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Question
if line segment bc is considered the base of triangle abc, what is the corresponding height of the triangle? 0.8 units 1.25 units 0.625 units 1.6 units
Step1: Calculate the area of the triangle using the formula \(A=\frac{1}{2}\times base\times height\) with another base - height pair
First, find the length of \(AC\). Using the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), for \(A(-1,1)\) and \(C(-1,-1)\), \(AC=\vert1-(-1)\vert = 2\) units.
Let's take \(AC\) as the base. The coordinates of \(B(3,2)\). The horizontal distance from \(B\) to the line \(x=-1\) (the line containing \(AC\)) is \(\vert3-(-1)\vert = 4\) units. So the area \(A=\frac{1}{2}\times AC\times\) (horizontal distance from \(B\) to \(AC\)) \(=\frac{1}{2}\times2\times4 = 4\) square units.
Step2: Calculate the length of \(BC\)
Using the distance formula for \(B(3,2)\) and \(C(-1,-1)\), \(BC=\sqrt{(3 + 1)^2+(2+1)^2}=\sqrt{16 + 9}=\sqrt{25}=5\) units.
Step3: Find the height corresponding to base \(BC\)
Since \(A=\frac{1}{2}\times BC\times h\), and \(A = 4\), \(BC = 5\). Then \(4=\frac{1}{2}\times5\times h\). Solving for \(h\), we get \(h=\frac{4\times2}{5}=1.6\) units.
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\(1.6\) units