Table of Contents

Last modified on September 6th, 2022

Base meaning bottom, it refers to any side of a triangle, which is perpendicular to its height or altitude. For scalene or equilateral triangle, any side can be the base, while for an isosceles triangle it is the single, unequal side.

The formula for base 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 = height

=> 2A = b x h

=> **b = 2A/h**

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

Let us solve some problems to derive the base of the given triangles.

**Problem: Finding the base of a triangle, when the HEIGHT and the AREA are known.**

**Find the base of a triangle with area 12 cm ^{2 }and height 8 cm**.

Solution:

As we know,

b = 2A/h, here area = 12cm^{2}, h = 8 cm

= 2 x 12/8

= 3 cm

**Problem: Finding the base of a right triangle, when the HEIGHT and the HYPOTENUSE are known.**

**Identify the base of a right triangle having height 6 cm and hypotenuse 9 cm.**

Solution:

Here, we will use the Pythagorean Theorem,

(Hypotenuse)^{2 }= (Base)^{2} + (Height)^{2}, here height = 6 cm and hypotenuse = 9 cm

(6)^{ 2} = (b)^{2} + (3)^{2}

(b)^{2} = (6)^{ 2} – (3)^{2}

b^{2} = 36- 9

b = **√**25 = 5 cm.

**Problem: Finding the base of an isosceles triangle, when the SIDES and the HEIGHT are known.**

**Find the missing base of the given isosceles triangle with two sides measuring 5 cm each and having height of 3 cm.**

Solution:

Here, we will use the Pythagorean Theorem,

In △ABC,

Since, the height AE divides the △ABC into two similar right triangles, △ABE and △ACE and the base BC into BE and EC equally

Such that,

BC = BE + EC

Now,

According to the Pythagoras theorem,

(Hypotenuse)^{2 }= (Base)^{2} + (Height)^{2}

In △ABE,

(AB)^{2} = (BE)^{2} + (AE)^{2}, here AB = 3 cm, AE = 5 cm

(BE)^{2}= (AB)^{2} – (AE)^{2}

(BE)^{2 }= (5)^{ 2} – (3)^{2}

(BE)^{2 }= 25 – 9

(BE)^{2 }= 16

BE = 4 cm

Similarly, EC = 4 cm

Thus,

BC = BE + EC = 8 cm

**Find the base of the given isosceles triangle with an area of 60 cm ^{2 }and the length of one of the equal sides is 13 cm.**

Solution:

Let the base = 2x and height = h

As we know,

Area of △ABC = ½ (b x h) = 60 cm^{2}

⇒½(b x h) = 60, here b = 2x

⇒ ½ x 2x x h = 60

⇒ xh = 60 …… (1)

Now, using Pythagorean theorem in △ABD, we get

AB^{2} = AD^{2} + BD^{2}

13^{2} = h^{2} + x^{2}

⇒ h^{2} + x^{2} = 169

Adding 2hx on both sides we get,

⇒ h^{2} + x^{2} + 2hx = 169 + 2hx

⇒ (h + x)^{2} = 169 + 2 x 60

⇒ (h + x)^{2} = 289

⇒ h + x = 17

⇒ h = 17 – x …… (2)

Substituting (2) in (1), we get,

x (17 – x) = 60

17x – x^{2} = 60

⇒ -x^{2} + 17x – 60 = 0

⇒ x^{2}– 17x + 60 = 0

⇒ x^{2}– 12x – 5x +60 = 0

⇒ x(x-12) – 5(x + 60) =0

⇒ (x-12) (x – 5) = 0

Either,

x -12 = 0

x = 12

Or,

x – 5 =0

x = 5

Hence, base of the given isosceles triangle is,

b = 2x = 2 x 5 =10 cm (if x = 5) or, 2x = 2 x 12 = 24 cm (if x = 12)

Last modified on September 6th, 2022