Page 1 CONSTRUCTION 1) Draw a line segment of length 6.4cm and divide in the ratio 3:2. Solution: STEPS OF CONSTRUCTION: 1) Draw AB= 6.4cm 2) Draw any ray AX , making an acute angle with AB. 3) Mark 5(=m+n, m =3, n =2) points A 1 , A2, A3, A4 and A 5 on the ray AX such that A A 1 = A 1 A 2 , =A 2 A 3 = A 3 A 4 = A 4 A 5 4) Join A 5 to B 5) Through the point A 3 draw a line parallel to A 5 B intersecting the line AB at the point C. Then AC:CB=3:2 Proof: Since 5 AB is parallel to 3 AC we can say that 3 35 AA AC CB A A ? (Basic Proportionality Theorem) By Construction 3 35 AA 3 A A 2 ? AC 3 CB 2 ?? Hence C divides AB in the Ratio 3:2. 2) Draw a line segment of length 7.8cm and divide in the ratio 5:8. Solution: Steps of Construction: 1) Draw AB=7.8cm. 2) Draw any ray AX making an acute angle with AB. 3) Draw a ray BY parallel to AX making ABY ? equal to BAX ? . 4) Mark the points 1 A , 2 A, 3 A, 4 A, and 5 A on AX and 1 B, 2 B, 3 B, ...... 8 B on BY such that 1 AA = 12 AA = 23 AA = 34 AA = 45 AA = 1 BB = 12 BB = 23 BB =......= 78 BB Page 2 CONSTRUCTION 1) Draw a line segment of length 6.4cm and divide in the ratio 3:2. Solution: STEPS OF CONSTRUCTION: 1) Draw AB= 6.4cm 2) Draw any ray AX , making an acute angle with AB. 3) Mark 5(=m+n, m =3, n =2) points A 1 , A2, A3, A4 and A 5 on the ray AX such that A A 1 = A 1 A 2 , =A 2 A 3 = A 3 A 4 = A 4 A 5 4) Join A 5 to B 5) Through the point A 3 draw a line parallel to A 5 B intersecting the line AB at the point C. Then AC:CB=3:2 Proof: Since 5 AB is parallel to 3 AC we can say that 3 35 AA AC CB A A ? (Basic Proportionality Theorem) By Construction 3 35 AA 3 A A 2 ? AC 3 CB 2 ?? Hence C divides AB in the Ratio 3:2. 2) Draw a line segment of length 7.8cm and divide in the ratio 5:8. Solution: Steps of Construction: 1) Draw AB=7.8cm. 2) Draw any ray AX making an acute angle with AB. 3) Draw a ray BY parallel to AX making ABY ? equal to BAX ? . 4) Mark the points 1 A , 2 A, 3 A, 4 A, and 5 A on AX and 1 B, 2 B, 3 B, ...... 8 B on BY such that 1 AA = 12 AA = 23 AA = 34 AA = 45 AA = 1 BB = 12 BB = 23 BB =......= 78 BB CONSTRUCTION 5) Join 5 A to 8 B . Name the point it intersects AB as C. Then AC: CB=5:8. Proof: ? 5 AA C ~? 8 BB C ? 5 8 AA AC 5 BB CB 8 ?? 3) Construct a triangle similar to a given triangle ABC with its sides equal to 3 4 of the corresponding sides of ?ABC. Solution: Page 3 CONSTRUCTION 1) Draw a line segment of length 6.4cm and divide in the ratio 3:2. Solution: STEPS OF CONSTRUCTION: 1) Draw AB= 6.4cm 2) Draw any ray AX , making an acute angle with AB. 3) Mark 5(=m+n, m =3, n =2) points A 1 , A2, A3, A4 and A 5 on the ray AX such that A A 1 = A 1 A 2 , =A 2 A 3 = A 3 A 4 = A 4 A 5 4) Join A 5 to B 5) Through the point A 3 draw a line parallel to A 5 B intersecting the line AB at the point C. Then AC:CB=3:2 Proof: Since 5 AB is parallel to 3 AC we can say that 3 35 AA AC CB A A ? (Basic Proportionality Theorem) By Construction 3 35 AA 3 A A 2 ? AC 3 CB 2 ?? Hence C divides AB in the Ratio 3:2. 2) Draw a line segment of length 7.8cm and divide in the ratio 5:8. Solution: Steps of Construction: 1) Draw AB=7.8cm. 2) Draw any ray AX making an acute angle with AB. 3) Draw a ray BY parallel to AX making ABY ? equal to BAX ? . 4) Mark the points 1 A , 2 A, 3 A, 4 A, and 5 A on AX and 1 B, 2 B, 3 B, ...... 