Since the ratio of area of two similar triangles is equal to the ratio of the squares of any two corresponding sides, we have
\(\frac { Area(\Delta ABC) }{ Area(\Delta DEF) } =\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } \) gives \(\frac { 54 }{ Area(\Delta DEF) } =\frac { { 3 }^{ 2 } }{ { 4 }^{ 2 } } \)
\(Area(\Delta DEF)=\frac { 16\times 54 }{ 9 } =96{ cm }^{ 2 }\)

In Is \(\triangle\)ABC ~ \(\triangle\)PQR
\(\frac { PQ }{ AB } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } ;\frac { QR }{ BC } =\frac { 4 }{ 10 } =\frac { 2 }{ 5 } \)
Since \(\frac { 1 }{ 2 } \neq \frac { 2 }{ 5 } ,\frac { PQ }{ AB } \neq \frac { QR }{ BC } \)
The corresponding sides are not proportional.
Therefore \(\triangle\)ABC  is not similar to \(\triangle\)PQR.

Given a triangle PQR, we are required to construct another triangle whose sides are \(\frac { 7 }{ 4 } \) of the corresponding sides of the triangle PQR.
Steps of construction
1. Construct a DPQR with any measurement.
2. Draw a ray QX making an acute angle with QR on the side opposite to vertex P.
3. Locate 7 points (the greater of 7 and 4 in \(\frac { 7 }{ 4 } \))
Q_1,Q_2,Q_3,Q_4,Q_5,Q₆ and Q₇ on QX so that 
QQ₁ = Q₁Q₂ = Q₂Q₃ = Q₄Q₅ = Q₅Q₆ = Q₆Q₇
4. Join Q₄ (the 4th point, 4 being smaller of 4 and 7 in \(\frac { 7 }{ 4 } \)) to R and draw a line through Q₇ parallel to Q₄R, intersecting the extended line segment QR at R'.
5. Draw a line through R' parallel to RP intersecting the extended line segment QP at P'.
Then \(\triangle\)P'QR' is the required triangle each of whose sides is seven-fourths of the corresponding sides of \(\triangle\)PQR.

Construction
Step 1: Draw a line segment PQ = 8cm.
Step 2: At P, draw PE such that \(\angle\)QPE = 60^o.
Step3: At P, draw PF such that\(\angle\)EPF = 90^o.
Step 4: Draw the perpendicular bisector to PQ, which intersects PF at O and PQ at G.
Step 5: With O as centre and OP as radius draw a circle.
Step 6: From G mark arcs of radius 5.8 cm on the circle. Mark them as R and S.
Step 7: Join PR and RQ. Then \(\triangle\)PQR is the required triangle.
Step 8: From R draw a line RN perpendicular to \(\angle\)Q. \(\angle\)Q meets RN at M
Step 9: The length of the altitude is RM = 3.8 cm.

In \(\triangle\) ABC we have DE || BC.
By Thales theorem, we have \(\frac { AD }{ DB } =\frac { AE }{ EC } \)
\(\frac { x }{ x-2 } =\frac { x+2 }{ x-1 } \) gives x(x - 1) = (x - 2)(x + 2)
When x = 4, AD = 4, DB = x - 2, AE + x + 2 = 6, EC = x - 1 = 3 
Hence, AB = AD + DB = 4 + 2 = 6, AC = AE + EC = 6 + 3 = 9
Therefore, AB = 6, AC = 9

Let AC is the lamp post and ED is the girl.
From the triangles ABC and DBE
o
By AA criteria
Their sides are Proportional
\(\frac{A C}{D E}=\frac{B C}{B E} \)
\( \frac{A C}{12.5}=\frac{6.6}{2.5} \)
\(A C= \frac{6.6 \times 12.5}{2.5}=\frac{6.6 \times 12.5^{5}}{2.5}=33 \mathrm{~m} \)
Height of the lamp post = 33 m