diagonals AE and BC bisect each other at point D.

Therefore,

Quadrilateral ABEC is a parallelogram.

AC = BE and AB = EC (Opposite sides of a parallelogram are equal)

Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ = LR

It was given in the question that,

ΔABE ΔPQL (By SSS similarity criterion)

We know that corresponding angles of similar triangles are equal.

∠BAE = ∠QPL ..... (i)

Similarly, it can be proved that

ΔAEC ΔPLR and

∠CAE = ∠RPL ..... (ii)

Adding equation (i) and (ii), we obtain

∠BAE + ∠CAE = ∠QPL + ∠RPL

⇒∠CAB = ∠RPQ .... (iii)

In ΔABC and ΔPQR,

(Given)

∠CAB = ∠RPQ [Using equation (iii)]

ΔABC ΔPQR (By SAS similarity criterion)
Hence, Proved.

Let AB and CD be a tower and a pole respectively

And, the shadow of BE and DF be the shadow of AB and CD respectively

At the same time, the light rays from the sun will fall on the tower and the pole at the same angle

Therefore,

∠DCF = ∠BAE

And,

∠DFC = ∠BEA

∠CDF = ∠ABE (Tower and pole are vertical to the ground)

ΔABE ΔCDF (AAA similarity)

AB = 42 m
Height of the Tower = 42 m

It is given that ΔABC is similar to ΔPQR

Given: Δ ABC ~ Δ PQR

AD and PM are medians
We know that the corresponding sides of similar triangles are in proportion

..........eq(i)

∠A = ∠P

∠B = ∠Q

∠C = ∠R ..........eq(ii)

Since AD and PM are medians, they divide their opposite sides in two equal parts

BD = and,

QM = .......eq(iii)

From (i) and (iii), we get

(iv)

In ΔABD and ΔPQM,

∠B = ∠Q [Using (ii)]

[Using (iv)]

ΔABD ΔPQM (Since two sides are proportional and one angle is equal then by SAS similarity)

Hence, Proved

It is given that,

Δ ABC ~ Δ DEF

Therefore,

Given:

EF = 15.4 cm

ar (ΔABC) = 64cm²

ar (ΔDEF) = 121 cm²

Hence,

Taking square root on both of the sides

BC = (8 × 15.4) / 11

BC = 8 × 1.4 = 11.2 cm

Since AB || CD,

∴∠OAB = ∠OCD and ∠OBA = ∠ODC (Alternate interior angles)

In ΔAOB and ΔCOD,

∠AOB = ∠COD (Vertically opposite angles)

∠OAB = ∠OCD (Alternate interior angles)

∠OBA = ∠ODC (Alternate interior angles)

ΔAOB ~ΔCOD (By AAA similarity)
When two triangles are similar, the reduced ratio of any two corresponding sides is called the scale factor of the similar triangles. If two similar triangles have a scale factor of a : b, then the ratio of their areas is a² : b².

Since, AB = 2 CD (Given)

Therefore,

Therefore, the ratio of the areas of triangles AOB and COD is 4:1

Construction: Draw two perpendiculars AP and DM on line BC and AB

Proof:

Area of a triangle = 1/2 x Base x Height

Therefore,

=

In ΔAPO and ΔDMO,

∠APO = ∠DMO (Each = 90°)

∠AOP = ∠DOM (Vertically opposite angles)

∴ ΔAPO ΔDMO (By AA similarity)

Hence, proved.

Let us consider two similar triangles as ΔABC ΔPQR (Given)
When two triangles are similar, the reduced ratio of any two corresponding sides is called the scale factor of the similar triangles.
If two similar triangles have a scale factor of a : b, then the ratio of their areas is a² : b².

Given that,
ar(Δ ABC) = ar(Δ PQR)

Therefore putting in equation (i) we get,

AB = PQ

BC = QR

And,

AC = PR

Therefore,

ΔABC ΔPQR (By SSS rule)
Hence, Proved.

