Question

10 a rectangle has a length of \\(\sqrt{171}\\) and a width of \\(\sqrt{51}\\). what is the best estimate for its area? \\(\bigcirc\\) a. 98 \\(\bigcirc\\) b. 91 \\(\bigcirc\\) c. 112 \\(\bigcirc\\) d. 104

Explanation:

Step1: Recall the formula for the area of a rectangle

The area formula for a rectangle is \(A = l\times w\), where \(l\) is the length and \(w\) is the width. Here, \(l=\sqrt{171}\) and \(w = \sqrt{51}\), so \(A=\sqrt{171}\times\sqrt{51}\). Using the property \(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\), we get \(A=\sqrt{171\times51}=\sqrt{8721}\).

Step2: Estimate the square - roots

Estimate \(\sqrt{171}\):
Since \(13^2=169\) and \(14^2 = 196\), \(\sqrt{171}\approx13.1\).
Estimate \(\sqrt{51}\):
Since \(7^2 = 49\) and \(8^2=64\), \(\sqrt{51}\approx7.1\).
Then \(A=\sqrt{171}\times\sqrt{51}\approx13.1\times7.1=(13 + 0.1)\times(7+0.1)=13\times7+13\times0.1+0.1\times7 + 0.1\times0.1=91+1.3 + 0.7+0.01=93.01\).
Another way:
\(\sqrt{8721}\), we know that \(93^2=(90 + 3)^2=90^2+2\times90\times3+3^2=8100+540 + 9=8649\), \(94^2=(90+4)^2=90^2+2\times90\times4 + 4^2=8100+720+16=8836\).
\(8721\) is closer to \(8649\) than to \(8836\).

Answer:

B. 91