Mensuration Test-2

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Mensuration Test-2

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Question 1
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A 4 cm cube is cut into 1 cm cubes. What is the percentage increase in the surface area after cutting?

A
4 %

B
300 %

C
75 %

D
400 %

Solution

Volume of 4 cm cube = 64 cc. When it is cut into 1 cm cube, the volume of each of the cubes = 1cc
Hence, there will be 64 such cubes. Surface area of small cubes = 6 (1² ) = 6 sqcm.
Therefore, the surface area of 64 such cubes = 64 * 6 = 384 sqcm.
The surface area of the large cube = 6(4² ) = 6*16 = 96.
% increase
Question 2
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A regular hexagon is inscribed in a circle of radius r cm. What is the perimeter of the regular hexagon?

A
3r

B
6r

C
√72 r

D
9 r

Solution

A regular hexagon comprises 6 equilateral triangles, each of them having one of their vertices at the center of the hexagon. The sides of the equilateral triangle are equal to the radius of the smallest circle inscribing the hexagon. Hence, each of the side of the hexagon is equal to the radius of the hexagon and the perimeter of the hexagon is 6r.
Question 3
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If x units are added to the length of the radius of a circle, what is the number of units by which the circumference of the circle is increased?

A
x

B
2x

C
2 πx

D
x²

Solution

Let the radius of the circle be 'r 'units.
The circumference of the circle will therefore be 2 πr units.
If the radius is increased by 'x 'units, the new radius will be (r + x) units.

The new circumference will be
Or the circumference increases by 2 πx units .
Question 4
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ABCD has area equal to 28. BC is parallel to AD. BA is perpendicular to AD. If BC is 6 and AD is 8, then what is CD?

A
2 √2

B
2 √3

C
4

D
2 √5

Solution

The given shape is a trapezium.
Area of a trapezium
28 =
Height = 4.

BA is perpendicular to BC and AD.

So, drop a line parallel to BA from C to meet AD at E.
CED is a right triangle with side CE measuring 4 and ED measuring 2 units.

Hence, CD, the measure of the hypotenuse =
Question 5
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Which of the following figures has the largest area?
A circle of radius √2
An equilateral triangle whose sides each have length 4
A triangle whose sides have lengths 3, 4 and 5

A
I

B
II

C
III

D
I and III

Solution

Area of circle of radius
Area of an equilate ral triangle whose sides each have length

Area of a triangle whose sides measure 3, 4 and 5. The given triangle is a right triangle.
Hence, its area
We need to compare these three numbers.

Area of the circle,   6.28 sq units

Area of the equilateral triangle
Area of the right triangle = 6 sq units.

The area of the equilateral triangle is the largest amongst the 3 objects given in this question.
Question 6
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The hexagon ABCDEF is regular. That means all its sides are of the same length and all its interior angles are of the same size. Each side of the hexagon is 2 m. What is the area of the rectangle BCEF?

A
4 sq.m

B
4 √3 sq.m

C
8 sq.m

D
4 + 4 √3 sq.m

Solution

A regular hexagon comprises six equilateral triangles - each of side 2 m, the measure of the side of the regular hexagon - as shown above. The 6 triangles are numbered 1 to 6 in the figure shown above.

BX is the altitude of triangle 1 and XF is the altitude of triangle 2.
Both triangle 1 and triangle 2 are equilateral triangles.

Hence, BX = XF =
Therefore, BF, the length of the rectangle = 2 √3 M
Hence, the area of the rectangle BCEF = length * width = 2 √3 * 2 = 4 √3 sq. m
Question 7
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Ram a farmer, managed to grow shaped-watermelons inside glass cases of different shapes. The shapes he used were: a perfect cube, hemi-spherical, cuboid, cylindrical along with the normal spherical shaped watermelons. Thickness of the skin was same for all the shapes. Each of the glass cases was so designed that the total volume and the weight of the all the water-melons would be equal irrespective of the shape.
A customer wants to buy water-melon for making juice, for which the skin of the water-melon has to be peeled off, and therefore is a waste. Which shape should the customer buy?

A
Cube

B
Hemi - sphere

C
Cuboid

D
Normal spherical

Solution

This is the kind of question that is either very easy or very difficult depending on whether you know the concept behind the question.

For a given surface area, the volume contained increases with increasing symmetry of the object. For instance, if we are to make water melons of different shapes of the same surface area, the volume will be maximum when it is made into a sphere.

The corollary is that for a given volume, the surface area will be minimum when the object is a sphere. So, the customer should opt for spherical shaped water melons if she has to minimize wastage.

For 2-dimensional object, for a given perimeter, the area increases with increasing symmetry.

Among different triangles of a given perimeter, an equilateral triangle has the largest area.

The area increases with increasing number of sides - i.e., for a given perimeter the area of a square will larger than that of an equilateral triangle; the area of a regular pentagon of a given perimeter will be larger than that of a square and so on.

Among different regular polygons of a given perimeter / circumference a circle has the largest area.
Question 8
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The figure below has been obtained by folding a rectangle. The total area of the figure (as visible) is 144 square meters. Had the rectangle not been folded, the current overlapping part would have been a square. What would have been the total area of the original unfolded rectangle?

A
128 square meters

B
154 square meters

C
162 square meters

D
172 square meters

Solution

Folded part as shown in the first figure is a triangle - a right triangle.

The two perpendicular sides of the right triangle measure 6m each. So, the triangle is a right isosceles triangle.

When unfolded the folded area becomes a square as shown in the following figure.

The side of the square will be the width of the larger rectangle and is therefore, 6m.

Area of the square = 6 * 6 = 36 sq.m

When folded, only the area of the right triangle gets counted.
However, when unfolded the area of square gets counted.
The square comprises two congruent right triangles.

In essence, when folded only half a square is counted. When unfolded the entire square gets counted.

The area of the rectangle when unfolded = area of the rectangle when folded + area of half a square.
So area after unfolding= 144 + 18 = 162 sq.m.
Question 9
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A solid metal cylinder of 10 cm height and 14 cm diameter is melted and re-cast into two cones in the proportion of 3 : 4 (volume), keeping the height 10 cm. What would be the percentage change in the flat surface area before and after?

A
9 %

B
16 %

C
25%

D
50 %

Solution

Change in flat surface area = 21k –14k = 7k
% change in flat surface area 7k/14k * 100 = 50%.
Question 10
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A circular road is constructed outside a square field. The perimeter of the square field is 200 ft. If the width of the road is 7 √2 ft. and cost of construction is Rs. 100 per sq. ft. Find the lowest possible cost to construct 50% of the total road.

A
Rs.70,400

B
Rs.1,25,400

C
Rs.1,40,800

D
Rs.2,35,400

Solution

The area of the circular road
= πr_o² - πr_i² , where r_o is the outer radius and r_i is the inner radius.

25= 2508 sq. ft.

If per sq. ft. cost is Rs. 100, then cost of constructing the road = 2508 ×100 = Rs.2,50,800.

Cost of constructing 50% of the road = 50% of the total cost = 250800/2 =Rs.1,25,400