Perimeter and Area of a Triangle

Here we will discuss about the perimeter and area of a triangle and some of its geometrical properties.

Perimeter, Area and Altitude of a Triangle:

Perimeter, Area and Altitude of a Triangle

Perimeter of a triangle (P) = Sum of the sides = a + b + c





Semiperimeter of a triangle (s) = \(\frac{1}{2}\)(a + b + c)

Area of a triangle (A) = \(\frac{1}{2}\) × base × altitude = \(\frac{1}{2}\)ah

Here any side can be taken as base; the length of the perpendicular from the corresponding vertex to this side is the altitude.

Area = \(\sqrt{\textrm{s(s - a)(s - b)(s - c)}}\)  (Heron’s formula)

Altitude (h) = \(\frac{\textrm{area}}{\frac{1}{2} \times \textrm{base}}\) = \(\frac{2\triangle}{a}\)


Solved Example on Finding the Perimeter, Semiperimeter and Area

 of a Triangle: 

The sides of a triangle are 4 cm, 5 cm and 7 cm. Find its perimeter, semiperimeter and area.

Solution:

Perimeter of a triangle (P) = Sum of the sides

                                      = a + b + c

                                      = 4 cm + 5 cm + 7 cm

                                      = (4 + 5 + 7) cm

                                      = 16 cm


Semiperimeter of a triangle (s) = \(\frac{1}{2}\)(a + b + c)

                                             = \(\frac{1}{2}\)(4 cm + 5 cm + 7 cm)

                                             = \(\frac{1}{2}\)(4 + 5 + 7) cm

                                             = \(\frac{1}{2}\) × 16 cm

                                             = 8 cm

Area of a triangle = \(\sqrt{\textrm{s(s - a)(s - b)(s - c)}}\) 

                          = \(\sqrt{\textrm{8(8 - 4)(8 - 5)(8 - 7)}}\) cm\(^{2}\)

                          = \(\sqrt{\textrm{8 × 4 × 3 × 1}}\) cm\(^{2}\)

                          = \(\sqrt{96}\) cm\(^{2}\)

                          = \(\sqrt{16 × 6}\) cm\(^{2}\)

                          = 4\(\sqrt{6}\) cm\(^{2}\)

                          = 4 × 2.45 cm\(^{2}\)

                          = 9.8 cm\(^{2}\)


Perimeter, Area and Altitude of an Equilateral Triangle:

Perimeter, Area and Altitude of an Equilateral Triangle

Perimeter of an equilateral triangle (P) = 3 × side = 3a

Area of an equilateral triangle (A) = \(\frac{√3}{4}\) × (side)\(^{2}\) = \(\frac{√3}{4}\) a\(^{2}\)

Altitude of an equilateral triangle (h) = \(\frac{√3}{4}\) a


Trigonometric formula for area of a triangle:

Trigonometric Formula for Area of a Triangle

Area of ∆ABC = \(\frac{1}{2}\) × ca sin B

                    = \(\frac{1}{2}\) × ab sin C

                    = \(\frac{1}{2}\) × bc sin A

(since, ∆ = \(\frac{1}{2}\) ah = \(\frac{1}{2}\) ca ∙ \(\frac{h}{c}\) = \(\frac{1}{2}\) ca sin B, etc.)


Solved Example on Finding the Area of a Triangle: 

In a ∆ABC, BC = 6 cm, AB = 4 cm and ∠ABC = 60°. Find its area.

Solution:

Area of ∆ABC = \(\frac{1}{2}\) ac sin B = \(\frac{1}{2}\) × 6 × 4 sin 60° cm\(^{2}\)

                    = \(\frac{1}{2}\) × 6 × 4 × \(\frac{√3}{2}\) cm\(^{2}\)

                    = 6√3 cm\(^{2}\)

                    = 6 × 1.73 cm\(^{2}\)

                    = 10.38 cm\(^{2}\)


Some geometrical properties of an isosceles triangle:

Geometrical Properties of an Isosceles Triangle

In the isosceles ∆PQR, PQ = PR, QR is the base, and PT is the altitude.

Then, ∠PTR = 90°, QT = TR, PT\(^{2}\) + TR\(^{2}\) = PR\(^{2}\) (by Pythagoras’ theorem)

 ∠PQR = ∠PRQ, ∠QPT = ∠RPT.


Some geometrical properties of a right-angled triangle:

In the right-angled ∆PQR, ∠PQR = 90°; PQ, QR are the sides (forming the right angle) and PR is the hypotenuse.

Geometrical Properties of a Right-angled Triangle

Then, PQ ⊥ QR (therefore, if QR is the base, PQ is the altitude).