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

Because D, E and F are respectively the mid-points of sides AB, BC and CA of Δ ABC,

Midpoint Theorem: The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is congruent to one half of the third side.
Therefore, From mid-point theorem,

DE || AC and DE = AC [1]

DF || BC and DF = BC [2]

EF || AB and EF = AB [3]

From [1], [2] and [3]


By SSS Similarity Criterion, we can see that

ΔDEF ∼ ΔABC

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

=

Let us assume two similar triangles as ΔABC ~ ΔPQR.

Let AD and PS be the medians of these triangles

Then, because ΔABC ~ΔPQR

.....(i)

∠A = ∠P, ∠B = ∠Q, ∠C = ∠R .......(ii)

Since AD and PS are medians,

BD = DC =BC/2

And, QS = SR =QR/2

Equation (i) becomes,

.......(iii)

In ΔABD and ΔPQS,

∠B = ∠Q [From (ii)]

And,

[From (iii)]

ΔABD ~ ΔPQS (SAS similarity)

Therefore, it can be said that

.......(iv)

From (i) and (iv), we get

And hence,

Let ABCD be a square of side a

Two desired equilateral triangles are formed as ΔABE and ΔDBF

Side of an equilateral triangle, ΔABE, described on one of its sides = a

Side of an equilateral triangle, ΔDBF, described on one of its diagonals =

Therefore, Area of equilateral triangle on one of the side of the square is half of the area of equilateral triangle on diagonal.

Given: D is mid point of BC

We know:

Equilateral triangles have all its angles as 60° and all its sides are of the same length. Therefore, all equilateral triangles are similar to each other.

Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.

Let side of ΔABC = x

Therefore,

Side of ΔBDE =

Therefore,

Hence, the correct answer is C

If two triangles are similar to each other, then the ratio of the areas of these triangles will be equal to the square of theratio of the corresponding sides of these triangles.

It is given that the sides are in the ratio 4:9

Therefore,

Ratio between areas of these triangles =²

=

Hence, the correct answer is (D)

(i) Given: sides of the triangle are 7 cm, 24 cm, and 25 cm

Squaring the lengths of these sides, we get: 49, 576, and 625.

49 + 576 = 625

Or, 7² + 24² = 25²

The sides of the given triangle satisfy Pythagoras theorem

Hence, it is a right triangle

We know that the longest side of a right triangle is the hypotenuse

Therefore, the length of the hypotenuse of this triangle is 25 cm

(ii) It is given that the sides of the triangle are 3 cm, 8 cm, and 6 cm

Squaring the lengths of these sides, we will obtain 9, 64, and 36

However, 9 + 36 ≠ 64

Or, 32 + 62 ≠ 82

Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side

Therefore, the given triangle is not satisfying Pythagoras theorem

Hence, it is not a right triangle

(iii)Given that sides are 50 cm, 80 cm, and 100 cm.

Squaring the lengths of these sides, we will obtain 2500, 6400, and 10000.

And, 2500 + 6400 ≠ 10000

Or, 502 + 802 ≠ 1002

Now, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side

Therefore, the given triangle is not satisfying Pythagoras theorem

Hence, it is not a right triangle

(iv)Given: Sides are 13 cm, 12 cm, and 5 cm

Squaring the lengths of these sides, we get 169, 144, and 25.

Clearly, 144 +25 = 169

Or, 12² + 5² = 13²

The sides of the given triangle are satisfying Pythagoras theorem

Therefore, it is a right triangle

We know that the longest side of a right triangle is the hypotenuse

Therefore, the length of the hypotenuse of this triangle is 13 cm

⇒ Let ∠MPR = x

⇒ In Δ MPR, ∠MRP = 180-90-x

⇒ ∠MRP = 90-x

Similarly in Δ MPQ,

∠MPQ = 90-∠MPR = 90-x

⇒ ∠MQP = 180-90-(90-x)

⇒ ∠MQP = x

In Δ QMP and Δ PMR

⇒ ∠MPQ = ∠MRP

⇒ ∠PMQ = ∠RMP

⇒ ∠MQP = ∠MPR

⇒ Δ QMP ∼ Δ PMR

⇒ =

⇒ PM² = MR × QM

Hence proved

(i) In ΔADB and ΔCAB, we have

∠DAB = ∠ACB (Each of 90^o)

∠ABD = ∠CBA (Common angle)

Therefore,

ΔADB ΔCAB (AA similarity)

