8th grade school geometry

8/14/2019 8th Grade School Geometry

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(5 1) The Circle

Bashar designed water fountain in his

house with horizontal range 10 meters in

all directions. He wants to cultivate the

area which can be irrigated through the

fountain, how can you help him?

Example 1:

Let O, P two points on the plane, draw a circle with center P and passes through O.

Determine to this circle:

1. Three radii. 2. Diameter. 3. Chord.

Solution:

Compasses is opened slot equal to

the distance between points O, P. focusing

the cape horns of compass at the point P

and then draw a circle. You will notice

that the circle passes through the point O

as in figure (5 - 1).

1. PO, PD, PS, Three radii of the circle.

2. SO, Diameter of the circle..

3. AB, Chord of the circle

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Remember :* A CIRCLE is the set of all points in a

plane that are equidistance from a given

point in the plane. The given point is the

CENTER , and the given distance is the

RADIUS .

* A CHORD is the segment that joins two

points on a circle .


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Training 1:

1. With reference to example 1 draw a circle with

center O, and passes through P.

2. If the distance between any two points A, B equal

to 6 cm, using the compass and a straightedge to

draw a circle with chord AB.

Training 2:

Solve the issue problem at the beginning of the lesson 5-1 specified for a

certain area of land that could be planted and irrigated from the fountain.

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What is false?

The circle is the situated

area inside a closed

curve, All of its pointsare equidistance from

known point..

Activity 1

Carpenter wanted to a the surface of a circular table from wooden platform in

the form of rectangular with dimensions 122, 244 cm as in the figure (5 - 2).How can you help him in obtaining the largest possible area of the surface of

the table and more economical use of wood.

1) How many circular surfaces can be cut from a wooden board with the largest

possible area?

2) If a Carpenter chooses the circle centre at 70 cm from one of the edges piece

of wood. How many circular surfaces with largest possible area could be cut

from the wooden board?

3) If the Carpenter has not got a compass, what tools could be use instead?


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Training 3

Draw a circle with center M on a paper

and then draw the angles: HMB,

COM, where each measures 40 0

as in figure (5-4) cut the two sectors

and Congruent them,what is your note?

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Result 1:

when we draw congruent chords in

a circle, then their corresponding

arcs are congruent. And their

corresponding central angles are

congruent.

Remember:Two rays can be drawn from the

center of a circle to form an angle.If the rays and the circle are in the

same plane, such an angle is called

a CENTRAL ANGLE, and their

arc as a part of the circle.

Figure (5 - 4(

Activity 2Draw a circle with a radius 7 cm and center M.

Draw a chord A E: AE = 10 cm.Draw a chord CD which does not interrupt the chord AE: CD = 10 cm.Cut sectors MAB, MCD.Congruent sectors MAE, MCD.What is the relationship between themeasurements of the two central angels

AME, CMD?What is the relationship between the

two arcs AE, CD.


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Problems and Exercises

1) Draw a circle with diameter 10 cm.

2) Relied on the figure (5-5) to answer

the questions a d:

a. what does each of the

segments AB, MC, AC represents?

b. Where is the vertex of angels:

BMC, AMC?

And what are they called?

c. How many chords, diameters, radii

can be drawn in this circle?

d. If the length of the largest chord

is 8 cm, find the radius of this circle?

3) If a large group of students in grade 8 in your school coin at a distance

of 1 km from the classroom and homes scattered in all trends:

a. Draw a circle with a classroom center.

b. If Ahmed and Khalid are two students in this class, find the largest

possible distance between their homes?

4) a. The broadcast communications tower mobile phones covers a distance of 13

km Bashar has tried to contact with participants department in the network,

but the communication has not succeeded could you deviate reason? b. Circle with center M and radius 5 cm, find the length of MX if the X lies:

(1) On the circle. (2) Inside the circle. (3) Outside the circle.

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( 5 2) Triangle

Let X, Y are two coplanar points, then there is exactly one line

containing them. The length of the segment XY is the distance between point X

and point Y. figure (5-6).

Suppose X, Y, Z three coplanar points.

How many lines can be drawn containing any two of the points X, Y, Z?

What are the different situations for these points that affects on the number of

these lines, and the nature of the figures that produced of their intersections?

The first case:

Only one line passes through X, Y, Z and X, Y, Z are collinear.

To show that, use one of the following ways:

a)The fist way : Draw a line contain any two points of them, if the line passes through

the third point, then all points are collinear, otherwise it is not.

b)The second way: Find the measure of the distance between the three points twice. If the sum of the smallest distances equal to the largest distance, then the points are

collinear.

