Geometry and Trigonometry

Trigonometry

In a right-angled ∆ ABC, angle B is a right angle.

If sin C = 4/5, then find the value of cos A.

A) 4/5

B) -4/5

C) 5/4

D) -5/4

Solution:

In triangle ABC, angle B = 90°

Therefore, angle C + angle A = 90°

Angle C = 90° – A

Sin C = 4/5

Sin (90° – A) = 4/5 (sum of angle A and angle C is 90°)

Cos A = 4/5 (trigonometry formula)

Correct Answer: A




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2) Circle and Tangent

Circle with Tangent O(-4,-6) P(-7,-7) A B

2)A circle with center O (-4,6) and a point (-7,-7 )on the circle. A tangent through the point (-7, -7)

what is the slope of the tangent?

solution

The tangent is perpendicular to the radius of the circle (Theorem)

let N is the slope of the tangent

Let M is the slope of the radius

therefore M×N=-1 (properties of tangents of any two perpendicular lines)

slope of the radius M = y2 – y1/x 2-x1 (distance formula when the two coordinates of two points are given)

the slope of the radius M= -7- (-6)/-7-(-4) =-7+6/-7+4 =-1/-3=1/3

product of the slope =M×N=1/3N= -1

N/3=-1

N= -1×3slope of the tangent =-3

ANS -3

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3) Arc length and chord sum

Circle with Chord AB O A B stroke-dasharray=”0.1,0.1″ // 45°// Circle Chord AB

In the above circle with center O and chord AB it subtanen angle 45 at rhe center of the circle drawn in XY- plane

what is the Arc AB length?

Option

A)πr/4

B)πr/5

C)3πr/4

D)πr/6

solution

chord length = angle subtended at the center ×2πradius/360=θ2πr/360

chord length AB =45×2πr/360=45×2×πr/360 =2 ×πr/8 (simplify) =2πr/8=πr/4

correct ANS :A

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Circle with Radii O OA OB x

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equation of a circle

general equation of a circle with center with radius (h,K) and radius r

(x-h)2+(y-k) 2=r2

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1)The graph x2+x+ y2+y=31/2 in the XY -plane is a circle.

What is the length of the diameter?

Option:

A) 10

B) 6

C) 8

D)12

Solution:

The given equation x2+x+ y2+y=31/2

Write this equation in perfect squire

(x+1/2)2-1/4+(y+1/2)2-1/4=31/2

Add ¼+1/4 on both side

Therefore (x+1/2) 2-1/4+1/4+(y+1/2) 2-1/4+1/4=31/2+1/4+1/4

(x+1/2)2+(y+1/2)2=31/2+2×1/4

(x+1/2) 2+(y+1/2) 2= (31×2)/4+2×1/4=62/4+2/4 (by taking LCM)

(x+1/2)2+(y+1/2)2= (62+2)/4=64/4=16=42

Therefore radius =4 Units

The he diameter of the circle =2×r=2×4=8 unit

Correct ANS: C

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2)The equation of a circle in XY -plane is x2+5x+y2+3y=15/2

What is the center and radius of the above circle?

A)(-5/2,-3/2) and -4

B)(-5/2,-3/2) and 4

C)(5/2,-3/2) and 4

D)(-5/2,3/2) and 4

Solution:

The given equation = x2+5x+y2+3y=15/2

By perfect squaring method we write the above equation as

(x+5/2)2+(y+3/2)2-25/4-9/4=15/2

Adding both side by 25/4 and 9/4

We got (x+5/2)2-25/4+(y+3?2)2-9/4+25/4+9/4=15/2+25/4+9/4

Simply the Right-hand side by taking LCM

=(15×2+25+9)/4=(30+25+9)/4=(30+34)/4=64/4=16

The equation is (x+5/2)2+(y+3/2)2=16=42

Therefore by comparing with the standard form of equation (x-h) 2+(y-k) 2=r2

The center of the circle is (-5/2, -3/2) and the radius is 4

Correct ANS: B

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3) the function 3x2-6x+3y2+12y=12 in XY -plane is circle.

What is the radius of this circle.

Option:

A) (-2,1)

B) (2, -1)

C) (-1,2)

D (1, -2)

Solution:

The given of the circle is 3x2-6x+3y2+12y=12

By dividing by 3 on both side we got

X2-2x+y2+4y=4

By method of perfect square we got

(x-1)2+(y+2)2-1-4=4

By adding 1 and 4 on both side we got

(x-1)2+(y+2) 2-1-4+1+4=4+1+4=9

(x-1)2+(y+2)2=9=32

By comparing the standard form of the equation of the circle

(x-h)2+(y-K)2=r2

The center of the circle is (2)

Correct ANS : C

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4) (x+8)2+(y-6)2=16 is the equation of a circle A. another circle with the same center

and half of the radius of the circ A.

find the radius of the Circle B.

Solution:

The given circle is (x+8)2 + 9y-6)2=16

This can be written (x+8) 2+(y-6) 2=16=42

By comparing with the standard equation of the circle

(x-h)2+(y-6)2=r2

(x+8)2+(y-6) 2=42

Radius of the above circle is r=4

The radius of the circle B is = half of the radius of the circle A =1/2×4=2

The radius of the circle is 2.

