# Chapter 8 - Section 8.2 - Perimeter and Area of Polygons - Exercises - Page 371: 45a

No

#### Work Step by Step

(5,6,7) Using Heron's formula if three sides of a triangle have length a,b,c then area A of a triangle is A = $\sqrt s(s-a)(s-b)(s-c)$ where s = $\frac{a+b+c}{2}$ Lets take a=5 b=6 and c=7 s = $\frac{5+6+7}{2}$ = 9 A = $\sqrt 9(9-5)(9-6)(9-7)$ =$\sqrt 9 * 4*3*2$ = 6$\sqrt 6$ = 14.69 If a,b,c be the integer length of the sides of a triangle . If the area of the triangle is also a integer then (a,b,c) is known as a heron triplet. Since A = 14.69 not a integer (a,b,c) = (5,6,7) integer Therefore the triplet is not a heron triplet.

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