The second case:

The three points are noncollinear, in this

case we can draw three different lines, and

each of them passes through two pints asin figure (5-7).

Example 1:

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Let A, B, C are three points, so that AB = 5cm, BC = 7cm, BC = 10cm. Are

A, B, C collinear?

Solution :

Notice that the smallest distances are 5, 7, and the sum of 5+7 =12

not equal to the large distance

which is 10. Then the three points are

noncollinear, as in figure (5-8).

Training 1 :

Let the distance between A, B is equal to 12 cm; AC = 4 cm,

DC =8 cm. Are A, B, C collinear?

You must notice that a triangle form in the case that the three points are noncollinear,

i.e.

Example 2 :

Let CD =7 cm, DH= 4 cm, CH =5 cm. Determine whether it's possible to draw

a triangle CDH with sides of the given measures. If it's so, draw the triangle.

Solution:

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The sum of the lengths of any two sides of triangle is greater than

the length of the third side .


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You find the sum of the smallest two lengths from the length of the segments

CD, DH, CH and compare it by the largest segment, notice that:

4+5 > 7 i.e. DH + CH > CD. Then C, H, D vertices of triangle CDH. To draw the

triangle:

1. Draw one of the segments like CD with length 7cm.

2. Open the compasses aperture equals to the length of one of the other two

segments say 5 cm . Place the compasses' point on C and mark an arc.

3. Open the compasses aperture equals to the length of the third segment which is 4

cm. Place the compasses' point on D and mark another arc. The two arcs are

meeting at H .

4. Draw HC, HD, to form a triangle as in figure (5-9).

Training 2:

Draw a triangle for the following cases:

1) AB = 7 cm, BC= 6 cm, AC= 4 cm.

2) EF = 5 cm, DF= 2 cm, DE= 3 cm.

3) GH = 2cm, HI = 5 cm, GI = 8 cm.

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Example 3:

Draw a triangle XYZ on a paper such that XY =XZ, cut the triangle and fold it

on itself, XY congruent on XZ, draw the folding line? What do you notice?

Solution:

The triangle XYZ isosceles, it can be drawn as follows:

Draw a segment YZ.

The compass is open a suitable slot distance (greater than YZ).

Place the compass point firstly on Y and mark an arc, secondly on Z and mark

another arc intersecting the first arc in X.

Draw XY, XZ, to form a triangle as in figure (5-10) .

Cut the triangle and fold it on itself and draw

the folding line.

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Activity 1:

Draw AB where AB =12 cm.

Open the compasses aperture equals 5 cm, and draws a circle with

center A. Open the compasses aperture equals 6 cm, and draw a circle with

center B.

Do the circles intersect? Explain that.

Are the measurements 12 cm, 5 cm, 6 cm, forms a triangle sides?

What happened if the measurement of the segment is 11 cm?


What happened if the measurement AB = 8cm?



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You notice that: m Y = m Z.

m DXY = m DXZ.

DY = DZ. m XDZ = m XDY.= 90 0.

As in figure (5-11).


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Think

ow many isosceles

iangles can we drawn

o the segment YZ ise base of these

iangles?

Remember:

The sum of the measures

of a supplementary

angles equals 180(thetwo angels form a

straight angle).


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1) For each case of the following, show that if the three endpoints of the given

segments are collinear:

I. AB = 4 cm, BC = 7 cm, AC = 5 cm.

II. XY = 4 cm, YZ = 4cm, XZ= 4 cm.

III.EF = 6 cm, GF = 12cm, EG = 6 cm.

2) Draw a triangle ( if possible) in each case in exercise 1.

3) Use figure (5-12) to calculate the mesurment of ABD, AHC, if AB =

AC, AD =AH.

(5 3) The Exterior Angle of The Triangle

1. Draw three intersection lines to form triangle as in figure (5-13)

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WHD are called interior angles in the triangle

DHW.

What is the relationship between the degree

measures of an exterior angle of a triangle

and the degree measures of the remote

interior angle?

On the figure (5-14). Determine two sections

every one contain one of the remote interior

angle H, D.

Cut the two sections.

Make the vertex of every angle

on the vertex of the exterior

angle HWC as adjacent,

as in figure (5-15), what is your notice?

You notice that:

m HWC = m HDW + m WHD,

To prove result (1):

Given:

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esult1:

The measure of an exterior angle of a triangle, is equal to

e sum of the measures of the two remote interior angles .


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XYL exteriorangle of the triangle XYZ,

figure (5-16).

