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Question
The perimeter of a triangle is 72cm and its sides are in the ratio 3:4:5. Find its area and the length of the altitude corresponding to the longest side.
Solution
Perimeter P of the triangle = P = 72cm
Ratio of the sides = 3:4:5
Let the constant of proportionality be k
⇒ 3k + 4k + 5k = 72
⇒ 12k = 72
⇒ k = `(72)/(12)`
= 6
∴ the sides are: 3 x 6, 4 x 6 and 5 x 6
I.e. 18cm, 24cm and 30cm
We know that, Area of a Triangle whose sides are a, b, and c and semiperimeter is s given by `sqrt("s"("s" - "a")("s" - "b")("s" - "c")); "s" = ("a" + "b" + "c")/(2)`
For a triangle whose sides are 18cm, 24cm and 30cm
i.e. a = 18 b = 24 and c = 30, s = `(72)/(2)` = 36
Area
= `sqrt(36(36 - 18)(36 - 24)(36 - 30)`
= `sqrt(36(18)(12)(6)`
= 21.6cm2
Let the length of the perpendicular of the triangle to the side 15cm be h cm i.e. height = h cm
We also know that, Area of a Triangle = `(1)/(2)"b.h" "i.e." (1)/(2)("Base" xx "Height")`
Area of a Traingle with base = 30cm and height = h cm
⇒ `(1)/(2)15."h"` = 216cm2
⇒ h = `(216 xx 2)/(30)`
= 14.4cm.
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