AD²=BD x CD
\(\frac { AD }{ CD } =\frac { BD }{ AD } \)
\(\triangle\)ADC-\(\triangle\)BDA (by SAS D=90⁰)
\(\angle\)BAD=\(\angle\)ACD;
\(\angle\)DAC=\(\angle\)DBA  (Corresponding angles of similar triangles)
\(\angle\)BAD+\(\angle\)ACD+\(\angle\)DAC+\(\angle\)DBA=180⁰
\(\angle\)BAD+\(\angle\)DAC=90⁰
\(\angle\)A=90⁰
\(\angle\)BAD+2\(\angle\)DAC=180⁰

Given 2AB = DE and BC = 8 cm
\(\triangle\)ABC ~ \(\triangle\)DEF

So \(\frac { AB }{ BC } =\frac { DE }{ EF } \)
\(\Rightarrow d \frac { AB }{ 8 } =\frac { 2AB }{ EF } \)
\(\therefore EF=2\times 8=16 cm.\)

PQ || MN
So, \(\frac { KP }{ PM } =\frac { KQ }{ QN } \)
\(\Rightarrow \quad \frac { KP }{ PM } =\frac { KQ }{ KN-KQ } \)
\(\Rightarrow \quad \frac { 4 }{ 13 } =\frac { KQ }{ 20.4-KQ } \)
\(\Rightarrow \) 4 x 20.4-4KQ=13 KQ
\(\Rightarrow \)17 KQ=4 x 20.4
\(\therefore \quad KQ=\frac { 20.4\times 4 }{ 17 } =4.8\quad cm,\)

Since DE || BC
\(\because \quad \frac { AD }{ AB } =\frac { AE }{ EC } \)
\(\Rightarrow \frac { 3 }{ 4 } =\frac { 6 }{ EC } \)
\(\therefore\) EC=8 cm

Let the triangles be \(\triangle\)ABC and \(\triangle\)DEF
\(\frac { ar(\triangle ABC) }{ ar(\triangle DEF) } =\left( \frac { 5 }{ 6 } \right) ^{ 2 }=\frac { 25 }{ 36 } \)
25 : 36

In \(\triangle\)PAO and \(\triangle\)QBO
\(\angle\)A=\(\angle\)B=90⁰
\(\angle\)POA=\(\angle\)QOB   (Vertically Opposite Angle)
\(\triangle\)PAO~\(\triangle\)QBO,    (by AA)
\(\therefore \quad \frac { OA }{ OB } =\frac { PA }{ QB } \\ \Rightarrow \frac { 6 }{ 4.5 } =\frac { 4 }{ QB } \\ \Rightarrow \quad QB=\frac { 4\times 4.5 }{ 6 } \\ \therefore \quad QB=3\quad cm\)

Let AB be the lamp-post and ED be the position of girl after 4s.
Given, height of the girl, ED = 100 cm
and height of the lamp-post, AB = 5 m = 500 cm
Distance of the girl from lamp-post after 4 s
= 1.9 x 4 = 7.6 m = 760 cm
[\(\because\)  distance = speed x time]
i.e. BD = 760 cm
Let DC = x cm
In \(\triangle CDE\) and \(\triangle CBA\),
\(\angle DCE=\angle BCA\) [common angle]
\(\angle CDE=\angle CBA\)  [each 90°]
\(\therefore \triangle CDE\sim \triangle CBA\) [by AA similarity criterion]
So, \(\frac { CD }{ CB } =\frac { DE }{ BA } \Rightarrow \frac { x }{ x+760 } =\frac { 100 }{ 500 } \)
\(\Rightarrow\)  5x = x + 760
\(\Rightarrow\)  4x = 760
\(\Rightarrow\)  x = 190 cm
Hence, the length of her shadow 4s is 190 cm.

