GRE Geometry: Circles

 A circle is a labeled by its center point: O means the circle with center point O

Diameter

A line segment, generally denoted by the variable, that connects two points on the circles and passes through the center of the circle

 

Radius:- A line segment, generally denoted by the variable r from the center of the circle to any point on the circle.

Chord: A line Segment joining two points on the segment of a circle is a chord.

Tangent:- A line that touches only one point on the circumference of the circle is called Tangent. A line drawn a tangent to a circle is perpendicular to the radius at the point of contact.

Circumference:- The distance around a circle is called the circumference.

An Arc is a portion of the circumference of a circle. The shorter distance between A an B along the circle is called the minor arc.

The longer distance between A and B is called the major arc.

Arc length:- For an arc with a central angle measuring θ Degree:

Length = ( θ / 360 )×(Circumference)

= (θ / 360 )×( π d )

 

Problem 1) What is the length of arc AC of the circle with center O as shown in the figure?

Sol. Arc length= (θ / 360 ) ( π d )= (60 / 360 ) (π (2))

12 π / 6    = 2 π units.

 

Area of circles: The area of circles is given by the formula A=  π r2

A sector is a portion of the circle that is bounded by two radii and an arc.

In a sector Whose central angle measures by Q degrees.

 

Area of sector= (Q / 360) x (Area of circle)

 

Problem 2) What is the area of sector AOC in the circle with the center O shown?

 

Sol. Area of sector AOC= ( 60 / 360 ) (36π )     = ( 6 π ) units.

 

PROPERTIES:

  1. The perpendicular from the center of a circle to a chord bisects the chord and vice versa.

 

If ∠OCB=90 °, THEN AC=BC

If AC=BC , then ∠OCB= 90°

 

2) Equal chords of a circle are equidistant from the center. Conversely, chords equidistant from the center are always equal.

3) Any two angles in the same segment are equal. Thus ∠ACB  = ∠ADB

 

 

4)The angle subtended by an arc at the center of the circle is twice the angle subtended by the same arc at any other point on the circle.

∠AOB= 2∠ACB

5)The angle subtended by a semicircle is a right angle. Conversely, the arc of a circle subtending a right angle at any point on the circle is a semi-circle.

 

If AB is a diameter, then ∠ACB = 90°

IF ∠ACB = 90° , then AB is a diameter.

6)Tangent drawn from common external point to a circle are equal.

 

Problem 3) A 5 by 12 rectangle is inscribed in a circle. What is the circumference of the circle?

Sol.

 

∠BCD= 90°

BD is a diameter (Because The arc of a circle subtending a right angle at any point on the circle is a semi-circle)

In BCD , BC2 + DC2 = BD2 (by Pythagoras theorem)

BD = 13

Circumference of the circle = 2π (Radius) = 13 π units

 

Problem 4) In the figure shown below, if the radius of a circle with centre  P is three times the radius of circle with centre A, ∠BAC=∠QPR, and the shaded area of the circle with centre A is 3 π SQUARE units, then what is the area of the shaded part of circle with centre P?

 

Sol. Let ∠BAC =∠QPR = Q°

The shaded area of circle with centre at A = 3 π sq. units = ( Q / 360° ) π (AC)2

PQ = 3AC

The shaded area of circle with centre at P

= ( Q / 360° ) π (PQ)2

= ( Q / 360° ) π (3AC)2

= (3π x 9)

= 27π sq. units

 

Problem 5) Two congruent, adjacent circles are cut out of a  16 by 8 rectangle. The circles have the maximum diameter possible. What is the area of the paper remaining after the circles have been cut out?

 

 

 

 

 

 

Sol. For the circles, the diameter of the circle is the same as the width of the rectangle.

Remaining area = Area of rectangle – 2 x area of circle

Area of rectangle = Length x Breadth = 8 x 16 = 128 Sq. units

radius of circle = 4 units

Area of circle = π (4)2      = 16 π Sq. units

Remaining area = (128- 2x(16π) ) = ( 128 – 32 π) sq. units

 

Problem 6) The Figure shows an equilateral triangle, where each vertex is the center of a circle. Each circle has a radius of 20. What is the area of the shaded region?

