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Answer 1

Answer:

$A) 14.89~ units$

Step-by-step explanation:

$AD=C=3.42~, AC=d=4.57, ~CD=a=4.39$

$S=\frac{C+d+a}{2} =\frac{3.42+4.57+4.39}{2} =6.19$

$Using ~ heron's~ formula:$

$Area~ of ADC~is:$

⇒ $\sqrt{s(s-c)(s-d)(s-a)}$

⇒ $\sqrt{6.19(6.19-3.42)(6.19-4.57)(6.19-4.39)}$

⇒ $\sqrt{6.19*2.77*1.62*1.8}$

⇒ $\sqrt{49.9986}$ ⇒ $7.071$

$Area ~of ~ ABD ~is:$

⇒ $\sqrt{s(s-d)(s-b)(s-a)}$

⇒ $\sqrt{8.35(8.35-7.84)(8.35-3.42)(8.35-5.44)}$

⇒ $\sqrt{8.35*0.51*4.93*2.91}$

⇒ $\sqrt{61.0937}$ ⇒ $7.816$

$So, Area ~ of ~quadrilateral~ ABCD ~ =$

$A=Area ~of ~ADC+Area ~ of ABD:$

$A=7.071+7.816$

$A=14.89 ~units^2$

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Answer 2

Answer:

A) 14.89 is your answer

Step-by-step explanation:

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