25 m

Yes, the pair of triangles is similar.
In \(\triangle \mathrm{MNL} \text { and } \triangle \mathrm{QPR}\),
\(\begin{aligned} \angle N M L & =\angle P Q R=70^{\circ} \\ \end{aligned}\)
\(\begin{aligned} \frac{M N}{P Q} & =\frac{2.5}{5}=\frac{1}{2} \\ \end{aligned}\)
and \(\begin{aligned} \frac{M L}{Q R} & =\frac{5}{10}=\frac{1}{2} \\ \end{aligned}\)
Then, \(\begin{aligned} \frac{M N}{P Q} & =\frac{M L}{Q R} \end{aligned}\)
\(\therefore\) \(\Delta\)MNL \(\sim\) \(\Delta\)QPR  [by SAS similarity criterion]

Given, AD and CE are altitudes which intersect each other at the point P.
In ΔABD and ΔCBE,
∠ADB = ∠CEB      [each 90°]
and ∠ABD = ∠CBE      [Common angle]
\(\therefore\) ΔABD ∼ ΔCBE    [by AA similarity criterion]

As, \(\triangle ABC\sim \triangle AMP\) [proved in part(i)]
\(\therefore d\frac{AC}{AP}=\frac{BC}{MP}\)
[since ratio of the corresponding sides of similar triangles are equal]
\(\Rightarrow \frac{CA}{PA}=\frac{BC}{MP}\)

Prove that \(\Delta A P B \sim \Delta C P D\) [byAA similarity criterion]
PD = 5 cm, CD = 2 cm

Given, in \(\triangle ABC\), \(DE\parallel BC\)
By Thales theorem, we get
\(\frac { AD }{ DB } =\frac { AE }{ EC } \)

\(\Rightarrow \frac { 4x-3 }{ 3x-1 } =\frac { 8x-7 }{ 5x-3 } \)
[\(\because \)  AD = 4x - 3, DB = 3x - 1, AE = 8x - 7, EC = 5x - 3]
\(\Rightarrow \) (4x - 3)(5x - 3) = (8x - 7)(3x - 1)
\(\Rightarrow \)  20x² - 12x + 9 - 15x = 24x² - 21x - 8x + 7
\(\Rightarrow \)  4x² - 2x - 2 = 0
\(\Rightarrow \)  2x² - x - 1 = 0     [dividing both sides by 2]
\(\Rightarrow \)  2x² - 2x + x - 1 = 0   [by splitting the middle term]
\(\Rightarrow \)  2x(x - 1) + 1 (x - 1) = 0
\(\Rightarrow \)  (2x + 1) (x - 1) = 0   \(\therefore x=-\frac { 1 }{ 2 } \) or x = 1
If \(x=-\frac { 1 }{ 2 } \), then AD = \(4\times -\frac { 1 }{ 2 } -3=-5<0\)  [not possible]
Hence, x = 1 is the required value.

Given \(\triangle PQR\) is a right angled triangle \(\angle Q\) = 90°, QS = SR.

To show PR²  = 4 PS²  - 3 PQ²
Proof \(\triangle PQR\),
PR²  = PQ² + OR²  [by Pythagoras theorem]  ... (i)
Now, QS = SR  [given]
QR = SR = \(\frac{1}{2}\) QR
QR = 2QS      ... (ii)
Put QR = 2QS in Eq.(i),
then, PR² = PQ² + (2QS)² = PQ² + 4QS²
In \(\triangle PQS\),
PS² = PQ² + QS²
\(\Rightarrow \)  QS² = PS² - PQ²    ... (iii)
\(\Rightarrow \)  PR² = PQ² + 4 (PS² - PQ²) [from Eq.(iii)]
= PQ² + 4 PS² - 4 PQ²
\(\therefore \)  PR²  = 4 PS²  - 3 PQ²

In \(\Delta\)ACB,    LM || CB         [given]
\(\Rightarrow \quad \frac{A M}{M B}=\frac{A L}{L C}\)    ....(i)
[ by basic proportionality theorem]
In \(\Delta\)ACD, LN ||CD    [given]
\(\Rightarrow \quad \frac{A N}{N D}=\frac{A L}{L C}\)    ....(ii)
[by basic proportionality theorem]
From Eqs. (i) and (ii), we get
\(\frac{A M}{M B}=\frac{A N}{N D} \Rightarrow \frac{M B}{A M}=\frac{N D}{A N}\)
[on taking reciprocal of the terms]
\(\Rightarrow \quad \frac{M B}{A M}+1=\frac{N D}{A N}\) + 1  [adding 1 on both sides]
\(\begin{aligned} & \Rightarrow \quad \frac{M B+A M}{A M}=\frac{N D+A N}{A N} \\ \end{aligned}\)

\(\begin{aligned} \Rightarrow \quad \frac{A M}{A M+M B}=\frac{A N}{A N+N D} \end{aligned}\)
[on taking reciprocal of the terms]
\(\therefore \quad \frac{A M}{A B}=\frac{A N}{A D} \quad\left[\begin{array}{c} \because A D=A N+N D \\ \text { and } A B=A M+M B \end{array}\right]\)
Hence proved.