ANS:2

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5) convert 1080 degrees in to Radians

A)5 π

B)3 π

C)7 π

D)6 π

Solution: radians

We know that 180 degree = π

Therefore 1080 degrees =π×1080/180=6π

Correct ANS:D

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Properties Of Triangle

!) In General Triangle has Three sides and three angles

.

sum of all three angles = 180 degrees,< p>

Sum of any two sides is greater than the third side

Example : In a triangle ABC

AB+BC>Ac

BC+AC>AB

AB+AC>BC

2) Area of a triangle = ½ ×base ×height

= 1/2×b×h squire units

3)Isosceles Triangle;

Any two sides are equal then it is called Isosceles Triangle

4) Equilateral Triangle

Let the side of the equilateral triangle is = a units

Then area of triangle = √3/4×a2 squire units

Side = a =√3/2×a units

4) In triangle any one of the angle is 90 degree then it is called Right angled Triangle

Exterior angle is equal to sum the interior opposite angles in the triangle

Right-Angled Triangle Height Base Hypotenuse

hypotenuse2=Base2+height22

Hypotenuse =√〖base〗2 )+√(〖height〗2

Base =√〖hypotenuse]2 – √〖height〗2

Height =√〖hypotenuse〗2 – √〖base〗2

Area of this triangle=1/2×base×height squire units

5) Similar triangle:

If two triangles are similar then their ANGLES are EQUAL.

Their sides are in the same ratio

6)Congruent Triangle

if two triangle are congruent then their ANGLES are equal

and their sides are equal

6)Trigonometry ratios

Sinθ = opposite side/Hypotenuse, Cosθ= adjacent side /Hypotenuse ,

Tanθ=Opposite side / adjacent side

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Right-Angled Similar Triangles A B C D E F

1) triangle ABC and DEF are similar.

Angle A corresponds to angle D

Angle C corresponds to Angele F

Angle B corresponds to angle E

Then SinA=1/2 and DE =100.

Find the Value of DF

Solution:

∆ ABC and ∆DEV are similar

Therefore corresponding angles are equal

SinA =SinD=1/2

SinD=(Opposite side )/Hypotenuse = FE/ED=FE/100=1/2

FE /100×100=1/2×100 (multiply by 100 on both side)

FE=1/2×100 = 50

DE2= DF2

+ EF2 ( Pythagoras theorem)

DF =√(DE2-EF2 ) = √(1002 -502)= 10√75=10×√(25×3)=10×5√3

Correct ANS:50√3

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3)

Right Angled Triangle ABC A B C D E

in the above diagram SinB=3/5.if BC=25

Find the length of the sides of the triangle?

Option:

A)15,20,25

B)20,25,30

C)15,25,30

D)10,15,15

solution:

consider the∆ BDE

it is given SinB=3/5 , it means opposite side = DE=3 and Hypotenuse=BE=5

therefore BD =√(52-32)=√(25-9)=√(16)=4

the side of the ∆ BDE are 3,4and 5

the sides of the ∆ BAC are BA,AC and 25

sides of the ∆s are similar

the ratio of the sides are BA/BD=AC/DE=BC/BE

BA/4=Ac/3=25/5

BA/4=5

BA=5×4=20 (multiply by 5 on both side)

AC/3=5

AC=5×3=15 (Multiply by 5 on both side)

the three sides are 15,20,25

correct ANS:A

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3)sum with similar triangle

Right Angled Triangle with Perpendicular A B C D E

In the above triangle ABC .

the line DE is parallel to side AC of the triangle

the following measurements are given for the diagram.

angle A is 90 degrees and side AC= 15 , DE =3 and DB=4 .

find the value of BC in the diagram?

Option:

A)35

B) 45

C) 25

D) 15

solution:

consider the Triangle DBE

it is a right angled triangle

in the triangle sides DB=4 and DE=3 the hypotenuse BE = √DB2+DE2=√42+32

That is BE = √16+9=√25=5

now consider ∆ABC and ∆ DBE

They are similar , therefore their sides are in the same ratio

AC/DE=BC/BE, 15/3=BC/5

BC/5=15/3

BC/5=5

BC=5×5 (multiply by 5 on both side)

BC= 25

Correct ANS:C

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Solution

4 In a right-angled triangle with two legs are 24 and 32 units what is the length of the hypotenuse ?

Option:

A)40

B)42

C)45

D)35

solution

By Pythagoras Theorem hypotenuse2=sum of the squires of the two sides

Hypotenuse2= 242+322

Hypotenuse=√(242+322 )

hypotenuse=√(576+1600) =40

Correct ANS:A

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5) Sum related to Angle property of a Triangle

Triangle ABC with Perpendicular AD A B 60 C D x y y+7 60

In figure AD is perpendicular to BC.

Measure Angle BAD =y degrees , mesure of angle CAD = x degrees

and Measure angele ABD = 60 degrees. Find the value of angle X.

Option:

A) 43

B)53

C) 37

D)63

Solution :

In ∆ABC Angle B = 60 degrees and AD is perdiclure to side BC of the Triangle.

Angle BCA = y+7 , Angle BAD = y degrees and Angle CAD = x degrees

Consider ∆ ABD

Sum of three angles in the ∆ ABD = 180 ( By theorem)

That is angle ABD + Angle BDA + angle BAD = 180

60+90+y =180 , 150+y=180

Add -150 0n both side of the equation

150+y-150=180-150, y= 30.