AB² = CB x BD

(ii) Let ∠CAB = x

In ΔCBA,

∠CBA = 180^o – 90^o – x

∠CBA = 90^o – x

Similarly, in ΔCAD

∠CAD = 90^o - ∠CBA

= 90^o – x

∠CDA = 180^o – 90^o – (90^o – x)

∠CDA = x

In triangle CBA and CAD, we have

∠CBA = ∠CAD

∠CAB = ∠CDA

∠ACB = ∠DCA (Each 90^o)

Therefore,

ΔCBA ΔCAD (By AAA similarity)

AC² = DC * BC

(iii) In ΔDCA and ΔDAB, we have

∠DCA = ∠DAB (Each 90^o)

∠CDA = ∠ADB (Common angle)

Therefore,

ΔDCA ΔDAB (By AA similarity)

AD² = BD x CD

To Prove: 2 AC² = AB²

Given: ΔABC is an isosceles triangle

Proof:

AC = CB ( Two sides of an isosceles triangle are equal, as the side opposite to right angle is largest, rest of the two sides are equal)

Pythagoras Theorem: It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Using Pythagoras theorem in ΔABC (i.e., right-angled at point C), we get

AC² + CB² = AB²

AC² + AC² = AB² (AC = CB)

2AC² = AB²

To Prove: ABC is a right angled triangle
Given: AB² = 2AC²

Now 2 AC² can be split into two parts

AB² = AC² + AC²

AB² = AC² + BC² (As, AC = BC)

Therefore, the given triangle is a right-angled triangle
Hence, Proved.

Let AD be the altitude of the given equilateral triangle, ΔABC

We know that altitude bisects the opposite side

BD = DC = a

In triangle ADB,

∠ADB = 90^O

Using Pythagoras theorem, we get

AD² + DB² = AB²

AD² + a² = (2a)²

AD² + a² = 4a²

AD² = 3a²

AD = a

In an equilateral triangle, all the altitudes are equal in length.

Hence, the length of each altitude will be

In Rhombus ABCD,
AB, BC, CD and AD are the sides of the rhombus.
BD and AC are the diagonals.

To prove: AB² + BC² + CD² +AD² = AC² + BD²
Proof:
The figure is shownbelow:

In ΔAOB, ΔBOC, ΔCOD, ΔAOD,

Applying Pythagoras theorem, we obtain

AB² = AO² + OB² .....................eq(i)

BC² = BO² + OC^{2 .......................eq} (ii)

CD² = CO² + OD² ....................eq(iii)

AD² = AO² + OD^{2.....................eq} (iv)

Now after adding all equations, we get,

AB² + BC² + CD² + AD² = 2 (AO² + OB² + OC² + OD²)

=(AC)² + (BD)²

Hence Sum of squares of sides of a rhombus equals to sum of squares of diagonals of rhombus.

(i)

To Prove : OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + EC²
Given: OD, OE and OF are perpendiculars on sides BC, AC and AB respectively
Construction : Join OA, OB and OC
Now according to pythagoras theorem, In a right angled triangle,
(hypotenuse)² = (altitude)² + (base)²

Applying Pythagoras theorem in ΔAOF, we obtain

OA² = OF² + AF² ..................eq(i)

Similarly, in ΔBOD,

OB² = OD² + BD² ..................eq(ii)

Similarly, in ΔCOE,

OC² = OE² + EC² .................eq(iii)

Adding these equations, we get

OA² + OB² + OC² = OF² + AF² + OD² + BD² + OE² + EC²
Rearranging the equations we get,

OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + EC²

(ii)

To Prove: AF² + BD² + EC² = AE² + CD² + BF²
From the above given result from (i),

AF² + BD² + EC² = (AO² – OE²) + (OC² – OD²) + (OB² – OF²)

AF² + BD² + EC² = AE² + CD² + BF²
Hence, Proved

Let OA be the wall and AB be the ladder

By Pythagoras theorem,

AB² = OA² + BO²

(10)² = (8)² + OB²

100 = 64 + OB²

OB² = 36

OB = 6 cm

Therefore, the distance of the foot of the ladder from the base of the wall is 6 m

To find: OA

Let OB be the pole and AB be the wire

By Pythagoras theorem,

AB² = OB² + OA²

(24)² = (18)² + OA²

OA² = (576 – 324)

OA² = 252

OA = 6√7 m

we know,

Distance = speed × time

Distance traveled by the plane flying towards north in = 1,000 x

= 1,500 km

Similarly, distance traveled by the plane flying towards west in = 1,200 x

= 1,800 km

Let these distances be represented by OA and OB respectively.

