Chapter-6, TRIANGLES

Explanation:

 From ∆ADB and ∆ADC,

According to the question, we have,

∠D = ∠D = 90° (∵ AD ⊥ BC)

∠DBA = ∠DAC [each angle = 90°- ∠C]

Using AAA similarity criteria,

∆ADB ∼ ∆CDA

BD/AD = AD/CD

BD.CD = AD2

Explanation:

A rhombus ABCD is there, the diagonals bisect each other at a point O. Let the side of the rhombus be AB.

1st diagonal = AC =16 cm

2nd diagonal = BD = 12 cm

AO = AC/2 = 16/2= 8 cm

OB = BD/2 = 12/2 = 6 cm

By Pythagoras' theorem,

AO^2+ OB^2 = AB^2

8^2+6^2 = AB^2

64+36 = AB^2

100 = AB^2

√100 = AB

10 cm = AB

So the length of the side is 10 cm. Hence option (B) is the right answer.

Explanation:

When two triangles are similar to each other then the ratio of corresponding sides will be equal

Given that

ΔABC is similar to ∆EDF  

The ratio of the corresponding sides is

=> AB/ED = BC/DF = AC/EF

Take AB/ED = AC/EF  

=> AB×EF = AC× ED  

∴ Option b) AB×EF = AC×DE is true  

From BC/DF = AC/EF

=> BC × EF = DF × AC

∴ Option a) BC×EF= AC ×FD is true

From AB/ED = BC/DF  

=> AB × DF = ED × BC  

∴ Option d) BC×DE = AB×FD is true  

∆ABC is not similar to ∆DEF  

The ratio of the corresponding sides is  

=> AB/DE ≠ BC/EF ≠ AC/DF    

From AB/DE ≠ BC/EF  

=> BC × DE ≠ AB × EF  

∴ Option c) BC×DE = AB×EF is false

Therefore,

The false option in the given options is Option c) BC×DE = AB×EF

Explanation:

(B) ΔCBA ≅ ΔPRQ

Explanation:

According to the question,

AB = QR, BC = PR and CA = PQ

Since, AB = QR, BC = PR and CA = PQ

We can say that,

A corresponds to Q, B corresponds to R, C corresponds to P.

Hence, (B) ΔCBA ≅ ΔPRQ

Hence, option B is the correct answer.

Explanation:

Given, two line segments AC and BD intersect each other at point P.

The length of the line segments

PA = 6 cm,PB = 3 cm,PC = 2.5 cm,PD = 5 cm

Given, ∠APB = 50° and ∠CDP = 30°

We have to find the value of ∠PBA.

SAS 

Checking for the ratio of the corresponding sides,

PA/PD = 6/5

PB/PC = 3/2.5 = 30/25 = 6/5

Thus, PA/PD = PB/PC

As the ratio of the corresponding sides are in the same proportion, the corresponding angles

∠APB = ∠DPC (Vertically opposite angles) are also equal.

Therefore, the triangles APB and DPC are similar.

So, ∠A = ∠D = 30°

In the triangle APB,

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

∠A + ∠B + ∠APB = 180°

30° + ∠B + 50° = 180°

80° + ∠B = 180°

∠B = 180° - 80°

∠B = 100°

Therefore, ∠PBA = 100°.

Explanation:

Given: Two triangles DEF and PQR, ∠D = ∠Q and ∠R = ∠E,

To Find: which of the following is not true?

Solution:

ΔDEF  and Δ PQR

∠D  = ∠ Q

∠E =  ∠R

=> ∠F  = ∠P=> ΔDEF ≈  ΔQRP=> DE/ QR  = EF/ RP   = DF / QP

option 1 : EF/PR = DF/PQ          PR = RP  & QP = PQ

Hence Correct

option 2:  DE/QR = EF/RP  Correct

option 3  : DE/QR  = DF/PQ      PQ = QP

Hence Correct

Option 4   EF/RP  = DE/QR   Correct

All options are correct

if option b is   DE/PQ = EF/RP then it is not true.

Explanation:

ΔABC  & ΔDEF

∠B = ∠E

∠C  = ∠F

=> ΔABC ≈  ΔDEF       ( AA )

AB/DE = BC/EF  = AC/DF

AB = 3DE​

=> AB/DE = 3

=> AB/DE = BC/EF  = AC/DF  = 3

for triangles to be congruent side ratios must be 1

Triangles are similar but not congruent.

Explanation:

Given BC/QR = 1/3

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

Thus ar(∆ABC)/ar(∆PQR) = (BC/QR)²

=> ar(∆ABC)/ar(∆PQR) = (1/3)²

=> ar(∆ABC)/ar(∆PQR) = (1/9).

Explanation:

Given, the triangles ABC and DEF are similar.

Also, ∠A =30° and ∠C = 50°

The length of the sides

AB = 5 cm,AC = 8 cm,DF = 7.5 cm

We have to determine the correct solution from the given options.