Now consider the∆ ACD

In ∆ ACD , sum of three angles = 180 ( by theorem)

That is angle CAD +angle ACD+ angle ADC =180

90+ x+y+7=180, substitute the value of y in the equation

90+ x+30+7=180, x+37+90=180

X+127=180

X+127-127=180-127

(Adding both side by -127)

so x= 53

Correct ANS: B

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5) sum with rectangle and diagonal

Rectangle with Diagonal

In the above rectangle ABCD ,the sides are in the ratio 3:4 and diagonal 25 units

Find the sides of the rectangle

Option:

A)15,20

B) 20.25

C)10.15

D)16,18

Solution:

Consider the rectangle ∆ADC

it is given sides are in the ratio 3:4 and, diagonal AC =25 units

let Breath of the rectangle AD=3x and DC=4x and AC=25 units

Diagonal AC 2= (3x)2+(4x)2

AC=√(3x)2+(4x)2 =√(9x2+162 )

AC =√25x

AC = 5x= 25

x5/5=25/5 (divide both side by 5

x= 5/p>

birth – 3x= 3×5=15, length =5×5=25

Correct ANS:A

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Formulas for Area and volumes of geometrical diagram

1) Area of Rectangle:

Area of rectangle is =length breath =lb squire units.

Perimeter of the rectangle =2(a+b) units

2)Area of squire:

Area of squire = side ×side=a×asquire units

Perimeter of the squire =4 a units

3)Area of Parallelogram:

Area of Parallelogram= base ×height= bh squire units

Perimeter of the parallelogram=2 (l+b)

4) Area of Triangle:

Area of Triangle = base ×height.= b×h/2 squire units

Area of triangle =√(s(s-a)(s-b)(s-c)) where s = a+b+c/2 and a,b and c are the sides of the triangle

Area of equilateral Triangle =√3 a2/4 squire units, where a is the side of the triangle.

Perimeter of the triangle =( a+b+c ) units.

5) Area of circle

Area of the circle = πr2 squire units , where r is the radius.

Perimeter of the circle = 2πr= πd squire units ,where r is the radius of the circle and d is the diameter and d=2r.

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Volume of geometrical diagrams:

1) Volume of a Right circular cylinder:

Volume of right circular cylinder; πr2h cubic units where r = radius of the circle and h = height of the cylinder)

Surface area of the cylinder:2πrl squire units

Total surface area of the cylinder:(2πrh+2πr2)=2πr(r+h) squire units

2)Volume of the cone:

Volume of cone: πr2h/3 cubic units r= radius and h = height of the cone.

Surface Area of the cone:2πrl squire units. Where r= radius of the cone.

and l=√(r2+h2 )

Totalsurface area of the cone = πr2+2πrl= πr(r+2l)squire units

3)Volume of the Cube:

Volume of the Cube: a3 cubic units, where a is the length of the cube

Surface area of the cube : 4a2 squire units

Total surface area of the cube : 6a2 squire units

4) volume of the cuboid :

Volume of the cuboid = lbh cubic units ,where l= length of the cuboid

And b = breath of the cuboid and h = height of the cuboid

surface area of the cuboid = 2(lh+bh) squire units

total surface area = 2(lb+bh+lh) squire units

5) volume of sphere :

Volume of the sphere = 4πr3 /3 cubic units.

Surface area of the sphere = 4πr2 squire units

6)Volume of the Hemi sphere :

Volume of the hemi sphere = 2πr3/3 cubic units.

Surface Area of the hemi sphere: 2πr2 squire units

Total surface Area of the hemi sphere = 3πr2 squire units.

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1)Sides of a right angled triangle are3 √5,4√5 which form the right angle in the right angled triangle. Find the third

side and area of the triangle.

Option:

A)5√5 and area of the Triangle =30 squire units

B) 7√5 and area of the triangle = 30 squire units.

C) 5√5 and area of the triangle = 35 squire units

D)7√5 and area of the triangle =30 squire units

Solution:

Area of triangle = base ×height/2

Here base and height are 3√(5 )and 4√5

Area of the triangle = 3√5 ×4√5/2=12×5/2=60/2=30 squire units

third side =√((3√5) 2+(4√5)2)

=√(45+80)= √125=5√5

Correct ANS: A

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2) Area of a given circle is 45π squire units. what is the circumference of a circle.?

Option :

A)6√5 π

B) 9√5 π

C)6√3 π

D) 5√6 π

Solution:

It is given Area of the circle = πr2 ( r- radius of the circle)

πr2=45π ( divide both side by π)

πr2/π=45π /π

r2=45=5×9

Therefore r =√9×5= 3√5 ( considering squire root of both side)

Now the perimeter of the circle = 2πr=2π3√5=6√5 π units

Correct ANS: A

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3) triangle XYZ and PQR are similar Triangle.

And it is given XY corresponds to PQ

.

XY = 3PQ .What is the ratio of the area of the triangle .?

Option :

A) 3

B) 9

C12

D)√3

Solution:

By theorem we know that the ratio area of similar triangle = squire or their sides Ratio

Here it is given sides are in the ratio 3.

Therefore ratio pof their area = 32=9

Correct ANS: B

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>

4 area of triangle sum

Circle with Diameter and Points O A B C D

In the above diagram AD is the diameter.

If AD and BD is equal to x and AC = x- 2, BD= x+1.

