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- These tests focus on geometry and mensuration and are meant to indicate your preparation level for the subject.
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Geometry and Mensuration: Test 18
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Question 1 |
An angle is equal to 1/3rd of its supplement. Find its measure.
60o | |
80o | |
90o | |
45o |
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 = 90o and D is mid-point of AC. The value of BC2 – BD2 is equal to
AD2 | |
2AD2 | |
3AD2 | |
4AD2 |
Question 2 Explanation:
BC2 = AB2 + AC2
BD2 = AB2 + AD2
BC2 – BD2 = AB2 + AC2 – AB2 –AD2
= AC2 – AD2
= (AC– AD) (AC + AD)
=(2AD – AD) (2AD + AD)
= AD × 3AD = 3AD2
BD2 = AB2 + AD2
BC2 – BD2 = AB2 + AC2 – AB2 –AD2
= AC2 – AD2
= (AC– AD) (AC + AD)
=(2AD – AD) (2AD + AD)
= AD × 3AD = 3AD2
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 3 Explanation:
In equilateral triangle centroid, incentre,
orthocenter coincide at the same points
Height/3 = inradius
Therefore Height= Median= 3×3= 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 4 Explanation:
Let AB = 2 unit.
As the triangle is an equilateral triangle, the perpendicular from A will bisect BC. (in this case at D) BD = 1 unit.
Required ratio AB: BD= 2: 1.
Correct option is (c)
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 = x2 + 64
=> x2 + 4 + 4x = x2 + 64
=> 4x = 60 => x = 15m.
Hence the length of the ladder = 15 + 2 = 17m.
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