Find the area (correct to three significant digits) of quadrilateral ABCD with angle BCA=90°, AB=26cm and ACD as an equilateral triangle of side 24 cm.
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Since ACD is an equilateral triangle
ar(ACD) = √3/4*a^2
√3 = 1.73
ar(ACD) = 1.73*24*24/4
= 1.73*24*6
= 249.12
=249 (correct to 3 significant digits)
since ABC is a right angled triangle
(BC)^2=(AB)^2-(AC)^2
(BC)^2=(26*26)-(24*24)
(BC)^2=676-576
(BC)^2=100
BC=10
ar(ABC)=1/2*24*10
=120
ar(ABCD) = ar(ABC)ar(ACD)
ar(ABCD) = 120+249
ar(ABCD) = 369
ar(ACD) = √3/4*a^2
√3 = 1.73
ar(ACD) = 1.73*24*24/4
= 1.73*24*6
= 249.12
=249 (correct to 3 significant digits)
since ABC is a right angled triangle
(BC)^2=(AB)^2-(AC)^2
(BC)^2=(26*26)-(24*24)
(BC)^2=676-576
(BC)^2=100
BC=10
ar(ABC)=1/2*24*10
=120
ar(ABCD) = ar(ABC)ar(ACD)
ar(ABCD) = 120+249
ar(ABCD) = 369
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