Question

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Explanation:

Step1: Calculate the area of the rectangle

The formula for the area of a rectangle is \(A = l\times w\). Here, the length \(l = 5\) and the width \(w = 4\).

$$A_{rectangle}=5\times4 = 20$$

Step2: Calculate the area of the triangle

The formula for the area of a triangle is \(A=\frac{1}{2}\times b\times h\). Here, the base \(b = 5\) and the height \(h = 4\).

$$A_{triangle}=\frac{1}{2}\times5\times4=10$$

Step3: Calculate the area of the shaded region

The area of the shaded region \(A = A_{rectangle}-A_{triangle}\)

$$A = 20 - 10=10$$

Wait, no, looking at the figure again, if we assume it's a rectangle divided by a diagonal (the dotted line). The area of the rectangle with sides \(l\) and \(w\) is \(A = l\times w\). If we consider the two - triangle parts formed by the diagonal of the rectangle, they are congruent. But if we assume the figure is a rectangle with length \(l = 4\) (vertical side) and width \(w\) (horizontal side) and the diagonal divides it. Wait, no, re - checking.
Another approach: If we assume the figure is a rectangle. Let's use the formula for the perimeter of a rectangle \(P=2(l + w)\). But no, the options are 14,16,18,20. Wait, if it's a rectangle with length \(l = 5\) (the diagonal of a right - triangle with legs \(3\) and \(4\) (since \(3 - 4-5\) is a Pythagorean triple) and the other side of the rectangle is \(4\). Wait, no.
Wait, if we consider the figure as a rectangle. Let's assume the vertical side is \(4\) and the horizontal side is \(x\). Using the Pythagorean theorem for the right - triangle formed by half of the rectangle (if the diagonal is \(5\)). But if it's a rectangle, and the diagonal \(d = 5\), one side \(a = 4\), then using \(d^{2}=a^{2}+b^{2}\), \(b=\sqrt{d^{2}-a^{2}}=\sqrt{25 - 16}=3\). The area of the rectangle \(A=(3 + 4)\times2=14\) (wait, no).
Wait, no, another way: If we consider the figure as composed of two congruent right - triangles. Each right - triangle has legs \(3\) and \(4\) (since \(3^{2}+4^{2}=5^{2}\)). The perimeter of the figure (assuming it's a polygon). If it's a rectangle - like figure (a parallelogram with right angles) with sides \(3 + 4\) and \(2\) (no, no). Wait, no, the options are 14,16,18,20.
Wait, if we consider the sum of the lengths of the sides. If the figure is a quadrilateral (a rectangle divided by a diagonal). The vertical side is \(4\), the horizontal side (assuming the base of the right - triangle is \(3\) (from \(3 - 4-5\) triangle)), then the perimeter \(P=3 + 4+5+2\) (no). Wait, no, re - looking at the options.
Wait, if we assume it's a rectangle with length \(l = 5\) (the diagonal of a \(3 - 4-5\) triangle) and width \(w = 4\). No, the perimeter of a rectangle \(P = 2(l + w)\). If \(l = 5\) (wrong, \(l\) should be the side of the rectangle). Wait, no, the correct approach:
If we consider the figure as a rectangle. Let the sides of the rectangle be \(a\) and \(b\). The diagonal \(d = 5\), and one side \(b = 4\). Then \(a=\sqrt{5^{2}-4^{2}}=3\). The perimeter \(P=2(a + b)=2(3 + 4)=14\)

Answer:

14