The sides of the triangle are 8cm, 10cm and 12cm. find the perimeter and area of a triangle whose

The sides of the triangle are 8cm, 10cm and 12cm. find the perimeter and area of a triangle whose vertices are the midpoints of the sides of this triangle.

Let in △ ABC AB = 8 cm, BC = 10 cm, AC = 12 cm. Points M, N and K are the midpoints of sides AB, BC and AC, respectively. The sides of △ MNK (MN, NK, MK) are the middle lines of △ ABC.
The midline of the two sides of the triangle is equal to half of the third side of the triangle:
MN = AC / 2 = 12/2 = 6 (cm);
NK = AB / 2 = 8/2 = 4 (cm);
MK = BC / 2 = 10/2 = 5 (cm).
1. The perimeter of △ MNK is:
P = MN + NK + MK = 6 + 4 + 5 = 15 (cm).
2. The area △ MNK is found by Heron’s formula:
S = √ (p * (p – a) * (p – b) * (p – c)),
where p is the semi-perimeter of the triangle, a, b and c are the sides of the triangle.
Semi-perimeter:
p = P / 2 = 15/2 = 7.5.
a = 6, b = 4, c = 5.
Find the area △ MNK:
S = √ (7.5 * (7.5 – 6) * (7.5 – 4) * (7.5 – 5)) = √ (7.5 * 1.5 * 3.5 * 2.5 ) = (let’s convert decimal fractions to common ones) = √ (75/10 * 15/10 * 35/10 * 25/10) = √ ((75 * 15 * 35 * 25) / (10 * 10 * 10 * 10) ) = (√ (75 * 15 * 35 * 25)) / 100 = (15 √ 7) / 4 (cm²).
Answer: P = 15 cm, S = (15 √ 7) / 4 cm².