Let rays BP and CQ intersect each other at A and rays ER and FS intersect each other at D. In the two triangles ABC and DEF, you can see that ∠ B = ∠ E, ∠ C = ∠ F and Ð A = ∠ D. That is, corresponding angles of these two
triangles are equal. What can you say about their corresponding sides. Note that \(\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{3}{5}=0.6\) What about \(\frac{\mathrm{AB}}{\mathrm{DE}} \text { and } \frac{\mathrm{CA}}{\mathrm{FD}} \) On measuring AB, DE, CA and FD, you will find that \(\frac{\mathrm{AB}}{\mathrm{DE}} \text { and } \frac{\mathrm{CA}}{\mathrm{FD}}\) are also equal to 0.6 (or nearly equal to 0.6, if there is some error in the measurement). Thus, \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{CA}}{\mathrm{FD}}\) You can repeat this activity by constructing several pairs of triangles having their corresponding angles equal. Every time, you will find that their corresponding sides are in the same ratio (or proportion). This activity leads us to the following criterion for similarity of two triangles.

This criterion is referred to as the SSS (Side - Side - Side) similarity criterion for two triangles.
This theorem can be proved by taking two triangles ABC and DEF such that
\(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{CA}}{\mathrm{FD}}(<1)\)

Cut DP = AB and DQ = AC and join PQ.
It can be seen that \(\frac{\mathrm{DP}}{\mathrm{PE}}=\frac{\mathrm{DQ}}{\mathrm{QF}} \text { and } \mathrm{PQ} \text { II EF }\)
So, ∠ P = ∠ E and ∠ Q =  ∠ F.
Therefore, \(\frac{\mathrm{DP}}{\mathrm{DE}}=\frac{\mathrm{DQ}}{\mathrm{DF}}=\frac{\mathrm{PQ}}{\mathrm{EF}}\)
So, \(\frac{\mathrm{DP}}{\mathrm{DE}}=\frac{\mathrm{DQ}}{\mathrm{DF}}=\frac{\mathrm{BC}}{\mathrm{EF}}\)
So, BC = PQ
Thus, \(\Delta\) ABC ≅ \(\Delta\) DPQ
So, ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F

It is given that AD = 1cm, BD = 2 cm and DE II BC.
\(\text { In } \Delta A D E \text { and } \Delta A B C\)
\(\angle A D E=\angle A B C\) [corresponding angles]
\(\angle A=\angle A\) [common]
Therefore, by AA similar condition
\(\Delta A D E-\Delta A B C\)
Ratio of areas of similar triangles is equal to the square of the ratio of the corresponding sides.
\( \therefore \frac{\operatorname{ar}(\Delta A B C)}{\operatorname{ar}(\Delta A D E)}=\frac{A B^{2}}{A D^{2}} \Rightarrow \frac{\operatorname{ar}(\Delta A B C)}{\operatorname{ar}(\Delta A D E)}=\left(\frac{3}{1}\right)^{2} \)
\(\Rightarrow \frac{\operatorname{ar}(\Delta A B C)}{\operatorname{ar}(\Delta A D E)}=\frac{9}{1} \)
Therefore, the ratio of the area of MBC to the area of ΔADE is 9: 1.

Quadrilateral PQRS and ABCD are not similar as their corresponding sides are proportional, i.e. 1:2, but their corresponding angles are not equal.

Consider a trapezium ABCD such that \(AB\parallel DC\) and draw its diagonals AC and BD which intersect at point O.

In \(\triangle OCD\) and \(\triangle OAB\),
\(AB\parallel DC\)        [given]
\(\therefore \angle 1=\angle 3,\angle 2=\angle 4\)    [alternate interior angles]
Also, \(\triangle DOC=\triangle BOA\)     [vertically opposite angles]
\(\therefore \triangle OCD\sim \triangle OAB\)     [by AAA similarity criterion]
\(\Rightarrow \frac { OC }{ OA } =\frac { OD }{ OB } \)
[since, ratios of the corresponding sides of the similar triangles are equal]
\(\therefore \frac { OA }{ OC } =\frac { OB }{ OD } \)    [on taking reciprocal of the terms]
Hence proved.

