**Pythagoras Theorem:**

__Case-I (At Right Angle)__

__Case-II (At Acute Angle)__

__Case-III (At ObtuseAngle)__**a**

^{2}= b^{2}+ c^{2 }**a**

^{2}< b^{2}+ c^{2 }**a**

^{2}> b^{2}+ c^{2}

__The possible quant-comparison Q’s for GRE:__

Each question from Q1 to Q5 consists of two quantities, one in column A and one in column B.

You are to compare
the two quantities and choose

A if the quantity in column A is
greater;

B if the quantity in column B is
greater;

C if the quantities are equal;

D if the relationship cannot be
determined from the information given;

A 20 foot ladder
leaning against a vertical wall with the base of the ladder 10 feet

from wall is pulled 2
feet farther from the wall causing the top of the ladder to

drop x feet.

__Column A__

__Column B__

X 2

Solution:

Consider right
triangle ABC. Let BE = y

Applying Pythagoras
theorem at angle B,

(x+y)

^{2}+ 10^{2}= 20^{2}
(x+y)

^{2}= 400 – 100 = 300 - -> (1)
Consider right
triangle ABC. Let BE = y

Applying Pythagoras
theorem at angle B,

y

^{2}+ 12^{2}= 20^{2}
y

^{2}= 400 – 144 = 256 => y = √256 = 16 - -> (2)
Substituting value of
y from (2) in (1),

(x+16)

^{2}= 300 => x+16 = √300 => x = √300 – 16
√300 value lies
between 17 and 18 (as 17

^{2}= 289 and 18^{2}= 324)
Hence x value lies between
(17-16) and (18-16) ie., 1 and 2

and must be less than
2. Hence answer option is “A”

**Q2)**

__Column A__

__Column B__

AB

^{2}+ BC^{2}AC^{2}**Solution:**

This is a tricky
question. Don’t take it granted that ÐB is 90.

As we have no clues
about ÐB or lengths of the sides,

we could not find the
relationship between the given columns.

Hence answer option
is “D”

**Q3)**

__Column A__

__Column B__

X z

**Solution:**

Observe that ÐB = 90,

Applying Pythagoras theorem
at B, z

^{2}= x^{2}+ y^{2}=> z > x
And hence answer
option is “B”

**Q4)**

__Column A__

__Column B__

ÐB 90

^{0}
Solution:

AC

^{2}= 25^{2}= 625; AB^{2}= 20^{2}= 400; BC^{2}= 15^{2}= 225
Observe that AC

And hence answer option is “C”^{2}= AB^{2}+ BC^{2}, hence ÐB = 90

ÐB 90

^{0}

**Solution:**

AC

^{2}= 25^{2}= 625; AB^{2}= 22^{2}= 484; BC^{2}= 15^{2}= 225
AB

^{2}+ BC^{2 }= 484+225 = 709
Observe that AC

^{2}< AB^{2}+ BC^{2}, hence ÐB < 90^{0}
And hence answer
option is “A”

__The possible DS section Q’s for GMAT:__
Q6 consists
of a mathematical problem, and 2 statements containing data.

Then you
have to choose one of the 5 different answer choices regarding

which of the
2 statements is sufficient for you to answer the problem.

The answers
will be:

A. Statement 1 ALONE is sufficient but statement 2 alone
is not sufficient.

B. Statement 2 ALONE is sufficient but statement 1 alone
is not sufficient.

C. BOTH statements TOGETHER are
sufficient, but NEITHER statement alone is sufficient.

D. EACH statement ALONE is
sufficient.

E. Statements 1
and 2 TOGETHER are NOT sufficient.

**Q6)**In the adjacent figure , what is the length of segment BC?

(1)

*x*= 90^{0}
(2)The perimeter of △ABC is 40.

**Solution:**

Using clue (1), applying
Pythagoras theorem at angle A,

BC

^{2}= 12^{2}+ 16^{2}= 144 + 256 = 400 => BC = 20
Using clue (2),

Perimeter = AB + AC +
BC = 40 =>12 + 16 + BC = 40

=>BC = 40 - 28 = 12

As we could find the
solution independently with the clues (1) and (2),

Answer option is “D”

__The Possible Problem solving Q’s for all exams:__**Q7)**The diagonals of a rhombus measure 24 and 10 units. Find the length of sides

of rhombus?

Solution:

In rhombus, diagonals
are at right angles and bisect each other.

AB

^{2}= OA^{2}+ OB^{2}= 12^{2}+ 5^{2 }
= 144 + 25 =
169

=>AB = 13

**Q8)**The length and width of a rectangle measure 24 and 18 units respectively.

Find radius of circle
circumscribing this rectangle?

**Solution:**

AC

^{2}= AB^{2}+ BC^{2}= 24^{2}+ 18^{2}= 576 + 324 = 900
AC = 30 units

AC is diameter of the
circle.

=> radius of circle = 30/2 = 15
units

**Q9)**

Find perimeter of
the given parallelogram?

**Solution:**

Applying Pythagoras theorem
at right angle ‘E’,

AD

^{2}= DE^{2}+ AE^{2}= 3^{2}+ 4^{2}= 9 + 16 = 25
AD = 5; BC = AD = 5

AB = CD = DE + EC =
3+5 = 8

Perimeter = AB + BC + CD + AD =
8+5+8+5 =26
Q10) John walked 6 m east
and then 8 m north. How far is he from his starting point?

**solution:**

Assume that John
started at A and walked 6 m east up to B and

then 8 m North up to
C.

Distance from the starting
point = AC

Applying Pythagoras at
right angle B,

AC

^{2}= AB^{2}+ BC^{2}= 6^{2}+ 8^{2}= 36+64 = 100
AC = 10 m

**Q11)**

Radius of the circle is 5 m and OC = 3 m. Find length of the chord AB?

**Solution:**

AO = radius = 5

Applying Pythagoras at
right angle C,

OA

=>AC = 4 =>AB = 2 AC = 8 m^{2}= OC^{2}+ AC^{2}=>AC^{2}= OA^{2}– OC^{2}= 5^{2}- 3^{2}= 25-9 = 16
Very good examples. Thanks Vemuri

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