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) x = 5 cm and y = 12 cm
(b) x = 6.5 cm and y = 6.5 cm
(c) x = 12 cm and y = 5 cm
(d) x = 0 cm and y = 13 cm
The correct answers to the above question in:
Answer: (b)
P does not lie on the circle at x = 6.5cm and
y = 6.5 cm because, as <APB = 90°
then, $AB^2 ∼ ∼ = AP^2 ∼ ∼ + BP^2$
$(13)^2 ≠ (6.5)^2 ∼ ∼ + (6.5)^2$
Hence, point P does not lies on the circle ∼
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Read more model questions set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
In the figure given above, ABCD is a trapezium. EF is parallel to AD and BG. ∠y is equal to
a) 60°
b) 30°
c) 45°
d) 65°
Answer »Answer: (a)
From figure.
BC || EF || AD
∴ x° = z° = 50° (corresponding interior angle
∴ θ + z° = 180° (linear pair)
∴ ∼ θ = 180° – 50° = 130°
In quadrilateral
AQFD, x° + y° + 120° + θ = 360°
50° + y° + 120° + 130° = 360°
y = 360° – 300° = 60°
Question : 2
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 = ${16}/9$
Question : 3
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 : 4
ABCD is a quadrilateral with AB = 9 cm, BC = 40 cm, CD = 28 cm, DA = 15 cm and angle ABC is a right-angle.
What is the area of quadrilateral ABCD ?
a) 312 $cm^2$
b) 300 $cm^2$
c) 306 $cm^2$
d) 316 $cm^2$
Answer »Answer: (c)
In right triangle ABC,
AC = $√{(AB)^2 + (BC)^2} = √{(40)^2 + (9)^2}$ = 41 cm
Area of ΔABC = $1/2 × 9 × 40 = 180 cm^2$
∵ Area of quadrilateral ABCD
= 126 + 180 = 306 $cm^2$
Question : 5
The dimensions of a field are 15 m by 12 m. A pit 8 m long, 2.5 m wide and 2 m deep is dug in one corner of the field and the earth removed is evenly spread over the remaining area of the field. The level of the field is raised by
a) 25 cm
b) 15 cm
c) 20 cm
d) ${200}/9$ cm
Answer »Answer: (a)
Volume of pit = l b h = 8 × 2.5 × 2 = 40 $m^3$.
Let the label of the earth spread over remaining area = h.
Volume of the earth spread = Volume of a pit
⇒ [(12 × 15) – (8 × 2.5)] × h = 40
∴ h = ${40}/{180 - 20} = {40}/{160} = 1/4$m = 25 cm
Question : 6
In the figure given above, LM is parallel to QR. If LM divides the ΔPQR such that area of trapezium LMRQ is two times the area of ΔPLM, then what is ${PL}/{PO}$ equal to?
a) $1/2$
b) $1/√2$
c) $1/√3$
d) $1/3$
Answer »Answer: (c)
In the given figure.
ar MRQL = 2 ar ΔPLM
Let area of ΔPLM be x, then
∴ the area of trapezium = 2x
∴ ar ΔPQR = 2x + x = 3x
Here it is clear from the given figure that ΔPQR ∼ ΔPLM
∴ ${\text"ar ΔPQR"}/{\text"ar ΔPLM"} = {3x}/x$
${PL^2}/{PQ^2} = 1/3 ∴ = {PL}/{PQ} = 1/√3$
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