Required :

Prove that m 4 = m 1 + m 2

Proof:

m 1+ m 2 + m 3 = 180 0 (Triangle sum theorem )

m 3 + m 4 = 180 0 (The angles in a linear pair are supplementary )

Then m 3 + m 4 = m 1 + m 2 + m 3 (Substitution )

m 4 = m 1 + m 2 (Subtraction Property of Equality )

Example1:

Use the information in the figure (5-17) to find the measure of the angle ABC

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Think Is it possible that measure of

more than one external angle of the same triangle less than 90 ?


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Solution :

HAD is an exterior angle of the triangle ABD, then

m HAD = x + m DBA

100 0 = X +X ( ABC is isosceles )

100 0 = 2X

X=50 0

m ABC =180 0 50 0 = 130 0.

Traning 2:

In the triangle ABC; m BAC =45 0,

ABD is an exterior angle and

m ABD =120 0, find the measure of A,

B, C in the triangle ABC

Traning3:

Use the information in figure (5-18), to calculate m DYZ

Given that XY=ZY.


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Think:Solve in other waywithout using theresult.


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1) If HAB, CBO, ACF are an exterior angles of the triangle ABC,

using figure (5-19) to find the sum of measurements of these angles.

2) Using the information in figure (5-20) below to find the measures of:

1, 2, 3.

3) In the triangle EFG, EFD is an exterior angle of the triangle;

m EFD = 110 0, and the angle EGC is another exterior angle of the

triangle ; m EGC = 140 0.

Find the measure of each angle in the triangle EFG.

(5-4) Right Triangle

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* Draw the triangle XYZ; XY = 6 cm,

XZ = 10 cm, YZ =8 cm.

* Draw the triangle ABC; AB = 5 cm,

AC = 13 cm, BC =12 cm.

* Measure XYZ, ABC,

* you notice that the

measures of each of them is 90 0,

as in figure (5-21) and:

82 + 6 2 = 10 2 , 5 2 + 12 2 = 13 2.

Training 1:

Draw a triangle with sides X, Y, X 2 +Y 2, and take a different values of X

and Y. Use the protractor to find the measure of the largest angle in each triangles.

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Result1:

If the sum of the squares of the measures of two sides of a triangle equals the

square of the measure of the longest side, then the triangle is a RIGHT ANGLE.


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?=

Example 1:Determine whether a triangle with sides measuring 8 cm, 15 cm, 17 cm, is a

right angle.

Solution:

It is only necessary to check one equation: (17) 2 (15) 2 + (8) 2

289 = 225 + 64. Since the equation is true, the triangle is a right angle.

Training 2:

Determine whether a triangle with sides measuring, 12 cm, 7cm, 193 cm, is a

right angle.

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Activity 1

Draw triangles with angles measuring 30 0 ,60 0 ,90 0 .

Find the length of its sides.

Draw another 30 0 ,60 0 ,90 0 triangles and measure the

length of their sides.

Result2:

In a 30 0 , 60 0 , 90 0 triangle, the hypotenuse is twice as a

long as the shorter leg.


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180 0 - 100 0

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Example 2 :

ABC is a right triangle in B , D is the midpoint of the segment AC;

m ADB = 100 0. Find m BAC

m ACB as in figure (5-23).

Solution:

D is the midpoint of AC, then

BD = AC = AD = DC,

Triangles ADB, BDC are isosceles,

m BAC = = 40 0,

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Activity 2

Draw a right angle on a paper like ABC.

Cut the triangle.

Bisect the hypotenuse BC in a point O.

Folding the triangle two

times, the first one, put

vertex C on vertex A,

and the second one, put

vertex B on vertex A, as

in figure (5-22).

What your notice.

esult 3 :

In a right triangle, the length of the hypotenuse is twice as a

ng as the length of the segment from the vertex of the right

ngle to the midpoint of the hypotenuse.

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Then m CDB = 180 0- 100 0 = 80 0,

m ACB = 50 0.

Training 3:

In a triangle AXY it is known that m X = 90 0 , D is the midpoint of AY, H

is the midpoint of XY, prove that DH XY.

i.e. (XY) 2 = (XL) 2 + (YL) 2

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Activity 3

Draw a right angle YLX; L is a right angle.

Build a square on each side of the triangle

shown in figure (5-24).

What is the relationship between the

quare of the hypotenuse and the squares

the other triangle legs?

Draw more right triangles, and search

n the relationship between the square

the hypotenuse and the squares of the

ther triangles legs?

