Question

If one side of an isosceles triangle is given by \(y=2\) and the base is provided by the points \((2,0)\) and \((0,2)\), then its area (in sq. units) is:

• A. \(2\sqrt{2}\)
• B. \(1\)
• C. \(2\)
• D. \(4\)

Hint:

For an isosceles triangle, the altitude from the vertex passes through the midpoint of the base and lies along the perpendicular bisector of the base.

Answer 1

The Correct Option is C

Step 1: Identify the base of the triangle.
The endpoints of the base are \[ A(2,0),\qquad B(0,2) \] The length of the base is \[ AB = \sqrt{(2-0)^2+(0-2)^2} \] \[ = \sqrt{4+4} \] \[ = 2\sqrt2 \]

Step 2: Find the equation of the base.
The line through \((2,0)\) and \((0,2)\) is \[ x+y=2 \] Since the triangle is isosceles with base \(AB\), the vertex lies on the perpendicular bisector of \(AB\). The midpoint of \(AB\) is \[ M\left(1,1\right) \] The slope of \(AB\) is \[ -1 \] Hence the perpendicular bisector has slope \(1\). Its equation is \[ y-1=x-1 \] or \[ y=x \]

Step 3: Use the condition \(y=2\).
One side of the triangle lies on \[ y=2 \] The vertex of the isosceles triangle is the intersection of \[ y=x \] and \[ y=2 \] Thus, \[ V=(2,2) \]

Step 4: Find the perpendicular distance from \(V\) to the base.
Distance from \((2,2)\) to the line \[ x+y-2=0 \] is \[ d= \frac{|2+2-2|} {\sqrt{1^2+1^2}} \] \[ = \frac{2}{\sqrt2} \] \[ = \sqrt2 \]

Step 5: Find the area.
Area of the triangle is \[ \frac12 \times \text{base} \times \text{height} \] \[ = \frac12(2\sqrt2)(\sqrt2) \] \[ = \frac12(4) \] \[ =2 \]

Step 6: Final conclusion.
Therefore, \[ \boxed{2} \]