Applying Pythagoras theorem,

Distances between planes is 300√61 km

Let CD and AB be the poles of height 11 m and 6 m

Therefore, CP = 11 - 6 = 5 m

From the figure, it can be observed that AP = 12m

Applying Pythagoras theorem for ΔAPC, we obtain

AP² + PC² = AC²

(12)² + (5)² = AC²

AC² = (144 + 25)

AC² = 169

AC = 13 m

Therefore, the distance between their tops is 13 m

To Prove: AE² + BD² = AB² + DE²

Given: D and E are midpoints of AD and CB and ABC is right angled at C
Applying Pythagoras theorem in ΔACE, we obtain

AC² + CE² = AE² ..........eqn(i)

Applying Pythagoras theorem in triangle BCD, we get

BC² + CD² = BD² .........eqn(ii)

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

AC² + CE² + BC² + CD² = AE² + BD² .................eqn (iii)

Applying Pythagoras theorem in triangle CDE, we get

DE² = CD² + CE²

Applying Pythagoras in triangle ABC, we get

AB² = AC² + CB²

Putting these values in eqn(iii), we get

DE² + AB² = AE² + BD²

We have two right angled triangles now ΔACD and ΔABD

Applying Pythagoras theorem for ΔACD, we obtain

AC² = AD² + DC²

AD² = AC² – DC² ......................eq(i)

Applying Pythagoras theorem in ΔABD, we obtain

AB² = AD² + DB²

AD² = AB² – DB² ........................eq(ii)

From (i) and (ii), we get

AC² – DC² = AB² – DB² (iii)

It is given that 3DC = DB

Therefore,
DC + DB = BC

DC = and DB =

Putting these values in (iii), we get

AC² – = AB² –

16AC² – BC² = 16AB² – 9BC²

16AB² – 16AC² = 8BC²

2AB² = 2AC² + BC²

The figure is given below:

Given: BD = BC/3

To Prove: 9 AD² = 7 AB²

Proof:
Let the side of the equilateral triangle be a, and AM be the altitude of ΔABC

BM = MC = BC/2 = a/2 [ Altitude of an equilateral triangle bisect the side]

And, then, in ΔABM, by pythagoras theorem we write,
Pythagoras Theorem : Square of the Hypotenuse equals to the sum of the squares of other two sides.

AM² = AB² - BM²

or AM² = a² - a²/4

BD = a/3 [ BC = a]

DM = BM – BD

= a/2 – a/3

= a/6

According to pythagoras theorem in a right angled triangle,
(hypotenuse)² = (altitude)² + (base)²

Applying Pythagoras theorem in ΔADM, we obtain

AD² = AM² + DM²

Now, a = AB or a² = AB²

36 AD² = 28 AB²

9 AD² = 7 AB²
Hence, Proved

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC

To Prove : 4 × (Square of altitude) = 3 × (Square of one side)
Proof:
Altitude of equilateral triangle divides the side in two equal parts. Therefore,

Pythagoras Theorem : It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Applying Pythagoras theorem in ΔABE, we obtain

AB² = AE² + BE²

4 × (Square of altitude) = 3 × (Square of one side)

Given that, AB = 6 cm,

AC = 12 cm,

And BC = 6 cm

It can be observed that

AB² = 108

AC² = 144

And, BC² = 36

AB² +BC² = AC²

ΔABC, is satisfying Pythagoras theorem.

Therefore, the triangle is a right triangle, right-angled at B

∠B = 90°

Hence, the correct answer is (C)

Construct a line segment RT parallel to SP which intersects the extended line segment QP at point T

Given: PS is the angle bisector of ∠QPR.