Theorem 1: In a right triangle, the square length of the

hypotenuse is equal to the sum of the squares of lengths of the other trian le le s.


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Example 3:

Represent the real number 2 on the real numbers line.

Solution:

We can use the Pythagorean Theorem to find the point which opposite to 2 on the

line, because: ( 2 ) 2 = 2, i.e. ( 2 ) 2 = 1 2 +1 2 . Open the compass with distance equal to the hypotenuse in a triangle ABC.

Place the compass point on 0 on the numbers line and mark two arcs intersect

the line in two points, the first one, on the right of point 0 which represent the

location of( 2 ). And the other one on the left of the point 0 which represent

the location of (- 2 ), as shown in figure (5-25).

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Add to your knowledge.

Theorem 1 is called

Pythagorean Theorem according

to Pythagoras, a Greek

mathematician and philosopher

from the sixth century (582 -500

BC ), he was built the Pythagorean

school in Italy, many different

basic theorems in planer geometry

has been roved since that.

Think :

If the shape which was built on every side of the right

triangle is equilateral triangle, is theorem 1 still true?

Is the theorem still true if the built shapes are:

1(Another polygon such as pentagon, hexagon. ..

2(Circles with diameter the sides of the triangle.

3(Irregular polygon.

?What is false

;In a triangle ABC

AC =AB + BC

AB))2 = (AC) 2 + (BC )2

BC))2 = (AC) 2 + (AB) 2

C B


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Example 4:

Given XYZ aright triangle, such that XY = 4 cm, XZ =3 cm, find the length of

YZ?

Solution:

According to Pythagorean theorem XYZ is a

right triangle, then:

(YZ) 2 = (XY) 2 + (XZ) 2

= (4) 2 + (3) 2

= 16 + 9 =25, then YZ = 5

(- 5 is neglected)

Then YZ = 5 cm.

Training 4:

Given ABC a right triangle, such that AC =10 cm, BC = 6cm, find the length of

AB?


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+

Activity 4

.Represent the real number 3 on the real line numbers


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1) Calculate the length of the third side of the triangle in each of the following

case:

I. XYZ, so that XY = YZ = 1 cm, Y = 90 0.

II. ABC, so that AB = 1 cm, AC = 3 cm, B is a right angle.

III. ZLH, so that ZL = 3 cm, LZ = 7 cm, L is a right angle.

IV.KLM, so that KM =25 cm, kl = 24 cm, k is a right angle.

2) For each of the following, show that the three given legs are sides of a right

angle.

A. 15 cm, 20 cm, 25 cm.

B. 9 cm, 40 cm, 41 cm.

C. 2 cm, 2 cm, 2 2 cm.

D. 3 cm, 7 cm, 4 cm.

3) A person wants to measure an angle a room to the tile, and he wanted to make

sure it has a right angle. How can you help him by using Pythagorean

Theorem?

4) Kareem is standing next to an electric trunk; he walked 10 meters towards

south, and 6 meters east. How fare is Kareem from the standing point?

5) A paper is wanted to be made as square with

diameter 12 cm. What are the dimensions of

the paper?

6) A ladder 2.5 meters is leaning against a

vertical wall; calculate the distance of the top

of the ladder from the ground if the distance

of the bottom of the ladder from the wall is

0.7 meters, as in figure (5-27).

(5 5)Angles Transferring

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Architect Nermeen was designed

sketch for a children garden

fence containing angels, Ahmad,

one of the 8 th grade students, wanted

to transfer these angles into another copy.

How can you help him?

Example 1:

Figure (5-28A) represents the angle

BHO; draw an angle congruent to the angle

BHO.

Solution:

Place a compasses on H, and draw a

circle intersects the sides of the angle

in C, D as in figure (5-28B).

Draw XY.

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With the same compass setting, draw

a circle center X cutting XY in two

points, one of them is Z as in figure

(5-28C).

Open the compass slot equal to the

length of the chord CD, focusing the

cape horns of the compass at Y and

then draw an arc cutting the circle in

L as in figure (5-28D).

Draw XL, then the angle .

LXZ is formed with the same

measurement of the angle

BHO as in figure (5-28E).

Training 1

Draw an angle with mesure 60 using

A protractor , then transfere this angle

using the ruler and the compasses only.

Then verify the mesure of the result angle

Using

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Think :Is it enough to sketch part of

the circle ( arc) to transfere a

known angle?


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Training 2:

Draw an obtuse angle on a paper, using the straightedge and the compass to

transfer it to another paper, then cut the two angles and place the vertex and the side

of one to the vertex and the side of the other. What is your notice?