Find the area of ∆ACDS

Option:

A)7/8

,

B)8/7

,

C)14/8

,

D)16/17

Solution:

Consider ∆ ACB and ∆ADB in this triangle

Angle C and angle D are 90 degrees

By Pythagoras theorem AB2 =AD2+BD2=AC2+CB2

=x2 +x2=(x-2)2+(x+1)2

2x2=X2-4x+4 +x2+2x+1 ( By Expanding squires)

=2x2-2x+5

2x2=2x2-2x+5 (Add -2x2 on both side}

2x2-2x2=2x2-2x+5-2x2

0=-2x+5

By adding 2x on both side we get

2x= -2x +2x+5

2x = 5,

x= 5/2

(divide 2 on both side)

X=5/2

Area of the ∆ ACB = ½ Base ×height= ½ (x-2)(x+1)

=1/2(5/2-2)(5/2+1)=1/2((5-4)/2)((5+2)/2) By Taking LCM)

= 1/2×1/2×7/2=7/(2×2××2)=7/8

Correct ANS:A

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6 Problem with circle and similar Triangle

Circle with Diameter, Chord, and Radius O X Y !– Label P and Q –> P Q S T

In the above figure ,If the radius is 20 and PQ = 32,

What is the Area of the triangle OPT.

Option:

A) 106

B) 86

C) 96

D) 116

Solution;

Consider ∆ OPT and∆ OTQ

OP=OQ = Radius=20

OS is common

Therefore ∆OPT and ∆OTS are Isosceles and similar

So The perpendicular OS bisects OP

OT =√(20^2-16^2 ) =√(400-256)= √144= 12

So Area of the ∆OPT =1/2 ×base × height

=1/2×12×16=6×16=96

Correct ANS:C

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7 circle and similar triangle problems

Congruent Circles O P A B

In the above diagram O and P are the centers of two congruent circle.

A and B are the point of inter section of the congruent circles

OA and OB are radius of the circle with center O

PA and PB are the Radius of the congruent circle with the center P

Find the ratio of angle BOA and the angle OAP

A) 3:1

B) 1:3

C) 1:2

D) 2:1

Solution:

In the above diagram O and P are centers of the two congruent Circle.<.p>

A and B are the point of intersection of the two congruent circles

Consider ∆ OPA and ∆OPB,

Op is common

AP = PB radius of the circle with center P.

Therefore ∆ OPA ≅∆ OPB (by Theorem)

And∆ OPA and∆ OPB arr Equilateral Triangles

That is all sides are qual and all the Angles are equal.

Therefor each angle is 60

In ∆OPA, Angle OAP =60 ,angle APO = 60 and AOP = 60.

In ∆ OPB , Angle OBP =60 ,angle OPB = 60 and angle POB =60

Now consider AOB =angle AOP + angle POB = 60 + 60 =120

Angle OAP = 60

Now the ratio of Angle AOB and OAP = angle AOB: angle OAP = 120 :60=2:1

Correct ANS:D

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8) sum related Tangents of a Circle from a point outside the circle

Circle with Tangents O P 4 4 4 A B PA PB

In the diagram P is point out side the circle with center O at distance of 80 units

and radius r= 48 units and in PA and PB are the tangents from the point O

.

the length of the tangents are 64 units A and B are points are on the circle.

A and OB are Radius and it is equal to =48 units. Angle BPA is 60. Find Arc length AB (Smaller) .

OPTION:

A)32π

B) 23 π

C)43π

D)25π

Solution:

In above diagram P is point outside the circle at a distance of 80 units from The center

That is OP= 80 units.

PA and PS are tangents from the point P and PA =PB (properties tangents)

Let r = radius=48 units

It is also given angle BPA = 60

Consider the ∆ POA

Angle PAO = 90 (By the property of tangent and radius)

Angle OBP =90 (by the property of Tangents and Radius)

Now consider the Quadrilateral OAPB

Sum of four angles are 360 (Properties of quadrilateral)

Therefore ∠AOB + ∠OBP + ∠ OAP +∠ APB = 360

∠ AOB +90+90 +60 = 360

∠ AOB+240 =360

∠AOB+ 240-240 = 360 – 240 ( By adding -240 on both side)

∠AOB= 120.

ARC Length AB( smaiier)=θ/360×2πr

Where r= radius of the circle = 48 units

θ=Angle subtended at the center by Arc AB.=120

Are length(Smaller)= 120/360× 2π ×48 ( By substituting the vales in the above formula)

= 1/3×2π48=2π×16=32π

Correct ANS A

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9) sum related to equilateral Triangles

Equilateral Triangle with Midpoints P Q R X Y Z

In above diagram ∆ PQR is a Equilateral . X .Y and Z are midpoint of sides PQ ,PR and RQ .

The Area of the∆ PQR is equal to 81√3 squire units. Find the perimeter of the ∆XYZ.

OPTION:

A)25/2

B)37/2

C)35/2

D)27/2

Solution:

Area of the Equilateral ∆PQR= 81√3 squire units

That is √3/4a^2 = 81 ( formula for Area of Equilateral ∆)

(Multiply by 4 on both sides on both side)

√3/4a2×4=81√3×4

(√3a2= 81√3×4,

√3 a2/√3= 81√3/√(3 ) (divide by √3)

a2 =81,

√a2=√81,

a=9

The sides of the Equilateral ∆PQR =9 units

The point X,Y and Z are the mid points of PQ ,PR .

