Question

1 The lengths of two sides of a triangle are 8 centimetres and 10 centimetres
and the angle between them is 140°. Calculate its area. What is the area
of the triangle with sides of the same length, but angle between them140°?​

Answer 1

$\huge{\boxed{\fcolorbox{cyan}{pink} {Answer}}}$

GIVEN ;!

• $\sf{Two \:sides \:of \:a \:triangle \:are \:8cm \:and \:10cm}$
• $\sf{\:Angle \:btw \:them = 140°}$

TO FIND

• $\sf{\:Calculate \:the \:area = ?¿}$
• $\sf{What \:is \:the \:area</li><li>\:of \:the \:triangle \:with \:sides \:of \:the \:same \:length, \:but \;angle \:between \:them \:140°}$

LET'S DO SOME CONSTRUCTION ;

• $\sf{\:Consider \:ABC \:is \:a \:triangle}$
• $\sf{AB = 10cm \:and \:AC = 8cm \:and \:∠BAC = 40°}$
• $\sf{Draw \:a \:perpendicular \:from \:C \:on \:AB = CH}$

$~~~~~~~~~~~~~~~$(refer a figure)

$\sf{\:From \:triangle \:ABC}$

$\sf{⇝CH = AC × \sin 40°}$

$\sf{⇝CH = 8 × \sin 40°}$

$\sf{(We \:know \sin 40 \:value = 0.642 \:from \:the \:table}$

$\sf{⇝CH = 8 × 0.642}$

$\sf\boxed{⇝CH = 5.136}$

$\sf{Time \:to \:find \:area ; }$

We know the formula ;

${\green{\overline{\green{\underline{\blue{\boxed{\orange{\mathtt{Area \:of \:triangle = \frac{1}{2} × base × height}}}}}}}}}$

$\sf{⇝Area \:(∆ABC) = \frac{1}{2} × AB × CH}$

$\sf{⇝Area \:(∆ABC) = \frac{1}{2} × 10 × 5.136}$

$\sf{⇝Area \:(∆ABC) = \frac{1}{\cancel{2}_{1}} × \cancel{10}^{5} × 5.136}$

$\sf{⇝Area \:(∆ABC) = 5 × 5.136}$

$\sf\boxed{⇝Area \:(∆ABC) = 25.68cm²}$

It's given angle is 140° i.e ∠BAC = 140°

$~~~~~~~~~~~~~~~$(refer b figure)

LET'S DO SOME CONSTRUCTION ;

• $\sf{Draw \:perpendicular \:from \:C \:on \:AB \:and \:extend \:AB \:to \:meet \:it \:at \:H}$

$~~~~~~~~\sf{\therefore CH ⊥ HB}$

$~~~~~~~~~~\sf{HB \:is \:the \:straight \:line}$

$\sf{⟹ ∠HAB = ∠HAC + ∠BAC =180°}$

$\sf{⟹ ∠HAC + 140° = 180°}$

$\sf{⟹ ∠HAC = 180° - 140°}$

$\sf\boxed{⟹ ∠HAC = 40°}$

$~~~~~~~~~~\sf{In \:right \:angle \:triangle CHA}$

• $\sf{∠HAC = 40° and AC = 8cm}$

$\sf{⇝CH = AC × \sin 40°}$

$\sf{⇝CH = 8 × \sin 40°}$

$\sf{(We \:know \sin 40 \:value = 0.642 \:from \:the \:table}$

$\sf{⇝CH = 8 × 0.642}$

$\sf\boxed{⇝CH = 5.136}$

Hence if we find the area we get the same value

____________________________________

Hence Area is same only the position of perpendicular line was changed if the angle is considered as 40°

So ;

Area of triangle ( angle 40° ) = 25.68 cm²

Area of triangle ( angle 140° ) = 25.68 cm²

___________________________________

$\sf\red{Here \:we \:get \:all \:your \:answer :)}$

Answer 2

${\ddagger{\mathrm {\green{Solution}}}}$⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\mathrm{CASE~1}$⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\sf{Given~:}$

• $\rm{Length~of~two~sides~of~∆~=~ 8~and~10~cm.}$

⠀⠀⠀⠀$\bf{AB~=~8cm ; BC~=~10cm}$

• $\rm{\angle{AC}~=~40°}$⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\sf{To~Find~:}$⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• $\rm{are~of~∆~with~\angle~=~40°.}$

$\sf{Construction~:}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• $\rm{Draw~bisector~AD}$ ( refer fig. 1 )

$\sf{Solution~:}$⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\bf{angle{BDC}~=~90°~(AD~\perp~BC)}$

$\bf{BD~=~BC~=~5cm~( AD~bisect~BC)}$

$\bf{In~∆ABD~,}$

$\bf{ sin\theta~=~\dfrac{Perpendicular}{Height}}$

$\bf{ sin40°~=~\dfrac{AD}{AB}}$

$\bf{ 0.75~=~\dfrac{AD}{8}}$

$\bf{ 6 cm~=~AD}$

$\bf{ Area~of~∆ABC, \angle{A}~=~40°}$

$\bf{ ar(∆ABC) ~=~ \frac{1}{2}~×~base~×~height}$

$\bf{ ar(∆ABC) ~=~ \frac{1}{2}~×~10~×~6}$

${\green {\boxed {\bf {\blue{ ar(∆ABC) ~=~ 30{cm}^{2}}}}}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\mathrm{CASE~2}$⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\sf{Given~:}$⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• $\rm{Length~of~two~sides~of~∆~=~ 8~and~10~cm.}$

⠀⠀⠀⠀$\bf{AB~=~8cm ; AC~=~10cm}$

• $\rm{\angle{ABC}~=~140°}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\sf{To~Find~:}$⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• $\rm{area~of~∆~with~\angle~=~140°.}$

$\sf{Construction~:}$⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• $\rm{Draw~bisector~BD}$ ( refer fig. 2)

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\sf{Solution~:}$⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\bf{angle{ABD}~=~70°~(BD~\perp~AC)}$

$\bf{AD~=~DC~=~5cm~( BD~bisect~AC)}$

$\bf{In~∆ABD~,}$

$\bf{ sin\theta~=~\dfrac{Perpendicular}{Height}}$

$\bf{ sin70°~=~\dfrac{BD}{AB}}$

$\bf{ 0.78~=~\dfrac{AD}{8}}$

$\bf{ 6.24 cm~=~AD}$

$\bf{ Area~of~∆ABC, \angle{A}~=~140°}$

$\bf{ ar(∆ABC) ~=~ \frac{1}{2}~×~base~×~height}$

$\bf{ ar(∆ABC) ~=~ \frac{1}{2}~×~10~×~6.24}$

${\green {\boxed {\bf {\blue{ ar(∆ABC) ~=~ 31.2{cm}^{2}}}}}}$

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