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
What is the area of quadrilateral ABCD ?
(a) 312 $cm^2$
(b) 300 $cm^2$
(c) 306 $cm^2$
(d) 316 $cm^2$
The correct answers to the above question in:
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 $cm^2$
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Read more model questions set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
A line segment AB is the diameter of a circle with centre at O having radius 6.5 cm. Point P is in the plane of the circle such that AP = x and BP = y. In which one of the following cases the point P does not lie on the circle ?
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
Answer »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 ∼
Question : 2
In the figure given above, ABCD is a trapezium. EF is parallel to AD and BG. ∠y is equal to
![mensuration area and volume aptitude mcq 23 38](https://careericons.com/adminicon/bunch/images/mensuration-area-and-volume-aptitude-mcq-23-38.png)
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 : 3
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?
![mensuration aptitude mcq a 12](https://careericons.com/adminicon/bunch/images/mensuration-aptitude-mcq-a-12.png)
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 : 4
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 : 5
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
∴ ${\text"ar ΔPQR"}/{\text"ar ΔPLM"} = {3x}/x$
${PL^2}/{PQ^2} = 1/3 ∴ = {PL}/{PQ} = 1/√3$
Question : 6
Two straight lines AB and AC include an angle. A circle is drawn in this angle which touches both these lines. One more circle is drawn which touches both these lines as well as the previous circle. If the area of the bigger circle is 9 times the area of the smaller circle, then what must be the angle A?
a) 75°
b) 45°
c) 60°
d) 90°
Answer »Answer: (c)
Let the radius of the bigger circle be a and radius of the smaller circle be b.
Then the angle made by direct common tangents when two circles of radius a and b touch externally is given by θ = $2sin^{- 1}({a - b}/{a + b})$
We are given that area of the bigger circle = 9 area of the smaller circle
⇒ $πa^2 = 9πb^2 ⇒ a^2 = 9b^2 ⇒ a = 3b$
Let us consider ∠BAC = θ
Thus,
θ = $2 sin^{-1} ({a - b}/{a + b}) = 2 sin^{-1} ({3b - b}/{3b + b}) = 2sin^{-1} ({2b}/{4b})$
= $2 sin^{-1} (1/2) = 2 sin^{-1}$ (sin 30°) = 2 × 30° = 60°
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