- 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 18

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*Geometry and Mensuration: Test 18*.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 |

An angle is equal to 1/3

^{rd}of its supplement. Find its measure.60 ^{o} | |

80 ^{o} | |

90 ^{o} | |

45 ^{o} |

Question 1 Explanation:

$\displaystyle \begin{array}{l}\begin{array}{*{35}{l}}
Let\text{ }that\text{ }angle\text{ }be\text{ }p \\
Supplement\text{ }of\text{ }this\text{ }angle\text{ }=\text{ }180-p \\
\end{array}\\Therefore\,\,\,p=\frac{1}{3}\,\left( 180-p \right)\\\Rightarrow 3p=180-p\\\Rightarrow p={{45}^{o}}\end{array}$

Question 2 |

In a triangle ABC, ∠A = 90

^{o}and D is mid-point of AC. The value of BC^{2}– BD^{2}is equal toAD ^{2} | |

2AD ^{2} | |

3AD ^{2} | |

4AD ^{2} |

Question 2 Explanation:

BC

BD

BC

= AC

= (AC– AD) (AC + AD)

=(2AD – AD) (2AD + AD)

= AD × 3AD = 3AD

^{2}= AB^{2}+ AC^{2}BD

^{2}= AB^{2}+ AD^{2}BC

^{2}– BD^{2}= AB^{2}+ AC^{2}– AB^{2}–AD^{2}= AC

^{2}– AD^{2}= (AC– AD) (AC + AD)

=(2AD – AD) (2AD + AD)

= AD × 3AD = 3AD

^{2}Question 3 |

The in-radius of an equilateral triangle is of length 3 cm. Then the length of each of its medians is

12 cm | |

9/2 cm | |

4 cm | |

9 cm |

Question 4 |

If ABC is an equilateral triangle and D is a point on BC such that AD⊥BC ,then

AB: BD= 1: 1 | |

AB: BD= 1: 2 | |

AB: BD= 2: 1 | |

AB: BD= 3: 2 |

Question 5 |

A ladder leans against a vertical wall. The top of the ladder is 8m above the ground. When the bottom of the ladder is moved 2 m farther away from the wall, the top of the ladder rests against the foot of the wall. What is the length of the ladder?

10 m | |

15 m | |

20 m | |

17 m |

Question 5 Explanation:

Let AB be the wall and AC is the ladder with C as the foot of the ladder.

Let BC = x and CD = 2. So the length of the ladder is x + 2 i.e. AC = x + 2.

Now in ΔABC, we have (x + 2)

^{2}= x

^{2}+ 64

=> x

^{2}+ 4 + 4x = x

^{2}+ 64

=> 4x = 60 => x = 15m.

Hence the length of the ladder = 15 + 2 = 17m.

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