Mensuration is a pure formula-based topic and tricks/shortcuts are seldom applied here. So in this series I will try to solve all the mensuration problems that have appeared in CGL lately and in the process I will share the important concepts/formulas.

Lateral Surface Area = Height * Perimeter of the Base

Volume = Height * Area of the Base

In this question the Total surface area is being asked

Height of the prism = 10 cm

Perimeter of the base = 5 + 12 + 13 (Calculate the hypotenuse with Pythagoras Theorem) = 30 cm

So Lateral Surface Area = 10 * 30 = 300 cm

Area of the base = 1/2 * base * height = 1/2 * 5 * 12 = 30 cm

So Total Surface Area = 300 + 2*30 = 360 cm

Put the value of r/R from equation (1)

Area of base/quadrilateral = 84 + 54 = 138 cm

Volume = Height * Area of the Base

2070 = Height * 138

So, Height of the prism = 15 cm

Lateral Surface Area = Height * Perimeter of the Base

Perimeter of the base = AB + BC + CD + DA = 48 cm

Lateral Surface Area = 48 * 15 = 720 cm^2

Volume of the prism = Area of the base * Height =√3/4 * a^2 * h ... (1)

Lateral surface Area of the prism = Perimeter of the base * Height = 3a * h ...(2)

Divide equation (1) by (2)

Volume/Area = (1/4√3) * a

40√3/120 = a/4√3 [Since Volume = 40√3 and Lateral surface Area = 120]

a = 160 * 3/120

__For Prism and Calendar (figures with uniform girth) -__Lateral Surface Area = Height * Perimeter of the Base

Volume = Height * Area of the Base

In this question the Total surface area is being asked

**Total Surface Area of a Prism = Lateral Surface Area + Area of the two bases**Height of the prism = 10 cm

Perimeter of the base = 5 + 12 + 13 (Calculate the hypotenuse with Pythagoras Theorem) = 30 cm

So Lateral Surface Area = 10 * 30 = 300 cm

Area of the base = 1/2 * base * height = 1/2 * 5 * 12 = 30 cm

So Total Surface Area = 300 + 2*30 = 360 cm

**Answer : (A)**
In such questions, remember one thing SIMILARITY

r/R = h/H ... (1)

where r = radius of small cone

R = radius of Big cone

h = height of small cone

H = height of big cone

Volume of cone = 1/3 * π* r

^{2 }* h
Given, Volume of big cone = 27 * Volume of small cone

1/3 * π * R

27 * (r/R)^2=H/h^{2 }* H = 27 * 1/3 * π * r^{2 }* hPut the value of r/R from equation (1)

27 * (h/H)^2 = H/h

27 * h

Put H = 30 cm^{3}= H^{3}
So h = 10 cm

The question asks us the height above the base, which is (30 - h) = 30 - 10 = 20 cm

**Answer : (B)**

The base of the prism looks like the figure above.

AD = 12 cm, AB = 9 cm

Hence BD = 15 cm (Pythagoras Theorem)

Area of base = Area of triangle ABD + Area of triangle BDC

Area of triangle ABD = 1/2 * 9 * 12 = 54 cm

Area of triangle BDC = 84 cm (Apply Heron's formula)Area of base/quadrilateral = 84 + 54 = 138 cm

Volume = Height * Area of the Base

2070 = Height * 138

So, Height of the prism = 15 cm

Lateral Surface Area = Height * Perimeter of the Base

Perimeter of the base = AB + BC + CD + DA = 48 cm

Lateral Surface Area = 48 * 15 = 720 cm^2

**Answer : (A)**

Area of the base = √3/4 * a^2, where a is the side of the equilateral triangle

Perimeter of the base = 3aVolume of the prism = Area of the base * Height =√3/4 * a^2 * h ... (1)

Lateral surface Area of the prism = Perimeter of the base * Height = 3a * h ...(2)

Divide equation (1) by (2)

Volume/Area = (1/4√3) * a

40√3/120 = a/4√3 [Since Volume = 40√3 and Lateral surface Area = 120]

a = 160 * 3/120

a = 4 cm

**Answer : (A)**

This question is about 'Pyramid'. So let me just give you some basic understanding of Pyramids. CGL can ask questions about two types of Pyramids - Pyramid with a Triangular Base and Pyramid with a square base. Both these pyramids have different formulas. Look at the below figure and understand the labellings, i.e.,

**Slant edge**and**Slant Height**
In the below image, I have written formulas for both types of Pyramids. The formula for Volume is same for both the Pyramids-

**V = 1/3 * A * h**

where A = Area of the base (calculation of A will be different for both)

h = Height of the Pyramid

When lateral surface area is asked, you will first calculate the 'Slant Height'. Then with the help of slant height you will find the area of one lateral face (let's call this area M). If the pyramid is having a triangular base then multiply M with 3, to get the lateral surface area of the pyramid. And if the pyramid is having a square base, then multiply M with 4.

The area of the square is 324, hence its side is 18 cm

Volume of the pyramid = 1/3 * Area of the base * Height

1296 = 1/3 * 324 * Height

So Height = 12 cm

Slant Height of the pyramid with square base = √ (h^2 + a^2/4) = √(12^2 + 18^2/4)

Slant Height = 15cm

Area of the lateral face = 1/2 * Base * Height = 1/2 * 18 * 15 = 135 cm^2

Pyramid with a square base has 4 lateral faces, so lateral surface area of the pyramid = 4 * 135 = 540 cm^2

**Answer : (D)**

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I seen this blog just now sir .. i like ur blog.. very helpful..

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteVery good and informative notes...highly appreciated...

ReplyDeletethnku sir....but dere is nt ny button to save pdfs

ReplyDeletePrint PDF is visible to me only after clicking on "Comments"

Deleteya..got it..thnku

ReplyDeleteThis comment has been removed by the author.

Deletevery helpful for us. thank u

ReplyDeletevery helpful for us. thank u

ReplyDeleteThere is no "print pdf" button on the blog???

ReplyDeleteYou will see the "Print PDF" button when you will explicitly open a post by clicking on its title...

DeleteHello sir u r doing gr8 job my request is that plz post ur valuable tricks on regular basis

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ReplyDeletethanx for the good work.In the second question ,the equation in the third line from the bottom should be H^3= 27*h^3.. thanx again

ReplyDeletefind volume of pyramid with right angled traingle base of sides 3,4,5 and slat edge 13

ReplyDeleteThank u sir

ReplyDeleteThank u sir

ReplyDeleteIn question 3, if i am calculating the area of BDC by the formula 1/2*DC*AD it is giving different result. However AD is the height of both the triangles. correct me please if I am wrong.

ReplyDeleteNo...AD is not the height of the triangle BDC...check the figure again...I have updated it :)

DeleteHello Prashant!

ReplyDeleteQ2 and Q5 type questions even if i knew all the basic approach formulae , still takes about 2- 2.5 min to solve to get answer ..

so in actual exam is it worth it to try these questions ?

If you take time in solving mensuration questions, attempt them at last...

Deletehi sir, i had one doubt while solving old paper 2008, kindly provide solution sir..please

Delete2008 prelims paper.

If S1 and S2 be the surface area of a sphere and the curved surface area of the circumscribed cylinder respectively, then S1 is equal to

(a) 2/3 S^2

(b) 1/3 S^2

(c) 1/2 S^2

(b) S^2

i got s1= s2, but in kiran publications he gave 3rd option

kindly provide solution as soon as you see this comment,please sir

@purna..s1=s2 is the correct answer

ReplyDelete