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Question

In a quadrilateral ABCD, if AB = 5 m, BC = 5 m, CD = 6 m, AD = 6 m and diagonal AC = 6 m, what is its area?


A

2(4+33)m2

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B

3(4+33)m2

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C

5(4+33)m2

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D

7(4+33)m2

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Solution

The correct option is B

3(4+33)m2


Given: AB = 5m, BC = 5m, CD = 6m, DA = 6m, AC = 6m

For this quadrilateral ABCD, we can obtain its area as follows:

Area (ABCD) = area (ΔABC) + area ( Δ ACD)

Also, Δ ABC is an isosceles triangle, while Δ ACD is an equilateral triangle.

(AB = BC = 5m) (AC = CD = DA = 6m)

So, area ( ΔACD)=(34)×(side)2=(34)×(6)2

=93sq.m

And, height of the isosceles ΔABC = h = BE

(BE is the altitude from vertex B on the side AC)

Applying Pythagoras theorem:

(BE)2+(EC)2=(BC)2

h2+32=52

(for an isosceles triangle, altitude from the vertex containing the equal sides is also the bisector of the third side: EC=(AC)2=(62)=3m )

h2 = (25 - 9) = 16

h = 4 m

Area (ΔABC)=(12)×h×(AC)=(12)×4×6

= 12 sq. m

So, area (ABCD) = (12 + 9 3)

= 3 (4 + 3 3 ) sq. m.


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