Question

The perimeter of a triangle is 50 cm. One side of a triangle is 4 cm longer than the smaller side and the third side is 6 cm less than twice the smaller side, then the area of the triangle is $20\sqrt{30}c{m}^2$
If true then enter $1$ and if false then enter $0$

Answer 1

$Lettheverticesofthe△beA,B,C\&thelengthofsmallersideBCisa.ThensideAB=a+4andsideAC=2a-6.Giventheperimeterof△ABCis50cm.Sowehavea+(a+4)+(2a-6)=50\Rightarrow 4a-2=50\Rightarrow 4a=52\Rightarrow a=13.\therefore ThesidesareBC=a=13cm,AB=(a+4)=17cmandAC=(2a-6)=20cm.NowaccordingtoHero{n}^{\prime }sformulatheareaoftriangleABCwithsidesa,b\&cisgivenasA=\sqrt[2]{2\left[s(s-a)(s-b)(s-c)\right]}where2s=(a+b+c).Herea=13cm,b=17,c=20,s=\frac{(13+17+20)}{2}=\frac{50}{2}=25Areaof△ABC=\sqrt[2]{2[25\times (25-13)(25-17)(25-20)]}=\sqrt[2]{2(25\times 12\times 8\times 5)}=\sqrt[2]{2(5\times 5\times 6\times 2\times 8\times 5)}=\sqrt[2]{2(5\times 5\times 6\times 16\times 5)}=5\times 4\times \sqrt[2]{2(6\times 5)}=20\sqrt[2]{230}$