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4 cm to you later then I will come to office
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Given :- (from image)
- ABCD is a square of side 6 cm.
- AE = 2 cm.
- DE = AD - AE = 6 - 2 = 4 cm.
- DF = 3 cm.
- FC = DC - DF = 6 - 3 = 3 cm.
- CB = AB = 6 cm.
To Find :-
- Area of the ∆BEF = ?
Solution :-
we know that,
- Area of square = (side)² .
- Area of right angled ∆ = (1/2) * Base * Perpendicular.
- All angles of a square is equal to 90° .
then,
→ Area ∆BEF = Area of Square ABCD - (Area of right angled ∆EDF + Area of right angled ∆FCB + Area of right angled ∆EAB).
→ Area ∆BEF = (CB)² - [ {(1/2) * DE * DF} + {(1/2) * FC * CB} + {(1/2) * EA * AB} ]
→ Area ∆BEF = (6)² - [ {(1/2) * 4 * 3} + {(1/2) * 3 * 6} + {(1/2) * 2 * 6} ]
→ Area ∆BEF = 36 - [6 + 9 + 6]
→ Area ∆BEF = 36 - 21
→ Area ∆BEF = 15 cm². (Ans.)
Hence, Area of the ∆BEF is equal to 15 cm².
Learn more :-
In ABC, AD is angle bisector,
angle BAC = 111 and AB+BD=AC find the value of angle ACB=?
https://brainly.in/question/16655884
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