Menu Close

In-a-triangle-with-one-angle-of-120-the-lengths-of-the-sides-form-an-AP-If-the-length-of-the-greatest-side-is-7-cm-the-area-of-the-triangle-is-

Tinku Tara 0 Comments
Pin




Question Number 43641 by gunawan last updated on 13/Sep/18
In a triangle with one angle of 120°, the lengths of the sides form an AP. If the length of the greatest side is 7 cm., the area of the triangle is
$$\mathrm{In}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{one}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{120}°, \\ $$$$\mathrm{the}\:\mathrm{lengths}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{form}\:\mathrm{an}\:\mathrm{AP}. \\ $$$$\mathrm{If}\:\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{side}\:\mathrm{is} \\ $$$$\mathrm{7}\:\mathrm{cm}.,\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{is} \\ $$
Answered by $@ty@m last updated on 13/Sep/18
Let c>b>a c^2 =a^2 +b^2 −2abcos C 7^2 =a^2 +b^2 −2abcos 120 49=a^2 +b^2 +ab (∵cos 120=−(1/2)) ......(1) ∵ a, b,c are in AP ∴2b=a+c ⇒2b=a+7 ⇒a=2b−7 ...(2) ∴from(1) 49=(2b−7)^2 +b^2 +(2b−7)b 49=4b^2 −28b+49+b^2 +2b^2 −7b 7b^2 −35b=0 7b(b−5)=0 ∴b=5 ⇒a=2×5−7=3 ∴Area of △=(√(s(s−a)(s−b)(s−c))) =(√(((15)/2)(((15)/2)−3)(((15)/2)−5)(((15)/2)−7))) =(1/4)×(√(15×9×5×1)) =(1/4)×5×3(√3) =((15)/4)(√3) sq. unit
$${Let}\:{c}>{b}>{a} \\ $$$${c}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\mathrm{cos}\:{C} \\ $$$$\mathrm{7}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\mathrm{cos}\:\mathrm{120} \\ $$$$\mathrm{49}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}\:\left(\because\mathrm{cos}\:\mathrm{120}=−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……\left(\mathrm{1}\right) \\ $$$$\because\:{a},\:{b},{c}\:{are}\:{in}\:{AP} \\ $$$$\therefore\mathrm{2}{b}={a}+{c} \\ $$$$\Rightarrow\mathrm{2}{b}={a}+\mathrm{7} \\ $$$$\Rightarrow{a}=\mathrm{2}{b}−\mathrm{7}\:…\left(\mathrm{2}\right) \\ $$$$\therefore{from}\left(\mathrm{1}\right) \\ $$$$\mathrm{49}=\left(\mathrm{2}{b}−\mathrm{7}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} +\left(\mathrm{2}{b}−\mathrm{7}\right){b} \\ $$$$\mathrm{49}=\mathrm{4}{b}^{\mathrm{2}} −\mathrm{28}{b}+\mathrm{49}+{b}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} −\mathrm{7}{b} \\ $$$$\mathrm{7}{b}^{\mathrm{2}} −\mathrm{35}{b}=\mathrm{0} \\ $$$$\mathrm{7}{b}\left({b}−\mathrm{5}\right)=\mathrm{0} \\ $$$$\therefore{b}=\mathrm{5} \\ $$$$\Rightarrow{a}=\mathrm{2}×\mathrm{5}−\mathrm{7}=\mathrm{3} \\ $$$$\therefore{Area}\:{of}\:\bigtriangleup=\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)} \\ $$$$=\sqrt{\frac{\mathrm{15}}{\mathrm{2}}\left(\frac{\mathrm{15}}{\mathrm{2}}−\mathrm{3}\right)\left(\frac{\mathrm{15}}{\mathrm{2}}−\mathrm{5}\right)\left(\frac{\mathrm{15}}{\mathrm{2}}−\mathrm{7}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\sqrt{\mathrm{15}×\mathrm{9}×\mathrm{5}×\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\mathrm{5}×\mathrm{3}\sqrt{\mathrm{3}} \\ $$$$=\frac{\mathrm{15}}{\mathrm{4}}\sqrt{\mathrm{3}}\:\:{sq}.\:{unit} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *