<span><span><span><span><span>(<span>5+4</span>)</span><span>(2)</span></span>+6</span>−<span><span>(2)</span><span>(2)</span></span></span>−1</span><span>=<span><span><span><span><span>(9)</span><span>(2)</span></span>+6</span>−<span><span>(2)</span><span>(2)</span></span></span>−1</span></span><span>=<span><span><span>18+6</span>−<span><span>(2)</span><span>(2)</span></span></span>−1</span></span><span>=<span><span>24−<span><span>(2)</span><span>(2)</span></span></span>−1</span></span><span>=<span><span>24−4</span>−1</span></span><span>=<span>20−1</span></span><span>=<span>19</span></span>
Since triangle ABC is similar to triangle DEF then the ratio of the corresponding sides is constant.
The ratio of the corresponding lengths is referred to as the linear scale factor.
Considering the heights of the two triangles;
L.S.F = 14/6
= 7/3
The ratio in area (A.S.F) is given by (L.S.F)²
Therefore, A.S.F = (7/3)² = 49/9
Thus te ratio of the area of triangle ABC to DEF is 49:9
when a data consists of hundreds of values
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The law of cosines for this particular situation is b^2 = a^2 + c^2 - 2ac cosB.
Filling in what you know, you have b^2 = 25 + 49 - [2(5)(7)-.1908], which simplifies to b^2 = 74 - 69.8092 which gives you a b^2 value of 4.1908, but you have to take the square root of that so you get a side value for b of 2.05.