Proof:

∠QPS = ∠SPR (i)

By construction,

∠SPR = ∠PRT (As PS || TR, By interior alternate angles) (ii)

∠QPS = ∠QTR (As PS || TR, By interior alternate angles) (iii)

Using these equations, we get:

∠PRT = ∠QTR

PT = PR

By construction,

PS || TR



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.

By using basic proportionality theorem for ΔQTR,

Hence, Proved

(i) To Prove: DM² = DN. MC

Construction:join DB

We have, DN || CB,

DM || AB,

And ∠B = 90° (Given)

As opposite sides are parallel and equal and also each angle is 90°, DMBN is a rectangle

DN = MB and DM = NB

The condition to be proved is the case when D is the foot of the perpendicular drawn from B to AC

∠CDB = 90°

Now from the figure we can say that

∠2 + ∠3 = 90° ........eq(i)

In ΔCDM,

∠1 + ∠2 + ∠DMC = 180° [ Sum of angles of a triangle = 180°]

∠1 + ∠2 = 90° ...........eq(ii)

In Δ DMB,

∠3 + ∠DMB + ∠4 = 180° [ Sum of angles of a triangle = 180°]

⇒∠3 + ∠4 = 90° ........eq(iii)

From (i) and (ii), we get

∠1 = ∠3

From (i) and (iii), we get

∠2 = ∠4

In ΔDCM and ΔBDM,

∠1 = ∠3 (Proved above)

∠2 = ∠4 (Proved above)

ΔDCM similar to ΔBDM (AA similarity)

(AA Similarity : When you have two triangles where one is a smaller version of the other, you are looking at two similar triangles.)

BM/DM = DM/MC

Cross multiplying we get,

DN/DM = DM/MC (BM = DN)

DM² = DN × MC

Hence, Proved.


(ii)
To Prove: DN²= AN x DM

In right triangle DBN,

∠5 + ∠7 = 90° (iv)

In right triangle DAN,

∠6 + ∠8 = 90° (v)

D is the foot of the perpendicular drawn from B to AC

∠ADB = 90°

∠5 + ∠6 = 90° (vi)

From equation (iv) and (vi), we obtain

∠6 = ∠7

From equation (v) and (vi), we obtain

∠8 = ∠5

In ΔDNA and ΔBND,

∠6 = ∠7 (Proved above)

∠8 = ∠5 (Proved above)

Hence,

ΔDNA similar to ΔBND (AA similarity criterion)

( AA similarity Criterion: When you have two triangles where one is a smaller version of the other, you are looking at two similar triangles.)

AN/DN = DN/NB

DN² = AN x NB

DN²= AN x DM (As NB = DM)

Hence, Proved

Using Pythagoras theorem in ΔADB, we get:

AB² = AD² + DB² (i)

Applying Pythagoras theorem in ΔACD, we obtain

AC² = AD² + DC²

AC² = AD² + (DB + BC)²

AC² = AD² + DB² + BC² + 2DB x BC

AC² = AB² + BC²+ 2DB x BC [Using equation (i)]

To Prove: AC² = AB² + BC² - 2BC x BD

Given: AD is Perpendicular on BC and angle ABC < 90°

Proof:

Pythagoras Theorem : It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Applying Pythagoras theorem in Δ ADB, we obtain

AD² + DB² = AB²

AD² = AB² – DB² .....eq(i)

Applying Pythagoras theorem in Δ ADC, we obtain

AD² + DC² = AC²

AB²– BD² + DC² = AC² [Using equation (i)]

AB²– BD² + (BC - BD)² = AC² [ DC = BC - BD]

AC² = AB²– BD² + BC² + BD² -2BC x BD

AC² = AB² + BC² - 2BC x BD

Hence, Proved.