8 B on BY such that 1 AA = 12 AA = 23 AA = 34 AA = 45 AA = 1 BB = 12 BB = 23 BB =......= 78 BB CONSTRUCTION 5) Join 5 A to 8 B . Name the point it intersects AB as C. Then AC: CB=5:8. Proof: ? 5 AA C ~? 8 BB C ? 5 8 AA AC 5 BB CB 8 ?? 3) Construct a triangle similar to a given triangle ABC with its sides equal to 3 4 of the corresponding sides of ?ABC. Solution: CONSTRUCTION Steps of Construction 1) Given ?ABC 2) Draw and ray BX making an acute angle with BC on the side opposite to the vertex A. 3) Mark four points 1 B, 2 B, 3 B and 4 B on BX such that 1 BB = 12 BB = 23 BB = 34 BB 4) Join 4 B to C and draw a line parallel to 4 BC through 3 B ( smaller of 3 & 4 in 3/4)intersecting BC at C’ 5) Draw a line C’A’ through C’ parallel to AC intersecting AB at A’. A’BC’ is the required triangle. 4) Construct a triangle with sides 7 cm, 8 cm and 9 cm. And then another triangle whose sides are 7 5 of the corresponding sides of the first triangle. Solution: Steps Of Construction 1) Draw a line segment BC of length 9 cm. 2) With B as centre and radius 7 cm draw an arc. 3) With C as centre and radius 8 cm draw another arc on the same side and name the point of intersection as A. 4) Join AB and AC to get first triangle ABC. 5) Now draw a ray BX such that it makes an acute angle with BC. Page 4 CONSTRUCTION 1) Draw a line segment of length 6.4cm and divide in the ratio 3:2. Solution: STEPS OF CONSTRUCTION: 1) Draw AB= 6.4cm 2) Draw any ray AX , making an acute angle with AB. 3) Mark 5(=m+n, m =3, n =2) points A 1 , A2, A3, A4 and A 5 on the ray AX such that A A 1 = A 1 A 2 , =A 2 A 3 = A 3 A 4 = A 4 A 5 4) Join A 5 to B 5) Through the point A 3 draw a line parallel to A 5 B intersecting the line AB at the point C. Then AC:CB=3:2 Proof: Since 5 AB is parallel to 3 AC we can say that 3 35 AA AC CB A A ? (Basic Proportionality Theorem) By Construction 3 35 AA 3 A A 2 ? AC 3 CB 2 ?? Hence C divides AB in the Ratio 3:2. 2) Draw a line segment of length 7.8cm and divide in the ratio 5:8. Solution: Steps of Construction: 1) Draw AB=7.8cm. 2) Draw any ray AX making an acute angle with AB. 3) Draw a ray BY parallel to AX making ABY ? equal to BAX ? . 4) Mark the points 1 A , 2 A, 3 A, 4 A, and 5 A on AX and 1 B, 2 B, 3 B, ...... 8 B on BY such that 1 AA = 12 AA = 23 AA = 34 AA = 45 AA = 1 BB = 12 BB = 23 BB =......= 78 BB CONSTRUCTION 5) Join 5 A to 8 B . Name the point it intersects AB as C. Then AC: CB=5:8. Proof: ? 5 AA C ~? 8 BB C ? 5 8 AA AC 5 BB CB 8 ?? 3) Construct a triangle similar to a given triangle ABC with its sides equal to 3 4 of the corresponding sides of ?ABC. Solution: CONSTRUCTION Steps of Construction 1) Given ?ABC 2) Draw and ray BX making an acute angle with BC on the side opposite to the vertex A. 3) Mark four points 1 B, 2 B, 3 B and 4 B on BX such that 1 BB = 12 BB = 23 BB = 34 BB 4) Join 4 B to C and draw a line parallel to 4 BC through 3 B ( smaller of 3 & 4 in 3/4)intersecting BC at C’ 5) Draw a line C’A’ through C’ parallel to AC intersecting AB at A’. A’BC’ is the required triangle. 4) Construct a triangle with sides 7 cm, 8 cm and 9 cm. And then another triangle whose sides are 7 5 of the corresponding sides of the first triangle. Solution: Steps Of Construction 1) Draw a line segment BC of length 9 cm. 2) With B as centre and radius 7 cm draw an arc. 3) With C as centre and radius 8 cm draw another arc on the same side and name the point of intersection as A. 4) Join AB and AC to get first triangle ABC. 5) Now draw a ray BX such that it makes an acute angle with BC. CONSTRUCTION 6) Mark 7 points on BX 1 B, 2 B, ...... 