Let AB be the ladder and B be the window, then
BC = 12 m and AC = 5m
Since \(\Delta\)ABC is right triangle; right angled at C
AB² = AC²+BC²
AB² = 5²+12² = 25+144
AB² = 169
AB = 13m
Hence the length of the ladder is 13 m

(i) We have a = 6cm,
b = 5 cm and c = 10cm
Here the larger side is c = 10 cm
Therefore a² + b² = 6² + 8²
= 36 + 64 = 100
= (10)² = C²
Therefore The triangle with the given sides is a right triangle.
(ii) We have a = 5 cm,
b = 8 cm and c = 11 cm
Here the larger side is c = 11 cm
a² + b² = 5² + 8² = 25 + 64
a² + b² = 89
But C² = 11² = 121
\(\therefore a^{2}+b^{2} \neq c^{2}\)
The triangle with the given sides are not a right angled triangle.

Given: In \(\Delta\)ABC , the bisector AD of \(\angle\)A bisects the side BC.
To prove: AB = AC
Proof: In \(\Delta\)ABC , AD is the bisector of \(\angle\)A
\(\frac{A B}{A C}=\frac{B D}{D C}\)
\(\frac{A B}{A C}=1\)
[\(\because\) D is the mid point of BC, BD = DC]
AB = AC.
\(\therefore\)The triangle ABC is isosceles.

Given BC = 24 cm and BD = 6 cm
Therefore DC = BC - BD
= 24 - 6 = 18 cm
\(\text { Now, } \frac{A B}{A C}=\frac{8}{24}=\frac{1}{3}\)
\(\frac{B D}{D C}=\frac{6}{18}=\frac{1}{3}\)
\(\frac{A B}{A C}=\frac{B D}{D C}\)
Therefore By the converse of angular bisector theorem. AD is the bisector of \(\angle A \text { of } \triangle A B C\)

\(\text { Since } \mathrm{AD} \text { is the bisector of } \angle A\)
\(\frac{A B}{A C}=\frac{B D}{D C} \Rightarrow \frac{3.5}{4.2}=\frac{B D}{2.8}\)
\(\mathrm{BD}=\frac{3.5 \times 2.8}{4.2}\)
\(=\frac{0.70}{0.3}=\frac{7}{3}=2.33\)
\(BD = 2.33 cm\)

Let BD = x cm,
Then DC = (12 - x) cm
Since AD is the bisector of  \(\angle\)A
\(\frac{A B}{A C} =\frac{B D}{D C} \Rightarrow \frac{20}{28}=\frac{x}{12-x} \)
\(\frac{5}{7} =\frac{x}{12-x} \Rightarrow 5(12-x)=7 x \)
\(60-5 x=7 x\)
\(12 \mathrm{x}=60\)
\(x=\frac{60}{12}=5\)
BD = 5cm
DC = 12 - 5 = 7cm

\(\text { In } \triangle A B C, A D \text { is the bisector of } \angle A\)
\(\therefore \frac{B D}{D C} =\frac{A B}{A C} \)
\(\frac{6}{4.5} =\frac{9}{A C} \)
\(A C =\frac{9}{6} \times 4.5 \)
\(A C =6.75 \mathrm{~cm} \)

We have a trapezium ABCD in which AB || DC.
The diagonals AC and BD intersect at O 
In \(\Delta\)OAB and \(\Delta\)OCD 
AB || DC 
and BD intersects them
\(\therefore\) \(\angle\)OBA = \(\angle\)ODC (Alternate interior angles)
Also \(\angle\)OAB = \(\angle\)OCD (Alternate interior angles)
Using AA similarity criteria
\(\triangle O A B \sim \triangle O C D\)

\(\text { In } \triangle M N L \text { and } \Delta Q P R\)
\(\frac{M N}{Q P}=\frac{2.5}{5}=\frac{1}{2} \)
\(\frac{M L}{Q R}=\frac{5}{10}=\frac{1}{2} \)
\(\frac{M L}{Q R} =\frac{M N}{Q P} \)
\(\angle N M L =\angle P Q R \)
Using SAS criteria of similarity, we have,
\(\triangle M N L \sim \triangle Q P R\)

\(\text { In } \triangle A B C \text { and } \triangle P Q R\)
\(\text { we have } \angle A=\angle P=60^{\circ}\)
\(\angle B=\angle Q=80^{\circ} \)
\(\angle C=\angle R=40^{\circ} \)
The corresponding angles are equal
Using AAA similarity rule
\(\triangle A B C \sim \triangle P Q R\)

\(\text { Given } \triangle P Q R \sim \Delta L M N\)
\(\frac{\text { area }(\Delta \mathrm{PQR})}{\text { area of }(\Delta \mathrm{LMN})}=\frac{P Q^{2}}{L M^{2}}\)
\(\frac{64}{\operatorname{area}(\Delta L M N)}=\left(\frac{4}{5}\right)^{2}=\frac{16}{25}\)
\(\therefore\ \text { Area of } \Delta L M N=\frac{64 \times 25}{16}\)
\(=100 \mathrm{~m}^{2}\)
\(\text { Area of } \Delta L M N=100 \mathrm{~m}^{2}\)

\(​​\text { Given } \ \triangle A B C \sim \Delta D E F\)
\(\frac{\operatorname{area}(\triangle A B C)}{\operatorname{area}(\Delta D E F)}=\frac{B C^{2}}{E F^{2}}\)
\(\frac{81}{100} =\frac{B C^{2}}{(5)^{2}} \)
\(\frac{B C}{5} =\frac{9}{10} \)
\(B C =\frac{9}{10} \times 5=\frac{9}{2} \mathrm{~cm} \)
\(=4.5 \mathrm{~cm} \)
\(\therefore \ B C =4.5 \mathrm{~cm} \)

Given XY IIA C

So,

\(\therefore \Delta ABC\sim \Delta XbY\) (AAA similarity criterion)
So, \(\cfrac { ar(ABC) }{ ar(XBY) } =\left( \cfrac { AB }{ XB } \right) ^{ 2 }\)         ...(1)
ar(ABC) = 2ar(XBY)
\(\cfrac { ar(ABC) }{ ar(ABC) } =\cfrac { 2 }{ 1 } \)           ...(2)
From (1) and (2),
\(\left( \cfrac { AB }{ XB } \right) ^{ 2 }=\cfrac { 2 }{ 1 } i.e.,\cfrac { AB }{ XB } =\cfrac { \sqrt { 2 } }{ 1 } \)
\(\cfrac { XB }{ AB } =\cfrac { 1 }{ \sqrt { 2 } } \)
\(1-\frac{X B}{A B}=1-\frac{1}{\sqrt{2}}\)
\(\cfrac { AB-XB }{ AB } =\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } \)
\(\cfrac { AX }{ AB } =\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } =\cfrac { 2-\sqrt { 2 } }{ 2 } \)

PQ II RS


Also

\(\angle \therefore \Delta POQ\sim \Delta SOR\) (AAA similarity criterion)

Given AB = 5.6 cm; AD = 1.4 cm;
AC = 7.2 cm; AE = 1.8 cm
DB = AB - AD
= 5.6 - L.4 = 4.2 cm
EC = AC - AE
= 7.2 - 1.8 = 5.4 cm
Now \(\frac{A D}{D B}=\frac{1.4}{4.2} \)
\(\frac{A D}{D B}=\frac{1}{3} \)
\(\frac{A E}{E C}=\frac{1.8}{5.4} \)
\(\frac{A E}{E C}=\frac{1}{3} \)
From (1) and (2)
\(\frac{A D}{D B}=\frac{A E}{E C}\)
DE || BC