Question

9. In the adjoining figure, ABCD is a
rectangle in which AB = 12 cm and
BC = 8 cm. E is the point on BC such
that BE = 5 cm. AE is produced to meet
DC at F.

(i) Calculate the length AE.
(ii) Prove that AABE ~ AFCE.
Hence, find the lengths of EF
and CF.​

Answer 1

i. In ∆ABE, AB=12cm

BE=5cm

as ABCD is a triangle

angle B is 90°

therefore, ∆ABE is a right angle triangle

acc to Pythagoras theorem

AB+BE=AE

12+5=AE

17=AE

ii. In ∆ABE and ∆FCE

angle B=angle C........(angles of a triangle)

anglle BEA =angle CEF.......(vertically opp angle)

therefore, ∆ABE~∆CEF...... (by AA test of similarity)

iii. CE=BC-BE

CE=8-5

CE=3

by 30°-60°-90° theorem

CE=1/2 ×EF

EF=2×CE

EF=2×3

EF=6