# 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$$^{2}$$

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