 

Sol.   Side of equilateral triangle =40

Area of equilateral triangle= √3/4    (40)2

=√3/4 (1600) = (400 √3)

Area of three sectors subtending an angle of 60° = 3 x (60/360) (π) (r2)

=(1/2) π (400) = 200π

Area of shaded Region= Area of equilateral triangle- area of three sectors subtending an angle of 60°

= (400√3  – 200 π ) sq. units

 

Problem 7)

 

 

 

 

 

 

 

 

 

If the diameter of the circle is 36, what is the length of arc ABC?

Sol.

 

 

 

 

 

 

 

 

Diameter =36, radius =18

Note that a minor arc is the “short way around” the circle from one point to another, & a major arc is the “long way around”. Arc is thus the same as major arc AC

∠AOC = 2∠ABC = 2(40°) = (80°)

The angle subtended by an arc at the centre of the circle is twice the angle subtended by the same arc at any other point on the circle.

Length of minor arc AC =

\(=\quad (\frac { 80 }{ 360 } )\times (2\pi r)\\ =\quad (\frac { 4 }{ 18 } )(2\pi )(18)\\ =\quad (8\pi )\)

Circumference of circle =

\(=2\pi (18)=36\pi\)

Length of major Arc AC = Circumference – Length of minor Arc AC =

\(=36\pi -8\pi ={ (28\pi ) }^{ Ans }\)

 

Problem 8)

 

 

 

 

 

 

 

 

AB is not a diameter of the circle.

Quantity A:         The area of the circle

Quantity B:         9

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given.

Sol.  Since, a diameter is the longest straight line, you can draw from one point on a circle to another (that is, a diameter is the longest chord in a circle), the actual diameter must be greater than 6.

If the diameter were exactly 6, the radius would be 3, & the area would be:

Area \(={ (3) }^{ 2 }\pi =9\pi\)

However, since the diameter must actually be greater than 6, the area must be greater than \(9\pi \). Therefore, Quantity A is greater.

 

Problem 9) In the following figure, O is the centre of the circle & \(\angle ABO={ 30 }^{ 0 }\). Find \(\angle ACB\).

 

 

 

 

 

 

 

 

 

 

Sol.

 

 

 

 

 

 

 

 

 

 

In \(\triangle \quad ABO,\quad AO=OB\) (radii of circle)

\(\angle OBA=\angle BAO={ 30 }^{ 0 }\)

(Because Angles opposite to equal sides are equal)

\(In\quad \triangle ABO,\quad \angle ABO+\angle BAO+\angle AOB={ 180 }^{ 0 }\\ \angle AOB={ 120 }^{ 0 }\\ \angle AOB=2\angle ADB\)

(Because the angle subtended by an arc at the canter of the circle is twice the angle subtended by the same arc at any other point on the circle)

\({ 120 }^{ 0 }=2\angle ADB,\quad \angle ADB={ 60 }^{ 0 }\) \(\angle ADB+\angle ACB={ 180 }^{ 0 }\\\)

(Opposite angles of cyclic quadrilateral are supplementary)

\(\angle ACB={ 120 }^{ 0 }\\\)

 

Directions for questions 10 to 12: In the figure below, X & Y are circles with centres O & O’ respectively. MAB is a common target. The radii of X & Y are in the ratio 4:3 & OM=28cm.

 

 

 

 

 

 

 

 

 

 

 

 

Problem: 10) What is that ratio of the length of OO’ to that of O’M?

a) 1:4

b) 1:3

c) 3:8

d) 3:4

Sol.

 

 

 

 

 

 

 

 

 

 

 

 

\(\triangle MAO’\sim \triangle MBO\\ (\angle OBM=\angle O’AM={ 90 }^{ 0 }(Radius\quad is\quad always\quad perpendicular\quad to\quad the\quad tangent\quad at\quad the\quad point\quad of\quad contact)\\ \angle BMO=\angle AMO'(Common))\)

The radii of X and Y are in the ratio 4:3.

