Question

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consider a quadrilateral with vertices ( a(2,2) ). ( b(2,5) ). ( c(7,2) ), and ( d(7,5) ). what is the sha area of quadrilateral ( abcd )?
quadrilateral ( abcd ) is a with an area of square units.

Explanation:

Step1: Determine the type of quadrilateral

• For points \(A(2,2)\) and \(B(2,5)\), the \(x\) - coordinates are the same (\(x = 2\)). The length \(AB=\vert5 - 2\vert=3\) (using the distance formula for vertical points \(d=\vert y_2 - y_1\vert\)).
• For points \(A(2,2)\) and \(C(7,2)\), the \(y\) - coordinates are the same (\(y = 2\)). The length \(AC=\vert7 - 2\vert=5\) (using the distance formula for horizontal points \(d=\vert x_2 - x_1\vert\)).
• Since \(AB\perp AC\) (one is vertical and one is horizontal), and \(AB = 3\), \(AC=5\), \(BC=\sqrt{(7 - 2)^2+(2 - 5)^2}=\sqrt{25 + 9}=\sqrt{34}\), \(AD=\sqrt{(7 - 2)^2+(5 - 2)^2}=\sqrt{25+9}=\sqrt{34}\), \(CD = 3\), \(BD = 5\). Opposite sides are equal (\(AB = CD = 3\) and \(AC=BD = 5\)) and adjacent sides are perpendicular. So, it is a rectangle.

Step2: Calculate the area of the rectangle

The area formula for a rectangle is \(A=\text{length}\times\text{width}\). Here, length \(l = 5\) (distance between \(A(2,2)\) and \(C(7,2)\)) and width \(w = 3\) (distance between \(A(2,2)\) and \(B(2,5)\)).
So, \(A=5\times3 = 15\).

Answer:

Quadrilateral \(ABCD\) is a rectangle with an area of \(15\) square units.