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# Find The area of of 4th Part in Rectangle

A Rectangle is ABCD is given and line DN and CM intersect at O.
As shown in the Figure.

Area of △MON, □AMOD and △DOC are 1 , 3 and 4 respectively.

Find the area of □BNOC.

Figure is not drawn to scale.

Answer: 4

Explanation:
∠MON = ∠DOC (∵ Vertically Opposite angles)
∠MND = ∠NDC (∵ Alternate Interior angles)

Therefore, we can say that both the triangles (MON and DOC) are similar.

And we know that, Area is proportional to square of side length in similar triangles.

Areas are in ration 4:1 => Length are in ration 2:1

Let us assume MN = a
we get DC = 2a

and also assume height of triangle MON is h => height of triangle DOC becomes 2h

Consider the area of top triangle MON
△MON = ah/2 = 1 (Area of △MON is given)
Therefore ah = 2 .........................eq(1)

Now,lenght AD = height of △MON + height of △DOC
AD = 3h

Also, we found that DC = 2a

We know the formula for the area of rectangle i.e. Height x Base
□ABCD = (3h)(2a) = 6ah
from eq(1) let us substitute value of ah as 2

□ABCD = 6(2) = 12

□ABCD = △MON + □AMOD + △DOC + □BNOC
12 = 1 + 3 + 4 + □BNOC
□BNOC = 4