You can help Ahmad to transfer the angles of the sketch to another copy (the

question in the beginning of the lesson), by using angles transferring steps.

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1) Using figure (5-29) and straightedge to transfer the following angles:

A. BAH.

B. BZH.

C. ZHO.

D. AHO

2)Transfer to your notebook the right

angle in figure (5-30) by using the

straightedge and the compass, then

build a right triangle their legs

has lengths 8 cm , 6 cm.

2) Let ABC a right triangle; B =90 0,

as in figure (5-31). Using straightedge

and compass to:

A. Transfer the angle ABC on a paper with vertex Y.

B. On one of the angle legs

which their vertex Y, determine

a point Z, such that YZ = BA.

C. On the other leg of the angle

which vertex Y, determine a point X,

such that YX = BC.

D.Cut the triangle XYZ and congruent it to triangle ABC. What is your

notice?

(5 6) Bisection the angle

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Dear student, how you can determine the position of water

container to be build around the meeting

point of the bisectors of the base angels of a farm

which has an isosceles triangle shape to irrigate

the farm ?

Example (1)

Bisect the angle XYH figure (5-32A) using

unmarked ruler and the compasses .

Solution:* Open the compasses with suitable aperture

And fix its sharpen point at the vertex of the angle Y, draw

arc crosses the sides of XYH in D and W

figure (5-32B)

* Open the compasses with suitable aperture

And fix its sharpen point at d and draw an

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arc inside XYH then fix the compasses'

sharpen point at w and by the same aperture

draw another arc meeting the first arc

in l figure (5-32C).

* Join YL with a ruler, this be bisector of the

angle XYH that is

measure XYL = measure HYL figure (5-32D).

Training 1

Copy to your notebook each of the

two angles XAB , YOL figure (5-33).

No doubt you have gotten an ability to identify the meeting point of the two bisectorsof the farm's base, in the beginning of the item (5- 6).

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Think

In

how much method can you

determined that the two angels

x l & h l have the same


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Activity 1

Look for steps to sketch two multiples of a given angle.

Activity 2 Sketch a triangle DHB on a sheet , then bisect

the two angels DHB , HBD the two bisectors

meet at M.

Join the point m with d figure ( 5 - 34).

Fold the angle hdb around MD after

cutting it .

Note that MD bisector to the angle d from

congruence the legs of the angle HDB ( DH , DB ).

Repeat the test for different triangles and write

your notes about the meeting point of the bisectors

Result 1

Bisectors of the triangle angels meeting at one point.


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1) AB , CD two straight lines meeting at the point h figure ( 5- 35)

use the ruler and the compasses to do the following :

Bisect the angle AHC by the bisector HW .

Bisect the angle AHD by the bisector HX .

Bisect the angle AHB by the bisector HO .

Find measure WHX without using the protractor.

2) Sketch angels with measures = 30 , 60 , 15, 7.5 using

unmarked ruler and the compasses .

3) ABC a triangle m the meeting point of its angels bisectors,measure C =40 , measure B = 30 , prove thatmeasure AMB = measure BAC .

4) Sketch a straight angle, and then divide it into six equal angels without using

the protractor.

(57) Composing Perpendicular from Assumed Point on a StraightLine

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An engineer in Great Amman Municipality

sketch a design for suspension bridge joins

between two parallel streets. To reach the

least costs he decide to make the bridge's

length short as possible . The position

of the bridge is known on one of the two

streets . how can this is done using the

ruler and the compasses ?

Example (1)

Let (M) be assumed point on the

line L, figure (5 36A) .

Raise perpendicular from the point (M) on the straight line ( L) using the ruler and the compasses .

Solution

Open the compasses with suitable aperture

and fix its sharpen point in (M) and drawtwo arcs crossing the straight line ( L)

At the two points W , H figure (5-36b).

Open the compasses a aperture bigger

than the previous one and fix its sharpen

point in ( W) and draw an arc , do the same

thing from (H) and draw another arc .The two arcs meeting at (X) figure (5 36C) .

Join (X) with (M) to get XM perpendicular

on the straight line (L) figure (5 36D)

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Activity (1)

Arrange with your friends and your school a scientific trip to Amman to see

A set of designs which is really implemented or will implemented.

Training (1)

Raise perpendicular from the assumed

point on each of the two straight lines

L , M figure ( 5 37)

Training (2)

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Think :

a) How you can sure that x m

perpendicular on the straight line (L)?

b) Is it possible to sketch another

perpendicular from the opposite side?