Since the two triangles are similar, the corresponding sides will be

AB/DF = AC/DE = BC/EF

5/7.5 = 8/DE = BC/EF

Taking 5/7.5 = 8/DE

On cross multiplication,

5(DE) = 8(7.5)

DE = 60/5

DE = 12 cm

In triangle ABC,

∠A + ∠B + ∠C = 180°

30° + ∠B + 50° = 180°

∠B = 180° - 80°

∠B = 100°

As the triangles are similar, the corresponding angles will be equal.

So, ∠B = ∠F = 100°

Therefore, DE = 12 cm and ∠F = 100°.

Explanation:

Now if the angle made by both the sides of each triangle are equal to each other then triangles ΔABC and ΔDEF will become similar.

Hence ∠B should be equal to ∠D 

means ∠B=∠D.

Explanation:

Given, the triangles ABC and PQR are similar.

Area of ABC/Area of PQR = 9/4

The length of the sides

AB = 18 cm

BC = 15 cm

We have to find the length of PR.

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

So, the area of ABC/area of PQR = BC^2/RP^2

9/4 = (15)^2/RP^2

Taking square root,

3/2 = 15/RP

On cross multiplication,

3(RP) = 15(2)

RP = 30/3

RP = 10 cm

Therefore, the length of PR is 10 cm.

Explanation:

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, the 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

Explanation:

According to Pythagoras' Theorem, a triangle is right-angled if

(Longest Side)² = Sum of square of other two sides

Here AB = 24 cm, BC = 10 cm and AC = 26 cm, if it is right-angled then

AC² = AB² + BC²

LHS               RHS

AC²               AB² + BC²

26²                24² + 10²

676                576 + 100

676                676

LHS      =       RHS

Sides are satisfying Pythagoras' Theorem, given triangle ABC is a right-angled triangle, and

∵ AC is the longest side ∴ angle opposite to AC (∠B) will be 90°

Hence, ΔABC is a right-angled triangle where ∠B = 90°.

Explanation:

Given, the points P and Q lie on the sides DE and DF of a triangle DEF.

Also, DP = 5 cm

DE = 15 cm,DQ = 6 cm,QF = 18 cm

We have to check if PQ is parallel to EF.

The basic Proportionality Theorem(BPT) 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.

By BPT,

DP/PE = DQ/QF

PE = DE - DP = 15 - 5 = 10 cm

LHS: DP/PE

= 5/10= 1/2

RHS: DQ/QF

= 6/18= 1/3

LHS ≠ RHS

BPT is not satisfied.

Therefore, PQ is not parallel to EF.

Explanation:

Given, the triangles FED and STU are similar.

We have to determine if DE/ST = EF/TU is true.

By the property of similarity,

F ↔ S,E ↔ T,D ↔ U

So, the corresponding sides will be

EF/ST = DE/TU = FD/SU

Therefore, the relation DE/ST = EF/TU is not true.

Explanation:

According to the question,

Let us assume that,

A = 25 cm,B = 5 cm,C = 24 cm

Now, Using Pythagoras' Theorem,

We have,

A² = B^2 + C^2

B² + C^2 = (5)^2 + (24)^2

B² + C^2 = 25 + 576

B² + C^2 = 601

A² = 600

600 ≠ 601

A² ≠ B² + C²

Since the sides do not satisfy the property of Pythagoras' theorem, a triangle with sides 25cm, 5cm, and 24cm is not a right triangle.

Explanation:

GIVEN:

∆ DEF ~  ∆RPQ

∠ D = ∠ R

∠ E = ∠ P

∠ F = ∠ Q

[Similarity of triangles should be expressed symbolically using correct correspondence of their vertices]

Hence , ∠ D = ∠ R = TRUE

∠ F = ∠ P = FALSE.

Explanation:

PB + BR = PR

PR= 10 cm

PA/PQ = 5/12.5

= 2/5

= PB/PQ = 4/10

= 2/5

as PA/PQ = PB/PQ BY Basic Proportionality Theorem, they are parallel.

Explanation:

PB = 5cm,PC = 6 cm,PD = 10 cm,PE = 12 cm

PB/PD  = 5/10  = 1/2

PC/PE = 6/12 = 1/2

=> PB/PD = PC/PE = 1/2

∠BPC = ∠DPE   ( Vertically opposite angles)

Δ PBC and ΔPDE  

PB/PD = PC/PE

∠BPC = ∠DPE

=> Δ PBC ≈  ΔPDE   ( Using SAS)  

Yes ΔPBC ~ ΔPDE  based on SAS similarity criteria.

Explanation:

We know that the sum of three angles of a triangle is 180°.

In ΔPQR, ∠P + ∠Q + ∠R = 180°

⇒ 55° + 25° + ∠R = 180°

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

In ΔTSM, ∠T + ∠S + ∠M = 180°

⇒ ∠T + ∠25°+ 100° = 180°

⇒ ∠T = 180°-(25° +100°)

=180°-125°= 55°

In ΔPQR and A TSM, and

∠P = ∠T, ∠Q = ∠S and ∠R = ∠M

ΔPQR ~ ΔTSM [since all corresponding angles are equal]

Hence, ΔQPR is not similar to ΔTSM, since the correct correspondence is P ↔ T, Q < r→ S and R ↔M

Explanation:

Given, the corresponding angles of two quadrilaterals are equal.

We have to determine if the two quadrilaterals are similar.

