For-a-triangle-with-perpandicular-height-h-and-base-length-b-the-area-of-the-triangle-is-given-by-A-1-2-hb-Why-is-this-the-case-I-understand-that-two-identicle-triangles-can-construct-a-rectangl

Question Number 3280 by Filup last updated on 09/Dec/15

For a triangle with perpandicular

height h and base length b, the

area of the triangle is given by :

A = \frac{1}{2} h b

Why is this the case ?

I understand that two identicle triangles

can construct a r e c t a n g l e, so the area

is half of the area of its rectangle with

lengths and height b and h

Is there any other reasoning ?

Answered by 123456 last updated on 09/Dec/15

from right angled triangle you can construct any

triangle, so generating a formula to a

right angled triangle will be sulficient

Answered by prakash jain last updated on 09/Dec/15

I think all formulas for area are derived from

unit square \to rectange \to integral .

Answered by Filup last updated on 09/Dec/15

y = m x + b

let y_{1} and y_{2} constuct a triangle with

the x - axis .

y_{1} = m_{1} x t h r o u g h (0, 0)

y_{2} = m_{2} x + b

y_{1} and y_{2} intersect at :

x = \frac{b}{m_{1} - m_{2}} = t

y_{1} intersects at x = 0

y_{2} intersects at x = - \frac{b}{m_{2}} = j

∴ A = \int_{0}^{t} y_{1} d x + \int_{t}^{j} y_{2} d x

can this explain the solution ?

Commented by prakash jain last updated on 09/Dec/15

a . While computing area of traingle with y_{1}, y_{2}

and y = 0 You need to compute area under two

lines and subtract second from the first .

b . Also signed are important if you are

interest in calculating area . Other y = x

\int_{- 3}^{3} y d x = 0 . Depending upon where you want

to use this result . It maynot be what you

are expecting .

Integrals can also give - ve result .

Your formula for A will not give area of

the triangle y_{1}, y_{2} and x = 0 .

Also integrals as area is also based and

rectangle area formula .

Commented by 123456 last updated on 09/Dec/15

this is because actualy area in the ontegral

take the sign

for y = x

at x \in [0, 3] the function is positive

at x \in [- 3, 0] the function is negative

so the area in the [0, 3] is positive and

in [- 3, 0] the area is negative

them the integral givd the diference of

area over axis x and under axis x

if actualy you want the total are just

take integral of ∣ y ∣ instead, wich solve

the problem of sign

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