Question

1.23 quiz: special right triangles: mastery test rounded to the nearest tenth, what is the perimeter of rectangle abcd? a. 48.3 cm b. 54.6 cm c. 60.0 cm

Explanation:

Step1: Find side - lengths using 30 - 60 - 90 triangle rules

In a 30 - 60 - 90 triangle, if the hypotenuse $AC = 20$ cm, the side opposite the 30 - degree angle (let's say $AB$) is half of the hypotenuse. So $AB=\frac{1}{2}\times20 = 10$ cm. The side opposite the 60 - degree angle (let's say $AD$) is $\sqrt{3}$ times the side opposite the 30 - degree angle. So $AD = 10\sqrt{3}$ cm.

Step2: Calculate the perimeter of the rectangle

The perimeter $P$ of a rectangle with length $l$ and width $w$ is $P = 2(l + w)$. Here, $l=10\sqrt{3}$ cm and $w = 10$ cm. So $P=2(10\sqrt{3}+10)=20(\sqrt{3}+1)$.

Step3: Approximate the value

We know that $\sqrt{3}\approx1.732$. Then $P=20(1.732 + 1)=20\times2.732=54.64\approx54.6$ cm.

Answer:

B. 54.6 cm