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Question Number 113092 by bobhans last updated on 11/Sep/20

What is the area of a tringle where the sides of triangle are 91 cm, 98 cm, and 105 cm

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{tringle}\:\mathrm{where}\:\mathrm{the} \\ $$$$\mathrm{sides}\:\mathrm{of}\:\mathrm{triangle}\:\mathrm{are}\:\mathrm{91}\:\mathrm{cm},\:\mathrm{98}\:\mathrm{cm},\:\mathrm{and}\:\mathrm{105}\:\mathrm{cm} \\ $$

Answered by bemath last updated on 11/Sep/20

question. What is the area of triangle where the sides of triangle are 91 cm, 98 cm and 105 cm? area = ((√(4a^2 b^2 −(a^2 +b^2 −c^2 )^2 ))/4) where a,b,c are the length sides of △. hence area = ((√(4×91^2 ×98^2 −(91^2 +98^2 −105^2 )^2 ))/4) = 4116 cm^2

$$\mathrm{question}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of} \\ $$$$\mathrm{triangle}\:\mathrm{where}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{triangle} \\ $$$$\mathrm{are}\:\mathrm{91}\:\mathrm{cm},\:\mathrm{98}\:\mathrm{cm}\:\mathrm{and}\:\mathrm{105}\:\mathrm{cm}? \\ $$$$\mathrm{area}\:=\:\frac{\sqrt{\mathrm{4a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} −\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\mathrm{4}} \\ $$$$\mathrm{where}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{are}\:\mathrm{the}\:\mathrm{length}\:\mathrm{sides}\:\mathrm{of}\:\bigtriangleup. \\ $$$$\mathrm{hence}\:\mathrm{area}\:=\:\frac{\sqrt{\mathrm{4}×\mathrm{91}^{\mathrm{2}} ×\mathrm{98}^{\mathrm{2}} −\left(\mathrm{91}^{\mathrm{2}} +\mathrm{98}^{\mathrm{2}} −\mathrm{105}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{4116}\:\mathrm{cm}^{\mathrm{2}} \\ $$

Commented by bobhans last updated on 11/Sep/20

waw....the new formula

$$\mathrm{waw}....\mathrm{the}\:\mathrm{new}\:\mathrm{formula} \\ $$

Commented by MJS_new last updated on 11/Sep/20

the formula is not new, it′s Heron′s Formula usually we write it Area=((√((a+b+c)(a+b−c)(a+c−b)(b+c−a)))/4)

$$\mathrm{the}\:\mathrm{formula}\:\mathrm{is}\:\mathrm{not}\:\mathrm{new},\:\mathrm{it}'\mathrm{s}\:\mathrm{Heron}'\mathrm{s}\:\mathrm{Formula} \\ $$$$\mathrm{usually}\:\mathrm{we}\:\mathrm{write}\:\mathrm{it} \\ $$$${Area}=\frac{\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}+{c}−{b}\right)\left({b}+{c}−{a}\right)}}{\mathrm{4}} \\ $$

Commented by bobhans last updated on 11/Sep/20

Archimedes Theorem 16(area)^2 =4a^2 b^2 −(c^2 −a^2 −b^2 )^2 area = (√((4a^2 b^2 −(c^2 −a^2 −b^2 )^2 )/(16)))

$$\mathrm{Archimedes}\:\mathrm{Theorem} \\ $$$$\mathrm{16}\left(\mathrm{area}\right)^{\mathrm{2}} =\mathrm{4a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} −\left(\mathrm{c}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\mathrm{area}\:=\:\sqrt{\frac{\mathrm{4a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} −\left(\mathrm{c}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{16}}} \\ $$

Commented by bemath last updated on 11/Sep/20

not same sir. i think it from Archimedes theorem. sorry if i wrong. hahaha

$$\mathrm{not}\:\mathrm{same}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{think}\:\mathrm{it}\:\mathrm{from}\: \\ $$$$\mathrm{Archimedes}\:\mathrm{theorem}. \\ $$$$\mathrm{sorry}\:\mathrm{if}\:\mathrm{i}\:\mathrm{wrong}.\:\mathrm{hahaha} \\ $$

Commented by MJS_new last updated on 11/Sep/20

(a+b+c)(a+b−c))a+c−b)(b+c−a)= =−(a^4 +b^4 +c^4 )+2(a^2 b^2 +a^2 c^2 +b^2 c^2 )= 4a^2 b^2 −(a^2 +b^2 −c^2 )^2 = =4a^2 b^2 −(a^4 +b^4 +c^4 +2(a^2 b^2 −a^2 c^2 −b^2 c^2 ))= =−(a^4 +b^4 +c^4 )+2(a^2 b^2 +a^2 c^2 +b^2 c^2 )

$$\left.\left(\left.{a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\right){a}+{c}−{b}\right)\left({b}+{c}−{a}\right)= \\ $$$$=−\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} \right)+\mathrm{2}\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} \right)= \\ $$$$\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} = \\ $$$$=\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +\mathrm{2}\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} −{a}^{\mathrm{2}} {c}^{\mathrm{2}} −{b}^{\mathrm{2}} {c}^{\mathrm{2}} \right)\right)= \\ $$$$=−\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} \right)+\mathrm{2}\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} \right) \\ $$

Answered by malwaan last updated on 11/Sep/20

((91+98+105)/2) = ((294)/2) = 147 so the area = (√(147(147−91)(147−98)(147−105))) = (√(147×56×49×42)) = (√(16941456)) = 4116

$$\frac{\mathrm{91}+\mathrm{98}+\mathrm{105}}{\mathrm{2}}\:=\:\frac{\mathrm{294}}{\mathrm{2}}\:=\:\mathrm{147} \\ $$$${so}\:{the}\:{area}\:= \\ $$$$\sqrt{\mathrm{147}\left(\mathrm{147}−\mathrm{91}\right)\left(\mathrm{147}−\mathrm{98}\right)\left(\mathrm{147}−\mathrm{105}\right)} \\ $$$$=\:\sqrt{\mathrm{147}×\mathrm{56}×\mathrm{49}×\mathrm{42}} \\ $$$$=\:\sqrt{\mathrm{16941456}}\:=\:\mathrm{4116} \\ $$

Commented by bemath last updated on 11/Sep/20

same sir with my answer. please check

$$\mathrm{same}\:\mathrm{sir}\:\mathrm{with}\:\mathrm{my}\:\mathrm{answer}. \\ $$$$\mathrm{please}\:\mathrm{check} \\ $$

Commented by malwaan last updated on 11/Sep/20

yes and thanks for this new formula

$${yes} \\ $$$${and}\:{thanks}\:{for}\:{this}\:{new}\:{formula} \\ $$

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