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 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 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:

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:

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.

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.

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

From Perimeter and Area of a Triangle to HOME PAGE