Question

a frequent problem in surveying city lots and rural lands adjacent to curves of highways and railways is that of finding the area when one or more of the boundary lines is the arc of a circle. find the area of the lot shown in the figure.

Explanation:

Step1: Divide the figure into a right - angled triangle and a sector.

The right - angled triangle has base \(b = 40\) yd and height \(h=30\) yd. The sector has a radius \(r\) which is the hypotenuse of the right - angled triangle.

Step2: Calculate the area of the right - angled triangle.

The area formula for a triangle is \(A_{triangle}=\frac{1}{2}bh\). Substituting \(b = 40\) yd and \(h = 30\) yd, we get \(A_{triangle}=\frac{1}{2}\times40\times30=600\) square yards.

Step3: Calculate the hypotenuse of the right - angled triangle (radius of the sector).

Using the Pythagorean theorem \(r=\sqrt{30^{2}+40^{2}}=\sqrt{900 + 1600}=\sqrt{2500}=50\) yd.

Step4: Calculate the area of the sector.

The area formula for a sector of a circle with central angle \(\theta\) (in degrees) and radius \(r\) is \(A_{sector}=\frac{\theta}{360}\times\pi r^{2}\). Here \(\theta = 60^{\circ}\) and \(r = 50\) yd. So \(A_{sector}=\frac{60}{360}\times\pi\times50^{2}=\frac{1}{6}\times\pi\times2500=\frac{1250\pi}{3}\approx1309\) square yards.

Step5: Calculate the total area of the figure.

\(A = A_{triangle}+A_{sector}=600+\frac{1250\pi}{3}\approx600 + 1309=1909\) square yards.

Answer:

\(600+\frac{1250\pi}{3}\text{ yd}^2\approx1909\text{ yd}^2\)