\(\frac { OB }{ O’A } =\frac { 4 }{ 3 } \\ \frac { OB }{ O’A } =\frac { OM }{ O’M } =\frac { BM }{ AM } \\ \frac { 4 }{ 3 } =\frac { OM }{ O’M } (Subtract\quad 1\quad from\quad both\quad sides)\\ \frac { 4 }{ 3 } -1=\frac { OM }{ O’M } -1=\frac { OM-O’M }{ O’M } =\frac { OO’ }{ O’M } \\ { (\frac { 1 }{ 3 } =\frac { OO’ }{ O’M } ) }^{ Ans }\)

d) is correct.

 

Problem 11) What is the radius of circle with centre O?

a) 2cm

b) 3cm

c) 4cm

d) 5cm

Sol.

\(\frac { OB }{ O’A } =\frac { OM }{ O’M } =\frac { BA }{ AM } \\ \frac { 4 }{ 3 } =\frac { OM }{ O’M } =\frac { 28 }{ O’M } ,\quad O’M=21\\ OO’=OM-O’M=28-21=7cm\\\)

7cm = (radius of circle with centre O) + (radius of circle with centre O’)

Let radius of circle with centre O be \(4x\) and radius of circle with centre O’ be \(3x\).

\(7cm=4x+3x\\ x=1\)

Therefore, radius of circle with centre O = 4cm

c) is correct.

 

Problem 12) The length of AM is

a) \(8\sqrt { 3 }\)

b) 10cm

c) 12cm

d) 14cm

Sol.

Radius of circle X=OB=4cm

OM=28cm

\(In\quad \triangle OBM,\quad \angle OBM={ 90 }^{ 0 }\\ { OB }^{ 2 }+{ BM }^{ 2 }={ OM }^{ 2 }\\ OB=4cm,\quad OM=28cm\\ { 4 }^{ 2 }+{ BM }^{ 2 }={ 28 }^{ 2 },\quad { BM }^{ 2 }=768\\ BM=16\sqrt { 3 } \\ \frac { 4 }{ 3 } =\frac { BM }{ AM } =\frac { 16\sqrt { 3 } }{ AM } \\ AM=12\sqrt { 3 }\)

c) is correct.

 

Problem 13) O is the centre of a circle of radius 5 units. the chord AB subtends an angle of \({ 60 }^{ 0 }\) at the centre. Find the area of the shaded portion (approximate value).

 

 

 

 

 

 

 

 

a) 50 Sq. units

b) 75 Sq. units

c) 88 Sq. units

d) 67 Sq. units

Sol. 

Area of circle = \(\pi { (5) }^{ 2 }=25\pi \quad Sq.units\)

In\quad \triangle AOB,\quad \angle AOB={ 60 }^{ 0 }\\ \angle OAB=\angle OBA

(Angles opposite to equal sides are equal)

\(\angle AOB+\angle OBA+\angle BAO={ 180 }^{ 0 }\\ \angle OAB=\angle OBA={ 60 }^{ 0 }\) \(\triangle AOB\) is an equilateral triangle with side 5 units.

Area of equilateral \(\triangle AOB\) =

\(\frac { \sqrt { 3 } }{ 4 } ({ side }^{ 2 })=\frac { \sqrt { 3 } }{ 4 } { (5) }^{ 2 }=\frac { 25\sqrt { 3 } }{ 4 }\)

Area of shaded portion =

\(25\pi -\frac { 25\sqrt { 3 } }{ 4 } =25(\pi -\frac { \sqrt { 3 } }{ 4 } )=67.67\quad Sq.units\)

GRE: Three Dimensional Figures (Uniform Solids)

Volume

The volume of a solid is the amount of space enclosed by the solid.

Surface area

In general, the surface area of a solid is equal to the sum of the areas of the solid’s faces.

Rectangular Solid

A solid with six rectangular faces ( all edges meet at right angles)

Volume= Area of Base x Height
= l x w x h

Surface Area = Sum of areas of faces
=2(lw + wh + hl )

Cube

A special rectangular solid with all edges equal (l=w=h) is called a cube. For Example:- Die

Volume = L3

Surface area = Sum of areas of faces
= 6 x L2

Cylinder

A uniform solid whose base is a circle is called a cylinder.