(i) Using, Pythagoras theorem in ΔAMD, we get

AM² + MD² = AD² (i)

Applying Pythagoras theorem in ΔAMC, we obtain

AM² + MC² = AC²

AM² + (MD + DC)² = AC²

(AM² + MD²) + DC² + 2MD.DC = AC²

AD² + DC² + 2MD.DC = AC²[Using equation (i)]

Using, DC = BC/2 we get

AD² + (BC/2)² + 2MD * (BC/2) = AC²

AD² + (BC/2)² + MD * BC = AC²

(ii) Using Pythagoras theorem in ΔABM, we obtain

AB² = AM² + MB²

= (AD²– DM²) + MB²

= (AD²– DM²) + (BD - MD)²

= AD²– DM² + BD² + MD² - 2BD × MD

= AD² + BD² - 2BD × MD

= AD² + (BC/2)² – 2 (BC/2) * MD

= AD² + (BC/2)² – BC * MD

(iii)Using Pythagoras theorem in ΔABM, we obtain

AM² + MB² = AB² (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM² + MC² = AC²(2)

Adding equations (1) and (2), we obtain

2AM² + MB² + MC² = AB² + AC²

2AM² + (BD - DM)² + (MD + DC)² = AB² + AC²

2AM²+BD² + DM² - 2BD.DM + MD² + DC² + 2MD.DC = AB² + AC²

2AM² + 2MD² + BD² + DC² + 2MD ( - BD + DC) = AB² + AC²

2 (AM² + MD²) + (BC/2)² + (BC/2)² + 2MD (-BC/2 + BC/2) = AB² + AC²

2AD² + BC²/2 = AB² + AC²

ABCD is a parallelogram in which AB = CD and AD = BC

Perpendicular AN is drawn on DC and perpendicular DM is drawn on AB extend up to M

In Δ AMD,

AD² = DM² + AM² ...............eq(i)

In Δ BMD,

BD² = DM² + (AM + AB)²

Or,

BD² = DM² + AM² + AB² + 2AM x AB ..................eq(ii)

Substituting the value of AM² from (i) in (ii), we get

BD² = AD² + AB² + 2 x AM x AB .............eq(iii)

In Δ AND,

AD² = AN² + DN² ......................eq(iv)

In Δ ANC,

AC² = AN² + (DC – DN)²

Or,

AC² = AN² + DN² + DC² – 2 x DC x DN ............eq(v)

Substituting the value of AD² from (iv) in (v), we get

AC² = AD² + DC² – 2 x DC x DN ...............eq(vi)

We also have,

AM = DN and AB = CD

Substituting these values in (vi), we get

AC² = AD² + DC² – 2 x AM x AB .................eq(vii)

Adding (iii) and (vii), we get

AC² + BD² = AD² + AB² + 2 x AM x AB + AD² + DC² – 2 x AM x AB

Or,

AC² + BD² = AB² + BC² + DC² + AD²

Hence, proved.

(i) In triangle APC and DPB,

∠CAP = ∠BDP (Angles on the same side of a chord are equal)

∠APC = ∠DPB (Opposite angles)

Hence,

Δ APC ~ Δ DPB (By AAA similarity)

(ii) Since, the two triangles are similar

Hence,

Or,

AP * PB = CP * DP

Hence, proved

(i) In triangle PAC and PDB

∠PAC + ∠CAB = 180^o (Linear pair)

∠CAB + ∠BDC = 180^O (Opposite angles of a cyclic quadrilateral are supplementary)

Hence,

∠PAC = ∠PDB

Similarly,

∠PCA = ∠PBD

Hence,

Δ PAC ~ Δ PDB

(ii) Since the two triangles are similar, so

Or,

PA * PB = PC * PD

Hence, proved

As per the question:

AD = 1.8 m
BD = 2.4 m
CD = 1.2 m
speed of string when she pulls in = 5 cm per second

In triangle ABD, length of string i.e. AB can be calculated as follows [By Pythagoras theorem]

AB² = AD² + BD²

= (1.8)² + (2.4)²
= 3.24 + 5.76
= 9

Or, AB = 3 m

Let us assume that the string reaches at point M after 12 seconds

Length of string pulled in, after 12 seconds = 5 * 12 = 60 cm

Remaining length, AM = 3 – 0.6 = 2.4 m

In triangle AMD, we can find MD by using Pythagoras theorem,

MD² = AM² – AD²
= 2.4² – 1.8²
= 5.76 – 3.24
= 2.52 m

Or, MD = 1.58 m

Also,

Horizontal distance between the girl and the fly = CD + MD
= 1.2 + 1.58 = 2.78 m