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) 3 and 9 units
(b) 5 and 9 units
(c) 5 and 6 units
(d) 3 and 6 units
The correct answers to the above question in:
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
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Read more model questions set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
If the length of the hypotenuse of a right angled triangle is 10 cm, then what is the maximum area of such a right angled triangle ?
a) 25 cm$^2$
b) 100 cm$^2$
c) 50 cm$^2$
d) 10 cm$^2$
Answer »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 = $1/2 BP = 1/2 P^2 = 1/2. H^2/2 = H^2/4$
= ${100}/4 = 25 cm^2$
Question : 2
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 : 3
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 : 4
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 : 5
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
Question : 6
ABCD is a rectangle of dimensions 6 cm × 8 cm. DE and BF are the perpendiculars drawn on the diagonal of the rectangle. What is the ratio of the shaded to that of unshaded region?
a) 4:3 $√2$
b) 7 : 3
c) 16 : 9
d) Data insufficient
Answer »Answer: (c)
${\text"Area of Δ DAE "}/{\text"Area of Δ DEC"} = {1/2 × DE × AE}/{1/2 × DE × CE}$
= ${AE}/{CE} = {(AD)^2}/{(DC)^2} = (6/8)^2 = {9/{16}}$
Similarly, in ΔABC,
${\text"Area of ΔBCF"}/{\text"Area of ΔBFA"} = 9/{16}$
∴ The area of shaded to unshaded region = ${16}/9$
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