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Question
In the following figure, 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. What is the area of ∆DAO, where O is the midpoint of DC?
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Solution
Given, Area of ΔAFB = Area of parallelogram ABCD
⇒ `1/2` × AB × EF = CD × (Corresponding height) ...`[(∵ "area of triangle" = "base" xx "height and area of"),("parallelogram" = "base" xx "corresponding height")]`
⇒ `1/2` × AB × EF = CD × EG
Let the corresponding height be h.
Then, `1/2` × 10 × 6 = 10 × h ...[∵ altitude, EF = 16 cm and base, AB = 10 cm, given] [∵ AB = CD]
⇒ h = 8 cm
In ΔDAO, DO = 5 cm ...[∵ O is the mid-point of CD]
∴ Area of ΔDAO = `1/2` × OD × h
= `1/2 xx 5 xx 8`
= 20 cm2
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