8th Grade School Geometry

Unit 5 : Geometry Preface: Do you know dear student how the houses and buildings built? Are you seen one day a design for building? Who sketches this design? What the architect use to reach high accuracy? are geometrical instruments as the ruler , compasses, protractor and triangles enough for that ? This unit will take you to the world of geometry, Mathematical methods and logical methods which are help the architect to sketch accurate designs .A set of methods will display to sketch geometrical constructions using simple instruments; as un marked ruler and the compasses depending on the properties of the circle with high accuracy and after studding it we expect from you to: 1) Explore that any two perpendicular bisectors on two chords in the same circle

meeting in its center. 2) Explore the fact which says: The bisectors of the three angels of a triangle meet

at one point, which is the center of the circle sketched inside the triangle. 3) Compose problems related to the measures of the triangle angels and solve it. 4) Construct a circle given its center, radius and one of its points or three points of

the circle . 5) Construct the segments and the angels using several methods ( as fold papers,

the ruler and the compasses) 6) Investigate Pythagorean Theorem and apply it in problems related to area and right triangles. 7) Prove geometrical facts included properties of the straight lines and the circles.

1

(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? :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.

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

2

Training 1: 1. With reference to example 1 draw a circle with

?What is false

center O, and passes through P.

The circle is the situated

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.

area inside a closed curve, All of its points are 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?

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.

3

Activity 2 Draw 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 Cut sectors MAB, MCD. Congruent sectors MAE, MCD. What is the relationship between the measurements of the two central angels

AE: CD = 10 cm.

AME, CMD? What is the relationship between the two arcs AE, CD.

:Remember

:Result 1

Two rays can be drawn from the

when we draw congruent chords in

center of a circle to form an angle.

a circle, then their corresponding

If the rays and the circle are in the

arcs are congruent. And their

same plane, such an angle is called

corresponding central angles are

a CENTRAL ANGLE, and their

.congruent

.arc as a part of the circle

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?

4 (Figure (5 - 4

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.

5

( 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 as in figure (5-7).

Example 1: 6

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. The sum of the lengths of any two sides of triangle is greater than the length of the third side.

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: 7

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.

8

: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?

•

Are the measurements 11 cm, 5 cm, 6 cm, forms a triangle sides?

•

What happened if the measurement AB = 8cm?

•

Are the measurements 8 cm, 5 cm, 6 cm, forms a triangle sides?

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.

9

You notice that: • m Y=m •

m

DXY = m

Z. DXZ.

• DY = DZ. •

m

XDZ = m

XDY.= 900.

As in figure (5-11). :Think

:Remember

How many isosceles

The sum of the measures

triangles can we drawn

of a supplementary

so the segment YZ is

angles equals 180º(the

the base of these

two angels form a

triangles?

straight angle).

Problems and Exercises 10

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)

11

2. What is the relationship between ZAC and the triangle ABC? 3. What is the relationship between GCA and the triangle ABC? 4. What is the relationship between EAB and the triangle ABC? You notice that every of the previous three angles are adjacent and supplementary with one of the angels of the triangle.

These angles are called exterior angle of triangle ABC? i.e. ZAC,

GCA,

XBC exterior angles of triangle ABC. i.e.

The exterior angle of triangle is the angle which is confined between one of the triangle's sides and the extension of the other.

Training 1: Name other exterior angles of the triangle ABC, in figure (5-13).

In figure (5-14),HWC called an exterior angle of the triangle DHW .HDWand

:Remember The sum of degree measures of two supplementary angles is 1800. 12

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,

Result1: The measure of an exterior angle of a triangle, is equal to the sum of the measures of the two remote interior angles.

To prove result (1): Given: 13

XYL exteriorangle of the triangle XYZ, figure (5-16).

Required : Prove that m

4=m

1+m

2+m

3 = 1800

2 Think Is it possible that measure of more than one external angle of the same triangle less than 90 º ?

Proof: m

1+ m m

Then m

3+m 3+m

m

4=m

4 = 1800 4=m 1+m

(Triangle sum theorem) (The angles in a linear pair are supplementary)

1+m 2

2+m

3

(Substitution)

(Subtraction Property of Equality)

Example1: Use the information in the figure (5-17) to find the measure of the angle

ABC

14

Solution : HAD is an exterior angle of the triangle ABD, then m

HAD = x + m

1000 = X +X

DBA

(ABC is isosceles)

1000 = 2X X=500 m

ABC =1800 – 500 = 1300.

:Think Solve in other way without using the result.

Traning 2: In the triangle ABC; m BAC =450, ABD is an exterior angle and m B,

ABD =1200, find the measure of

A,

C in the triangle ABC

Traning3: Use the information in figure (5-18), to calculate m

DYZ

Given that XY=ZY.

Problems and Exercises

15

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, m

EFD is an exterior angle of the triangle;

EFD = 1100, and the angle

triangle ; m

EGC is another exterior angle of the

EGC = 1400.

Find the measure of each angle in the triangle EFG.

(5-4) Right Triangle 16

* 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 900, as in figure (5-21) and: 82 + 62 = 102

,

52 + 122 = 132.