Suppose ABCD is a rhombus in which
AB = BC = CD = DA = a     [say]
Here, diagonals AC and BD are right angle bisectors of each other at O.
In \(\triangle AOB,\angle AOB={ 90 }^{ ° }\)
\(OA=\frac{1}{2}AC\)  \([\because OA=OC]\)   ... (i)
and \(OB=\frac{1}{2}BD\)  \([\because OB=OD]\)   ... (ii)
Using Pythagoras theorem, we get

OA² = OB² = AB²
\(\Rightarrow \left( \frac { 1 }{ 2 } AC \right) ^{ 2 }+\left( \frac { 1 }{ 2 } BD \right) ^{ 2 }={ AB }^{ 2 }\) [from Eqs.(i) and (ii)]
\(\Rightarrow\) AC² + BD² = 4 AB²
\(\Rightarrow\) AB² + BC² + CD² + DA² = AC² + BD²
\([\because AB = BC = CD = DA]\)

Join BX
In ΔABX, U is mid-Point of AB and Y is mid-Point AX
(given)
∴ UY || BX(using mid-Point theorem) ..(i)

In ΔBCX, V is mid-Point of BC and Z is mid-Point of XC
∴ VZ || BX   ....(ii)
From (i) and (ii)
UY || VZ
In ΔABC, U is mid-point of AB and V is mid-point of BE
∴ UV || AC 
⇒ UV || YZ
Hence Proved

(i) (a): Speed = 1200 km/hr
\(\text { Time }=1 \frac{1}{2} \mathrm{hr}=\frac{3}{2} \mathrm{hr}\)
\(\therefore\) Required distance = Speed x Time
\(=1200 \times \frac{3}{2}=1800 \mathrm{~km}\)
(ii) (c): Speed = 1500 km/hr
Time = \(\frac{3}{2}\) hr.
\(\therefore\) Required distance = Speed x Time
\(=1500 \times \frac{3}{2}=2250 \mathrm{~km}\)
(iii) (b): Clearly, directions are always perpendicular to each other.
\(\therefore \quad \angle P O Q=90^{\circ}\)
(iv) (a): Distance between aeroplanes after \(1\frac{1}{2}\) hour 
\(\begin{array}{l} =\sqrt{(1800)^{2}+(2250)^{2}}=\sqrt{3240000+5062500} \\ =\sqrt{8302500}=450 \sqrt{41} \mathrm{~km} \end{array}\)
(v) (d): Area of \(\Delta\)POQ= \(\frac{1}{2}\)x base x height
\(=\frac{1}{2} \times 2250 \times 1800=2250 \times 900=2025000 \mathrm{~km}^{2}\)

Given OB = x, OD = x + 3, OC = 3x + 19 and OA = 3x + 4

and AC || BD
To prove \(\triangle O A C \sim \triangle O B D\)
Proof In \(\triangle O A C \text { and } \triangle O B D\)
\(\angle D O B=\angle C O A\)
                         [vertically opposite angles]
As, AC is parallel to BD, AB and DC are transversal
\(\begin{aligned} & \angle B A C=\angle A B D \end{aligned}\)
and   \(\begin{aligned} & \angle C D B=\angle D C A \end{aligned}\)
                              [\(\because\)alternate interior angles]
\(\triangle O A C \sim \triangle O B D\)
[by AA similarity criteria]
                               Hence proved.
(ii) To prove \(\frac{O A}{A C}=\frac{O B}{B D}\)
Proof From part (i) we know that
\(\triangle O A C \sim \triangle O B D\)
Corresponding sides of \(\Delta\)OAC and \(\Delta\)OBD must be proportional
\(\begin{aligned} \frac{O A}{O B} & =\frac{A C}{B D} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad & \frac{O A}{A C}=\frac{O B}{B D} \end{aligned}\)        Hence proved.
(iii) (a) From part (i)
\(\begin{aligned} \triangle O A C & \sim \triangle O B D \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{O A}{O B} & =\frac{O C}{O D} \Rightarrow \frac{3 x+4}{x}=\frac{3 x+19}{x+3} \end{aligned}\)
[given OA = 3x +4, OB = x, OC = 3x+19, OD = x+3]
\(\Rightarrow\) (3x + 4)(x + 3) = (3x + 19)x
\(\Rightarrow\) 3x² + 9x + 4x + 12 = 3x² + 19x
\(\Rightarrow\)      13x + 12 = 19x \(\Rightarrow\) x = 2
Given,       OC = 3x + 19
On substituting x = 2, we get
OC = 3 \(\times\) 2 + 19 = 6 + 19 = 25
Therefore, OC = 25
(b) From part (ii), we have
\(\begin{aligned} & \frac{O A}{A C}=\frac{O B}{B D} \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad \frac{B D}{A C}=\frac{O B}{O A} \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad \frac{B D}{A C}=\frac{x}{3 x+4} \end{aligned}\)
                       [given, OB = x and OA = 3x + 4]
On substituting x = 2, we get
\(\frac{B D}{A C}=\frac{2}{3 \times 2+4}=\frac{2}{10} \Rightarrow \frac{B D}{A C}=\frac{1}{5}\)
Therefore, \(\frac{B D}{A C}=\frac{1}{5}\)

(i) Let the length of roads connecting X to Y, Y to Z and Z to X be 5x, 3x and 4x, respectively.

Mayank, Biju and Shanti had drawn similar triangles as angles in them are of same measure and sides in them are of same proportion.
(ii) Hint Distance of Village Y from the meeting point of the roads = 3/2 x x
Ans : 22.5 km
(iii) The length of the road would be the same in all three scale drawings as the line parallel to a side in a triangle and cutting the other two sides in a specific ratio is unique.