Mensuration Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 2 EXERCISES
The following question based on Mensuration topic of quantitative aptitude
(a) 25 cm$^2$
(b) 100 cm$^2$
(c) 50 cm$^2$
(d) 10 cm$^2$
The correct answers to the above question in:
Answer: (a)
Area will be maximum when P and B will be same
So $P^2 + P^2 = H^2 ⇒ P^2 = H^2/2$
⇒ P = $H/√2$
Area = $1/2 BP = 1/2 P^2 = 1/2. H^2/2 = H^2/4$
= ${100}/4 = 25 cm^2$
Discuss Form
Read more model questions set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
If area of a circle and a square are same, then what is the ratio of their perimeters ?
a) $√π/2$
b) 2 $√π$
c) $√π$
d) $√π/4$
Answer »Answer: (a)
Let radius of cirlce is ‘r’ and side of the square is a ∼ then, from question
$π r^2 = a^2 ⇒ r/a = 1/√π$
Now ∼ ${2πr}/{4a} = {2π}/{4√π} = √π/2$
Question : 2
Which one of the following is a Pythagorean triple in which one side differs from the hypotenuse by two units?
a) ($2n^2$ , 2n, 2n + 1)
b) (2n + 1, 4n, $2n^2$ + 2n)
c) (2n, 4n, $n^2$ + 1)
d) (2n, $n^2$ – 1, $n^2$ + 1) Where, n is a positive real number.
Answer »Answer: (d)
According to Pythagorean triplet.
The sum of square of base and perpendicular equal to square of hypotenuse.
By hit and trial method:—
$(2n)^2 + (n^2 – 1)^2 = (n^2 + 1)^2$
$4n^2 + n^4 + 1 – 2n^2 = n^4 + 2n^2$ + 1
$n^4 + 2n^2 + 1 = n^4 + 2n^2$ + 1
LHS = RHS
Question : 3
A circle of radius r is inscribed in a regular polygon with n sides (the circle touches all sides of the polygon). If the perimeter of the polygon is p, then the area of the polygon is
a) ${pr}/2$
b) (p + n) r
c) (2p – n) r
d) None of the above
Answer »Answer: (a)
The n-sided polygon can be dinded into 'n' triangle with O, the Centre of the circle as one vertex for each triangle. The altitude of each triangle is r. Let the sides of the polygon be '$a_1 ', a_2 ----- a_n$.(Given $a_1 = a_2 = ------ a_n$)
∴ The area of polygon is ${nr}/2 = {pr}/2$
Area of polygon = ${a_1r}/2 + {a_2r}/2 + --- {a_nr}/2 = {pr}/2$
So, option (a) is correct.
Question : 4
The hypotenuse of a right triangle is 3$√10$ unit. If the smaller side is tripled and the longer side is doubled, new hypotenuse becomes 9$√5$ unit. What are the lengths of the smaller and longer sides of the right triangle, respectively?
a) 3 and 9 units
b) 5 and 9 units
c) 5 and 6 units
d) 3 and 6 units
Answer »Answer: (a)
Suppose the smaller and larger sides of a right triangle be x and y, respectively.
By given condition,
$x^2 + y^2 = (3 √{10})^2$
⇒ $x^2 + y^2$ = 90 ... (i)
and $9x^2 + 4y^2$ = 405 ... (ii)
On solving Eqs. (i) and (ii), we get
x = 3 units and y = 9 units
Question : 5
PQ is a common chord of two circles. APB is a secant line joining points A and B on the two circles. Two tangents AC and BC are drawn. If ∠ACB = 45°, then what is ∠AQB equal to?
a) 120°
b) 75°
c) 90°
d) 135°
Answer »Answer: (d)
The tangents drawn from an outer point on a circle are always equal = ∠CBA.
Therefore, ∠CAB = ∠CBA
∴ 45° + x + x = 180°
⇒ 2x = 180° – 45°
⇒ x = 67 ${1°}/2$
∠AQP = ∠x = ∠BQP
= 67 ${1°}/2$
(alternate interior segments properties)
⇒ ∠AQB = ∠AQP + ∠BQP
= $67 {1°}/2 + 67{1°}/2$ = 135°
Question : 6
The diagonals of a trapezium are at right angles, and the slant sides, if produced, form an equilateral triangle with the greater of the two parallel sides. If the area of the trapezium is 16 square cm, then the distance between the parallel sides is
a) 8 cm
b) 2 cm
c) 4 cm
d) Cannot be determined due to insufficient data
Answer »Answer: (c)
ΔEAB is equilateral
ΔEDC is also equilateral
Area of trapezium ABCD
= $(1/2 × DB × OA) + 1/2 (DB × OC)$
= $1/2$ × DB × AC
Let AO = OB = x and DO = OC = y
Area (ABCD) = $1/2 (x + y)^2$ = 16(given)
⇒ x + y = 4 $√2$ ... (i)
ΔAOB is a right angled isosceles triangle.
So, AB = $√{x^2 + x^2} = √2x$
Similarly, DC = $√2$y
Now, FG = EF – EG
⇒ FG = AB sin 60° – DC sin 60°
= $√3/2 (AB - DC) = √6/2$ (x - y) ... (ii)
Area of trapezium
= Area ΔEAB – Area ΔEDC
= $√3/4 (AB^2 - DC^2)$
= $√3/4 [(x√2)^2 - (y√2)^2]$
⇒ Area = $√3/2$ (x + y) (x – y)
Now, $√3/2$ (x + y)(x - y) = 16
⇒ x – y = ${32}/{√3 (x + y)} ⇒ x - y = 8/√6 (∵ x + y = 4√2)$
Height = $√6/2 (x - y) = √6/2 × 8/√6$ = 4 cm
GET Mensuration PRACTICE TEST EXERCISES
Mensuration Model Questions Set 1
Mensuration Model Questions Set 2
Mensuration Shortcuts and Techniques with Examples
Verbal Reasoning
Question & Answer Quiz
Non Verbal Reasoning
Question & Answer Quiz
Quantitative Aptitude
Question & Answer Quiz
Computer MCQ
Question & Answer Quiz
General English
Question & Answer Quiz
History GK
Question & Answer Quiz
Polity GK
Question & Answer Quiz
Geography GK
Question & Answer Quiz
Economy GK
Question & Answer Quiz
General Awareness GK
Question & Answer Quiz
Recently Added Subject & Categories For All Competitive Exams
100+ Quadratic Equation Questions Answers PDF for Bank
Quadratic Equation multiple choice questions with detailed answers for IBPS RRB SO. more than 250 Attitude practice test exercises for all competitive exams
Continue Reading »
IBPS Aptitude Linear Equations MCQ Questions Answers PDF
Linear equations multiple choice questions with detailed answers for IBPS RRB SO. more than 250 Attitude practice test exercises for all competitive exams
Continue Reading »
New 100+ Compound Interest MCQ with Answers PDF for IBPS
Compound Interest verbal ability questions and answers solutions with PDF for IBPS RRB PO. Aptitude Objective MCQ Practice Exercises all competitive exams
Continue Reading »
100+ Mixture and Alligation MCQ Questions PDF for IBPS
Most importantly Mixture and Alligation multiple choice questions and answers with PDF for IBPS RRB PO. Aptitude MCQ Practice Exercises all Bank Exams
Continue Reading »