To become familiar with the fact that geometry (similar triangles) can be Description
In this project I tried to find situations in daily life where geometrical notions can be effectively used, I selected the following examples: 2. To find height of a tower 1. To find the width of a river
iC
BS E
.co
used in real life to find height of certain things and width of many others.
m
Objective
iC BS E.c om
To find the width of a river
Walked along the river, fixed another pole at R at a distance of 9 metres. Walked another 3 metres to S, from here walked at right angles to the river till the point T is reached such that T is directly in line with R and P. width of the river was determined. Measured the distance ST. Using the property of similarity of triangles the
iC
BS
E.c
om
Fixed a pole at Q directly opposite to a tree P on the other side of the river.
iC BS E.c om
In right triangle RQP and RST Angle PRQ = angle TRS (vertically opposite angles) ൌ ͻ ൌ ͵
୕ ୗ
Therefore triangle RQP ~ triangle RST by AA corollary
ൌ
ଷ ଵ
------- (i)
Now ST = 4m, substituting its value in (i) ͵ ൌ Ͷ ͳ
Therefore width of river = 12m
iC
QP = 12m
BS
E.c
om
Angle PQR = angle RST = 90°
iC BS E.c om
To find the height of a tower shadow is at the same place as the ends of the shadow of the tower. Knowing the relevant distance, the height of the tower can be estimated.
iC
BS
E.c
om
Placed the ruler upright in the shadow of the tower, so that the ends of its
iC BS E.c om
Solution :
In ∆ABE and ∆CDE angle E = angle E (common) angle B = angle D = 90°
by AA corollary
େୈ
=
ୈ ଶ ଶହ
=
………
(i)
On measuring CD we get CD = 40cm Substituting value of CD in (i)
ܤܣ ൌ ͶͲ ଶ ଶହ
=
= 320cm = 3.2m
Height of tower = 3.2m
iC
ൌ AB =
ଶൈସ ଶହ
BS
E.c
(Corresponding parts