- This is an assessment test.
- To draw maximum benefit, study the concepts for the topic concerned.
- Kindly take the tests in this series with a pre-defined schedule.

## Geometry and Mensuration: Level 2 Test 1

Congratulations - you have completed

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

Question 1 |

The area of a square is 1444 square metre. The breadth of a rectangle is 1/4th of the side of the square and the length of the rectangle is thrice the breadth. What is the difference between the area of the square and the area of the rectangle?

1152.38 sq. mtr. | |

1169.33 sq. mtr. | |

1181.21 sq. mtr. | |

1173.25 sq. mtr. |

Question 1 Explanation:

The side of the square = 38 m.

The breath of the rectangle = 38/4 m

Length = 3x38/4 m

Difference between area of square and rectangle = 1444 –(3x38/4 x 38/4) = 1173.25 sq.m

The breath of the rectangle = 38/4 m

Length = 3x38/4 m

Difference between area of square and rectangle = 1444 –(3x38/4 x 38/4) = 1173.25 sq.m

Question 2 |

If area of an equilateral triangle is a and height b, then value of b

^{2}/a is: 3 | |

1/3 | |

√3 | |

1/√3 |

Question 2 Explanation:

$ \begin{array}{l}Let\text{ }the\text{ }side\text{ }of\text{ }the\text{ }equilateral\text{ }triangle\text{ }be\text{ }x\text{ }units.\\Area\text{ }=\text{ }a\text{ }=\surd 3/4\text{ }{{x}^{2}}~\\and\text{ }Height=b=\text{ }\surd 3/2\\\frac{{{b}^{2}}}{a}=\frac{3}{4}{{x}^{2}}\times \frac{4}{\sqrt{3}}\times \frac{1}{{{x}^{{}}}}=\sqrt{3}\end{array}$

Question 3 |

Circumference of a circle-A is 11/7 times perimeter of a square. Area of the square is 784 sq. cm. What is the area of another circle-B whose diameter is half the radius of the circle-A?

38.5 sq. cm | |

156 sq. cm | |

35.8 sq. cm | |

None of these |

Question 3 Explanation:

The side of the square = 28 cm.

Perimeter = 4 X 28 = 112 cm

The circumference of the circle = 11/7 x 112 = 22/7 x 2 x 28

Radius = 28 cm

Half the radius of Circle-A = 14 cm.

Radius of Circle-B = 14/2 = 7 cm

The area = 22/7 x 7 x 7 = 54 cm

Perimeter = 4 X 28 = 112 cm

The circumference of the circle = 11/7 x 112 = 22/7 x 2 x 28

Radius = 28 cm

Half the radius of Circle-A = 14 cm.

Radius of Circle-B = 14/2 = 7 cm

The area = 22/7 x 7 x 7 = 54 cm

^{2}Question 4 |

An isosceles triangle ABC is right-angled at B.D is a point inside the triangle ABC. P and Q are the feet of the perpendiculars drawn from D on the sides AB and AC respectively of ΔABC. If AP= a cm, AQ= b cm and ∠BAD= 15

^{o}, sin 75^{o}=$ \displaystyle \frac{2b}{\sqrt{3}\,\,a}$ | |

$ \displaystyle \frac{a}{2b}$ | |

$ \displaystyle \frac{\sqrt{3}\,\,a}{2b}$ | |

$ \displaystyle \frac{2\,\,a}{\sqrt{3}\,b}$ |

Question 4 Explanation:

$ \displaystyle \begin{array}{l}To\text{ }find\text{ }Sin\text{ }{{75}^{0}}we\text{ }need\text{ }AD\\So\text{ }\\From\,\,\,\,\Delta AQD\\\sin {{60}^{o}}\,\,\,=\frac{AQ}{AD}\\\frac{\sqrt{3}}{2}=\frac{b}{AD}\\\Rightarrow AD=\frac{2b}{\sqrt{3}}\\Now\,\,Sin\theta =\frac{perpendicular}{hypotenuse}\\so\,\,Sin{{75}^{0}}=\frac{a}{AD}\\OR\\\sin {{75}^{o}}=\frac{AP}{AD}=\frac{a}{\frac{2b}{\sqrt{3}}}=\frac{\sqrt{3}a}{2b}\end{array}$

Question 5 |

A horse is tethered to a peg with a 14 metre long rope at the corner of a 40 metre long and 24 metre wide rectangular grass field. What area of the field will the horse graze?

154 m ^{2} | |

308 m ^{2} | |

240 m ^{2} | |

480 m ^{2} |

Once you are finished, click the button below. Any items you have not completed will be marked incorrect.

There are 5 questions to complete.

List |