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CAT 2019 Slot 2 — QA Question 11

PolygonsEasy

Let A and B be two regular polygons having a and b sides, respectively. If b = 2a and each interior angle of B is 3/2 times each interior angle of A, then each interior angle, in degrees, of a regular polygon with a + b sides is

Answer & solution

Correct answer: 150

Answer: 150

Solution

Easy

Each interior angle of a regular nn-gon is (n−2)n×180∘\dfrac{(n-2)}{n}\times 180^\circ. Substitute aa and b=2ab=2a, set BB's angle equal to 32\tfrac32 of AA's, solve for aa, then compute the interior angle of the (a+b)(a+b)-gon.

1

Write both interior angles. With b=2ab=2a:

Angle(A)=a−2a×180∘Angle(B)=2a−22a×180∘\begin{aligned} &\text{Angle}(A)=\frac{a-2}{a}\times 180^\circ\\ &\text{Angle}(B)=\frac{2a-2}{2a}\times 180^\circ \end{aligned}
2

Apply the 32\tfrac32 ratio. Set Angle(B)=32 Angle(A)\text{Angle}(B)=\tfrac32\,\text{Angle}(A); the 180∘180^\circ cancels.

2a−22a=32⋅a−2a⇒ a−1a=3(a−2)2a⇒ 2(a−1)=3(a−2)(×2a)\begin{aligned} &\frac{2a-2}{2a}=\frac{3}{2}\cdot\frac{a-2}{a}\\ &\Rightarrow\ \frac{a-1}{a}=\frac{3(a-2)}{2a}\\ &\Rightarrow\ 2(a-1)=3(a-2) \quad\text{(}\times 2a\text{)} \end{aligned}
3

Solve for aa, then bb and a+ba+b.

2a−2=3a−6⇒ a=4,b=2a=8⇒ a+b=12\begin{aligned} &2a-2=3a-6\\ &\Rightarrow\ a=4,\quad b=2a=8\\ &\Rightarrow\ a+b=12 \end{aligned}
4

Interior angle of the 1212-gon.

12−212×180∘=1012×180∘=150∘\begin{aligned} &\frac{12-2}{12}\times 180^\circ=\frac{10}{12}\times 180^\circ=150^\circ \end{aligned}
Each interior angle=150∘\text{Each interior angle}=150^\circ

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CAT 2019 Slot 2 QA Q11: Let A and B be two regular polygons having a and b sides, respectively. If b = 2a and each interior angle of B — Solution | TheCATExam