The area of a triangle having sides a, b, c and s as semi-perimeter is given by, `A = sqrt( s(s-a)(s-b)(s-c)`, where `s = (a+b+c)/2` a = 3 cm ; b = 4 cm ; c = 5 cm `s = (a+b+c)/2` `s = (3+4+5)/2` `s =12/2` s = 6 cm Now, area `A = sqrt(s(s-a)(s-b)(s-c)` \[= \sqrt{6(6 - 3)(6 - 4)(6 - 5)}\]\[ = \sqrt{6 \times 3 \times 2 \times 1}\]\[ = \sqrt{36}\] \[ = 6 c m^2\]
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This triangle is a right-angled triangle,because [tex]3^2+4^2=5^2[/tex] Area of the triangle [tex]= \frac1{2}.3.4=6[/tex]  dduclam Posts: 36Joined: Sat Dec 29, 2007 10:42 amLocation: HUCE-Vietnam Reputation: 4
is there a formula using only the value of the sides? i mean like of its not a right triangle? how do we solve for this? donaldhall
Hi! YES, there is. It's called the "Law of cosines" and can be aplied in each and every triangle.Law of cosines: In the ABC triangle,[tex]BC^{2}=AB^{2}+AC^{2}-2XABXAC[/tex]Let's consider that our ABC triangle has the measures in the following way:AB=3; BC=5; AC=4Applying the Law of cosines:[tex]5^{2}=3^{2}+4^{2}-2X3X4XcosAcos A=0 => A=90^\circ => sinA=1[/tex] An area formula is ABXACXsinA / 2.So, the ABC's area is 3X4X1/2=6 Hope I helped you. stefantudose Posts: 2Joined: Mon Dec 19, 2011 3:52 pm Reputation: 2
Area of a triangle = A=hb.b/2 => 3.4/2 = 6 leesajohnson
Wrongly solving ..... It's a simple problem ....S=(a+b+c)/2Consider a=3 b=3 and c=5So S= (3+3+5)/2=11/2 or 5.5We know A=√s(s-a)(s-b)(s-c)=√ 5.5(5.5-3)(5.5-3)(5.5-5)=√5.5 x2.5x 2.5x .5=√17.1875 =4.146 scm Montramugdha Posts: 1Joined: Fri Mar 03, 2017 8:38 am Reputation: 1
The easy way is to use this online calculator: https://ezcalc.me/triangle-calculator . You can calculate any three missing parameters given any three known parameters (say three sides) as well as the area and perimeter. There you can find all relevant formulas. Guest
Montramugdha is using "Heron's formula" (https://en.wikipedia.org/wiki/Heron%27s_formula): Given sides of length a, b, and c, let s be the "semi-perimeter", [tex]s= \frac{a+ b+ c}{2}[/tex].Then the area is given by [tex]A= \sqrt{s(s-a)(s-b)(s-c)}[/tex].In the example initially given, a= 3, b= 4, c= 5, [tex]s= \frac{3+ 4+ 5}{2}= 6[/tex] so the area is [tex]A= \sqrt{6(6-3)(6-4)(6-5)}= \sqrt{6(3)(2)(1)}= \sqrt{36}= 6[/tex] which is, of course, the same as (3)(4)/2 since this is a right triangle. If we were given an equilateral triangle with side lengths all 4, we could find the area by dividing it into two right triangles with hypotenuse of length 4, one leg of length 2, so the other leg of length [tex]\sqrt{4^2- 2^2}= \sqrt{12}= 2\sqrt{3}[/tex]. Each right triangle has area [tex]\frac{2(2\sqrt{3}}{2}= 2\sqrt{3}[/tex] and the original equilateral triangle has area [tex]4\sqrt{3}[/tex]. By Heron's formula, [tex]s= \frac{4+ 4+ 4}{2}= 6[/tex] so the area is [tex]A= \sqrt{6(6-4)(6-4)(6-4)}= \sqrt{6(2)(2)(2)}= \sqrt{48}= \sqrt{3(16)}= 4\sqrt{3}[/tex]. HallsofIvy Posts: 341Joined: Sat Mar 02, 2019 9:45 am Reputation: 125
As CK=4, then BK=4
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