# By heron's formula area of triangle is?

**Asked by: Kimberly Richards**| Last update: 18 June 2021

Score: 4.3/5 (46 votes)

Another is **Heron's formula** which gives the **area** in terms of the three sides of the **triangle**, specifically, as the square root of the product s(s – a)(s – b)(s – c) where s is the semiperimeter of the **triangle**, that is, s = (a + b + c)/2.

Regarding this, Which of the following is the Heron's formula to find the area of a triangle of side a B and C and semi-perimeter s?

Hence,

**area of a triangle**by

**heron's formula**is A=

**s**(

**s**−a)(

**s**−

**b**)(

**s**−

**c**) .

Simply so, How do you use Heron's area formula?.

**Use Heron's formula**to find the

**area**of triangle ABC, if AB=3,BC=2,CA=4 . Substitute S into the

**formula**. Round answer to nearest tenth. Since

**Heron's formula**relates the side lengths, perimeter and

**area**of a triangle, you might need to answer more challenging question types like the following example.

Moreover, What is Heron's formula?

In geometry,

**Heron's formula**(sometimes called Hero's

**formula**), named after Hero of Alexandria, gives the area of a triangle when the length of all three sides are known. Unlike other triangle area

**formulae**, there is no need to calculate angles or other distances in the triangle first.

Why we use Heron's formula?

**Heron's formula**is

**used**to find the area of a triangle when

**we**know the length of all its sides. It is also termed as Hero's

**Formula**.

**We**don't have to

**need**to know the angle measurement of a triangle to calculate its area.

**31 related questions found**

### Does Heron's formula work for all triangles?

**Heron's formula** is used to determine the area of **triangles** when lengths of **all** their sides are given or the area of quadrilaterals. We also know it as Hero's **formula**. This **formula** for finding the area **does** not depend on the angles of a **triangle**. It solely depends on the lengths of **all** sides of **triangles**.

### Is Heron's Formula accurate?

**Heron's formula** computes the area of a triangle given the length of each side. If you have a very thin triangle, one where two of the sides approximately equal s and the third side is much shorter, a direct implementation **Heron's formula** may not be **accurate**.

### Who found Heron's formula?

**Heron's formula**, **formula** credited to **Heron** of Alexandria (c. 62 ce) for finding the area of a triangle in terms of the lengths of its sides.

### What is the area of the triangle?

The **area** of a **triangle** is defined as the total region that is enclosed by the three sides of any particular **triangle**. Basically, it is equal to half of the base times height, i.e. A = 1/2 × b × h.

### How do you prove Heron's formula using trigonometry?

Proof. The following proof is **trigonometric**, and basically uses the cosine rule. First we compute the cosine squared in terms of the sides, and then the sine squared which we use in the **formula** A=1/2bc·sinA to derive the area of the triangle in terms of its sides, and thus **prove Heron's formula**.

### Can you use Heron's formula for right angle triangle?

Area of the **right angled triangle can** be calculated by two ways that id with the help of half of height multiplied by base **formula** and **we can** also **calculate** the value of the area **using heron's formula**. In **heron's formula**, Area =√s(s−a)(s−b)(s−c) where s is the sum of all the sides of a **triangle** divided by 2.

### What is the area of the right triangle?

**Area** of **Right Triangle** Formula

So the **area** of a **right triangle** is obtained by multiplying its base and height and then making the product half. The **area** of a **right triangle** with base 6 cm and height 4 cm is 1/2 × 6 × 4 = 12 cm^{2}.

### Who invented area?

In the 5th century BCE, Hippocrates of Chios was the first to show that the **area** of a disk (the region enclosed by a circle) is proportional to the square of its diameter, as part of his quadrature of the lune of Hippocrates, but did not identify the constant of proportionality.