Lateral Surface area (or area of rest of shell)
= 2 π r h
Volume = Area of base X Height = π r2h
Total Surface area = (2 X Area of base) + Lateral Surface Area

= 2 π r+ 2 π r h = 2 π r ( r + h)

Sphere

A sphere is made up of all the points in space at a certain distance from a center point, it is also called a three-dimensional circle.
The difference from the center to a point on the sphere is the radius of the sphere.

Volume of sphere = 4/3 π r3

Total Surface area = 4 π r2

 

Problem 1) The solid shown is half a rectangular solid. What is the volume of the solid shown?

 

 

 

 

 

 

 

Sol. In Triangle ABC,

AB2 + BC2 = AC2

AB+ 9= 25

AB=4
Area of base = (1/2) x 4 x 3 = 6 Sq. units
Volume = 6 x 4 = 24 Cubic units

 

Problem 2)  The height of a cylinder is twice its radius. If the volume of the cylinder is 128 π. What is the radius?

Sol. Let radius be r
Height of cylinder (h) = 2 r
Volume = π r2 h
128 π = π r2 ( 2 r )
64= r3
r = 4 units.

 

Problem 3) A cube of ice has edges of length 10. What is the volume of the largest cylinder that can be carved from the cube?

Sol. The largest cylinder would have diameter 10 and height 10 ( each equal to the edge of the cube)

Volume = π r2 h
= π (100/4) (10)

=1000 π / 4
= 250 π Cubic units

 

Problem 4) A solid metal cylinder with a radius of 6 and a height of 3 is melted down and all of the metal is used to recast a new solid cylinder with a radius of 3. What is the height of the new cylinder?

Sol. Volume of solid metal cylinder
=Volume of new solid cylinder

Let radius and height of solid metal cylinder be and new solid cylinder be

π R12H1 = π R22H2

62. (3) = 32. h

H = (36 x 3) / (3 x 3) = 12 units

Lines, Angles and Triangles

Conditions of similarity:-

 

  • AAA Similarity:- If in two triangles, corresponding angles are equal, then triangles are similar.

 

 

 

 

 

 

If ∠A = ∠P , ∠B = ∠Q , ∠C = ∠R

Then △ABC ~ △ PQR

  • SSS Similarity:- If the corresponding sides of two triangles are proportional then they are similar.

  

 

If AB/PQ = BC/QR = AC/PR

                     Then   △ABC ~ △ PQR

  • SAS Similarity: If in two triangles, one pair of corresponding sides are proportional and the included angles are equal then the two triangles are similar.

      

If AB/PQ = BC/QR  &  ∠B = ∠Q

Then △ABC ~ △ PQR

The similarity of Triangles:- Similarity of triangles is a special case where if either of the conditions of similarity holds, The other will hold automatically.

Congruency of Triangles:-If two triangles are congruent, then corresponding angles are equal.

Conditions of Congruency:-

1.SAS Congruency:- If two sides and an included angle of one triangle are equal to two sides and an included angle of another, the two triangles are congruent.

     

In  △ABC and △ PQR

IF AB=PQ,

BC=QR,

∠ABC = ∠PQR

Then △ABC~△ PQR

2.ASA Congruency:- If two angles and the included side of one triangle is equal to two angles and the included side of another, The two triangles are congruent.

In  △ABC and △ DEF

If  ∠A =∠D

AB = DE

∠B = ∠E, Then △ABC ≅ △ PQR

3. AAS Congruency:- If two angles and the side opposite to one of the angles is equal to the corresponding angles and the side of another triangle, the triangles are congruent.

If  In △ABC and △ DEF

∠A =∠D

∠B=∠E

AC=DF

Then △ABC≅△ DEF

4.SSS Congruency:- If three sides if one triangle is equal to three sides of another triangle, the two triangles are congruent.

