- This is an assessment test.
- These tests focus on geometry and mensuration and are meant to indicate your preparation level for the subject.
- Kindly take the tests in this series with a pre-defined schedule.

## Geometry and Mensuration: Test 28

Congratulations - you have completed

*Geometry and Mensuration: Test 28*. 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 |

ABCD is a rhombus. AB is produced to F and BA is produced to E such that AB= AE= BF. Also ED and CF are produced to meet at G. Then:

ED > CF | |

EGâŠ¥GF | |

ED ^{2} +CF^{2}= EF^{2} | |

ED||CF |

Question 1 Explanation:

height="246" />

Here AB= AE= BF.......(1) Given ABCD is a rhombus.

Therefore, AB = BC = CD = AD ... (2)

On equating (1) and (2), we get

BC = BF

â‡’âˆ 4 = âˆ 3 [Angles opposite to equal sides are equal]

Again, âˆ B is the exterior angle of triangle BFC.

Therefore, âˆ 2 = âˆ 3 + âˆ 4 = 2âˆ 4 ... (3)

Similarly, AE = AD â‡’âˆ 5 = âˆ 6

Also, âˆ A is the exterior angle of triangle ADE

â‡’âˆ 1 = 2âˆ 6 ... (4)

Also, âˆ 1 +âˆ 2 = 180Â° [consecutive interior angles]

âˆ´ 2âˆ 6 + 2âˆ 4 = 180Â°

â‡’âˆ 6 + âˆ 4 = 90Â° ... (5)

Now, in triangle EGF, by angle sum property of triangle

â‡’ 90Â°+ âˆ G = 180Â° [using (5)]

Thus, EG âŠ¥ FG.

Here AB= AE= BF.......(1) Given ABCD is a rhombus.

Therefore, AB = BC = CD = AD ... (2)

On equating (1) and (2), we get

BC = BF

â‡’âˆ 4 = âˆ 3 [Angles opposite to equal sides are equal]

Again, âˆ B is the exterior angle of triangle BFC.

Therefore, âˆ 2 = âˆ 3 + âˆ 4 = 2âˆ 4 ... (3)

Similarly, AE = AD â‡’âˆ 5 = âˆ 6

Also, âˆ A is the exterior angle of triangle ADE

â‡’âˆ 1 = 2âˆ 6 ... (4)

Also, âˆ 1 +âˆ 2 = 180Â° [consecutive interior angles]

âˆ´ 2âˆ 6 + 2âˆ 4 = 180Â°

â‡’âˆ 6 + âˆ 4 = 90Â° ... (5)

Now, in triangle EGF, by angle sum property of triangle

â‡’ 90Â°+ âˆ G = 180Â° [using (5)]

Thus, EG âŠ¥ FG.

Question 2 |

Two equal circles of radius 4 cm intersect each other such that each passes through the centre of the other. The length of the common chord is:

$ \displaystyle 2\sqrt{3}$ | |

$ \displaystyle 4\sqrt{3}$ | |

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

$ \displaystyle 8$ |

Question 2 Explanation:

$ \begin{array}{l}The\text{ }length\text{ }of\text{ }the\text{ }common\text{ }chord\text{ }=\text{ }\\2\text{ }\times \text{ }length\text{ }of\text{ }the\text{ }altitude\text{ }of\text{ }the\text{ }equilateral\text{ }triangle\text{ }of\text{ }side\text{ }4\text{ }cm.\\Thus\text{ }the\text{ }length\text{ }of\text{ }common\text{ }chord\text{ }is\\2\times \frac{\sqrt{3}}{2}\times 4\\=4\sqrt{3}\end{array}$

Question 3 |

The length of two chords AB and AC of a circle are 8 cm and 6 cm and âˆ BAC= 90

^{o}, then the radius of circle is25 cm | |

20 cm | |

4 cm | |

5 cm |

Question 4 |

The tangents are drawn at the extremities of a diameter AB of a circle with centre P. If a tangent to the circle at the point C intersects the other two tangents at Q and R, then the measure of the âˆ QPR is

45 ^{o} | |

60 ^{o} | |

90 ^{o} | |

180 ^{o} |

Question 5 |

The circumcentre of a triangle Î”ABC is O. If âˆ BAC= 85

^{o}and BCA= 75^{o}, then the value of âˆ OAC is40 ^{o} | |

60 ^{o} | |

70 ^{o} | |

90 ^{o} |

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 |