Training 1: Draw a triangle with sides X, Y,

X2 +Y2, and take a different values of X

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

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. 17

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. Activity 1 •

Draw triangles with angles measuring 300,600,900.

•

Find the length of its sides.

•

Draw another 300,600,900 triangles and measure the length of their sides.

•

What is your notice?

Result2: In a 300, 600, 900 triangle, the hypotenuse is twice as a long as the shorter leg.

18

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.

Result 3: In a right triangle, the length of the hypotenuse is twice as a long as the length of the segment from the vertex of the right angle to the midpoint of the hypotenuse.

Example 2: ABC is a right triangle in B , D is the midpoint of the segment AC; ADB = 1000. Find m

m m

BAC

ACB as in figure (5-23).

Solution: D is the midpoint of AC, then 1 BD = 2

AC = AD = DC,

Triangles ADB, BDC are isosceles, m

1800 - 1000 BAC 2 =

= 400,

19

Then m m

CDB = 1800- 1000 = 800,

ACB = 500.

Training 3: In a triangle AXY it is known that m is the midpoint of XY, prove that DH

X = 900 , D is the midpoint of AY, H

XY.

Activity 3 * Draw a right angle YLX; L is a right angle. * Build a square on each side of the triangle as shown in figure (5-24). * What is the relationship between the square of the hypotenuse and the squares of the other triangle legs? * Draw more right triangles, and search in the relationship between the square of the hypotenuse and the squares of the other 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 triangle legs

i.e. (XY)2 = (XL)2 + (YL)2 20

.……Add to your knowledge Theorem 1 is called

?What is false

Pythagorean Theorem according ;In a triangle ABC

to Pythagoras, a Greek

AC =AB + BC

mathematician and philosopher

2

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

from the sixth century (582 -500

2

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

BC

), he was built the Pythagorean

school in Italy, many different B

C

basic theorems in planer geometry has been proved 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 . ..…Another polygon such as pentagon, hexagon (1 .Circles with diameter the sides of the triangle (2 .Irregular polygon (3

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 line, because: (

2 on the

2 )2 = 2, i.e. ( 2 )2 = 12 +12 .

•

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).

21

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?

Activity 4 .Represent the real number

3 on the real line numbers

Problems and Exercises 22

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

case: I.

Y = 900.

XYZ, so that XY = YZ = 1 cm,

II.

ABC, so that AB = 1 cm, AC =

III.

ZLH, so that ZL = 3 cm, LZ =

3 cm, 7 cm,

IV. KLM, so that KM =25 cm, kl = 24 cm,

B is a right angle. L is a right angle. 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 23

Architect Nermeen was designed sketch for a children garden fence containing angels, Ahmad, one of the 8th 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. 24

• 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 º using A protractor , then transfere this angle using the ruler and the compasses only.

Think :

Then verify the mesure of the result angle Is it enough to sketch part of Using

the circle ( arc) to transfere a known angle?

25

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.

26

Problems and Exercises 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 =900,

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 27

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 28

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).

Think In how much method can you Training 1

determined that the two angels

Copy to your notebook each of the

x y l & h y l have the same

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

measure?

No doubt you have gotten an ability to identify the meeting point of the two bisectors of the farm's base, in the beginning of the item (5- 6). 29

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.

Problems and Exercises 30

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 = 30º, 60º, 15º, 7.5º using unmarked ruler and the compasses .

3) ABC a triangle m the meeting point of its angels bisectors, measure C =40º , measure B = 30º , prove that measure AMB = measure BAC . 4) Sketch a straight angle, and then divide it into six equal angels without using the protractor.

(5–7) Composing Perpendicular from Assumed Point on a Straight

Line 31

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 draw two 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) 32

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?

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)

33

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 3 Determine the center of the circle in Figure (5-39) by sketching two diameters to the circle

34

Problems and Exercises 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 outside the line. 35

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.

36

* 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) 37

* 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.

Problems and Exercises 38

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 39

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 draw an arc crossing the first arc at the two points H , L figure (5 – 45C).

Think *Join h l using the ruler to cut the segment ZW at X If you complete drawing figure (5-45D). the circle when you draw the two * Let one of the two segments HL , ZW the core of fold . What is your observation

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 . 41

* 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.

Problems and Exercises 42

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) Applications First: The circle sketched inside triangle and tangent to its sides. 43

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, 44

see figure ( 5 – 48 C).

* Open the compasses aperture equal BA. * Fix the compasses at (B) and sketch a 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 45

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 piece of 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 equal to 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)

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?

Activity (1) 47

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?

Problems and Exercises 48

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 the internal.

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 the compasses.

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 º. 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 the right 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 c) ∆ ABC is acute angels

b) A , B , C are collinear points d)

BAC is obtuse angle

If (X) a set contains of 1000 points of the plane where each of them is 5 cm apart (2) :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). 52

(3) If ∆ ABC is acute triangle then the measure of one of its external angels is: a) Less than 90 º

b) 90 º

c) More than 90 º

d) 180 º

(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 º and an angle of measure 22.5 º . 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 triangle and the center of the circle which tangent to the sides of the triangle from internal lies on the straight line HD.

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