Question

Measures of sides of a triangle are in Arithmetic progression. Its perimeter is 30 cm, and the
between the longest and shortest side is 4 cm, then find the meqasure of the sides​

Answer 1

Correct Question :-

Measures of sides of a triangle are in Arithmetic progression. Its perimeter is 30 cm, and thedifference between the longest and shortest side is 4 cm, then find the measure of the sides.

Solution :-

Measures of sides of a triangle are in AP

So, let the measures of the sides of a triangle be a, (a + d), (a + 2d) [ General form of AP ]

Perimeter of the triangle = 30 cm

⇒ Sum of measures of all sides = 30 cm

⇒ a + (a + d) + (a + 2d) = 30

⇒ a + a + d + a + 2d = 30

⇒ 3a + 3d = 30

⇒ 3(a + d) = 30

⇒ a + d = 30/3

⇒ a + d = 10

Difference between longest side and shortest side = 4 cm

⇒ Measure of longest side - Measure of shortest side = 4

⇒ (a + 2d) - a = 4

⇒ a + 2d - a = 4

⇒ 2d = 4

⇒ d = 4/2

⇒ d = 2

Substituting d = 2 in a + d = 10 we get

⇒ a + 2 = 10

⇒ a = 10 - 2

⇒ a = 8

→ Measure of one of the side = a = 8 cm

→ Measure of second side = a + d = (8 + 2) = 10 cm

→ Measure of third side = a + 2d = 8 + 2(2) = 8 + 4 = 12 cm

Hence, the measures of the sides of the triangle are 8 cm, 10 cm and 12 cm.

Answer 2

Given, Sides of triangle are in arithmetic progression

Let sides be (a−d)cms, a  cm and (a+d)cms,

Now given perimeter is 30cm , that gives 3a=30 , therefore a=10cms.

and also given difference between the longest and shortest side is 4cm, so (a+d)−(a−d)=4, which gives d=2cms

Therefore sides of triangle are 8cms, 10cmsand 12cms