What is the relation between them?


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Consider the question in the beginning of item ( 5-7) and show how the engineer

in Great Amman Municipality will act since the position of the bridge on

one of the streets is known.

Result ( 1 )The least distance between two parallel lines equals the length of the

perpendicular joins between them.

Activity (2)

*The straight line dh intersects a circle whose center (M) at the points A, B.

Construct a perpendicular from the point (A) on the straight line DH to meet the

circle at (W.(*Join the two points B, W What is your notes at the hypotenuse W B figure (5-38(

* Repeat the process by composing a perpendicular from the point (B) on the straight

line DH to meet the circle at O.

* Join the two points A, O what are your notes on the hypotenuse WB?

Training 3Determine the center of the circle in

Figure (5-39) by sketching two

diameters to the circle

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1) Compose a perpendicular on the straight line XY from the point (A) which is

located on it using the ruler and the compasses.

2) A , B two points on straight line ,the distance between the two points are 6 cm,

use the ruler and the compasses to sketch a square where AB is one of its sides.

3) If AB is a diameter in a circle whose center is (M), compose a perpendicular on

it from (M).

4) X , L are two points on straight line , use the ruler and the compasses to draw

a right triangle in (X) so it has equal leg

(5-8) Download a perpendicular on a straight line from point outsidethe line.

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Abed Alkareem wants to join his

house with the main water line

which passes throw the opposite

side. He must dig the street

with least possible distance to

reduce the cost. How can this be

done?

Example (1)

a is a point outside the straight line (M) figure ( 5 40A) , download a perpendicular

on (M) from (A) using the ruler and the compasses.

Solution :

Open the compasses a suitable aperture and fix its sharpen point in (A) and

draw an arc crossing the straight line (M)at the two points D, H figure (5

40B).

Open the compasses aperture different from the first one and fix its sharpen

point in (D) and draw an arc.

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* Fix the sharpen point of the compasses

in H , and with the same previous aperture

draw another arc to meet the first arc in (E)

figure ( 5 40D )

* sketch the straight line e a to meet

the straight line (M) at the point (W) figure

(5 40H). So w a is perpendicular on

the straight line (M) .

Verification

* Cut the figure and fold the angle

to get congruence between DH and WH.

Note that the fold line is WA and

the angle (DWA) is congruent to the

angle HWA.

Because they are on the same straight line,so (WA) is perpendicular on the straight line (M).

Training (1)Download a perpendicular on the line ( L )

from the point (M) figure (5 41) and

verify that your sketch is true.Training (2)

Tell Abed Alkareem how to dig the street to join his house with the main water line

(see the beginning of item ( 5- 8)).

Activity (1)

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* Sketch an angle like XYA , then bisect it

by the bisector YC.

* let the point (C) on the bisector .

* Download a perpendicular from

the point (C)on the line (XY)

and another one on (YA )

figure (5 42) and.

*Fold the angle xya around its

bisector core then compare

the distance of ( C )from xy

with the distance of ( C )from

YA . What is your observation?.

result (1)

A point on a bisector if an angle has equidistance from its legs.


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1) Use the ruler and the compasses to sketch a rectangle whose one side is XY and its

opposite side lies on the straight line CD, knowing that XY // CD, figure (5 43).

2) Download a perpendicular from the point (Y) outside the straight line AB to meet

it at (O) using unmarked ruler and the compasses. Then select a point ( L) on AB

such that YL = LO , see figure ( 5 44 ).

(5 9) Bisecting a segment

Abed Al Azez wants to fix scouts lighting in

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the mid of the top side of the front interface

of his building . how can he determine

the fixing point of the scouts using the

design ?

Example (1)

Determine the mid point of the segment ZW ,

figure (5 45 A).

Solution* Open the compasses an aperture greater than

half length of ZW (by estimation) ,fix its sharpen point at one of ZW ends

and draw an arc figure (5 45B).

* Fix the sharpen point of the compasses at

the other end of the segment and drawan arc crossing the first arc at the

two points H , L figure (5 45C).

*Join h l using the ruler to cut the segment ZW at X

figure (5-45D).

* Let one of the two segments HL , ZWthe core of fold . What is your observation

40

Think If you complete drawing

the circle when you draw the two

arcs to bisect the segment z w,

what is the relation between the

composed perpendicular from the

mid point of the hypotenuse in

the circle and its center?


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Training (1)

Help Abed Al Azez to determine

the position of the scouts lighting on

his building ( see the beginning of item ( 5 9 )).

