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Angles and Triangles Construction

Angles and Triangles Construction. Konstruksi Sudut dan Segitiga. Duplicating Angles. Steps: Construct a line QR ( buatlah garis QR )

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Angles and Triangles Construction

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  1. Angles and Triangles Construction Konstruksi Sudut dan Segitiga

  2. Duplicating Angles Steps: • Construct a line QR (buatlah garis QR) • 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) • 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) • With DE as the radius construct an arc centered at S and intersect the arc that maked in step 3 at point T. • Construct a line passing through Q and T and named it QP. • The construction of ∠PQR which has the same measure ∠ABC is done. Construct ∠PQR which is equal to ∠BAC

  3. Bisecting Angles Steps: • 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) • 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) • 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 ) Bisect ∠RPQ

  4. 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

  5. Constructing a 45º Angle We know that: 45 = ½ . 90 So, to construct an angle of 45º, first construct a 90º angle and then bisect it. Step 1:  Construct ∠BAC = 90o Step 2:  Construct the bisector of ∠BAC, and call it AR, this way we have ∠BAR = 45o

  6. 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 Pso that intersect the arc in step 2 at point R Step 4:  Join the points P and R, now you have ∠RPQ = 60o

  7. Constructing a 30º Angle We know that: 30 = ½ . 60 So, to construct an angle of 30º, first construct a 60º angle and then bisect it. Step 1:  Construct ∠BAC = 60o Step 2:  Construct the bisector of ∠BAC, and call it AR, this way we have ∠BAR = 30o

  8. Constructing a 150º Angle Construct∠ABC = 150º We know that: 150 = 90 + 60 So, to construct an angle of 150º, first construct a 90º angle and then add with a 60o angle. Step 1:  Construct ∠ABP = 90o Step 2:  Construct ∠PBC = 60o

  9. Constructing Special Lines of Triangles There are 4 special lines in a triangle. • Altitude (garis tinggi) • Angle Bisector (garis bagi) • Median (garis berat) • Perpendicular Bisector (garis sumbu)

  10. Constructing a Perpendicular • 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: • construct a circular arc centered at P so that intersect line k at A and B • Make two circular arcs having same radius,centered at A and B, so that intersect at point C • Join the points P and C, so PC perpendicular to k, or PC⊥ k

  11. 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: • construct a circular arc centered at M so that intersect line k at A and B • Make two circular arcs having same radius (radius should be greater than MA),centered at A and B, so that intersect at point C • Join the points M and C, so MC perpendicular to k, or MC⊥ k

  12. Altitude of Triangle DC, EA, FB are the altitudes of triangle ABC DC, EA and FB intersect at single point. C E F T A B D An altitude of a triangle is a line passing through a vertex of the triangle and perpendicular to opposite side.

  13. Constructing Altitude of Triangle 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. A B

  14. Angle Bisectors in a Triangle(Garis Bagi Segitiga) DC, EA, FB are angle bisectors of triangle ABC DC, EA and FB intersect at single point. C E F Z ● * ● * A D B 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.

  15. Constructing Angle Bisectorin a Triangle 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. A B

  16. Perpendicular Bisectors of a Triangle (Garis Sumbu) C A B A perpendicular bisector of a triangle is a line passing through the midpoint of a side and perpendicular to the side

  17. Constructing a PerpendicularBisector of a line Segment Construct a perpendicular bisectors of line segment AB STEPS: • construct a circular arc centered at A so that intersect segment AB • 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 • Join the points P and Q, so PC perpendicular to AB, or PQ⊥ AB

  18. Median of a Triangle(Garis Berat) DC, EA, FB are medians of triangle ABC DC, EA and FB intersect at single point. C E F A B D A median of a triangle is a line passing through a vertex and the midpoint of the opposite side.

  19. Constructing Medianin a Triangle 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 S Step 3: the segment PS is the median through point P P Q

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