i think your answer to the altitude of an isosceles triangle is wrong. kindly check so it will not confused your reader. thank you and godbless

Altitude or height of a triangle is the perpendicular line drawn from the vertex of a triangle to its opposite side. It makes a right angle with the base of the triangle.

Each triangle has three possible altitudes. Different triangles have different types of altitudes. They can be found either inside a triangle (as in acute triangles) or outside (as in obtuse triangles) or can be one of the three sides (as in right triangles).

The altitude of a triangle can be calculated using the formula given below:

Derivation

The formula to calculate the altitude of a triangle can be derived from the standard formula of area of a triangle as shown below:

As we know,

Area (A) = ½ (b x h), here b = base, h = altitude

=> 2A = b x h

=> h = 2A/b

Hence, mathematically, altitude of a triangle can also be defined as twice the area divided by the base of the triangle.

Although we can use the above formula to determine the altitude for all types of triangles, some specific triangles such as equilateral triangle, isosceles triangle, and right triangle have specific formulas to determine their altitude.

The altitude of an equilateral triangle can be determined using the formula given below:

Derivation

To derive the formula of altitude of an equilateral triangle, two different methods can be used. They are given below:

Using Trigonometric Function

In the given △ABC, AB = BC = AC, and AE is the altitude that divides the base BC equally into BE and EC.

In △ABC,

sin60° = Perpendicular/Hypotenuse

√3/2 = h/a [∵ sin60 = √3/2]

h = √3/2a

Using Heron’s Formula

We can also find the altitude of an equilateral triangle by using the Heron’s formula for finding area of a triangle:

Area (A) = √s(s–a)(s–b)(s–c), here semiperimeter(s) = ½(a+b+c)

In an equilateral triangle, all the three sides are equal, thus s = 3a/2

Putting the value of ‘s’ in the standard formula h = 2A/b, we get,

h = 2√s(s–a)(s–b)(s–c)/b …..(1)

Since in equilateral triangle, a = b

Eqn (1) can be written as,

h = 2√s(s–a)(s–b)(s–c)/a …..(2)

Substituting the value of ‘s’ in (2), we get

h = 2/a√3a/2(3a/2 – a) (3a/2 – a) (3a/2 – a)

h = 2/a√3a/2x a/2xa/2xa/2

h = 2/axa2√3/4

h = a√3/2

The altitude of a scalene triangle can be determined using the formula given below:

Derivation

To derive the formula for altitude of a scalene triangle, we use the Heron’s formula for finding area of a triangle:

Area (A) = √s(s–a)(s–b)(s–c) ……(1), here semiperimeter(s) = ½ (a+b+c)

Now, the general formula for area of triangle is given as:

Area (A) = ½ (b x h) …… (2), here b = base, h = altitude

Equating (1) and (2) we get,

½ (b x h) = √s(s–a)(s–b)(s–c)

h = 2√s(s–a)(s–b)(s–c)/b

The altitude of an isosceles triangle can be determined using the formula given below:

Derivation

The altitude of an isosceles triangle bisects its base. This property is used to drive the formula for calculating the altitude of an isosceles triangle.

Use our isosceles triangle image and see attachment

By Pythagoras theorem in △AEB, we get

AE2= AB2 – BE2 …..(1)

Since AE is the bisector of side BC, dividing BC equally into BE and EC

Thus,

BE = ½ X BC (b)…… (2)

Substituting the value of BE in equation (1) we get,

AE2= AB2 – BE2

h2 = a2 – (1/2 x b)2

h = √a2 – b2/4

The altitude of a right triangle can be determined using the formula given below:

Derivation

The altitude of a right triangle divides the existing triangle into two similar triangles. To derive the formula for altitude of a right triangle, we use the ‘Right Triangle Altitude Theorem’ which states that: ‘the measure of the altitude drawn from the vertex to its hypotenuse is the geometric mean of the measure of the two segments formed when the altitude divides the hypotenuse’.

If two triangles are similar, then

i) Corresponding angles are equal

ii) Corresponding sides are equal

In the given figure, the altitude AE is drawn from the vertex A to the hypotenuse BC, forming two similar triangles △BEA and △CEA

∠BEA = ∠CEA = 90°

As △ABC is right angle at A

∠A = α + β = 90° …… (1)

In △BEA,

∠ABE + ∠EAB + ∠ABE = 180° (Angle Sum Property)

∠ABE = 180° – (90° + β) = 90° – β = α (From Eqn 1)

Similarly,

In △CEA,

∠CEA + ∠EAC + ∠ACE = 180° (Angle Sum Property)

∠ACE = 180° – (90° + α) = 90° – α = β (From Eqn 1)

By similarity theorem,

△BEA ≅ △CEA

AE/CE = BE/AE

h/y = x/h

h2 = xy

h = √xy

In an obtuse triangle, the altitude lies outside the triangle. The base is extended and the altitude is drawn from the opposite vertex to its base. The altitude of an obtuse triangle can be determined using the formula given below:

To find the altitude of a triangle, we first need to identify the type of triangle. After the identification of the type, we use the specific formulas given above for each type to find the value of the altitude.

Let us solve some examples to understand the concepts better.

Find the altitude of an equilateral triangle with sides measuring 6 cm.

As we know,
h = √3/2a, here a = 6 cm
= √3/2 x 6
= 3√3 [∵√3 ~ 1.732]
= 5.196

Calculate the altitude of a scalene triangle having sides 5 cm, 6 cm, and 7 cm.

As we know,
h = 2√s(s–a)(s–b)(s–c)/b, here a = 5 cm, b = 6 cm, and c = 7 cm
In the given triangle,
s = (5+6+7)/2 = 9 cm
Thus,
h = 2√9(9-5)(9-6)(9-7)/6
= 2√216/6
= 2√36
=2√6 x 6
= 12 cm

Find the altitude of an isosceles triangle with sides measuring 6 cm, 4 cm and 6 cm.

As we know,
h = √a2 – b2/4, here a = 6 cm, b = 4 cm
= √62 – ¼(4)2
= √32 cm

Find the altitude of a right triangle that is dividing its base of 12 cm into 8 cm and 4 cm.

As we know,
h = √xy, here x = 8 cm, y = 4 cm
= √8 x 4
= 5.65 cm

Calculate the length of the altitude of the given obtuse triangle with sides measuring 4 m, 5m, and 6m.

As we know,
h = 2√s(s–a)(s–b)(s–c)/b, here a = 4 cm, b = 5 cm, and c = 6 cm
In the given triangle,
s = (4+5+6)/2 = 7.5 cm
Thus,
h = 2√7.5(7.5-4)(7.5-5)(7.5-6)/5
= 2√(7.5 x 3.5 x 2.5 x 1.5)/5
= 2√19.68
= 8.87 cm

An altitude is the perpendicular distance from the base to the opposite vertex. It can be found either outside or inside a triangle. In contrast the median of a triangle is the line segment drawn from the vertex to the opposite side that divides a triangle into two equal parts. The median bisects the triangle formed at the vertex from where it is drawn and the base of the triangle. It always lies inside the triangle.