Find the height of the triangle whose vertices are (2,0),(-2,0),(4,0)
Answers
Answer:
Step-by-step explanation:
As the vertices are given, we need to find the length of each sides. Using “The distance between the two points (x1,y1) and (x2,y2) is:
(x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−−√”
Length between the vertices (−2,0) and (4,0) or AB is:
AB=(4−−2)2+(0−0)2−−−−−−−−−−−−−−−−√
AB=62+02−−−−−−√=36+0−−−−−√
AB=6
Length between the vertices (4,0) and (3,3) or BC is:
BC=(3−4)2+(3−0)2−−−−−−−−−−−−−−−√
BC=(−1)2+32−−−−−−−−−√=1+9−−−−√
BC=10−−√
Length between the vertices (−2,0) and (3,3) or AC is:
AC=(3−−2)2+(3−0)2−−−−−−−−−−−−−−−−√
AC=52+32−−−−−−√=25+9−−−−−√
AC=34−−√
Now as we get the three sides of a Triangle we can easily find it's area using the Heron's Formula.
Perimeter (P)=AB+BC+AC=6+10−−√+34−−√
Semiperimeter (s)=P2=6+10√+34√2
Area (A)=s(s−AB)(s−BC)(s−AC)−−−−−−−−−−−−−−−−−−−−−−−√
A=6+10−−√+34−−√2(6+10−−√+34−−√2−6)(6+10−−√+34−−√2−10−−√)(6+10−−√+34−−√2−34−−√)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√
A=6+10−−√+34−−√2(−6+10−−√+34−−√2)(6−10−−√+34−−√2)(6+10−−√−34−−√2)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√
A=81−−√
A=9units2
Answer:
8 units
2units
3units
4units