In △ABC and △ DEF

If AB=DE, BC=EF and AC=DF

THEN  △ABC ≅ △ DEF

5.SSA Congruency:- If two sides and the angle opposite to any of the sides are equal to corresponding sides and the angle opposite to any of the sides, Then the triangles are congruent.

      

 

 

 

 

 

In △ABC and △ DEF

AB=DF

BC= EF

∠BAC = ∠EDF

Then △ABC ≅ △ DEF

Right Angled Triangle:

Pythagoras theorem: In the case of a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In right △ABC, AB2 + BC2 = AC2

Special figure:

Note: A lot of questions are based on the figure

In this figure, △ABC ~△ ADB ~ △BDC

  • △ABC ~△ ADB

 

AB/AD   =  BC/DB  =  AC/AB

 

  • △ADB ~△ BDC

AD/BD   =  DB/DC  = AB/BC

 

  • △ABC ~△ BDC

AB/BD = BC/DC = AC/BC

Area Theorem:

According to area theorem:The ratio of areas of two similar triangles is the square of the ratio of corresponding sides.

If △ABC ~ △ DEF

Ar(ABC): Ar(DEF) = (AB/DE)2 = (BC/EF)2 = (AC/DF)2

Problem 1) For similar triangles, the ratio of their corresponding sides is 2:3. What is the ratio of their areas?

Sol. According to area theorem :The ratio of areas of two similar triangles is the square of the ratio of corresponding sides.

Ratio of areas= (2/3)2  =  4/9

 

Problem 2) What is the area of the triangle shown?

 

 

 

 

 

 

 

 

 

 

Sol.

 

 

 

 

 

 

 

 

 

 

Construction: Draw AD⊥BC

∠B = ∠C = 60

∠A= 60

(Sum of angles of a triangle is equal to 180°)

 

∠A = ∠B = ∠C = 60°

(Angles of an equilateral triangle are equal to 60° & triangle having each angle equal to 60° is an equilateral triangle )

:. △ABC is an equilateral △

In △ABD

BD2 +AD2 = AB2 (Pythagoras theorem)

AD2 =75, ad = 5√3

Area of Triangle = ½ × Base × Height

=1/2 × 10 × 5√3

= 25√3  sq. unit

 

Problem 3) What is the area of the triangle Shown?

Sol.

x2 +( x + 1)2 =25 (by Pythagoras theorem)

2x2 +2x =24

2x2 + 2x -24 =0

x2 + x – 12 =0

x = 3 or -4

x= 3 (Because length cannot be negative)

Area = ½ (base) x (height) = ½(7)(8)   = 28 sq. units

 

Problem 4) What is the value of x in the triangle shown?

 

 

 

 

 

 

 

 

 

 

 

Sol.      In △ABC

4 + 3 = AC  (BY Pythagoras theorem)

AC = 5

△ABC~ △BDC

 

AB/BD =BC/DC = AC/BC

 

4/x  =  5/3

12 = 5x          ,      x =(12/5)  = Ans

 

Polygons

A polygon is a closed figure whose sides are straight line segments.

The perimeter of the polygon is the sum of the lengths of the sides. A vertex of a polygon is the point where two adjacent sides meet.

A diagonal of a polygon is a line segment connecting two nonadjacent vertices.

A regular polygon has sides of equal length and interior angles of equal measure.

 

Sum of all angles of a polygon with n sides

= (n-2) π radians = (n-2) (180°)

 Area of Regular polygon = (ns2/4) x Cot (180/n)   

Where     s = length of side

      n = No. of sides

 

Parallelogram (IIgm

A quadrilateral with two pairs of parallel sides.

Properties of parallelogram:

Area of Parallelogram = Base x Height

Diagonals of Parallelogram bisect each other.

The opposite angles in a Parallelogram are equal.

 

Rectangle

A Parallelogram with four equal angles, each at right angle is a rectangle.

 

 AB = CD, BC =AD

∠ABC = ∠BCD = ∠CDA = ∠BAD

 Properties of a Rectangle:

 Diagonals of a rectangle are equal and bisect each other.

 

Rhombus

A Parallelogram having all the sides equal is a rhombus.

