Question

In rectangle ABCD, ZC is trisected by CE and CF where E
lies on AB and Fon AD. If BE = 6 cm and AF = 2 cm, which
of the following integers is nearest to the area of the
rectangle ABCD in sq.cm.:
( )
190
( )
170
( )
150
(
)
130​

Answer 1

E is the midpoint of side BC of a rectangle ABCD and F is the midpoint of CD . The area of triangle AEF is 3 unit square. What is the area of the rectangle?

Find, evaluate and source engineering materials online.

Let ABCD is a rectangle in which AB= CD = 2a units. BC=AD= 2b units. E is the

mid point of BC therefore BE= CE= b units and F is the midpoint of CD so that

CF= DF = a units.

Area of rectangle ABCD =2a×2b = 4.a.b units ^2…………………….(1)

Area of ∆ AEF= area of rectangle ABCD-area of (∆ABE+∆ADF+∆FCE).

3 units ^2= 4.a.b units ^2 - (2a×b/2+a×2b/2+a×b/2)units ^2.

or. 3. = 4.a.b. - (5/2).a.b

or. 3. = 3.a.b/2

or. 3.a.b= 6. => a.b. = 2. ………………..(2)

Putting a.b. = 2 from eqn. (2) in (1)

Area of rectangle ABCD = 4.a.b units ^2. = 4×2. = 8. unit ^2. Answer