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Polygons — PYPs

4 solved Polygons previous year questions (PYQs) from past year papers — attempt each and check the answer.

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Q1:ipmat indore 2019QA › PolygonsEasySA · TITA

The sum of the interior angles of a convex

n

-sided polygon is less than

2019^\circ

. The maximum possible value of

n

Enter your answer to attempt

The Setup: A pure geometry fundamental. The sum of the interior angles of any convex

n

-sided polygon is the GOAT formula

(n-2)\times 180^\circ

. Set up the inequality, solve, and take the largest integer that survives. Step 1: Set up the inequality. The sum is strictly less than

2019^\circ

(n-2)\times 180 < 2019

Step 2: Isolate the variable. Divide both sides by 180 (positive, so the inequality direction is safe):

n-2<\frac{2019}{180}

Step 3: Crunch the division. Since

180\times 11=1980

and

180\times 12=2160

, the quotient sits between 11 and 12:

\frac{2019}{180}=11.216\ldots \implies n<13.216\ldots

Step 4: Lock in the maximum integer. A polygon has a whole number of sides, so the largest admissible

n

is 13. Step 5: Verify both sides of the boundary. A bound is only trustworthy if the next value up actually fails: *

n=13

: sum

=(13-2)\times 180=1980^\circ

, and

1980<2019

. Valid. *

n=14

: sum

=(14-2)\times 180=2160^\circ

, and

2160>2019

. Fails. So 13 works and 14 does not, which pins the maximum exactly. Final Answer: 13

Q2:jipmat 2025QA › PolygonsMediumQA · MCQ

The cost of fencing of an equilateral triangular park and a square park is the same. If the area of the triangular park is

16\sqrt{3}\text{ m}^2

then the length of the diagonal of the square park is:

A $6\sqrt{3}\text{ m}$
B $8\sqrt{3}\text{ m}$
C $8\sqrt{2}\text{ m}$
D $6\sqrt{2}\text{ m}$

Pick an option to attempt

The Setup: The fencing costs are identical, meaning their perimeters perfectly match each other's energy (they are exactly equal). We need to find the side of the triangle, use it to get the perimeter, transfer that perimeter to the square, and finally find the square's diagonal. Step 1: Find the side length of the equilateral triangle (

a

). Area formula:

\frac{\sqrt{3}}{4}a^2=16\sqrt{3}

. Cancel out

\sqrt{3}

\frac{a^2}{4}=16

a^2=64 \implies a=8\text{ m}

. Step 2: Calculate the perimeter of the triangle.

P=3 \times 8=24\text{ m}

Step 3: Find the side of the square (

s

). Since perimeters are equal:

4s=24 \implies s=6\text{ m}

Step 4: Find the diagonal of the square using the formula

d=s\sqrt{2}

d=6\sqrt{2}\text{ m}

Final Answer:

6\sqrt{2}

Q3:jipmat 2025QA › PolygonsEasyQA · MCQ

Arrange the following in descending order based on their perimeters: A. A square with an area of 36 sq. cm. B. An equilateral triangle with a side of 9 cm. C. A rectangle with 10 cm as length and 40 sq. cm as area. D. A circle with radius of 4 cm.

AC > A > B > D
BD > A > B > C
CC > B > A > D
DC > B > D > A

Pick an option to attempt

The Setup: We need to calculate the perimeter (or circumference) for each shape to see who is really built different, then rank them from largest to smallest. Step 1: Calculate A (Square).

\text{Area}=s^2=36 \implies s=6\text{ cm}

\text{Perimeter}=4s=4 \times 6=24\text{ cm}

. Step 2: Calculate B (Equilateral Triangle). Side is

9\text{ cm}

\text{Perimeter}=3s=3 \times 9=27\text{ cm}

. Step 3: Calculate C (Rectangle).

\text{Area}=L \times W \implies 40=10 \times W \implies W=4\text{ cm}

\text{Perimeter}=2(L+W)=2(10+4)=2(14)=28\text{ cm}

. Step 4: Calculate D (Circle). Radius

r=4\text{ cm}

\text{Circumference}=2\pi r=2 \times 3.1415 \times 4 \approx 25.13\text{ cm}

. Step 5: Rank them in descending order (biggest first).

C(28) > B(27) > D(25.13) > A(24)

. Final Answer: C > B > D > A

Q4:ipmat indore 2025QA › PolygonsEasyMCQ · MCQ

Area of a regular octagon inscribed in a circle of radius

1

unit is:

A $2\sqrt{2}$
B $2+\sqrt{2}$
C $\frac{9}{2\sqrt{2}}$
D $\sqrt{10}$

Pick an option to attempt

The Setup: We are dropping into a geometry raid involving a regular octagon inscribed inside a circle. Think of it like slicing up a pizza into 8 perfectly equal pieces. To find the total area, we can calculate the area of just one of those triangular pizza slices (sectors) branching out from the center, and then multiply it by 8. Step 1: Analyze the central angles of the octagon. A full circle is

360^\circ

(or

2\pi

radians). Since a regular octagon has

8

identical sides, it also has

8

identical triangles radiating out from the center point to each vertex. Let's find the central angle

\theta

for just one of these triangles:

\theta = \frac{360^\circ}{8} = 45^\circ

Step 2: Pull the stats for a single triangle. Each of these

8

triangles shares two sides that extend from the center to the circle's boundary. Since both of those lines are radii of the circle, their lengths are both equal to

r = 1

. So, we have an isosceles triangle with two side lengths of

1

and an included angle of

\theta = 45^\circ

. Step 3: Calculate the area of one triangle. The trigonometry cheat code for finding the area of a triangle when you know two sides (

a

and

b

) and the included angle (

\theta

) is:

\text{Area}_{\text{triangle}} = \frac{1}{2} a b \sin(\theta)

Plug in our stats (

a = 1

b = 1

, and

\theta = 45^\circ

\text{Area}_{\text{triangle}} = \frac{1}{2} (1)(1) \sin(45^\circ)

Since

\sin(45^\circ) = \frac{\sqrt{2}}{2}

\text{Area}_{\text{triangle}} = \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{4}

Step 4: Scale it up to the full octagon. An octagon consists of

8

of these exact identical triangles. To get the total area, just multiply the area of one triangle by

8

\text{Total Area} = 8 \cdot \left(\frac{\sqrt{2}}{4}\right)

Simplify the numbers:

\text{Total Area} = 2\sqrt{2}

The math is clean, fast, and flawless. Final Answer:

2\sqrt{2}

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