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babymother [125]
3 years ago
12

PLEASE HELP!! Proving the Parallelogram Side Theorem in Geometry B on edg.

Mathematics
1 answer:
RideAnS [48]3 years ago
9 0

Here are the answers :D

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Perimeter = 32.44 units

Area = 30 square units

Step-by-step explanation:

Given

Vertices

A(2,8), B(16,2) and C(6,2)

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d = \sqrt{(x_{2}- x_{1})^{2} +(y_{2}-y_{1})^{2}}\\AB=\sqrt{(16-2)^{2} +(2-8)^{2}}\\=\sqrt{(14)^{2} +(-6)^{2}}\\=\sqrt{196+36}\\ =\sqrt{232}\\=15.23\\\\BC=\sqrt{(6-16)^{2} +(2-2)^{2}}\\=\sqrt{(-10)^{2} +(0)^{2}}\\=\sqrt{100+0}\\ =\sqrt{100}\\=10\\\\AC=\sqrt{(6-2)^{2} +(2-8)^{2}}\\=\sqrt{(4)^{2} +(-6)^{2}}\\=\sqrt{16+36}\\ =\sqrt{52}\\=7.21\\\\

So the perimeter is:

Perimeter=AB+BC+AC\\=15.23+10+7.21\\=32.44\ units

Using hero's formula,

s=\frac{perimeter}{2}\\s=\frac{32.44}{2}\\ s=16.22\\Area=\sqrt{s(s-a)(s-b)(s-c)}\\=\sqrt{16.22(16.22-15.23)(16.22-10)(16.22-7.21)}\\=\sqrt{(16.22)(0.99)(6.22)(9.01)}\\=\sqrt{899.91}\\=29.99\ square\ units

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