Answer
(a), (b) , (c), (d)
Since these triangles have integer sides and areas, these triangles are Heron triangles.
Work Step by Step
(a) We can find the semi-perimeter $S$:
$S = \frac{11+13+20}{2} = 22$
We can find the area:
$A = \sqrt{S~(S-a)(S-b)(S-c)}$
$A = \sqrt{(22)~(22-11)(22-13)(22-20)}$
$A = \sqrt{(22)~(11)(9)(2)}$
$A = \sqrt{4356}$
$A = 66$
Since this triangle has integer sides and area, this triangle is a Heron triangle.
(b) We can find the semi-perimeter $S$:
$S = \frac{13+14+15}{2} = 21$
We can find the area:
$A = \sqrt{S~(S-a)(S-b)(S-c)}$
$A = \sqrt{(21)~(21-13)(21-14)(21-15)}$
$A = \sqrt{(21)~(8)(7)(6)}$
$A = \sqrt{7056}$
$A = 84$
Since this triangle has integer sides and area, this triangle is a Heron triangle.
(c) We can find the semi-perimeter $S$:
$S = \frac{7+15+20}{2} = 21$
We can find the area:
$A = \sqrt{S~(S-a)(S-b)(S-c)}$
$A = \sqrt{(21)~(21-7)(21-15)(21-20)}$
$A = \sqrt{(21)~(14)(6)(1)}$
$A = \sqrt{1764}$
$A = 42$
Since this triangle has integer sides and area, this triangle is a Heron triangle.
(d) We can find the semi-perimeter $S$:
$S = \frac{9+10+17}{2} = 18$
We can find the area:
$A = \sqrt{S~(S-a)(S-b)(S-c)}$
$A = \sqrt{(18)~(18-9)(18-10)(18-17)}$
$A = \sqrt{(18)~(9)(8)(1)}$
$A = \sqrt{1296}$
$A = 36$
Since this triangle has integer sides and area, this triangle is a Heron triangle.