Training (2)

Draw in your copybook a straight

Segment, then bisect it using the

compasses and unmarked ruler ,

verify that your sketch is true.

Activity (1) * Sketch a triangle WSE figure (5 -46)

* Compose the bisector perpendicular

of the side W S to meet the bisector

perpendicular of the side SE at the

point (M).

* Fold each of one of the triangles MWS,

MSE around the bisector perpendicular to

find

that WM = SM , SM = EM consequently :

WM = SM = EM

So WMS , SME are isosceles triangles* Bisect WE in D .

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* Join (M) with (D) then cut WME and fold it around MD then observe that

the two triangles WMD and EMD are completely congruent.

Result (1)a) The composed perpendiculars from the mid points of the triangle sides are meeting

at one point.

b) The vertices of a triangle are equidistance from the meeting point of the

perpendicular bisectors of its sides.


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1) Draw on your copybook a segment, then divide it into four equal parts .

2) Two persons have a rectangular plot of land , one of its sides lies on a straight

street ,they want to divide it into two equal parts each of them has rectangular shape

so that their sides from the street's side are equal . Use the ruler and the compasses to

divide the plot of land as they want .

3) Use the ruler and the compasses to sketch the segment joining between the mid

points of the triangle BXW.

4) Consider the figure ( 4- 47) to find the length of ( LO ) , given that( XYO) a

triangle and (M) the meeting point of the composed perpendiculars from the

midpoints of its sides ,(ML) one of these perpendiculars ,XM =13 cm and ML =5 cm.

(5 10) ApplicationsFirst: The circle sketched inside triangle and tangent to its sides.

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Example (1)

An ironsmith wants to cut a cover has

A circle shape from a triangle flat piece

of iron. How you help the ironsmith to

identify the center of the circular cover

and its radius to get the cover as large

as possible.

Solution

* The triangle DMH figure ( 5 48A)

represents the flat piece which

will be cut by the ironsmith to make the

cover.

* Bisect the angle (M) then bisect the

angle (H),the two bisectors will meet

at (B) , see figure ( 5 48 B).

* Download a perpendicular from the

point (B) to the side m hMH to meet it at A,

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see figure ( 5 48 C).

* Open the compasses aperture equal BA.

* Fix the compasses at (B) and sketcha circle ,note that the circle you sketched

tangent the sides of the triangle from inside.

figure ( 5 48 D) .The circle whose center

(B) is called the circle sketched inside the

triangle

Training (1)

* Sketch the circle which tangent

the sides of the triangle AXC from inside

figure ( 5 49 )

Training (2)

Sketch the circle which is tangent

to the sides of a right triangle

from inside.

Second: drawing a circle passing in three points

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Example (2)

Bnan cuts a triangular shaped paper and wants to lay it inside a circular ring . What

is the shortest diameter of the circular ring surrounds the triangle.

Solution:

*The triangle BHC represents the pieceof paper cut by Banan . Figure(5 50A)

*Sketch the perpendicular bisector

of the side BH Using the ruler and

the compasses. See figure (5 -50B)

* Sketch the perpendicular bisector of the side HC to meet the bisector

of BH in a See figure (5 -50C)

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* Open the compasses aperture equalto HA and fix its sharpen point at (A)

and sketch a circle .figure(5 50 D) .

Note that the circle passes the

triangle's vertices.

Training (1)Sketch the circle with center (A) and passes

through the points C,D,W figure (5 51)

Activity (1)

47

Think

If you asked to draw a circle

passing through the points w

, h , s and you don't find

a point with equal distances

from these points ,what does

it mean?


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Draw an equilateral triangle using the ruler and the compasses, then do the

following:

a) Bisect the side BC in D .

b) Join the point (D) with the vertices (A).

c) Cut the triangle ABC into two triangles ABD and ACD through the straight

line AD , then fold the two triangles such that the angle ABD congruent to the

angle ACD .

d) What is the measure of each of the two angels ADB and ADC ?

e) Calculate the length of AD using Pythagorean Theorem in terms of length side

of an equilateral triangle.

f) Find the measures of the two angels ABD and BAD . What is the relation

between the length of the side opposite to the angle BAD and the length of the

hypotenuse AB?


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ABC is a triangle, figure (5 52):

a) Sketch the two bisectors of the two angels B ,C to meet in (M).

b) Download a perpendicular from (M) on AB using the ruler and the

compasses.

c) Sketch the circle with center ( M ) tangent the triangle's sides from theinternal.