A trapezium ABCD in which \(PQ\parallel AB\), draw a line AC, which intersects PQ at O.
Join AC, which intersects PQ at O.
Given, \(AB\parallel DC\) and \(PQ\parallel AB\)
Then, \(AB\parallel PQ\parallel DC\)

In \(\triangle ADC\), \(PO\parallel DC\)
By basic proportionality theorem,
\(\frac{AP}{PD}=\frac{AO}{OC}\)  ... (i)
In \(\triangle CAB\), \(OQ\parallel AB\)
By basic proportionality theorem,
\(\frac{AO}{OC}=\frac{BQ}{QC}\) ... (ii)
On comparing Eqs. (i) and (ii), we get
\(\frac{AP}{PD}=\frac{BQ}{QC} \Rightarrow \frac{AP}{18}=\frac{35}{15}\)
[\(\because\)  PD = 18 cm, BQ = 35 cm and QC = 15 cm]
\(\Rightarrow AP=\frac { 18\times 35 }{ 15 } \) = 42 cm
\(\therefore\)  AD = AP + PO
= 42 + 18 = 60 cm

Let \(\Delta A B C\) is an equilateral triangle of each side '2a' inscribed in a circle of radius 6 ern and centre O.
Draw AD⊥BC. Then, BD = DC [∵ In an equilateral triangle, perpendicular bisects the base]
Also, O lies on AD [∵ In an equilateral triangle circumcentre and centroid coincides]

Thus, OA : OD = 2 : 1
\(\Rightarrow \frac{6}{O D}=\frac{2}{1} \Rightarrow O D=3 \mathrm{~cm}, \text { then } A D=9 \mathrm{~cm}\)
Now, use pythagoras theorem in ΔABD.
\(6 \sqrt{3} \mathrm{~cm}\)

12 cm, 16 cm

It is given that the sides of the triangle are 7 cm, 24 cm, and 25 cm.
Squaring the lengths of these sides, we will obtain 49, 576, and 625.
49 + 576 = 625
Or
72 + 242 = 252
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 25 cm.

Apply Pythagoras theorem, if the given three sides satisfies the theorem, then they form right angled \(\triangle \) . Not a right angle triangle.

9 : 7

\(\triangle ADB\sim \triangle ADC\Rightarrow \frac { AB }{ AC } =\frac { DB }{ DC } \Rightarrow \frac { 18 }{ p } =\frac { p }{ 2 } \)
p = 6 cm

Given, \(\triangle ABC\sim \triangle DEF\)
\(\therefore \frac { ar\left( \triangle ABC \right) }{ ar\left( \triangle DEF \right) } =\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } \)
[ using property of area of similar triangles]
\(\Rightarrow \frac { 64 }{ 121 } =\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } \Rightarrow \left( \frac { BC }{ EF } \right) ^{ 2 }=\left( \frac { 8 }{ 11 } \right) ^{ 2 }\)
\(\Rightarrow \frac { BC }{ EF } =\frac { 8 }{ 11 } \) [taking positive square root on both sides]
\(\Rightarrow BC=\frac { 8 }{ 11 } \times EF\)
\(\therefore BC=\frac { 8 }{ 11 } \times 15.4=11.2cm\quad \left[ \because EF=15.4cm,given \right] \)

Given A \(\triangle ABC\) in which \(\angle ABC=\) 90° and AB = BC.\(\triangle ABD\) and \(\triangle ACE\) are equilateral triangles.

To Prove \(ar\left( \triangle ABD \right) =\frac { 1 }{ 2 } ar\left( \triangle CAE \right) \)
Proof Let AB = BC = x units
Now, \(CA=\sqrt { { AB }^{ 2 }+{ AC }^{ 2 } } =\sqrt { { x }^{ 2 }+{ x }^{ 2 } } =x\sqrt { 2 } \) units.
In \(\triangle ABD\) and \(\triangle CAE\), each angle is 60° as they are equilateral triangle.
\(\therefore \triangle ABD\sim \triangle CAE\)
Since, the ratio of the area of two similar triangles is equal to the ratio of the squares of their corresponding sides.
\(\therefore \frac { ar\left( \triangle ABD \right) }{ ar\left( \triangle CAE \right) } =\frac { { AB }^{ 2 } }{ { CE }^{ 2 } } =\frac { { x }^{ 2 } }{ { \left( x\sqrt { 2 } \right) }^{ 2 } } =\frac { { x }^{ 2 } }{ 2{ x }^{ 2 } } =\frac { 1 }{ 2 } \)
Hence, \(ar\left( \triangle ABD \right) =\frac { 1 }{ 2 } ar\left( \triangle CAE \right) \)