Now Consider the ∆ PQR and ∆XZYZ

The ratio of sides are 2 :1

Therefore XY is also = Half of its opposite side

=1/2 ×9= 9/2 units

Similarly YZ and ZX are also = half of its opposite sides = 9/2 (∆ PQR is equilateral )

∆XYZ is also a equilateral ,each side =9/2 units

The perimeter of the∆ XYZ= 9/2+9/2+9/2=3× 9/2=27/2 units

Correct ANS: D

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10) Area of triangle and parallelogram

Right Angled Triangle A B C D E F

ABC is a Right angled ∆ ,right angled at A.

D ,E and F are the mid points of AB, AC and BD respectively.

DF and DE are joined . DECF is a parallelogram.

Find the ratio of the triangle and the parallelogram.

Option :

A)1:2

B) 2:1

C) 4:1

D)1:4

Solution:

Area of right Triangle = ½ ×base× height

Herer base = AC and height =AB

The Area of the right △ ABE =1/2×AC×AB

Area Of the parallelogram = Base ×height

Here Area of the Parallelogram DECF=Base ×height

Base = EC and Height = DA

Area of Parallelograms = Base ×Height of the parallelogram

= AD×EC

Here base = EC = ½ AC

Height = DA = ½ AB

Area of the parallelogram = ½ AC ×AB/2= 1/4AC×AB

Area of the triangle : Area of the Parallelogram= !/2AB×AC :1/4AC×AB

Ratio = ½ :1/4= 1/2×4:1/4×4=2:1

Correct ANS:2:1

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11) A right circular cylinder has volume 350π cubic units. If the radius of the cylinder 10 units.

Find the Height of the cylinder?

option:

A)3.5

B)4.5

C) 2.5

D)5.5

Soluton:

Volume of the cylinder = πr2h , where r= radius of the cylinder

and h= height of the cylinder

It is given the volume of the cylinder = πr2h= 350π

Substitute r=10 in the above volume

π×10×10×h=350π

100πh=350π

100h=350, h100/100= 350/100 (divide both side by100)

That is h=3.5

Correct ANS:A

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12) The surface area of a cone 360 π ,whose radius and height are in the ratio 3:4.

Find the radius and height of the cone. ( formula =2πrl where l=√(r2+h2 ) r is radius, h is height)

Solution:

Surface area of the cone = 2 πrl where r = radius of the cone and l=√(r2+h2)

It is given surface area= 360 π

That is 2 πrl=360 π

The ratio of radius and height = 3:4

Let r=3x and h=4x where x is a constant

Therefore l =√((〖3x)〗2+(〖4x)〗2 )= √(9(x)2+16(x)2 )=√25x2=5x

Substitute r= 3x and l= 5x in the surface area formula

That is 2πrl=360π

2π3x5x=360π, 30πx2=360π

30πx2/π=360π/π (divide both side by π)

30x2=360, 30 x2/30=360/30 (divide both side by 30)

x2=12 , therefore x=√12=2√3

The radius of the cone =3x=3×2√3=6√3 units

Tha height of the cone = 4x=4×2√3=8√3units

correct ANS: Radius of the cone=6√3 units

Height of the cone=8√3units

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13) The total surface area of the sphere is 196πsquire units . What is the diameter of the sphere?

Option:

A)14 units

B) 41 units/p>

C) 21 units

D)31 units

Solation:

Total surface area of a sphere is =4πr2 (formula and r= radius of sphere)

Therefore 4πr2 = 196π

4πr2 / 4π = 196π/ 4π ( divide by4π on both side)

r2 =49, r=√49=7

Diameter =2 r = 2×7=14 units

Correct ANS:A

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14) Radius of a hemisphere is 14 units.

What is the volume of the hemisphere?

Solution:

Volume of hemisphere is = 2/3πr3 ( formula for the volume of a hemi sphere, r is the radius)

V= 2×14×14π=2×14×14×22/7=1232 cubic units

ANS: Volume of the hemisphere is=1232 cubic units

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15)A cube has a volume is 3375 cubic units. What is the surface area of the cube?

A)1350 squire units

B)1530 squire units

C)1503 squire units

D)1305 squire units

Solution: Volume of the cube = a3 cubic units. (Formula for volume of the cube<

and a is Measure of the side of the cube)

It is given V= volume of the cube =3375 cubic units

That is a3 = 3375 cubic units

Therefore a =∛3375= 15 units.

Now surface (Total) area = 6a2 ( formula and a is the length of the side)

Here a =15 units

Therefore surface ( total) area = 6×152=6×15 ×15=1350 squire units.

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16)What is the volume of a right rectangular prism has the following dimensions length= 15 units

Width is 10 units and height is 20 units.

Option:

A)3000 cubic units

B)300cubic units

C)30000cubic units

D)4000 cubic units

Solution:

We know that volume of the rectangular prism = l× b× h cubic units

Where l = length, b= width and h =height

Volume =15× 10× 20 =3000- cubic units

Volume of the right rectangular prism=3000 cubic units

Correct ANS:A

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17) Find the area of the circle, whose radius is 21 units?

