Basic Engineering Mathematics, Fifth Edition

178 Basic Engineering Mathematics

Problem 29. A rectangular shed 2m wide and

3m high stands against a perpendicular building of

height 5.5m. A ladder is used to gain access to the

roof of the building. Determine the minimum

distance between the bottom of the ladder and the

shed

A side view is shown in Figure 20.43, whereAF

is the minimum length of the ladder. Since BD

and CF are parallel,∠ADB=∠DFE (correspond-

ing angles between parallel lines). Hence, triangles

BADandEDF are similar since their angles are the

same.

AB=AC−BC=AC−DE= 5. 5 − 3 = 2 .5m

By proportion:

AB

DE

=

BD

EF

i.e.

2. 5

3

=

2

EF

Hence,EF= 2

(

3

2. 5

)

= 2 .4m=minimum distance

from bottom of ladder to the shed.

3m

2m 5.5m

Shed

D

EC

B

A

F

Figure 20.43

Now try the following Practice Exercise

PracticeExercise 80 Similar triangles

(answers on page 349)

1. In Figure 20.44, find the lengthsxandy.

111 

32 

32 ^37 

25.69mm

4.74mm

7.36mm

14.58mm

x

y

Figure 20.44

2. PQRis an equilateral triangle of side 4cm.
   WhenPQandPRare produced toSandT,
   respectively,STis found to be parallel with
   QR.IfPSis 9cm, find the length ofST.X
   is a point onSTbetweenSandTsuch that
   the linePXis the bisector of∠SPT.Findthe
   length ofPX.


3. In Figure 20.45, find
   (a) the length of BC when AB=6cm,
   DE=8cm andDC=3cm,
   (b) the length of DE when EC=2cm,
   AC=5cmandAB=10cm.

D E

C

A B

Figure 20.45

4. In Figure 20.46,AF=8m,AB=5m and
   BC=3m. Find the length ofBD.

D

E

C

B

A F

Figure 20.46