By the property of similarity,

To prove two figures are similar we must prove that their corresponding angles are congruent and their corresponding sides are in proportion.

Here, the corresponding angles are equal but no details about the sides are mentioned.

For example: Consider a rectangle and a square

The corresponding angles are equal to 90 degrees.

The ratio of corresponding sides are

DA/PS = 5/2

DC/SR = 5/4

BC/RQ = 5/2

AB/PQ = 5/4

It is clear that the corresponding angles are equal but the corresponding sides are not in proportion.

So, the given quadrilaterals are not similar.

Therefore, the two quadrilaterals cannot be similar only if their corresponding angles are equal.

Explanation:

Find the two triangles are similar

According to the question,

AB=3PQ  .......(1)

AC=3PR  .......(2)

Also, the perimeter of ΔABC is three times the perimeter of ΔPQR

AB+BC+CA=3(PQ+QR+RP)

AB+BC+CA=3PQ+3QR+3RP

3PQ+BC+3PR=3PQ+3QR+3PR   

using eq (1) and (2)  

BC=3QR   .....(3)

from equations (1), (2), and (3) we conclude that the sides of both the

triangles are in the same ratio.

As we know that if the corresponding sides of two triangles are in the same ratio,

then the triangles are similar 

Hence, the Two triangles are similar.

Explanation:

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, 

the sum of the interior angles of a triangle is 180°.

 Angle O of the first triangle must be equal to angle T of the second triangle. 

So, ∠O = ∠T Hence, by AAA similarity criteria, 

We have, ∆LMO ∼ ∆RST.

Explanation:

Given, the ratio of the corresponding altitudes of two similar triangles is 3/5.

We have to determine if the ratio of their areas is 6/5.

We know that,

Similar triangles have congruent corresponding angles and the corresponding sides are in proportion.

Similar triangles have the same shape, but not the same size.

By the property of similar triangles,

(area1)/(area2)=(altitude1)/(altitude2)^2

LHS:area₁/area₂

= 6/5

RHS: (altitude1)/(altitude2)^2

= (35)^2

= 9/25

LHS ≠ RHS

The ratio of areas is not equal to the ratio of the squares of the altitude.

Therefore, the ratio of the areas is not equal to 6/5.

Explanation:

In ΔPDQΔPDQ

In ΔPDQ and ΔPDR 

PD⊥QR [Given] 

∴∠PDQ=∠PDR=90∘

  PD does not bisect ∠P. 

∴∠1 ≠∠2 

and ∠1 ≠∠R 

and ∠Q ≠∠2 

and ∠Q ≠∠R[∵PQ ≠PR] 

Any ratio of sides is also not equal .

So, ΔPQD is not similar to ΔRPD. Hence, the given statement is false. Hence, the given statement is false.

Explanation:

True

In ΔADE and ΔACB,

∠A=∠A

 [ common angle]

∠D=∠C

 [given]

ΔADE∼ΔACB [ by AAA similarity criterion].

Explanation:

False 

The given statement is not correct. Here, one angle and two sides of two triangles are equal but these sides do not including an 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.

Explanation:

See the diagram enclosed.  A square of the largest size is possible only if you coincide two edges of the triangle with the two edges of the square.

Now area of triangle ABC = 1/2 *8 * 16  cm² = 64 cm²

Area of square = a²

area of triangle BDE =  1/2 (8-a) a = 4a - a²/2

area of triangle EFC = 1/2 (16 -a) a =  8a - a²/2

ΔABC = 64  = a² + 4a -a²/2 + 8a -a²/2  = 12 a 

a = 16/3

Explanation:

Let one side be x cm

then another side =(x+5)cm

x²+(x+5)^2=(25)^2

x^2+x^2+10x+25=625

2x^2+10x-600=0

x^2+5x-300=0

x^2+20x-15x-300=0

x(x+20)+15(x+20)=0

(x+20)(x-15)=0

As we know that the length of the sides must be positive so 

one side =15cm

other side=(x+5)=(15+5)=20 cm.

Explanation:

Solution: AD/DB = AE/EC (Given) Therefore,

 DE ∥ BC (Converse of Basic Proportionality Theorem) 

So, ∠D = ∠B and ∠E = ∠C (Corresponding angles) (1) 

But ∠D = ∠E (Given) 

Therefore, ∠B = ∠C [ From (1)] 

So, AB = AC (Sides opposite to equal angles) 

i.e., BAC is an isosceles triangle.

Explanation:

Given:

In ∆ PQR, PR²-PQ²= QR² & QM ⊥ PR

To Prove: QM² = PM × MR

Proof:

Since, PR² - PQ²= QR²

PR² = PQ² + QR²

So, ∆ PQR is a right-angled triangle at Q.

In ∆ QMR & ∆PMQ

∠QMR = ∠PMQ [ Each 90°]

∠MQR = ∠QPM [each equal to (90°- ∠R)]

∆ QMR ~ ∆PMQ [ by AA similarity criterion]

By the property of an area of similar triangles,

ar(∆ QMR ) / ar(∆PMQ)= QM²/PM²

1/2× MR × QM / ½ × PM ×QM = QM²/PM²

[ Area of triangle= ½ base × height]

MR / PM = QM²/PM²

QM² × PM = PM² × MR

QM² =( PM² × MR)/ PM

QM² = PM × MR.

