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 rectangle, so the area is half of the area of its rectangle with lengths and height b and h Is there any other reasoning?


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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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