Given A quadrilateral ABCD, in which 
\(\angle A+\angle D=90^{\circ}\)
Toprove AC² + BD² = AD² + BC²
Construction Produce AB and CD to meet at E.
Also, join AC and BD 
Proof In ΔAED \(\angle A+\angle D=90^{\circ}\) [given]

\(\therefore \angle E=180^{\circ}-(\angle A+\angle D)=90^{\circ}\) [by angle sum property of a triangle]
By Pythagoras theorem,
AD² = AE² + DE²........(i)
In ΔBEC, by Pythagoras theorem,
BC² = BE² + CE² .........(ii)
On adding Eqs. (i) and (ii), we get
AD² + BC² = AE² + DE² + BE² + CE² .......(iii)
In ΔAEC, by Pythagoras theorem,
AC² = AE² + CE² ......(iv)
and in ΔBED, by Pythagoras theorem,
BD² = BE² + DE² .......(v)
On adding Eqs. (iv) and (v), we get
AC² + BD² = AE² + CE² + BE² + DE²........(vi)
From Eqs. (iii) and (vi), we get
AC² + BD² = AD² + BC²

\(P R^{2}-P Q^{2}=Q R^{2} \Rightarrow P R^{2}=P Q^{2}+Q R^{2}\)
\(\Rightarrow\) ΔPQR is right angled triangle, right angled at Q.

Altitude AD which is perpendicular to BC, then D is the mid-point of BC.
BD = CD = \(\frac{1}{2}\) BC = \(\frac{8}{2}\) = 4 cm.
Apply, Pythagoras theorem to find AD.
AD = \(4\sqrt{3}\) cm.

Given, PQRS is a trapezium in which \(PQ\parallel RS\) and PQ = 3RS.

\(\Rightarrow \frac { PQ }{ RS } =\frac { 3 }{ 1 } \) ...(i) 
In \(\triangle POQ\) and \(\triangle ROS\),
\(​​\angle SOR=\angle QOP \)  [vertically opposite angles]
\(​​\angle SRP=\angle RPQ\) [alternate angles]
\(\therefore \triangle POQ\sim \triangle ROS\)     [by AA similarity criterion]
By property of area of similar triangle,
\(\frac { ar\left( \triangle POQ \right) }{ ar\left( \triangle ROS \right) } =\frac { { \left( PQ \right) }^{ 2 } }{ { \left( RS \right) }^{ 2 } } =\left( \frac { 3 }{ 1 } \right) ^{ 2 }\)  [from Eq.(i)]
\(\Rightarrow \frac { ar\left( \triangle POQ \right) }{ ar\left( \triangle SOR \right) } =\frac { 9 }{ 1 } \)
Hence, the required ratio is 9 : 1.

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

Given, \(\triangle ABC\sim \triangle QRP\), AB = 18 cm, BC = 15 cm

We know that, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
\(\therefore \frac { ar\left( \triangle ABC \right) }{ ar\left( \triangle QRP \right) } =\frac { { \left( BC \right) }^{ 2 } }{ { \left( RP \right) }^{ 2 } } \)
But \(\frac { ar\left( \triangle ABC \right) }{ ar\left( \triangle QRP \right) } =\frac { 9 }{ 4 } \)  given]
\(\Rightarrow \frac { { \left( 15 \right) }^{ 2 } }{ { \left( RP \right) }^{ 2 } } =\frac { 9 }{ 4 } \)  [\(\because \) BC = 15 cm]
\(\Rightarrow { \left( RP \right) }^{ 2 }=\frac { 225\times 4 }{ 9 } =100\)
\(\therefore \)  RP = 10 cm
[taking positive square root]

Prove \(\triangle AOP\) and \(\triangle COQ\) are similar.
then \(\frac { OA }{ OC } =\frac { AP }{ CQ } \)   [corresponding sides are proportional]

Prove \(\triangle ABC\) and \(\triangle AMP\)  are similar.
then take ratio \(\frac{AC}{AP}=\frac{BC}{MP}\)

Diagonals of a rhombus bisect each other at right angles.

So, OA = OC = 8 cm
and OB = OD = 6 cm
Now use pythagoras theorem in  ΔAOB
10 cm.

Draw the figure according to the question and get two triangles. Then, show both triangles are similar by AAA similarity criterion and then calculate the required distance.
She is at 9 m from the base of the pole.