Question

In the given figure, ABCD is a vertical wall
2 m high. It casts a shadow ABEF on the
ground. ABEF is a parallelogram in which
the height EN is 12 times the height of the
wall.
(a) Find the height EN.
(b) If the area of the wall is 7.6 m², find the
area of the shadow.
D
С
2
ΑΕ
B
N
F
E​

Answer 1

Answer:

In △ABC and △DEF

∠A=∠D=90

∘

Since the tower and the stick cast their respective shadows at the same time on the ground,

So, ∠C=∠F

∴△ABC∼△DEF ....(by AA similarity)

DE

AB

=

DF

AC

⇒

x

12

=

40

8

⇒x=60 m

Answer 2

Answer:

Step-by-step explanation:

GIVEN BC = 2 M

    EN = 3/2 TIMES OF BC

a)  Height of EN =$\frac{3}{2}$ × 2 = 3 m

b) AREA OF  wall ABCD= 7.6 sq . m (given)

     Height of wall BC = 2 M ( GIVEN)

    BASE OF WALL   = ? (FIND IT )

   AREA = BASE X HEIGHT

⇒  7.6   =  BASE  X 2

       ⇒ BASE = 7.6÷ 2 = 3.8 M

Area of shadow ABEF =?(FIND)

 BASE = 3.8 M

HEIGHT = 3

AREA = BASE X HEIGHT

   = 3.8 X 3 = 11.4 SQ. M