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## Geometry and Mensuration: Level 2 Test 5

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Question 1 |

In a cyclic quadrilateral ABCD, ∠A is double its opposite angle and the difference between the other two angles is one-third of ∠A. The minimum difference between any two angles of this quadrilateral is

30 ^{o} | |

10 ^{o} | |

20 ^{o} | |

40 ^{o} |

Question 1 Explanation:

Let ∠A be equal to 2x.

Thus 2+x= 180,

=> x=60.

Thus, Two angles are 60 and 120.

Difference between the other two angles = 1/3 of 120 = 40

Sum of the two angles = 180.

Thus the two angles = 110 and 70 degrees.

Thus The difference between any two angles will be minimum for 110 and 120 = 10

Thus 2+x= 180,

=> x=60.

Thus, Two angles are 60 and 120.

Difference between the other two angles = 1/3 of 120 = 40

Sum of the two angles = 180.

Thus the two angles = 110 and 70 degrees.

Thus The difference between any two angles will be minimum for 110 and 120 = 10

^{o }Correct option is (b)Question 2 |

Inside a triangle ABC, a straight line parallel to BC intersects AB and AC at the points P and Q respectively. If AB= 3 PB, then PQ: BC is

1: 3 | |

3: 4 | |

1: 2 | |

2: 3 |

Question 2 Explanation:

Since AB= 3 PB. Therefore, AP:PB = 2:1.

Thus PQ:BC= 2/(2+1)=2:3

The correct option is (d)

Thus PQ:BC= 2/(2+1)=2:3

The correct option is (d)

Question 3 |

The area of a rectangular field is 460 square metres. If the length is 15 per cent more than the breadth, what is breadth of the rectangular field?

15 metres | |

26 metres | |

34.5 metres | |

none |

Question 3 Explanation:

Let the breadth be x m Thus the length = 1.15x m

Thus, 1.15x

x =√400

x = 20

Thus, 1.15x

^{2}= 460x =√400

x = 20

Question 4 |

ABCD is a square, F is mid-point of AB and E is a point on BC such that BE is one-third of BC. If area of ΔFBE = 108 m2, then the length of AC is

60 m | |

$ \displaystyle 36\sqrt{2}$ | |

$ \displaystyle 63\sqrt{2}$ | |

$ \displaystyle 72\sqrt{2}$ |

Question 4 Explanation:

Let the length of each side of square be 6x.

Thus FB = 3x and BE = 2x.

Area = ½ (3x)(2x) = 3x

x = 6

Thus the diagonal AC= 6 x 6√2 = 36√2 m

Thus FB = 3x and BE = 2x.

Area = ½ (3x)(2x) = 3x

^{2}= 108x = 6

Thus the diagonal AC= 6 x 6√2 = 36√2 m

Question 5 |

Inside a square plot a circular garden is developed which exactly fits in the square plot and the diameter of the garden is equal to the side of the square plot which is 28 metres. What is the area of the space left out in the square plot after developing the garden?

98 m ^{2} | |

146 cm ^{2} | |

84 m ^{2} | |

168 m ^{2} |

Question 5 Explanation:

The area of the circle = 22/7 x 14 x 14 = 22 x 28 = 616m

Area of the square = 28x28 = 784 m

Area left out = 784 - 616 = 168 m

The correct option is = 168 m

^{2}Area of the square = 28x28 = 784 m

^{2}Area left out = 784 - 616 = 168 m

^{2}The correct option is = 168 m

^{2}
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