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Q.

In the figure given below, the area of ∆ AFB is equal to the area of parallelogram ABCD. If altitude EF is 16 cm long, find the altitude of the parallelogram to the base AB of length 10 cm. Determine the area of ∆DAO, where O is the mid-point of DC.

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a

10

cm 2

b

20

cm 2

c

30

cm 2

d

40

cm 2

answer is B.

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Detailed Solution

We are given that the area of ΔAFB is equal to the area of parallelogram ABCD.
Question Image

Question Image

Now, we know that the area of a triangle is half of the product of its base and height and the area of the parallelogram is the product of its base and height.
Here, Area of

Δ

AFB=Area of parallelogram ABCD.
‘AB’ is the base of the triangle given as 10cm and ‘EF’ is the height of the triangle given as 16cm, ‘CD’ is the base of the parallelogram which is 10cm and ‘EG’ is the corresponding height of parallelogram.
Thus,

12 × AB × EF = CD × EG

Thus,

12 × AB × EF = CD × EG 12 × 10 × 16 = 10 × EG 10h = 80 h = 8010 h = 8cm

In ∆DAO,
DO is 5 cm and height is found to be 8 cm.

Area of ​ ​ Δ ​ ​ DAO = 12 × base × height = 12 × OD × 8 = 12 × 5 × 8 = 20 cm 2

So, the area of

△ D A O

is obtained as 20

cm 2

.
Hence, the correct option is (2).

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