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## Geometry and Mensuration: Level 2 Test 7

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*Geometry and Mensuration: Level 2 Test 7*.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%%
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Question 1 |

55 ^{o} | |

27.5 ^{o} | |

60 ^{o} | |

30 ^{o} |

Question 2 |

An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is 2 cm

^{2}, then the area of the hexagon is2 cm ^{2} | |

3 cm ^{2} | |

4 cm ^{2} | |

6 cm ^{2} |

Question 2 Explanation:

Let the individual sides of the hexagon be 1 cm

Thus the perimeter =6 and the side of the equilateral triangle is 2cm.

Thus the area of the equilateral triangle = √3/4 x 4 = √3cm

Thus the area of the hexagon= {(6 x √3)/4} x 1

Thus the ratio of the triangle to hexagon is 2:3.

Thus the area of the hexagon = 2 x 3/2 = 3 cm.

The correct option is (b)

Thus the perimeter =6 and the side of the equilateral triangle is 2cm.

Thus the area of the equilateral triangle = √3/4 x 4 = √3cm

Thus the area of the hexagon= {(6 x √3)/4} x 1

^{2}= (3/2)√3Thus the ratio of the triangle to hexagon is 2:3.

Thus the area of the hexagon = 2 x 3/2 = 3 cm.

The correct option is (b)

Question 3 |

In a right angled ABC, ABC= 90

^{o}; BN is perpendicular to AC, AB= 6cm, AC= 10 cm. Then AN: NC is3: 4 | |

9: 16 | |

3: 16 | |

1: 4 |

Question 3 Explanation:

Before starting calculation, one can eliminate two choices = a

and since 1 cannot be broken down in the required ratios.

By Pythagoras theorem the third side = 8 cm

The length of BN = 48/10 =4.8 cm

Thus the length of CN according to Pythagoras theorem is √40.96 =6.4 cm

The ratio = (10-6.4): 6.4 = 9:16

Correct option is (b)

Question 4 |

In a triangle PQR, R = 90

^{o}and Q is mid-point of RP. The value of PS^{2}– QS^{2}is equal toPQ ^{2} | |

2PQ ^{2} | |

3PR ^{2} | |

4RQ ^{2} |

Question 4 Explanation:

By Pythagoras Theorem

PS

QS

PS

= PR

= PR

= (PR– QR) (PR + QR)

=(2QR – QR) (2QR + QR)

= QR × 3QR = 3QR

PS

^{2}= PR^{2}+ RS^{2}QS

^{2}= QR^{2}+ RS^{2}PS

^{2}– QS^{2}= PR

^{2}+ RS^{2}– QR^{2}–RS^{2}= PR

^{2}+ QR^{2}= (PR– QR) (PR + QR)

=(2QR – QR) (2QR + QR)

= QR × 3QR = 3QR

^{2}Question 5 |

If the interior angle of a regular polygon is double the measure of exterior angle, then the number of sides of the polygon is

6 | |

8 | |

10 | |

12 |

Question 5 Explanation:

$ \displaystyle \begin{array}{l}\begin{array}{*{35}{l}}
Let\text{ }the\text{ }number\text{ }of\text{ }sides\text{ }of\text{ }the\text{ }polygon\text{ }=\text{ }n \\
We\text{ }know\text{ }that\text{ }the\text{ }interior\text{ }angle\text{ }= \\
\end{array}\\=\left( \frac{2n-4}{n} \right)\times {{90}^{o}}\\And\,\,the\,\,exterior\,\,angle\,\,=\frac{{{360}^{o}}}{n}\\\therefore \,\,from\,\,question\,\,\\\frac{2n-4}{n}\times {{90}^{o}}=\frac{2\times 360}{n}\\\Rightarrow 2n-4=8\\\Rightarrow 2n=12\\\Rightarrow n=6=Number\,\,\,of\,\,\,sides\end{array}$

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