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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

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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=(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?
$$\mathrm{For}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{perpandicular} \\ $$$$\mathrm{height}\:{h}\:\mathrm{and}\:\mathrm{base}\:\mathrm{length}\:{b},\:\mathrm{the}\: \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}: \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}{hb} \\ $$$$ \\ $$$$\mathrm{Why}\:\mathrm{is}\:\mathrm{this}\:\mathrm{the}\:\mathrm{case}? \\ $$$$\mathrm{I}\:\mathrm{understand}\:\mathrm{that}\:\mathrm{two}\:\mathrm{identicle}\:\mathrm{triangles} \\ $$$$\mathrm{can}\:\mathrm{construct}\:\mathrm{a}\:{rectangle},\:\mathrm{so}\:\mathrm{the}\:\mathrm{area} \\ $$$$\mathrm{is}\:\mathrm{half}\:\mathrm{of}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{its}\:\mathrm{rectangle}\:\mathrm{with} \\ $$$$\mathrm{lengths}\:\mathrm{and}\:\mathrm{height}\:{b}\:\mathrm{and}\:{h} \\ $$$$ \\ $$$$\mathrm{Is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{other}\:\mathrm{reasoning}? \\ $$
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
$$\mathrm{from}\:\mathrm{right}\:\mathrm{angled}\:\mathrm{triangle}\:\mathrm{you}\:\mathrm{can}\:\mathrm{construct}\:\mathrm{any} \\ $$$$\mathrm{triangle},\:\mathrm{so}\:\mathrm{generating}\:\mathrm{a}\:\mathrm{formula}\:\mathrm{to}\:\mathrm{a} \\ $$$$\mathrm{right}\:\mathrm{angled}\:\mathrm{triangle}\:\mathrm{will}\:\mathrm{be}\:\mathrm{sulficient} \\ $$
Answered by prakash jain last updated on 09/Dec/15
I think all formulas for area are derived from unit square→rectange → integral.
$$\mathrm{I}\:\mathrm{think}\:\mathrm{all}\:\mathrm{formulas}\:\mathrm{for}\:\mathrm{area}\:\mathrm{are}\:\mathrm{derived}\:\mathrm{from} \\ $$$$\mathrm{unit}\:\mathrm{square}\rightarrow\mathrm{rectange}\:\rightarrow\:\mathrm{integral}. \\ $$
Answered by Filup last updated on 09/Dec/15
y=mx+b let y_1 and y_2 constuct a triangle with the x−axis. y_1 =m_1 x through (0, 0) y_2 =m_2 x+b y_1 and y_2 intersect at: x=(b/(m_1 −m_2 ))=t y_1 intersects at x=0 y_2 intersects at x=−(b/m_2 )=j ∴A=∫_0 ^( t) y_1 dx+∫_t ^( j) y_2 dx can this explain the solution?
$${y}={mx}+{b} \\ $$$$ \\ $$$$\mathrm{let}\:{y}_{\mathrm{1}} \:\mathrm{and}\:{y}_{\mathrm{2}} \:\mathrm{constuct}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with} \\ $$$$\mathrm{the}\:{x}−\mathrm{axis}. \\ $$$$ \\ $$$${y}_{\mathrm{1}} ={m}_{\mathrm{1}} {x}\:\:\:\:\:\:{through}\:\left(\mathrm{0},\:\mathrm{0}\right) \\ $$$${y}_{\mathrm{2}} ={m}_{\mathrm{2}} {x}+{b} \\ $$$$ \\ $$$${y}_{\mathrm{1}} \:\mathrm{and}\:{y}_{\mathrm{2}} \:\mathrm{intersect}\:\mathrm{at}: \\ $$$${x}=\frac{{b}}{{m}_{\mathrm{1}} −{m}_{\mathrm{2}} }={t} \\ $$$${y}_{\mathrm{1}} \:\mathrm{intersects}\:\mathrm{at}\:{x}=\mathrm{0} \\ $$$${y}_{\mathrm{2}} \:\mathrm{intersects}\:\mathrm{at}\:{x}=−\frac{{b}}{{m}_{\mathrm{2}} }={j} \\ $$$$ \\ $$$$\therefore{A}=\int_{\mathrm{0}} ^{\:{t}} {y}_{\mathrm{1}} \:{dx}+\int_{{t}} ^{\:{j}} {y}_{\mathrm{2}} \:{dx} \\ $$$$ \\ $$$$\mathrm{can}\:\mathrm{this}\:\mathrm{explain}\:\mathrm{the}\:\mathrm{solution}? \\ $$
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 ∫_(−3) ^3 ydx=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.
