# Quiz

**The interior angles of a polygon are in AP. If the smallest angle is 120∘ and the common difference is 5∘, then find the possible number of sides of that polygon?**

__Question:__**Options:**

A. | 16 | |

B. | 9 | |

C. | Both a and b | |

D. | None of these |

**Answer: Option B**

: B

Sum of the interior angles of a polygon = (n-2) 180 n2 * (2a+ (n-1)d)= (n-2)180 n2* [2*(120) + (n-1)5]=(n-2)180 n[48+(n-1)] = (n-2)72 n2 -25n + 144=0 n=9 and n=16 when n= 16, the greatest angle will be equal to a+15d = 120 + 15 x 5 = 195 and no interior angle of a polygon can be equal to or greater than 180. hence answer =b

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### More Questions on This Topic :

__Question 1.__**Concentric circles are drawn with radii 1, 2, 3 … 100. The interior of the smallest circle is colored black and the annular regions are colored alternately red and black, so that no two adjacent regions are the same color. The total area of the red regions divided by the area of the largest circle is**

- 12
- 51100
- 101200
- 50101

**Answer: Option C**

: C

Areas of circles follow the sequence 1,4,9,16,…………1002 Areas of Black & Red annuals – 1,3,5,7,9……………199 – General term 2n – 1 Red annuals – 3, 7, 11, …………….. 199 Sum = (502)*(3 + 199) Largest circle’s area = 1002

^{ }Ratio = 101200 Note : Ignore pi as it is a ratio of areas Shortcut: Reverse gear approach – We know that the denominator is 1002 and num is an integer, hence the denominator cannot be 101 as it is in no-way a factor of 10000. Hence option d) has been eliminated. We need to find R(R+B). R + B = 10000. Each of the red annual is greater in area by 2 units than the subsequent black annual and there are 50 such cases. Hence R = 5050 and B = 4950. R(R+B). = 505010000 = 101200

__Question 2.__**Find the side of a piece of cloth in the shape of an equilateral triangle, whose area costs as much to paint at Rs 10/ square metre as it would cost to lay a border around the three sides at Rs 25 per metre?**

- 13.33 m
- 17.32 m
- 14.14 m
- 8.66 m

**Answer: Option B**

: B

Option (b) Perimeter = 3a Area = √34a2 10x√34a2= 25x3a a=30√3= 17.32 m

__Question 3.__**In Δ ABC, D,E and F are taken on AB, AC and BC respectively, so that EF=CF and DF=BF. If ∠A=40∘, then find ∠DFE?**

- 100∘
- 120∘
- 90∘
- 140∘

**Answer: Option A**

: A

Option (a) Take DBF = a∘ and ECF = b∘.

40 + a∘ + b∘ = 180 ⇒a∘ + b∘= 140

DFB + EFC + DFE = 180 ⇒ 180 – 2a∘ + 180 – 2b∘ + DFE = 180 ⇒ 180∘ – 2(a + b) = - DFE

DFE = 280 – 180 = 100∘

__Question 4.__**There is a circle of radius 10 units that circumscribes an equilateral triangle. A quadrilateral is drawn by joining mid-points of two adjacent sides of the triangle. Find the area of the quadrilateral**

- 225√32
- 225√34
- 225√38
- 225√312

**Answer: Option B**

: B

In an equilateral triangle circum radius = 23 h, 10 = 23 * √32S Side = 10\(\sqrt3\) units. Area of Triangle = √34(10√3)2 By graphical division, we need 34th of the area of the triangle =34×√34×(10√3)2 = 225√34

- 18
- 12.5
- 12
- 14

**Answer: Option B**

: B

Option b

It’s important that we note that it is a Pythagoras triplet 72 = 252+242 Now, In a right-angled triangle, the median to the hypotenuse is half the hypotenuse and is also the circum radius of the triangle. As the hypotenuse is 25, the circum radius is 12.5

__Question 6.__**What is the maximum number of times x circles of the same size can intersect each other?**

- 2x
- (x - 1)
- x(x-1)
- x(x+1)/2

**Answer: Option C**

: C

Deducing a Pattern: Taking two circles. The number of points of intersection is 2 at most. If u consider 3 circles, the number of points of intersection = 6 ( the third circles will have 4 points of intersection with the other 2 circles and those two circles will have 2 points together). Shortcut:- Assumption & Reverse Gear: Taking two circles. The number of points of intersection is 2 at most. From this itself, option a, b and d can be eliminated giving answer c.

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