Explanation:

BASIC PROPORTIONALITY THEOREM (BPT): 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. That is also known as the Thales theorem.

GIVEN:

In ∆ABC, DE || AB

AD= 3x+19, DC= x+3, BE= 3x+4, EC = x

AD /DC = BE/EC

[ By Thales theorem(BPT)]

3x+19/x+3 = 3x+4/x

(3x+19) × x = (3x+4) × (x+3)

3x² + 19x = 3x² + 9x + 4x +12

3x² + 19x = 3x² + 13x +12

3x² - 3x² +19x -13x = 12

6x = 12

x = 12/6= 2

x = 2

Hence, the value of x is 2.

Explanation: 

According to the question,

∆ 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 Equations (i) and (ii).

PS/SQ = PT/TR

By converse of basic proportionality theorem

⇒ ST || QR

∴ ∠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.

Explanation:

Since, PQ||RS and PQ = 3RS

PQ/RS = 3/1 = 3 -------------- (1)

In △POQ and △ROS,

The vertically opposite angles ∠SOR and ∠POQ are equal.

i.e, ∠SOR = ∠POQ

The alternate angles ∠SRP and ∠QPR are equal.

i.e., ∠SRP = ∠QPR

By AAA criterion, the third angle will be equal.

i.e., ∠RSO = ∠OQP

Therefore, the triangles POQ and ROS are similar.

By the property of an area of similar triangles,

Area of the triangle POQ/Area of the triangle ROS = PQ²/RS²

Substituting (1) in the above relation,

Area of the triangle POQ/Area of the triangle ROS = 3²/1²

= 9/1

Therefore, the ratio of the area of the triangles POQ and ROS is 9:1.

Explanation: 

in triangle APO and triangle CQO

angle PAO = angle OCQ ( alternate angle)

angle POA = angle COQ( vertical opposite angle )

by AA criteria triangle APO is similar to triangle CQO

so

(AP/ CQ )= ( AO/ CO)

cross multiplication

AP X OC = CQ X OA.

Explanation:

Let ABC be an equilateral triangle of sides 8 cm

AB = BC = CA = 8 cm. (all sides of an equilateral triangle is equal)

Draw altitude AD which is perpendicular to BC.

Then, D is the mid-point of BC.

∴ BD = CD = ½

BC = 8/2 = 4 cm

Now, by Pythagoras' theorem

AB2 = AD2 + BD2

⇒ (8)2 = AD2 + (4)2

⇒ 64 = AD2 + 16

⇒ AD = √48 = 4√3 cm.

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

Explanation:

As per the dimensions given in the questions,

Now, we have to find out the perimeter of △ABC

Let △ABC∼△DEF

So, AB/DE=AC/DF=BC/EF

Consider, AB/DE=AC/DE

4/6=AC/12

By cross multiplication we get,

AC=(4×12)/6

AC=48/6

AC=8cm

Then, consider AB/DE=BC/EF

4/6=BC/9

BC=(4×9)/6

BC=36/6

BC=6cm

Therefore, the perimeter of △ABC=AB+BC+AC

=4+6+8

=18cm.

Explanation:

Given: DE||BC

DE = 6cm. BC = 12cm

To find, at ∆ADE: ar DECB

In ∆ADE and ∆ABC

∆A is common

∆ADE = ∆ABC. corresponding angle

.:. ∆ADE ~ ∆ABC. by AA

since both triangles are the similar ratio of the area of two similar triangles is equal to the square of their corresponding sides.

.:. ar ∆ADE / ar ∆ABC = (DE/BC)²

= (6/12)²

= 1/4

.:. ar ∆ADE / ar DECB = 1/ar ∆ABC - ar ∆ADE

= 1 / 4 - 1

= 1:3.

Explanation:

Given, PD = 18 cm

BQ = 35 cm

QC = 15cm

Also, AB||PQ||DC

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.

In △ABD,

Given, AB||PQ

So, PO||AB

By BPT, DP/AP = OD/OB -------------- (1)

In △BDC,

Given, DC||PQ

So, OQ||DC

By BPT, OB/OD = BQ/QC

On rearranging,

OD/OB = QC/BQ --------------------------- (2)

Equating (1) and (2),

DP/AP = QC/BQ

So, 18/AP = 15/35

AP(15) = 35(18)

AP = 7(18)/3

AP = 42 cm

We know, AD = DP + AP

AD = 18 + 42

AD = 60 cm

Therefore, the length of AD is 60 cm.

Explanation:

The area of the larger ∆ is 108 cm²

Step-by-step explanation:

Given:

Let the Smaller triangle be ΔABC & the bigger triangle be ΔPQR and the corresponding sides be BC & QR  

ΔABC ~ ΔPQR.

Area of ΔABC = 48 cm².

BC : QR = 2 : 3  

ar(ΔABC)/ar( ΔPQR) = (BC/QR)²

[The ratio of the area of two similar triangles is equal to the ratio of squares of their corresponding sides.]