$$\mathrm{a}.\:\mathrm{While}\:\mathrm{computing}\:\mathrm{area}\:\mathrm{of}\:\mathrm{traingle}\:\mathrm{with}\:{y}_{\mathrm{1}} ,{y}_{\mathrm{2}} \\ $$$$\mathrm{and}\:{y}=\mathrm{0}\:\mathrm{You}\:\mathrm{need}\:\mathrm{to}\:\mathrm{compute}\:\mathrm{area}\:\mathrm{under}\:\mathrm{two} \\ $$$$\mathrm{lines}\:\mathrm{and}\:\mathrm{subtract}\:\mathrm{second}\:\mathrm{from}\:\mathrm{the}\:\mathrm{first}. \\ $$$$\mathrm{b}.\:\mathrm{Also}\:\mathrm{signed}\:\mathrm{are}\:\mathrm{important}\:\mathrm{if}\:\mathrm{you}\:\mathrm{are} \\ $$$$\mathrm{interest}\:\mathrm{in}\:\mathrm{calculating}\:\mathrm{area}.\:\mathrm{Other}\:{y}={x} \\ $$$$\int_{−\mathrm{3}} ^{\mathrm{3}} {ydx}=\mathrm{0}.\:\mathrm{Depending}\:\mathrm{upon}\:\mathrm{where}\:\mathrm{you}\:\mathrm{want} \\ $$$$\mathrm{to}\:\mathrm{use}\:\mathrm{this}\:\mathrm{result}.\:\mathrm{It}\:\mathrm{maynot}\:\mathrm{be}\:\:\mathrm{what}\:\mathrm{you} \\ $$$$\mathrm{are}\:\mathrm{expecting}. \\ $$$$\mathrm{Integrals}\:\mathrm{can}\:\mathrm{also}\:\mathrm{give}\:−\mathrm{ve}\:\mathrm{result}. \\ $$$$\mathrm{Your}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{A}\:\mathrm{will}\:\mathrm{not}\:\mathrm{give}\:\mathrm{area}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{triangle}\:{y}_{\mathrm{1}} ,\:{y}_{\mathrm{2}} \:\mathrm{and}\:{x}=\mathrm{0}. \\ $$$$\mathrm{Also}\:\mathrm{integrals}\:\mathrm{as}\:\mathrm{area}\:\mathrm{is}\:\mathrm{also}\:\mathrm{based}\:\mathrm{and} \\ $$$$\mathrm{rectangle}\:\mathrm{area}\:\mathrm{formula}. \\ $$
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∈[0,3] the function is positive at x∈[−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
$$\mathrm{this}\:\mathrm{is}\:\mathrm{because}\:\mathrm{actualy}\:\mathrm{area}\:\mathrm{in}\:\mathrm{the}\:\mathrm{ontegral} \\ $$$$\mathrm{take}\:\mathrm{the}\:\mathrm{sign} \\ $$$$\mathrm{for}\:{y}={x} \\ $$$$\mathrm{at}\:{x}\in\left[\mathrm{0},\mathrm{3}\right]\:\mathrm{the}\:\mathrm{function}\:\mathrm{is}\:\mathrm{positive} \\ $$$$\mathrm{at}\:{x}\in\left[−\mathrm{3},\mathrm{0}\right]\:\mathrm{the}\:\mathrm{function}\:\mathrm{is}\:\mathrm{negative} \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{area}\:\mathrm{in}\:\mathrm{the}\:\left[\mathrm{0},\mathrm{3}\right]\:\mathrm{is}\:\mathrm{positive}\:\mathrm{and} \\ $$$$\mathrm{in}\:\left[−\mathrm{3},\mathrm{0}\right]\:\mathrm{the}\:\mathrm{area}\:\mathrm{is}\:\mathrm{negative} \\ $$$$\mathrm{them}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{givd}\:\mathrm{the}\:\mathrm{diference}\:\mathrm{of} \\ $$$$\mathrm{area}\:\mathrm{over}\:\mathrm{axis}\:{x}\:\mathrm{and}\:\mathrm{under}\:\mathrm{axis}\:{x} \\ $$$$\mathrm{if}\:\mathrm{actualy}\:\mathrm{you}\:\mathrm{want}\:\mathrm{the}\:\mathrm{total}\:\mathrm{are}\:\mathrm{just} \\ $$$$\mathrm{take}\:\mathrm{integral}\:\mathrm{of}\:\mid{y}\mid\:\mathrm{instead},\:\mathrm{wich}\:\mathrm{solve} \\ $$$$\mathrm{the}\:\mathrm{problem}\:\mathrm{of}\:\mathrm{sign} \\ $$

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