CLASSIFYING TRIANGLES BY ANGLES. Classifying Triangles by Angles ACUTE OBTUSE RIGHT EQUIANGULAR.
Angles and Triangles Construction
description
Transcript of Angles and Triangles Construction
Angles and Triangles ConstructionKonstruksi Sudut dan Segitiga
Construct ∠PQR which is equal to ∠BACDuplicating Angles
Steps:1. Construct a line QR (buatlah garis QR)2. In ∠BAC construct a circular arc with A as the
center such that it intersects AB at D and AC at E (pada ∠BAC buatlah busur dengan pusat A sehingga busur tersebut memotong AB di D dan AC di E)
3. With AD as the radius, construct an arc with Q as the center and intersects QR at S. (dengan jari-jari AD buatlah busur yang berpusat di Q dan memotong QR di S)
4. With DE as the radius construct an arc centered at S and intersect the arc that maked in step 3 at point T.
5. Construct a line passing through Q and T and named it QP.
6. The construction of ∠PQR which has the same measure ∠ABC is done.
Bisect ∠RPQBisecting Angles
Steps:1. Construct a circular arc centerd at P such that it
intersect PQ at S and PR at T (buatlah busur yang berpusat di P sehingga memotong PQ di S dan PR di T)
2. With S and T as the center points, construct two circular arcs (with equal radius) whics intersect at point U(dengan S dan T sebagai pusatnya, buatlah busur (jari-jari sama) dan berpotongan di titik U)
3. Join the points P and U such that PU is the line segment which bisects ∠RPQ so that ∠RPU = ∠UPQ (hubungkan titik P dan U sehingga PU adalah ruas garis yang membagi ∠RPQ sehingga ∠RPU = ∠UPQ )
Constructing a 90º Angle
Step 1: Draw the arm AB.Step 2: Place the point of the compass at B and draw
an arc that passes through A and interect the extension of AB at C.
Step 3: With point A and C as the center, construct two arcs (radius greater than AB) so that intersect at point D.
Step 4: Join the points B and D, now you have ∠ABD = 90o
Constructing a 45º Angle
We know that: 45 = ½ . 90So, to construct an angle of 45º, first construct a 90º angle and then bisect it.
Step 1: Construct ∠BAC = 90oStep 2: Construct the bisector of ∠BAC, and call it AR, this way we have ∠BAR = 45o
Constructing a 60º Angle
Step 1: Draw the arm PQ.Step 2: Place the point of the compass at P and draw
an arc that passes through Q.Step 3: Place the point of the compass at Q and draw an
arc that passes through P so that intersect the arc in step 2 at point R
Step 4: Join the points P and R, now you have ∠RPQ = 60o
Constructing a 30º Angle
We know that: 30 = ½ . 60So, to construct an angle of 30º, first construct a 60º angle and then bisect it.
Step 1: Construct ∠BAC = 60oStep 2: Construct the bisector of ∠BAC, and call it AR, this way we have ∠BAR = 30o
Constructing a 150º Angle
Construct ∠ABC = 150º
We know that: 150 = 90 + 60So, to construct an angle of 150º, first construct a 90º angle and then add with a 60o angle.
Step 1: Construct ∠ABP = 90oStep 2: Construct ∠PBC = 60o
Constructing Special Lines of TrianglesThere are 4 special lines in a triangle.1. Altitude (garis tinggi)2. Angle Bisector (garis bagi)3. Median (garis berat)4. Perpendicular Bisector (garis
sumbu)
Constructing a Perpendicular
a. Constructing a Perpendicular to a line segment through a point outside the line segment.(Menggambar garis tegak lurus terhadap suatu ruas garis melalui suatu titik di luar ruas garis)
Construct a perpendicular to line k passing through P.
STEPS:1. construct a circular arc centered at P so that
intersect line k at A and B2. Make two circular arcs having same
radius,centered at A and B, so that intersect at point C
3. Join the points P and C, so PC perpendicular to k, or PC⊥ k
Constructing a Perpendicular
b. Constructing a Perpendicular to a line segment through a point located on the line segment.(Menggambar garis tegak lurus terhadap suatu ruas garis melalui suatu titik yang terletak di ruas garis tersebut)
Construct a perpendicular to line k passing through M.
STEPS:1. construct a circular arc centered at M so that
intersect line k at A and B2. Make two circular arcs having same radius (radius
should be greater than MA),centered at A and B, so that intersect at point C
3. Join the points M and C, so MC perpendicular to k, or MC⊥ k
A B
C
D
EF
Altitude of Triangle
DC, EA, FB are the altitudesof triangle ABC
DC, EA and FB intersect at single point.
An altitude of a triangle is a line passing through a vertex of the triangle and perpendicular to opposite side.
T
Constructing Altitude of Triangle
A B
C
On the triangle ABC on the left, construct an altitude through the point B
Step 1: with B as the center point, construct a circular arc which intersect AC at P and Q
Step 2: At P and Q as the center points, construct two arcs with same radius which intersect at point R
Step 3: joint the point B and R to construct a line which intersect AB at S
Step 4: the line BS is the altitude through point B.
Angle Bisectors in a Triangle(Garis Bagi Segitiga)
DC, EA, FB are angle bisectorsof triangle ABC
DC, EA and FB intersect at single point.
An angle bisector of a triangle is a line passing through a vertex of the triangle which bisect (divides into two equal parts) the corresponding angles.
A B
C
D
EF
Z●● **
Constructing Angle Bisectorin a Triangle
A B
C
On the triangle ABC on the left, construct an angle bisector through the
point B
Step 1: with B as the center point, construct a circular arc which intersect AB at P and BC at Q
Step 2: At P and Q as the center points, construct two arcs with same radius which intersect at point R
Step 3: joint the point B and R to construct a line which intersect AC at T
Step 4: the line BT is the angle bisector through point B.
Perpendicular Bisectors of a Triangle (Garis Sumbu)
A perpendicular bisector of a triangle is a line passing through the midpoint of a side and perpendicular to the side
A B
C
Constructing a PerpendicularBisector of a line Segment
Construct a perpendicular bisectors of line segment AB
STEPS:1. construct a circular arc centered at A so that
intersect segment AB2. With the same radius at step one, construct a
circular arc centered at B so that intersect the arc that made in step one in point P and Q
3. Join the points P and Q, so PC perpendicular to AB, or PQ⊥ AB
Median of a Triangle(Garis Berat)
A median of a triangle is a line passing through a vertex
and the midpoint of the opposite side.
A B
C
D
EF
DC, EA, FB are mediansof triangle ABC
DC, EA and FB intersect at single point.
Constructing Medianin a Triangle
P Q
R
On the triangle PQR on the left, construct a median through the point P
Step 1: construct a perpendicular bisector of side RQ which intersect RQ at point S
Step 2: Join point P and SStep 3: the segment PS is the median through point P