An equilateral triangle has sides of 20. What are the lengths of another equilateral triangle with half the area?

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Answer 1 · 2018-05-24T16:10:14

$10 \sqrt{2}$ $10sqrt2$

Area of triangle $= \frac{1}{2} a b sin C$ $=1/2a b sinC$

for this equilateral triangle

$A_{1} = \frac{1}{2} \times 20 \times 20 \times sin 60$ $A_1=1/2xx20xx20xxsin60$

$A_{1} = 100 \sqrt{3}$ $A_1=100sqrt3$

now we have the relation

$ratio of areas = {(ratio of lengths)}^{2}$ $"ratio of areas " =(" ratio of lengths ")^2$

let the second triangle have area

$A_{2}$ $A_2$

$A_{2} = \frac{1}{2} \times 100 \sqrt{3} = 50 \sqrt{3}$ $A_2=1/2xx100sqrt3=50sqrt3$

so $A_{2} : A_{1} : : {(l_{2} : 20)}_{2}^{2}$ $A_2:A_1::(l_2:20)^2$

$:. (50sqrt3)/(100sqrt3)=l_2^2/400$

$=>l_2^2=1/2xx400$

$l_2=sqrt(200)$

$l_2=10sqrt2$