 

 Area of Rhombus = ½ x product of diagonals

 

Properties:-

Diagonals of a rhombus bisect each other at right angle.

Square

 A Square is a rectangle with adjacent sides equal or rhombus with angles equal to 90°.

 Properties:

Diagonals of square are equal and bisect each other.

Area = ½ x (diagonal) 2

 

Diameter of circle circumscribing a square also acts as a diagonal of square.

 

BD=Diameter of Circle

= Diagonal of Square = a √2

 

Trapezium

A Trapezium is a quadrilateral with only two sides parallel to each other.

 

 

 Area of Trapezium = ½ (Sum of Parallel sides)x height

Rectangular Hexagon

A Rectangular Hexagon is a actually a combination of 6 equilateral triangles all of side “a”

 

Area of hexagon = 6(3/4 a2) = 3√3 / 2 a2

  Problem 5) Right △ABC and rectangle EFGH have the same perimeter. What is the value of x?

Sol. In △ABC

AB2 + BC2 = AC2 (BY Pythagoras theorem)

AC=5

Perimeter of △ABC = 3+4+5 = 12

Perimeter of rectangle EFGH =2(2+x)

12 = 2 (2+x)

2 + x = 6

x = 4

 

Problem 6) What is the area of the square in the figure above?


Sol. (Side2 + Side2 = Diagonal2) (by Pythagoras theorem)

2 Side2  = 10= 100

Side = 5 √2

Area = Side2 = 50 sq units

 

Problem 7) The area of a rectangle can be represented by the expression 2x2+9x+10. The length is 2x+5 and the width is 6. What is the value of the area of the rectangle?

Sol. Area = 2x2 +9x 10 = Length x Breadth

2x2 +9x 10 = (2x +5)6

2x2  – 3x – 20 = 0

x = 4 or -2.5

If x = -2.5, then length = 2(-2.5) + 5 = 0 (Length can’ t be zero,∴ Negative value of x is discarded)

x = 4

Area = 2 (16) +9 (4) +10

= 32 +36+ 10   = 78 sq Units

 

Problem 8) In the figure shown, ABCD is a rectangle. AB=8 and BC=6. R,S,T, AND Q are midpoints of the sides of rectangle ABCD. What is the perimeter of RSTQ?

 

Sol. AB = CD = 8, BC = AD = 6

AR = RB = DT = CT

AQ = QC = BS = SD = 3

AR2 + AQ2 = RQ2  (By Pythagoras theorem)

32 + 42 = RQ2

RQ = 5

RQ=RS=ST=QT=5

Perimeter of RSTQ = 20

 

Problem 9) In the figure shown ABCD is a square with a side length of 16. R,S,T AND Q are midpoints of the sides of ABCD .

What is the area of RSTQ?

 

Sol. AB=16

AR=RB=BS=SC=TC=DT=QD=AQ=8

AR2 + AQ2 = RQ2  (By Pythagoras Theorem)

RQ=8 √2

Similarly RQ=RS=QT=ST= 8 √2

In △ARQ, ∠QAR = 90°, AR = AQ

∠AQR = ∠DQT (Angles opposite to equal sides are equal)

∠AQR = ∠DQT = 45°

∠RQT=90(Linear pair)

Therefore, RSTQ is square

Area (8 √2)2= 128 sq units

 

Problem 10) In the given figure, AD||BC. Find the value of x.

a) x=8, \(\frac { 15 }{ 3 }\)

b) x=8, \(\frac { 11 }{ 3 }\)

c) x=8, \(\frac { 13 }{ 3 }\)

d) x=7, \(\frac { 13 }{ 3 }\)

 

 

 

 

 

 

 

 

 

 

Sol.

