Pembuktian Rumus Luas dan Tinggi Segitiga Sembarang

$AF=\frac{b^2+c^2-a^2}{2c}$

$tc^2=b^2-\left ( \frac{b^2+c^2-a^2}{2c} \right )^2$

$tc^2=\left [ b+\left ( \frac{b^2+c^2-a^2}{2c} \right )\right ]\left [ b-\left ( \frac{b^2+c^2-a^2}{2c} \right )\right ]$

$tc^2=\left [ \frac{2bc+b^2+c^2-a^2}{2c}\right ]\left [ \frac{2bc-b^2-c^2+a^2}{2c}\right ]$ $tc^2=\left [ \frac{\left ( b+c \right )^2-a^2}{2c}\right ]\left [ \frac{(b-c)^2+a^2}{2c}\right ]$

$tc^24c^2=\left [ (b+c)^2-a^2 \right ]\left [ a^2-(b-c)^2 \right ]$
$tc^24c^2=\left [ \left [ (b+c)+a \right ]\left [ (b+c)-a \right ] \right ]\left [ \left [ a+(b-c) \right ]\left [ a-(b-c) \right ] \right ]$
$tc^24c^2=\left [ (a+b+c)(-a+b+c) \right ]\left [ (a+b-c)(a-b+c) \right ]$

$tc^2=\frac{16s}{4c^2}(s-a)(s-c)(s-b)$
$tc=\frac{2}{c}\sqrt{(s-a)(s-c)(s-b)}$

Luas

$Luas \triangle ABC=\frac{1}{2} AB+c$
$=\frac{1}{2}AB.\frac{2}{c}\sqrt{(s-a)(s-b)(s-c)}$
$=\frac{1}{2}c.\frac{2}{c}\sqrt{(s-a)(s-b)(s-c)}$
$=\sqrt{s(s-a)(s-b)(s-c)}$

Wednesday, September 10, 2014

Pembuktian Rumus Luas dan Tinggi Segitiga Sembarang

No comments:

Post a Comment

Paling Banyak Dilihat

Facebook

Total Pageviews