48 /ar( ΔPQR) = (2/3)²

48/ar( ΔPQR) = 4/9  

ar( ΔPQR) = (9 × 48)/4

ar( ΔPQR) = 9 × 12

ar( ΔPQR) = 108 cm²

Hence, the area of the larger ∆ is 108 cm².

Explanation:

PN. NR = QN² can be written as

PN/QN = QN/NR

From △PNQ and △RNQ,

PN/QN = QN/NR

∠PNQ = ∠RNQ = 90°

By SAS criterion, the triangles PNQ and RNQ are similar.

By the property of similar triangles,

∠PQN = ∠ QRN -------------- (1)

∠RQN = ∠QPN --------------- (2)

Adding (1) and (2),

∠PQN + ∠RQN = ∠QRN + ∠QPN

From the figure,

∠PQR = ∠PQN + ∠RQN ----------- (3)

So, ∠PQR = ∠QRN + ∠QPN --- (4)

We know that the sum of all three interior angles of a triangle is always equal to 180°

In △QNR,

∠QNR + ∠NRQ + ∠RQN = 180°

90°+ ∠NRQ + ∠RQN= 180°

∠NRQ + ∠RQN = 180° - 90°

∠NRQ + ∠QPN = 90° --- [From (2)]

∠PQR = 90° [From (4)]

Therefore, it is proved that ∠PQR = 90°.

Explanation:

Area of smaller triangle = 36 cm^2

Area of larger triangle = 100cm^2

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

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

By the property of an area of a similar triangle

Area of (larger triangle)/area of (smaller triangle) = (side of the larger triangle) / (side of the smaller triangle)

100/36=20^2/x^2

X^2=20^2×36/100

X = root 144

X = 12 cm.

Explanation:

In △ADC and △ACB,

Given, ∠ACB = ∠CDA

∠CAD = ∠BAC = common angle

By AAA criterion, the third angle will be equal.

Therefore, the triangles ADC and ACB are similar.

By the property of similar triangles,

The corresponding sides are proportional.

So, AC/AD = AB/AC

8/3 = AB/8

On cross multiplication,

8(8) = 3AB

AB = 64/3

Now, BD = AB - AD

= 64/3 - 3

= (64 - 9)/3

= 55/3

BD = 18.33 cm

Therefore, the length of BD is 18.33 cm.

Explanation:

From the question it is given that,

Height of a tower PQ=15m

It’s shadow QR=24 m

Let us assume the height of a telephone pole MN=x

It’s shadow NO=16m

Given, at the same time,

△PQR∼△MNO

Therefore, PQ/MN=ON/RQ

15/x=24/16

By cross multiplication we get,

x=(15×16)/24

x=240/24

x=10

Therefore, the height of the pole =10m.

Explanation:

A ladder 10m long rests against a vertical wall. If the foot of the ladder is 6m away from the wall and the ladder just reaches the top of the wall, then the wall is 8m high.

We can simply find the height of the wall by using the Pythagoras theorem.

According to Pythagoras' theorem,

     (H)² = (B)² + (P)²

The length of the ladder is the hypotenuse of the right-angle triangle and the foot of the ladder is the base, we have to find the perpendicular distance.

  (10)² = (6)² + (P)²

  100 - 36 = (P)²

  64 =  (P)²

   8 = P

∴ The perpendicular distance is 8m.

Hence, A ladder 10m long rests against a vertical wall. If the foot of the ladder is 6m away from the wall and the ladder just reaches the top of the wall, then the wall is 8m high.

Explanation:

OB is the perpendicular bisector of line segment DE, FA is perpendicular to OB and FE intersects OB at the point C as shown in the figure.

now, ∆OAF and ∆ODB

∠OAF = ∠OBD = 90° {because OB is the perpendicular bisector of DE so, OB ⊥ DE and OB ⊥ AF }

∠FOA = ∠DOB { common angle }

from A - A similarity, ∆OAF ~ ∆ODB

so, OA/OB = AF/DB = OF/OD -------(1)

similarly, ∆AFC and ∆BEC

∠FCA = ∠BCE

∠FAC = ∠CBE = 90°

from A - A similarity, ∆AFC ~ ∆BEC

so, AF/BE = AC/CB = FC/CE -------(2)

we know, DB = BE { perpendicular bisector of DE is OB } put it in equation (2)

AF/DB = AC/CB =FC/CE -------(3)

now, from equations (1) and (3),

OA/OB = AC/CB = ( OC - OA)/(OB - OC)

OA/OB = (OC - OA)/(OB - OC)

OA(OB - OC) = OB(OC - OA)

OA.OB - OA.OC = OB.OC - OB.OA

2OA.OB = OB.OC + OA.OC

dividing by OA.OB.OC both sides,

2OA.OB/OA.OB.OC = OB.OC/OA.OB.OC + OA.OC/OA.OB.OC

2/OC = 1/OA + 1/OB

Explanation:

Given:- ABC is a triangle

AC^2 =AB^2 +BC^2

 To prove:- ∠B=90°

Construction:- Construct a triangle PQR right angled at Q such that, PQ=AB and QR=BC

Proof:-

In △PQR

PR^2 =PQ^2 +QR^2

 (By Pythagoras theorem)

⇒PR^2 =AB^2 +BC^2 .....(1)(∵AB=PQ and QR=BC)

AC^2 =AB^2 +BC^2 .....(2)(Given)

From equation (1)&(2), we have

AC^2 =PR^2

 ⇒AC=PR.....(3)

Now, in △ABC and △PQR

AB=PQ

BC=QR

AC=PR(From (3))

∴△ABC≅△PQR(By SSS congruency)

Therefore, by C.P.C.T.,

∠B=∠Q

∵∠Q=90°

∴∠B=90°

Hence proved.