\(In\quad \triangle ODA\quad and\quad \triangle OBC,\quad AD\parallel BC\\ \angle ODA=\angle OBC\) (Alternate interior Angles are equal)

\(\angle OAD=\angle OCB\) (Alternate interior angles are equal)

\(\triangle ODA\sim \triangle OBC\\ \frac { OD }{ OB } =\frac { DA }{ BC } =\frac { OA }{ OC } \\ \frac { OD }{ OB } =\frac { AO }{ OC } \\ \frac { x-5 }{ x-3 } =\frac { 3 }{ 3x-19 } \\ (x-5)(3x-19)=3x-9\\ 3{ x }^{ 2 }-19x-15x+95-3x+9=0\\ 3{ x }^{ 2 }-37x+104=0\\\)

(solving this quadratic equation)

\(x=\frac { 37\pm \sqrt { { 37 }^{ 2 }-4(3)(104) } }{ 6 } =\frac { 37\pm 11 }{ 6 } \\ =8,\frac { 11 }{ 3 }\)

c) is correct.

 

Problem 11) In the right angle \(\triangle PQR\), find RS?

 

 

 

 

 

 

 

 

 

 

Sol.

\(In\quad \triangle QRP\quad and\quad \triangle RSP\\ \angle QRP=\angle RSP\quad (both\quad are\quad right\quad angles)\\ \angle QPR=\angle SPR\quad (common)\\ Therefore,\quad \triangle QRP\sim \triangle RSP\\ \frac { QR }{ RS } =\frac { RP }{ SP } =\frac { QP }{ RP } \\ \frac { 5 }{ RS } =\frac { QP }{ 12 } \\ In\quad \triangle QRP\\ { QR }^{ 2 }+{ RP }^{ 2 }={ QP }^{ 2 }(By\quad pythagoras\quad theorem)\\ { 12 }^{ 2 }+{ 5 }^{ 2 }={ QP }^{ 2 }\\ QP=13\\ \frac { 5 }{ RS } =\frac { 13 }{ 12 } \\ RS={ (\frac { 60 }{ 13 } ) }^{ Ans }\)

 

Problem 12)

 

 

 

 

 

 

 

 

The area of semicircle O is \(16\pi\).

\(\angle CDE=\angle ABE={ 30 }^{ 0 }\\ AC=\sqrt { 2 }\)

Quantity A:  DE

Quantity B:  \(3\sqrt { 5 }\)

Sol. 

\(In\quad \triangle EDC\quad and\quad \triangle EBA\\ \angle EDC=\angle EBA\\ \angle CED=\angle AEB={ 90 }^{ 0 }\\ \triangle EDC\sim EBA\\ \frac { EC }{ AE } =\frac { DC }{ BA } \\ Let\quad EC\quad be\quad x\\ In\quad \triangle EDC\\ \angle CED={ 90 }^{ 0 },\quad \angle CDE={ 30 }^{ 0 },\quad \angle DCE={ 60 }^{ 0 }\)

Recall that the sides of a 30-60-90 triangle are in the proportion

\(x:x\sqrt { 3 } :2x\).

.Or if you are familiar with trigonometry, just recall that

\(\sin { { 30 }^{ 0 } } =\frac { 1 }{ 2 } =\frac { perpendicular }{ hypotenuse } \\ Therefore,\quad \frac { CE }{ CD } =\frac { 1 }{ 2 } =\frac { x }{ CD } ,\quad CD=2x\\ \frac { EC }{ AE } =\frac { DC }{ BA } \\ \frac { x }{ x+\sqrt { 2 } } =\frac { 2x }{ BA } \)

Area of semicircle = \(\frac { \pi { r }^{ 2 } }{ 2 } =16\pi\) \({ r }^{ 2 }=36,\quad r=4\sqrt { 2 } =OB\\ \frac { x }{ x+\sqrt { 2 } } =\frac { 2x }{ 8\sqrt { 2 } } \\ x+\sqrt { 2 } =4\sqrt { 2 } \\ x=3\sqrt { 2 } \\ Therefore,\quad CE=3\sqrt { 2 } ,\quad CD=6\sqrt { 2 } \)

Side of \(\triangle CDE\) are in ratio \(x:x\sqrt { 3 } :2x\) \(CE=3\sqrt { 2 } ,\quad CD=6\sqrt { 2 } \\ ED=3\sqrt { 6 } \\ Therefore,\quad DE=3\sqrt { 6 } \)

(Quantity A is greater than Quantity B).