Given, \(DE\parallel BC\) , DE = 6 cm and BC = 12 cm
In \(\triangle ABC\) and \(\triangle ADE\),
\(\angle ABC=\angle ADE\)  [corresponding angles]
\(\angle ACB=\angle AED\) [corresponding angles]
and \(\angle A=\angle A\)  [common angle]
\(\therefore \triangle ABC\sim \triangle ADE\)  [by AAA similarity criterion]
We know that, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
\(\therefore \frac { ar\left( \triangle ADE \right) }{ ar\left( \triangle ABC \right) } =\frac { { \left( DE \right) }^{ 2 } }{ { \left( BC \right) }^{ 2 } } =\frac { { \left( 6 \right) }^{ 2 } }{ { \left( 12 \right) }^{ 2 } } =\left( \frac { 1 }{ 2 } \right) ^{ 2 }\)
\(\Rightarrow \frac { ar\left( \triangle ADE \right) }{ ar\left( \triangle ABC \right) } =\left( \frac { 1 }{ 2 } \right) ^{ 2 }=\frac { 1 }{ 4 } \)
Let \(ar\left( \triangle ADE \right) =k\) , then \(ar\left( \triangle ABC \right) =4k\)
Now, \(ar\left( \triangle DECB \right) =ar\left( \triangle ADC \right) - ar\left( \triangle ADE \right) \)
= 4k - k = 3k
\(\therefore \)  Required ratio = \(ar\left( \triangle ADE \right) : ar\left( \triangle DECB \right) \)
= k : 3k = 1 : 3

Here, \(\frac { AB }{ EF } =\frac { 6 }{ 4.5 } =\frac { 60 }{ 45 } =\frac { 4 }{ 3 } ,\frac { BC }{ DE } =\frac { 4 }{ 3 } \)
 So, \(\frac { AB }{ EF } =\frac { BC }{ DE } \) and \(\angle ABC=\angle FED\)   [given]
So, \(\triangle ABC\sim \triangle FED\)  [by SAS similarity criterion]
Hence, figure (i) and (ii) are similar triangles, but no other pairs of triangles in the given figure are similar.

In \(\triangle ABC\), \(\frac { AD }{ DB } =\frac { 6 }{ 9 } =\frac { 2 }{ 3 } \)

and \(\frac { AE }{ EC } =\frac { 8 }{ 12 } =\frac { 2 }{ 3 } \)
So, \(\frac { AD }{ DB } =\frac { AE }{ EC } \)
Hence, \(DE\parallel BC\)
[ by converse of basic proportionallity theorem]

Given, \(\triangle ABC\sim \triangle EDF\)
Also, AB = 5 cm, AC = 7 cm, DF = 15 cm
and DE = 12 cm                         .... (i)
Since, \(\triangle ABC\sim \triangle EDF\)
\(\therefore \frac { AB }{ ED } =\frac { AC }{ EF } =\frac { BC }{ DF } \)
[ ∵ Corresponding sides of similar triangles are proportions]
\(\Rightarrow \frac { 5 }{ 12 } =\frac { 7 }{ EF } =\frac { BC }{ 15 } \)  [from Eq. (i)]

On taking first and second terms, we get
\(\frac { 5 }{ 12 } =\frac { 7 }{ EF } \Rightarrow EF=\frac { 7\times 12 }{ 5 } \) = 16.8 cm
On taking first and third terms, we get
\(\frac { 5 }{ 12 } =\frac { BC }{ 15 } \Rightarrow BC=\frac { 5\times 15 }{ 12 } \) = 6.25 cm
Hence, lengths of the remaining sides of the triangles are EF = 16.8 cm and BC = 6.25 cm.