Explanation:

Distance traveled by aeroplane in 1 1/2 = 3/2 hours towards north direction

=> 300 * 3/2=> 150*3=> 450 km

So AB = 450 m.

Distance traveled by airplane in 3/2 hours towards north direction =

=> 400*3/2=> 200*3=> 600 km

So BC = 600 km.

Now from Pythagoras' theorem

     AC^2 = AB^2 + BC^2

=> AC^2 = (450)^2 + (650)^2

=> AC^2 = 202500 + 360000

=> AC^2 = 562500

=> AC = √562500

=> AC = 750 m.

Hence, the distance between two planes after 3/2 hours = 750 km.

Explanation:

Given, △ABC∼△DEF

Therefore,

DE/AB=EF/BC=FD/CA

So,2x−1/18

=2x+2/3x+9

=3x/6x

Now taking 2x−1/18

=3x/6x, we have 2x−1/18

=1/2 or 4x−2=18 or x=5

Therefore, AB=2×5−1=9,

BC=2×5+2=12,CA=3×5=15,

DE=18,EF=3×5+9=24

and FD=6×5=30

Hence,

AB=9cm,BC=12cm,CA=15cm,DE=18cm,EF=24cm and FD=30cm.

Explanation:

In △ABP and △CDP,

Given, ∠A = ∠C

So, ∠BAP = ∠PCD

Vertically opposite angles are equal

∠BPA = ∠CPD

By AAA criterion, △ABP ⩬ △CDP.

By the property of similarity,

The corresponding sides are in proportion.

AB/DC = BP/PD = AP/CP

6/DC = 15/PD = 12/4

6/DC = 15/PD = 3

Considering 6/DC = 3

3DC = 6

DC = 6/3

DC = 2 cm

Considering 15/PD = 3

3PD = 15

PD = 15/3

PD = 5 cm

Therefore, the length of PD and CD are 5 cm and 2 cm respectively.

Explanation:

As per the dimensions given in the questions,

From the question it is given that,

△DEF∼△LMN

So, AB/ED=AC/EF=BC/DF

Consider AB/ED=AC/EF

5/12=7/EF

By cross multiplication,

EF=(7×12)/5

EF=16.8cm

Now, consider AB/ED=BC/DF

5/12=BC/15

BC=(5×15)/12

BC=75/12

BC=6.25.

Explanation:

let us draw a triangle ABC as shown in the figure

line DE is drawn parallel to side BC

now, in triangles ADE and ABC,

∠ADE = ∠ABC (corresponding angles)

∠AED = ∠ACB (corresponding angles)

and ∠A is the common angle

then, as per the AAA test criterion, triangles ADE and ABC are similar.

now, we know that the ratio of the corresponding sides of similar triangles is equal,

⇒ AD/AB = AE/AC

⇒ AD/DB = AE/EC

now, if point D is the midpoint of AB,  AD/DB = 1/2

⇒ AD/DB = AE/EC = 1/2

Hence, proved.

Explanation:

In △OPS and △OAB,

∠POS = ∠AOB = common angle

The corresponding angles are equal. i.e.,∠OSP = ∠OBA

By AAA criterion, △OPS ⩬ △OAB

By BPT, PS/AB = OS/OB ----------------- (1)

In △CQR and △CAB,

∠RCQ = ∠BCA = common angle

The corresponding angles are equal. i.e.,∠CRQ = ∠CBA

By AAA criterion, △CQR ⩬ △CAB

By BPT, QR/AB = CR/CB ----------------- (2)

Given, PQRS is a parallelogram PS = QR,

So (2) becomes PS/AB = CR/CB ------------ (3)

From (2) and (3),

OS/OB = CR/CB

On rearranging,

OB/OS = CB/CR

On subtracting 1 from both sides, we get,

OB/OS- 1 = CB/CR - 1

OB-OS / OS = CB-CR / CR

From the figure,

OB - OS = BS

CB - CR = BR

Now, BS/OS = BR/CR

By Converse of the Basic Proportionality Theorem,

SR||OC

Therefore, it is proven that OC||SR.

Explanation:

Let the length of the ladder = AC = 5 m

Let the height of the wall on which the ladder is placed = BC = 4m.

From right-angled ∆EBD,

Using the Pythagoras Theorem,

ED^2 = EB^2 + BD^2

(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, we have,

EC = EB – BC = 4.8 – 4 = 0.8

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

Explanation:

According to the question,

AC⊥CB,

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

Thus, we get ∆ ACB right angled at C.