PQ\(^{2}\) + QR\(^{2}\) = PR\(^{2}\) (by Pythagoras’ theorem)

Area of the ∆PQR = \(\frac{1}{2}\) ∙ PQ ∙ QR

⟹ PQ ∙ QR = 2 × area of the ∆PQR.

Again, area of the ∆PQR = \(\frac{1}{2}\) ∙ QT ∙ PR

⟹ QT ∙ PR = 2 × area of the ∆PQR.

Therefore, PQ ∙ QR = QT ∙ PR = 2 × Area of the ∆PQR.


Solved Examples on Perimeter and Area of a Triangle:

1. Find the perimeter of an equilateral triangle whose area is equal to that of a triangle with sides 21 cm, 16 cm and 13 cm.

Solution:

Let a side of the equilateral triangle = x.

Then, its area = \(\frac{√3}{4}\) x\(^{2}\)

Now, the area of the other triangle = \(\sqrt{\textrm{s(s - a)(s - b)(s - c)}}\) 

Here, s = \(\frac{1}{2}\) (a + b + c)

           = \(\frac{1}{2}\) (21 + 16 + 13) cm

           = \(\frac{1}{2}\) 50 cm

           = 25 cm

Therefore, area of the other triangle = \(\sqrt{\textrm{25(25 - 21)(25 - 16)(25 - 13)}}\) cm\(^{2}\)

                                                     = \(\sqrt{\textrm{25 ∙ 4 ∙ 9 ∙ 12}}\) cm\(^{2}\)

                                                     = 60\(\sqrt{\textrm{3}}\) cm\(^{2}\)

According to the question, \(\frac{√3}{4}\) x\(^{2}\) = 60\(\sqrt{\textrm{3}}\) cm\(^{2}\)

⟹ x\(^{2}\) = 240 cm\(^{2}\)

Therefore, x = 4√15 cm


2. PQR is an isosceles triangle whose equal sides PQ and PR are 10 cm each, and the base QR measures 8 cm. PM is the perpendicular from P to QR and X is a point on PM such that ∠QXR = 90°. Find the area of the shaded portion.

Solved Examples on Perimeter and Area of a Triangle

Solution:

Since PQR is an isosceles triangle and PM ⊥ QR, QR is bisected at M.

Therefore, QM = MR = \(\frac{1}{2}\) QR = \(\frac{1}{2}\) × 8 cm = 4 cm

Now, PQ\(^{2}\) = PM\(^{2}\) + QM\(^{2}\) (by Pythagoras’ theorem)

Therefore, 10\(^{2}\) cm\(^{2}\) = PM\(^{2}\) + 4\(^{2}\) cm\(^{2}\)

or, PM\(^{2}\) = 10\(^{2}\) cm\(^{2}\) - 4\(^{2}\) cm\(^{2}\)

                       = 100 cm\(^{2}\) - 16 cm\(^{2}\)

                       = (100 - 16) cm\(^{2}\)

                       = 84 cm\(^{2}\)

Therefore, PM\(^{2}\) = 2√21 cm

Therefore, area of the ∆PQR = \(\frac{1}{2}\) × base × altitude

                                         = \(\frac{1}{2}\) × QR × PM

                                         = (\(\frac{1}{2}\) × 8 × 2√21) cm\(^{2}\)

                                         = 8√21) cm\(^{2}\)

From geometry, ∆XMQ ≅ ∆XMR (SAS criterion)

We get, XQ =XR = a (say)

Therefore, from the right-angled ∆QXR, a\(^{2}\) + a\(^{2}\) = QR\(^{2}\)

or, 2a\(^{2}\) = 8\(^{2}\) cm\(^{2}\)

or, 2a\(^{2}\) = 64 cm\(^{2}\)

or, a\(^{2}\) = 32 cm\(^{2}\)

Therefore, a = 4√2 cm

Again, area of the ∆XQR = \(\frac{1}{2}\) × XQ × XR

                                    = \(\frac{1}{2}\) × a × a

                                    = \(\frac{1}{2}\) × 4√2 cm  × 4√2 cm

                                    = \(\frac{1}{2}\) × (4√2)\(^{2}\) cm\(^{2}\)

                                    = \(\frac{1}{2}\) × 32 cm\(^{2}\)  

                                    = 16 cm\(^{2}\)  

Therefore, area of the shaded portion = area of the ∆PQR - area of the ∆XQR

                                                      = (8√21) cm\(^{2}\) - 16 cm\(^{2}\)

                                                      = (8√21 - 16) cm\(^{2}\)  

                                                      = 8(√21 - 2) cm\(^{2}\)  

                                                      = 8 × 2.58 cm\(^{2}\)  

                                                      = 20.64 cm\(^{2}\)  










9th Grade Math

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