Solution:

Area of the circle =πr2 ( Formula for the Area of the circle),

Here radius =21 units

Area of the circle=π21^2=22/7×21×21

=1386 squire units

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18) diameter of a circle is 14 units, what is the circumference of the circle?

solution

Circumference of a circle = 2πr ( formula)

Here diameter of the circle =14 units

Therefore radius of the circle=diameter of the circle/2=14/2= 7 units

Circumference of the circle=2πr=2π×7=2×22/7×7=44 units

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!9) the length and breath ( width) of a rectangle are 10 units and 6 units.

What is the area and perimeter of the rectangle.

Solution:

Area of the rectangle = length×Breath=l×b=10×6=60 squire units

Perimeter of the Rectangle= 2(length+breath)=2(l+b)

=2(10+6)=2×16=32 units

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20) Find the area and perimeter of a squire whose length of the side is 12 units.?

Solution

area of the squire = side × side=12×12=144 squire unit, (formula)

Perimeter of the squire =4×side =4×12= 48units ,( formula)

ANS:Area of tbe squire =144 squire units and perimeter=48 units

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Stright lines ,Angles and Triangles

Line Segment with Points and Intersections X Y A B C D M K L

1)In the figure above X,A,B.C.D and Y lie on the line segment XY. And line segment KC and BM intersects at the point L

. and it is also give∠AKC=55 ,∠KAC=85∠KLM=80.and∠MDY=160

Find the measure of ∠BMD.

Option:

A)110

B)95

C)100

D)120

Solution:

Now consider ∆AKC

∠KAC+∠AKC+∠ACK=180 ( Sum of three angles in triangle)

85+55+ACK=180,

140 +∠ACK=180

Add -140 on both side of the equation

Now 140 – 149 +∠ACK=180-140

Therefore ∠ACK=40 degrees

It is given ∠KLM=80 degrees

Therefore ∠BLC=80 ( Vertically opposite angle )

Now consider ∆ BLC

∠BLC+∠BCL+∠LBC=180 ( sum of the angles in triangle is 180)

Substitute ∠BLC=80

80+∠BCL+∠LBC=180

80-80+∠ BCL +∠LBC=180-80 (Add -80 on both side)

∠BCL+∠LBC=100

But ∠ACK=∠BCL=40

So 40+∠LBC=100

40-40+∠LBC=100-40=60

∠LBC=60

Consider ∆BMD

∠MDY is the exterior angle

∠BMD+∠MBD=∠MDY=160

∠MBD=∠LBC=60

Therefore ∠BMD+60=160

∠BMD+60-60=160-60

∠BMD=100 degrees

Correct ANS:100

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hello Sujan

Right-Angled Triangle with Perpendicular O Q P R

In the above figure ∆POQ is right angled at O.and OR is perpendicular to PQ.

And it is also given PQ = 10 units and OQ = 8 units.

Find the value of OR.

OPTION:

A)4.8

B) 8.4

C)5.4

D)4.5

Solution:

Consider ∆POQ and ∆ROQ.

Corresponding angles are ∠POQ and ∠ORQ

Their corresponding sides are PQ and OQ.

∠PQO and ∠RQO are corresponding angle

Their corresponding sides are OP and QR

The ∆POQ and ROQ are simillar.

The corresponding sides are in the same ratio.

PQ/OQ=OP/OR;

That is OQ/PQ=OR/Op;

8/10=OR/OP

Now consider right angled trianle POQ

Op2+OQ2=PQ2 ( BY Phythagores theorem)

Op2+82=102

OP2=100-64(Adding-64 on both sides)

Therefore OP=√36=6

Supbstitute the value of OP =6 in

OR/OP=8/10; OR/6=8/10

OR=8/10×6

OR=48/10= 4.8

Correct ANS:A

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hello world

Intersecting Lines AB and CD E A B C D 30° 40°

In the above figure Find the value x and y

If the Value of ∠BKD=98 degrees.

Solution:

Consider the ∆ KAD

The exterior= 30 +y ( sum of exterior angle is equal to sum of inter angles)

30+y= 98

Y=98-30=68

Consider the The Triangle CKB

In the above figure Find the value x and y .

If the Value of ∠BKD=98 degrees.

Exterior angle ∠BKD= 40 + x=98

40+x=98 (add -40 on both side)

X=98-40=58

Correct ANS: x=58 and y=68

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Drawing Parallel Lines with a Transversal

Line Diagram

A B C D E F G H

Properties of parallel and transverse lines

1) Alternate Interior angles are equal

In the diagram ∠AGR=∠GHD and ∠BGH=∠GHC are alternate interior angles

2) Alternate exterior angles are equal

from the diagram ∠EGB=∠FHC and ∠EGA=∠DHF are alternate exterior angles 3)Included Angles on the same side of the transverse lines are supplementary or 180

from the diagram ∠AGH+∠GHC=180 and ∠BGH+∠DHG=180

4) Corresponding angle are equal (on the same side of the transverse)

from the diagram ∠EGB=∠GHD and ∠FHD=∠HGB

∠EGH=∠HGC and∠FHC=∠HGA

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Parallel Lines

Parallel Lines p and q

p q A B C D N M O P Q

solution to the parallel lines sum

In the above figure p and r parallel lines

Line AB intersecting the the lines p and q at A and P.

Line CD intersecting the lines P and Q at M and O.

AB and CD are intersecting at the point Q.

