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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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