Question

In the given figure, ABCD is a square of side 6 cm. If AE = 2 cm and DF = 3 cm, find the area of the
∆BEF. ​

Answer 1

here is your solution hope you understand.

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Answer 2

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=?

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