Mensuration Model Questions Set 2 Practice Questions Answers Test with Solutions & More Shortcuts
Mensuration PRACTICE TEST [2 - EXERCISES]
Mensuration Model Questions Set 1
Mensuration Model Questions Set 2
Question : 1
The lengths of three line segments (in cm) are given in each of the four cases. Which one of the following cases is not suitable to be the three sides of a triangle?
a) 2, 4, 5
b) 2, 3, 4
c) 2, 3, 5
d) 3, 4, 5
Answer »Answer: (c)
We know that, in any triangle the sum of two sides is always greater than its third side and the difference of two sides is always less than its third side.
Only option (c) is not satisfy the above conditions
(i) 2 + 3 ≯ 5 (ii) ∼ 5 – 2 ≮ 3
Question : 2
The areas of two circles are in the ratio 1 : 2. If the two circles are bent in the form of squares, then what is the ratio of their areas?
a) 1 : $√2$
b) 1 : 2
c) 1 : 3
d) 1 : 4
Answer »Answer: (b)
According to question
$A_1/A_2 = {π r_1^2}/{πr_2^2} = 1/2$
⇒ $(r_1/r_2)^2 = 1/2$ ...(i)
As, Circles are bent in the form of square then their perimeter will become equal.
$2πr_1 = 4a_1$
⇒ $a_1 = {πr_1}/2$
Similarly, $a_2 = {π r_2}/2$
∴ $A_1/A_2 = a_1^2/a_2^2 = {({π r_1}/2)^2}/{({π r_2}/2)^2} = r_1^2/r_2^2 = 1/2$[from Eq. (i)]
Question : 3
A hollow right circular cylindrical vessel of volume V whose diameter is equal to its height, is completely filled with water. A heavy sphere of maximum possible volume is then completely immersed in the vessel. What volume of water remains in the vessel ?
a) ${2V}/3$
b) $V/2$
c) $V/3$
d) $V/4$
Answer »Answer: (a)
As per the given condition, the radius of sphere and cylinder will be same.
Volume of cylinder = $πR^2H = πR^2 .D = 2πR^3$
(where R = Radius)
Volume of sphere $V_s = 4/3 π R^3$
⇒ = ${V_S}/V ={4/3 π R^3}/{2π R^3} = 2/3 ⇒ V_s = 2/3 V$
Question : 4
In the figure given below, what is ∠BCD equal to?
![mensuration area and volume aptitude mcq 24 122](https://careericons.com/adminicon/bunch/images/mensuration-area-and-volume-aptitude-mcq-24-122.png)
a) 80°
b) 70°
c) 75°
d) 90°
Answer »Answer: (a)
∠BAC = ∠BDC {Angle made by same chord BC in the same side}
Now, from ΔBCD, sum of angles = 180°
∠CBD + ∠BDC + ∠BCD = 180°
∴ ∠BCD = 180° – 100° = 80°
Question : 5
Let the incircle to a ΔABC touch BC, AC and AB respectively at the points X, Y and Z.
- Statement I
- If AB > BC, then AB + AZ < BC + XC
- Statement II
- AZ = AY
a) Statement I is correct and Statement II is incorrect
b) Statement I and II are correct and Statements II is the correct explanation of Statement I
c) Statement I and II are correct and Statement II is not the correct explanation of Statement I
d) Statement I is incorrect and Statement II is correct
Answer »Answer: (d)
In ΔAOZ and ΔAOY,
AO = OA [common]
∠OAZ = ∠OAY [Since, OA bisect ∠A]
and ∠AZO = ∠AYO [each 90°]
∴ ΔAZO ≅ ΔAYO
So, AZ = AY [by CPCT]
Similarly, CX = CY and BX = BZ
Now, AB > BC
∴ AZ + ZB > BX + XC
AZ > XC [∵ BX = ZB]
If AB > BC, then AB + AZ > BC + XC
So, Statement I is incorrect and Statement II is correct.
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