It is also given that ∠CMN=112 degrees and ∠OPQ=55 degrees.

Find the angle OQP.

OPTION

A)57

b)75

C)47

D)74

Solution:

Line p is parallel to q (given)

∠CMN=112, Therefore

∠MOP=∠CMN (Transverse Theorem)

Now consider ∆OQP

In this triangle

∠MOP=∠OPQ +∠QPO =112 (Exterior angle in a triangle is equal to sum of interior opposite angles)

∠OPQ +∠OQP=112

55+∠OPQ=112

∠OPQ=112-55=57 degrees.

Correct ANS: A

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Triangle with Parallel Line P Q R X Y

In the above figure PQR is a triangle is parallel to PQ.

It is also given ∠PRQ=60 degrees and ∠YXR=70 degrees.

Find the measure of angle RPQ.

Option

A)50

B)55

C)45

D)70

Solution:

it is given PQR is Triangle

Therefore ∠YXR+∠RYX+∠XRY=180 (Sum of three angles in Triangle)

70+60+∠RYX=180

130+∠RYX=180

∠RYX=180-130=50

Now consider PQ and XY

They are parallel (given)

RQ is a transverse

∠RYX=∠RQP ( Corresponding angle)

∠RQP=50

Correct ANS: A

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A B C E D F 140 55 x —– > Triangle with Extended Line A B C E F 140 D

In the diagram side AB and BC are equal and angle BAC is equal to 55

angle CDE is equal to 140,Find angle measure of AED.

Option:

A)35

B)40

C)45

D)30

In the diagram it is given Side AB=side BC and and angle BAC=55angle EDC=140

Solution:

consider the Triangle ABC

Sode AB= Side BC

therefore Angle BAC =Angle ACB =55,( angle opposite to equal side)

∠BAC+∠CBA+∠ACB=180 ,( sum of three angles in a triangle =180)

∠CBA+55+55=180

∠CBA+110=180

∠CBA+110-110=180-110 ( add -110 0n both sides)

∠CBA=70

∠CDE+∠EDB=180 ( properties’ of a Stright line )

substitute the value of ∠CDE=140

140+∠EDB=180

Add -140 on both sides of the equation

140-140+∠EDB=180-140=40

∠EDB=40

Now consider Triangle EBD

∠ABC=∠BED+∠EDB (Exterior angle is equal to sum of interior angles)

substitute the value of ∠ABC=70 and ∠BDE=40

then we will get 70=∠BED+40

Add -40 on both sides of the equation

70-40=∠BED+40-40

therefore ∠BED=30 =∠AED

Correct ANS:B

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Circle with Tangents C P T₁ T₂

The graph of the given(x-9)2+(y+7)2=196

equation of circle in XY-plane. The point (m,n)lies on the circle

which of the following is the possible value of m.

Options:

A)-4

B)27

C)22

D)29

Solution:

Consider the equation of the given circle

That is (x-9)2+(y+7)2=196

Here it is of the form (x-h)2+(y-k)2=r2

Where (h,K) and r are the center and radius of the given circle

(9,-7) and 14 are the coordinates of the center of the circle and radius is 14.

It is also given that (m,n) lies on the circle.

It satisfy the condition that mis equal to h±r

That m is h+r 0r h-r

Here 9+14 or 9-14

That is 23 or 15

That is the vale of m lies between -5 and 23

Therefore the value of m is -5

Correct ANS:C

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2) In a right angle triangle ABC ,Angle C is right angle.

TanA=9/16 what is the value of sinA.

Option:

A)5/3

B)5/4

C)4/5

D)3/5

solution:

It is given In ∆ABC, C =90.

TanA=12/16 , But by definition of Tanθ =Opposite side /Adjacent side

Here TanA= BC/AC=12/16

The sides BC=12 and AC=16By Pythagoras theorem Hypotenuse2=Opposite side2+adjacent side2

Hypotenuse=√(█(opposite side2+adjacent side2<))

=√(12*2+16*2) =√(144+256)

=√400=20

Now SinA= opposite side /Hypotenuse=12/20=3/5

Correct ANS:D

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In a right angle ∆ PQR right angled at Q and measure of angle R is 30 .Length of hypotenuse is 48 .

find the value of Cosine of angle R.

Solution:

It is given in the Right angle ∆ PQR ,∠R=30 and ∠Q=90

Therefore ∠p =60 (By the properties of triangle sum of three angle is 180)

The angles are in the ratio 30:60:90

Therefore sides are in the ratio 1:√3:2

Here iis given hypotenuse =48

That is in the ratio 2 is equal to 48

Therefore 1 =48/2=24

Therefore √3= 24√3

The sides are in the ratio 24:24√3:48

Now Cosin= Adjacent side/Hypotenuse=24√3/48=√3/2 (Dive by 24 on both side)

Correct ANS:B

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Equilateral Triangle with Inscribed Circle

Equilateral Triangle with Inscribed Circle

A B C O D E F Angle Bisectors

In above equilateral Triangle ABC with the side a units.

What is the radius of the circle inscribed in the above equilateral Triangle.

Solution:

Let length of the side of equilateral triangle is = a units

By properties of equilateral triangle, the height =√3/2a units

We know that the point of inter section of the bisectors of any equilateral triangle

Dive the bisector in the ratio 2:1

AD,BE and CD are bisectors of the angle intersect at O

O is the center of the Circle inscribed in the equilateral triangle ABC

That is O is point dived the AD in the ratio 2:1

2+1=3 parts is =√3a/2 units.