Now, from ∆ACB, Using Pythagoras' Theorem,

AB2 = AC2 + BC2

⇒ (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

Dividing the equation 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 or x = 5

Since the distance can’t be negative, we neglect x = -12

∴ x = 5

Now,

AC = 2x = 10km

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

Thus, the distance covered to city B from city A via city C = AC + BC

AC + BC = 10 + 24 = 34 km

Distance covered to city B from city A after the highway was constructed = BA = 26 km

Therefore, the distance saved = 34 – 26 = 8 km.

Explanation:

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, 

LN^2 = LM^2 + MN^2 [by Pythagoras theorem] 

⇒ LN^2 = (9.6)^2 + (18)^2 

⇒ LN^2 = 92.16 + 324 

⇒ LN^2 = 416.16 

∴ LN = √416.16 = 20.4 m 

Hence, the required distance is 20.4 m.

Explanation:

Let,∠BCA=θ

From  ΔEDC

tanθ=ED/CD

or, tanθ=1.5/3

or, tanθ=1/2

From ΔABC,

tanθ=BA/CA

or, 1/2=6/x+3

or, x+3=12

or, x=12-3

or, x=9 m

∴, She is 9m away from the base of the pole.

Explanation:

In triangles ABC & ADB

∠A = ∠A (common)

∠ABC = ∠ADB (each 90)

Therefore Triangle ABC similar to Triangle ADB

=> AB/AD = AC/AB

=> AB2= AD x AC

=> AB2= 4 x 9 (AC=AD+CD = 4+5 = 9 cm)

=> AB2= 36

=> AB = 6 cm

In Triangle ABD using Pythagoras' Theorem

AB2= AD2+ BD2

62 = 42 + BD2

BD2= 36 - 16 = 20

BD = 2√5.

Explanation:

 In △PSQ,

PQ2 = PS2 + SQ2

(6)2 = (4)2 + SQ2

36 = 16 + SQ2

SQ2 = 36 - 16

SQ2 = 20

Taking square root,

SQ = 2√5 cm

So, SQ2 = PS × SR

(2√5)2 = 4 × SR

4(5) = 4SR

SR = 5 cm

In △QSR,

Given, QS⊥PR

So, ∠QSR = 90°

By Pythagoras' theorem,

QR2 = SQ2 + RS2

QR2 = (2√5)2 + (5)2

QR2 = 4(5) + 25

QR2 = 20 + 25

QR2 = 45

Taking square root,

QR = 3√5 cm

Therefore, the length of QS, RS, and QR are 2√5 cm, 5 cm and 3√5 cm.

Explanation:

Given, 

In ∆PQR, 

PD⊥QR,

 PQ = a, PR = b, QD = c and DR = d

 In right-angled ∆PDQ,

 PQ^2 = PD^2 + QD^2 [by Pythagoras theorem] 

= a^2 = PD^2 + c^2 

= PD^2 = a^2 – c^2 …..(i)

In right-angled ∆PDR, 

PR^2 = PD^2 + DR^2 [by Pythagoras theorem] 

= b^2 = PD^2 + d^2

 = PD^2 = b^2 - d^2 ……(i) 

From Equations (i) and (ii),

 A^2 - c^2 = b^2 - d^2 

A^2 - b^2 = c^2 - d^2

 = (a – b) (a + b) = (c – d) (c + d) Hence proved.

Explanation:

We have, ∠A + ∠D = 90°

In ΔAPD, by angle sum property,

∠A + ∠D + ∠P = 180°

 90° + P = 180°

∠P = 180° – 90° = 90°

In ΔAPC, by Pythagoras' theorem,

AC2 = AP2 + PC2 ....(1)

In ΔBPD, by Pythagoras' theorem,

BD2 = BP2 + DP2 ....(2)

Adding equations (1) and (2),

AC2 + BD2 = AP2 + PC2 + BP2 + DP2

AC2 + BD2 = (AP2 + DP2) + (PC2 + BP2) = AD2 + BC2

Hence proved.

Explanation:

The points A, E and C lie on l.

The points D, F and B lie on m.

In △APC and △PDB,

Vertically opposite angles are equal. i.e.,∠APC = ∠DPB

Alternate angles are equal i.e., ∠PAC = ∠PBD

By AAA criterion, △APC ⩬ △PDB

By the property of similarity,The corresponding sides are proportional.

AP/PB = PC/PD = AC/DB -------------- (1)

In △APE and △PBF,

Vertically opposite angles are equal. i.e.,∠APE = ∠BPF

Alternate angles are equal i.e., ∠PAE = ∠PBF

By AAA criterion, △APE ⩬ △BPF

By the property of similarity, The corresponding sides are proportional.

AP/PB = AE/FB = EP/FP ------------------------ (2)

In △CPE and △PDF,

Vertically opposite angles are equal. i.e.,∠CPE = ∠DPF

Alternate angles are equal i.e., ∠PCE = ∠PDF

By AAA criterion, △CPE ⩬ △PDF

By the property of similarity, The corresponding sides are proportional.