2) Copy The triangles ABC, DHW and XYO to your copybook, then draw the

circle which passes through the vertices of each triangle using the ruler and thecompasses.

3) Find the length of the circle's diameter which passes through the vertices of

the triangle HWL given that HW=15 cm, WL = 8 cm and HL = 17 cm.

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Review

1) We want to fix four lamps in the ceiling of a room which has a rectangular

shape with dimensions 4 m and 9 m .Each lamp will be fixed in the meeting point

of the rectangle with the bisector of the opposite rectangular angle , see figure

( 5 53 ) , determine the positions of the points where the lamps will be fixed.

2) we want to fix the door of a farm which has

a trapezoid shape figure ( 5 - 54 ),

to the wall by two joints at the two

points C , D so that AC = BD =( 1/4) AB .

A handleof the door will be laid in the

middle of WH, determine the two

points C , D and the position of fixing

the handle using the ruler and the compasses.3) We want to cut a rectangular wooden plate with length equals two times its width

into two circular bases. Find the radius and the center of the circular base to make the

circular base as large as possible.

4) Use the ruler and the compasses to cut a circle from a triangular piece of cloth so

that you have the large possible area.

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5) A , B , C are three villages .we want to build a container of water to provide the

three villages with water such that AB = 4 km , BC = 3 km and AC = 5 km .

Determine the container's position so that it is equidistance from these villages.

6) Conceder the trapezoid figure ( 5 55 ) ,

to sketch the segment ( AB ) joins between

the middle points of the two sides HW

and MD using the ruler and the compasses

then describe a method to show that

AB = (1/2) ( HM + WD) .

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Self Exam

1) Verify that the following statements are true or falls:

a) If the distance between two points on a circle equals 10 , then the radius of the

circle equals 5 cm .

b) If the point (A)lays on the circle whose center is (M) and its radius 7 cm ,then (M)

lays on the circle whose center is a and its radius 7 cm.

c) There is a triangle the sum of the measures of each of its angels is 90 .

d) If ABC is an external angle of the triangle ABC then ABC an obtuse angle.

e) The bisector of the straight angle is the composed perpendicular from its vertices.

f) The two distances of any point lays on the bisector of the angle from the angel's

sides are equal.

g) In the right triangle, the area of the circle whose diameter the hypotenuse of the

triangle equals the sum of the areas of the two circles whom diameters the legs of theright angle.

2) This question contains of (5) multiple choice items, each item has four answers

only one of them is correct .Circle the correct answer for each item.

(1) If A , B , C are three points in the plane such that AB = 7 cm , BC = 11 cm and

AC = 4 cm . Then :

a) ABC is a right triangle b) A , B , C are collinear points

c) ABC is acute angels d) BAC is obtuse angle

)2(If (X) a set contains of 1000 points of the plane where each of them is 5 cm apart

from the point (M), then the circle whose center (M) and its radius 5 cm is:

a) The set (X.(

b) Contains no point of the set (X.(

c) Contains some of the points of the set (X) but not all of them.d) Contains other points beside the points of the set (X).

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(3) If ABC is acute triangle then the measure of one of its external angels is:

a) Less than 90 b) 90 c) More than 90 d) 180

(4) If the lengths of a triangle sides are 15 cm , 17 cm , 8 cm , then the triangle is :

a) Acute angels b) Obtuse angle

c) right angle d) Has two with the same measure

(5) If ABC is equilateral , with length side equals 11 cm and AD the bisector of the

angle BAC meets BC in d then the length of AD equals :

a) 11 cm b) 5.5 cm c) ( 11 ) / 2 cm d) 11 cm

3) Use the ruler and the compasses to draw an angle of measure 45 and an angle of

measure 22.5 .

4) Draw the triangle XYO where XY = YO = OX = 10 cm , then

a) Bisect the angle XY O with the bisector YL to meet X O at D . Calculate the length

of YD ( use Pythagorean Theorem ) .

b) Calculate the length of OH, where h the meeting point between the bisector of the

angle YOX and the side XY.

c) Let (M) be the meeting point of the two sides YD and OH , show that the point (M)

is the center of two circles : the first passes through the vertices of the triangle XYO

and the second tangent the sides of the triangle from internal .

5) Sketch the triangle HLW where HL = HW , then bisect the angle LHW by the

bisector HY to meet LW in D, fold the triangle LHW around the straight line HD to

get completely congruence between the two triangles HLD and HWD .What is your

notes about the two sides LD , WD and the two angels HDL, HDW.

Prove that the center of the circle which passes through the vertices of the isosceles

8th grade school geometry

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