Therefore 1 part of the bisector AD=√3a/2×1/3 units.

Measure of OD=√3a/2×1/3=√3a/6

That is the radius of the circle inscribed in the equilateral triangle is =√3/6a units.

Correct ANS:√3a/6 units.

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Numerical Example of the above sum;

Find the radius of the circle inscribed in The above Equilateral Triangle ABC of side measure of 18 inches .

Solutiopn:

Consider the side of the triangle = 18 inches,

AD ,BE and CE are the bisectors of the angles A ,B and C.

Intersects at O ,By the property of Equilateral triangle

O intersects the bisector in the ratio 2:1

Now Cosider the Triangle ADC ,

AD the bisector of the angle A and it is perpendicular to The side BC .

By theorem AD bisects the side BC

Therefore Measure of DC=1/2 of the side BC

DC=18/2=9 inches

And AC = 18 inches

By Pythagoras theorem AC,2=AD2+(1/2DC)2

182=AD2+92

AD2=182-92=324-81=243

AD=√243=9√3

The center of the circle O dives in the ratio 2:1.

So total number of portion of the of AD divide is 3

One portion of AD= 9√3/3=3√3 inches

That is radius of the circle =3√3 inches.

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Equilateral Triangle with Circumscribed Circle and Bisectors A B C

In the above diagram a ∆ABC is inscribed in a circle with center O and radius 20 units.

Find the length of the side of the Triangle ABC.

Option:

A)30√3units

B)10√3units

3)15√3units

D)20√3units

Solution:

Consider the ∆ ABC and circle with center O and radius 20 units.

∆ABC is a equilateral triangle.

AD ,BE and CF are bisector of the ∆ABC.

By the properties of Equilateral triangle , We know that bisector of angles of the triangle

Intersect at point O and it is divide th bisector in the ratio 2:1.

Consider the Right angle Triangle ADC

AD the bisector bisect the opposite side BC and perpendicular to BC

By Pythagoras theorem

AC2=AD2+DC2

Here DC =1/2BC

AD,BE and CE are the bisector of the angle A,B an AC .

Intersect at a point O . It divdes any bisector in the ratio 2:1

Here the Largest portion of the Bisector 2

Total length of the bisector is 2+1=3

2 portion =20

3 portion=20×3/2=30.

Substitute the AD=30 , AC2=AD2+1/2AC2

Ac2=302+1/4AC2

Add -1/4AC2

AC2-1/4AC2=900+1/4AC2-1/4AC2

Simplify

3/4AC2=900, Multiply by 4/3

3/4×4/3AC2=900×4/3=300×4

AC2=300×4

AC=√(300×4)

AC=2√300=2√3×100 =2×10√3 =20√3

Correct ANS:D

____________________________________________________- Square Inscribed in a Circle

Square Inscribed in a Circle

A B C D Diagonal < Inscribed Circle with Diameter

Circle Inscribed in a Square

Square Diameter E F A B C D

The Equation (x-5)2+(y+2)2=169 represent a Circle P. Circle Q is obtained by shifting

By shifting circle P by 3 units right in the XY- plane .Which of the following equation represent the circle Q.

Option[:

A)x2+y2=-5

B)x2+y2

C)(x-8)2+(y=2)2=169

D)x2-y2=5

Solution:

consider the given equation (x-5)2+(y+2)2=169 =132

It is of the of the circle equation

(x-h)2+(y-k)2=r2Here h and k are coefficient of the center of the

here (h,k) is the center and radius is r

for the Circle P has center (5,-2) and radius=13

for the Circle Q the center (5+3,-2) and radius 13

The equation for the circle q is (x-8)2+(y=2)2=169

Correct ANS:C

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Inscribed Circle with Diameter

Circle Inscribed in a parallelogram

A B C D O E

in the above diagram area of the circle is 80% of the area of the parallelogram.

Find the radius of the circle ,if the area of the parallelogram is 770 square units

Option

A)28

B)21

C)15

D)14

Solution:

it is given that area of the circle = 80% of the area of the parallelogram

area of the circle =80% of 770=80/100×770=.80×770=616

Area of the circle=πr2=22/7r2=616

r2=616×7/22 , Multiply by 7/22 on both side

r=√616×7/22=√28×7=√4x7x7=2×7=14 units

Correct ANS:D

Area Of right triangle ABC is 5 times of the area of tright triangle PQR.

Area Of the triangle ABC is 2250 squire units.

Find the plenteous of the triangle PQR, if the sides are in the ratio!:2 times.

Solution:

It is given area of triangle ABC =2250

So area of triangle PQR=2250/5=450squire units.

Sides of the triangle ABC are in the ration=1:2

That is the ratio sides=1/2

Therefore 1x and 2xtheoram are sides of the triangle.

The given triangle is right angle triangle

The area of triangle =1XBaseXheight/2=450

1x1xx2x2=450

x2=450

both side by 2

2x2/2=450/2

X2=225

X=√225

=15

The sides are 1x and 2x =1×15, 2×15=15,30

The hypotenuse=√(x2+〖(2x)〗2 )=√(152+302 )=√1125=15√5

Corrcet ANS=15√5units. —————————————————————-