 

Problem 13) What is the area of a triangle that has two sides that each have a length of 10, & whose perimeter is equal to that of a square whose area is 81?

a) 30

b) 48

c) 36

d) 60

e) 42

Sol.

 

Area of square = 81

Side of square = \(\sqrt { 81 } =9\)

Perimeter of square = 4×9 = 36

Perimeter of triangle = 36

Two sides of triangle have a length of 10.

Third side of triangle = 36-10-10 = 16

 

 

 

 

 

 

 

 

Any triangle that has equal sides is an isosceles triangle. Drop a perpendicular to divide it into two individual right triangles.

\(In\quad \triangle ABC,\quad AD\bot BC\\ In\quad \triangle ABD,\quad { AB }^{ 2 }={ BD }^{ 2 }+{ AD }^{ 2 }\quad (by\quad pythagoras\quad theorem)\\ { 10 }^{ 2 }={ 8 }^{ 2 }+{ AD }^{ 2 }\\ AD=6\\ Area\quad of\quad \triangle ABC=\frac { 1 }{ 2 } \times Base\times Height=\frac { 1 }{ 2 } \times 16\times 16=48\quad Sq.units\)

 

Problem 14) Michael wants to split his rhombus-shaped garden into two triangular plots, one for planting strawberries & one for planting vegetables, by erecting a fence from one corner of the garden to the opposite corner. If one corner of the garden measures \({ 60 }^{ 0 }\) & one side of the garden measures \(x\) meters, which of the following could be the area of the vegetable plot?

Indicate all such value

a) \(\frac { { x }^{ 2 }\sqrt { 3 } }{ 2 }\)

b) \(\frac { { x }^{ 2 }\sqrt { 3 } }{ 4 }\)

c) \(\frac { { x }^{ 2 } }{ 2 }\)

d) \(\frac { { x }^{ 2 }\sqrt { 3 } }{ 4 }\)

e) \(\frac { { x }^{ 2 } }{ 4 }\)

Sol.  Whenever a geometry problem comes without a figure, start by drawing one yourself:

 

 

 

 

 

 

 

 

 

 

Michael wants to split this garden into two triangles.

Try it this way first:

 

 

 

 

 

 

 

 

 

 

\(\triangle ACD\) is an equilateral. (Because all angles are equal to \({ 60 }^{ 0 }\))

Area of triangle ACD = Area of vegetable plot =

\(=\frac { \sqrt { 3 } { x }^{ 2 } }{ 4 }\)

This is one of the choices, but the garden might be split the other way:

 

 

 

 

 

 

 

 

 

 

\(Draw\quad AE\bot BD,\quad In\triangle AEB\\ \angle AEB={ 90 }^{ 0 },\quad \angle ABE={ 30 }^{ 0 },\quad \angle EAB={ 60 }^{ 0 }\)

Recall that the sides of a 30-60-90 triangle are in the proportion \(x:\sqrt { 3 } x:2x\)

Or if you are familiar with basic trigonometry

\(In\quad \triangle AEB,\quad \sin { { 30 }^{ 0 } } =\frac { 1 }{ 2 } =\frac { AE }{ AB } =\frac { AE }{ x } \\ AE=\frac { x }{ 2 } ,\quad EB=\frac { x\sqrt { 3 } }{ 2 }\)

Therefore, we can easily derive that

\(AE=\frac { x }{ 2 } ,\quad EB=\frac { x\sqrt { 3 } }{ 2 }\) (by using properties of triangle or by using basic trigonometry)

Similarly, we can find that \(ED=\frac { \sqrt { 3 } x }{ 2 } ,\quad BD=\sqrt { 3 } x\) \(Ar(\triangle ABD)=\frac { 1 }{ 2 } \times base\times height\\ =\frac { 1 }{ 2 } \times \sqrt { 3 } x\times \frac { x }{ 2 } \\ =\frac { \sqrt { 3 } { x }^{ 2 } }{ 4 }\)

This is same as the previous area, so only one answer choice is B.