So, EP/FP = PC/PD = CE/DF ----------------- (3)

Equating (1), (2) and (3), we get,

AP/PB = PC/PD = AC/DB = AP/PB = AE/FB = EP/FP = EP/FP = PC/PD = CE/DF

Canceling out common terms,

AC/DB = AP/PB = AE/FB = CE/DF = EP/FP ------------------------- (4)

From (4), AE/BF = AC/BD = CE/FD

Therefore, it is proved that AE/BF = AC/BD = CE/FD.

Explanation:

PA, QB, RC,& SD are perpendicular on line l.

To find: PQ, QR & RS

Construction: produce SP  & l to meet each other at E.

Proof: In ΔEDS,

AP║ BQ║DS║CR        [ given]

∴ PQ : QR: RS  = AB:BC:CD

PQ : QR: RS = 6:9:12

Let PQ= 6x  , QR= 9x    RS= 12x

PS = PQ+QR+RS

∴ PQ+QR+RS=36                  [PS=36cm]

6x + 9x + 12x = 36

27x = 36

x = 36/27 = 4/3

x = 4/3

∴ PQ = 6x = 6× 4/3

PQ = 8cm

QR = 9x = 9 × 4/3

QR= 12cm

RS = 12x = 12× 4/3

RS = 16 cm

Hence PQ = 8cm , QR = 12cm,   RS = 16 cm.

Explanation:

See the attached diagram for the description of the question.  

ABCD is a trapezoid with AB || CD

AC and BD are diagonals that will meet at point O.

A line PQ is drawn from O which is parallel to AB and CD.  PQ || AB || CD

In ∆ABD and ∆POD

AB || PO

∠ADB = ∠ PDO (Since they are the same angles)

∠ABD = ∠POD (Since they are corresponding angles on the parallel lines)

∠DAB = ∠DPO (Corresponding angles)

By AAA law, we can say that  ∆ABD ~ ∆POD

From above similar triangles, we can get  

OP/AB = PD/AD  -----------------------------------------------------------E1

In ∆ABC and ∆OQC

AB || OQ

∠ACB = ∠OCQ

∠BAC = ∠QOC

∠ABC = ∠OQC.

BY AAA law, we can say that ∆ABC ~ ∆OQC

We can derive  

OQ/AB = CQ/BC  ----------------------------------------------------------E2

BQ/CQ = OA/OC-----------------------------------------------------------E3

Similarly, we can prove that ∆ADC ~ ∆APO

We can derive  

AP/PD = OA/OC ---------------------------------------------------------E4

From E3 and E4, we can derive

AP/PD = BQ/CQ

Add + 1 on both sides, and we get

AP/PD + 1 = BQ/CQ + 1

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

AD/PD = BC/CQ

PD/AD = CQ/BC

From E1 and E2 and the above result, we can derive

OP/AB = OQ/AB  

Hence OP = OQ.

Explanation:

Given that E is the midpoint of AC, then

CE=AE...............................1

also given that,

angle AEF=angle AFE................2

then, AE=AF........................3

by eq1 and eq2 we get ,

CE=AE=AF........................4

now draw CG parallel DF

we know that corresponding angles are equal,

therefore,angleAEF=angle ACG........5

also, angle AFE=angle AGC...............6

by eq2,5,6, we get

angleAEF=AFE=ACG=angleAGC

Then, AC=AG.......................7. as sides opposite to equal angles

AE+CE=AF+GF

By eq4,7 we get,

AE=AF=GF=CE

CE=AF=GF.............................8

By Thales theorem we get,

BC/CD=BG/GF

By adding 1 on both sides we get,

BC/CD+1=BG/GF+1

(BC+CD)/CD=(BG+GF)/GF

BD/CD=BF/GF

by eq 8 we get

BD/CD=BF/CE

Hence proved.

Explanation:

By Pythagoras' theorem:

y^2 + z^2 = x^2

Area of semicircle = 1/2π x Radius^2

Area of A = 1/2π(y/2)^2 = 1/8πy^2

Area of B = 1/2π(z/2)^2 = 1/8πz^2

Area of C = 1/2π(x/2)^2 = 1/8πx^2

Area A + Area B = 1/8πy^2 + 1/8πz^2

  = 1/8π(y^2 + z^2)

But from the Pythagoras theorem: y^2 + z^2 = x^2

So, Area A + Area B = 1/8π(y^2 + z^2)= 1/8πx^2

Which is the same as area C = 1/8πx^2

Conclusion:

The area of a semicircle drawn on the hypotenuse of a right-angled triangle is equal to the sum of the areas of the semicircles drawn on the other two sides.

Explanation:

triangle PAB and triangle QBC and triangle RCR are similar since all the triangles are equilateral

Two triangle PNB similar triangle QBC similar triangle RCA

Hence,

Area of triangle PAV / area of triangle are RCA.= ab square upon ac square --------------(1)

Area of triangle ABC / area of triangle are = BC square + AC square------------(2)

Adding equation (1) and (2)

area of triangle PAB / area of triangle RCA + area of triangle QBC / area of triangle RCA = AB square /AC square + BC square / AC square - AB square - BC square / AC Square

area of triangle PAB + area of triangle QBC / area of triangle RCA = AC square / AC square = 1

area of triangle PAB+ area of triangle QBC = area of triangle RCA

Hence proved.