# AREA AND VOLUME (MENSURATION) PART-4

#### AREA AND VOLUME (MENSURATION) PART-4

Most of these questions are taken from the previous year’s examinations conducted by the Staff Selection Commission (SSC) on Area and Volume (Mensuration) of the Quantitative Aptitude (Maths) section of the following exams as well as other similar exams. They are all solved and supported by detailed explanation.

1. Combined Graduate Level (CGL) Exam Tier-I & Tier-II

2. Combined Higher Secondary (10+2) Exam (CHSL) Tier-I

3. SI in Delhi Police and CPO Exam Paper-I & Paper-II

4. Grade-II DASS Exam conducted by Delhi Staff Subordinate Services (DSSSB)

#### QUERY 31

**The length of a hollow thick cylindrical metallic pipe is 6 cm and its total surface area including the surfaces at the ends is 98π cm². If the outer diameter is 8 cm, then the inner diameter in cm is?**

A) 6.5

B) 7

C) 5

D) 6

**MAHA GUPTA**

The outer radius (R) = 8/2 = 4 cm

Height (h) = 6 cm

Now, Let the inner diameter be 2r cm

Therefore the inner radius (r) = r cm

The total surface area of the hollow thick cylinder = External curved surface + Internal curved surface + 2(area of the base of the ring) = 2πRh + 2πrh + 2(πR² – πr²)

Therefore, 2πRh + 2πrh + 2(πR² – πr²) = 98π

Solving r = 3 cm

Hence the inner diameter = 2*3 = 6 cm (option ‘D’)

QUERY 32

**In an isosceles triangle, the measure of each of equal sides is 10 cm and the angle between them is 45°. Area of the triangle is?**

**ROHIT SHARMA
**To make things easier make this triangle into a parallelogram with its equal sides being adjacent ones.

We know area of the parallelogram = product of adjacent sides × sine of included angle = 10*10*sin45 = 100*(1/√2) = 50√2 cm²

Hence, area of the given triangle = half area of the above parallelogram i.e. = (50√2)/2 = 25√2 cm² (option ‘C’)

QUERY 33

**A circle with radius 5 cm and an arc on circle 3.5 cm which joins centre O with OA and OB, then what is the area of that sector?**

A) 9 cm²

B) 8.75 cm²

C) 11 cm²

D) 11.25 cm²

**MAHA GUPTA**

Area of the sector when length of the arc is given = (Length of the arc*area of the circle)/Circumference of the circle

So, the area of the sector here = (3.5 x π5²)/2πr

= (3.5*5)/2 = 17.5/2 = 8.75 cm² (option ‘B’)

NOTE: Area of the sector if angle is given = (angle/360)*πr²

QUERY 34

**Area of an isosceles trapezium is 176 cm2 and height is 2/11 of the sum of its parallel sides. Ratio of parallel side is 4 : 7; then length of diagonal?**

A) 528 cm

B) √528 cm

C) 50 cm

D) 25 cm

**MAHA GUPTA**

Let parallel sides be a and b and height is h.

Area of trapezium = (1/2)*sum of parallel sides*height

So, 176 = = (1/2)(a + b)h

h = (a + b)*(2/11) (given)

Putting this in the above equation

176 = = (1/2)(a + b)(a + b)(2/11)

=> (a + b)² = 1936

=> a + b = 44

Now ratio of a and b = 4 : 7

Therefore, a = 16, b = 28 and h = 44*(2/11) = 8

Now the figure will become like this, where triangle BEC is a right triangle in which CE = 16 + 6 = 22, height (h) = 8 and BC the hypotenuse which will incidentally be one of the two equal diagonals of the given trapezium. Remember diagonals of an isosceles trapezium are equal.

So by Pythagoras the the diagonal of the trapezium = √(CE² + BE²) = √(22² + 8²) = √528 (option ‘B’)

#### QUERY 35

**A lotus is seen 5 cm above the water level of a lake. With the onset of the wind it sinks in the water 10 cm away from its place. How deep is the water in that place?**

A) 5 cm

B) 5√5 cm

C) 7.5 cm

D) 10 cm

**KANW@LJEET**

Assume the length of the lotus is in water x cm

The length of the lotus above water is 5 cm (given)

When it sinks, you’ll see a right angled triangle will be formed with its dimensions as x cm, (x+5) cm, 10 cm.

Using pythagoras we get

(x+5)² = x² + 10²

=> 25 + 10 x = 100

=> x = 7.5 cm, so option ‘C’ is correct.

QUERY 36

**A semi circular sheet of diameter 28 cm is converted into a cone. What is the volume of the cone?**

A) 525 cm³

B) 622.12 cm³

C) 575.15 cm³

D) 600 cm³

**RONNIE BANSAL**

Just think in what way a semi-circular sheet of anything can be converted into a cone. Obviously the mid-point of the diameter of that semi-circle will be the point of that cone and the radius will become the slant height of that; and the curved part of the circumference of the semi-circle as the circumference of the base of the cone.

So here the slant height (l) of the cone = 28/2 = 14

And the circumference of the base of the cone (2πr) = πr of the semi-circle = (22/7)*14 = 44

In other words 2πr = 44

=> r = 7

For volume of the cone to be found we need height (h) of the cone also

So h² = l² – r²

=> h = √(14² – 7²)

=> h = √147

= 12.12 (appox)

Now the volume of the cone = (1/3)πr²h

= (1/3)(22/7)(7²)12.12

= 622.12 cm³ (option ‘B’)