7 B such that 1 BB = 12 BB = 23 BB =......= 67 BB 7) Join 5 B to C and draw a line B 7 C’ parallel to 5 B C through 7 B intersecting BC extended at C’. 8) Draw a line C’A’ parallel to AC through C’ intersecting BA at A’. ?A’BC’ is the required triangle. ?ABC ~?A’BC’ AB AC BC A'B A'C' BC' ? ? ? 5 7 BB BC BC' BB ? 5 7 ? BC' 7 BC 5 ? A'B A'C' BC' 7 AB AC BC 5 ? ? ? ? 5) Construct an isosceles triangle whose base is 5 cm and altitude is 4 cm. Then another triangle whose sides are 1 1 2 times the corresponding sides of the isosceles triangle. Solution: Page 5 CONSTRUCTION 1) Draw a line segment of length 6.4cm and divide in the ratio 3:2. Solution: STEPS OF CONSTRUCTION: 1) Draw AB= 6.4cm 2) Draw any ray AX , making an acute angle with AB. 3) Mark 5(=m+n, m =3, n =2) points A 1 , A2, A3, A4 and A 5 on the ray AX such that A A 1 = A 1 A 2 , =A 2 A 3 = A 3 A 4 = A 4 A 5 4) Join A 5 to B 5) Through the point A 3 draw a line parallel to A 5 B intersecting the line AB at the point C. Then AC:CB=3:2 Proof: Since 5 AB is parallel to 3 AC we can say that 3 35 AA AC CB A A ? (Basic Proportionality Theorem) By Construction 3 35 AA 3 A A 2 ? AC 3 CB 2 ?? Hence C divides AB in the Ratio 3:2. 2) Draw a line segment of length 7.8cm and divide in the ratio 5:8. Solution: Steps of Construction: 1) Draw AB=7.8cm. 2) Draw any ray AX making an acute angle with AB. 3) Draw a ray BY parallel to AX making ABY ? equal to BAX ? . 4) Mark the points 1 A , 2 A, 3 A, 4 A, and 5 A on AX and 1 B, 2 B, 3 B, ...... 8 B on BY such that 1 AA = 12 AA = 23 AA = 34 AA = 45 AA = 1 BB = 12 BB = 23 BB =......= 78 BB CONSTRUCTION 5) Join 5 A to 8 B . Name the point it intersects AB as C. Then AC: CB=5:8. Proof: ? 5 AA C ~? 8 BB C ? 5 8 AA AC 5 BB CB 8 ?? 3) Construct a triangle similar to a given triangle ABC with its sides equal to 3 4 of the corresponding sides of ?ABC. Solution: CONSTRUCTION Steps of Construction 1) Given ?ABC 2) Draw and ray BX making an acute angle with BC on the side opposite to the vertex A. 3) Mark four points 1 B, 2 B, 3 B and 4 B on BX such that 1 BB = 12 BB = 23 BB = 34 BB 4) Join 4 B to C and draw a line parallel to 4 BC through 3 B ( smaller of 3 & 4 in 3/4)intersecting BC at C’ 5) Draw a line C’A’ through C’ parallel to AC intersecting AB at A’. A’BC’ is the required triangle. 4) Construct a triangle with sides 7 cm, 8 cm and 9 cm. And then another triangle whose sides are 7 5 of the corresponding sides of the first triangle. Solution: Steps Of Construction 1) Draw a line segment BC of length 9 cm. 2) With B as centre and radius 7 cm draw an arc. 3) With C as centre and radius 8 cm draw another arc on the same side and name the point of intersection as A. 4) Join AB and AC to get first triangle ABC. 5) Now draw a ray BX such that it makes an acute angle with BC. CONSTRUCTION 6) Mark 7 points on BX 1 B, 2 B, ...... 7 B such that 1 BB = 12 BB = 23 BB =......= 67 BB 7) Join 5 B to C and draw a line B 7 C’ parallel to 5 B C through 7 B intersecting BC extended at C’. 8) Draw a line C’A’ parallel to AC through C’ intersecting BA at A’. ?A’BC’ is the required triangle. ?ABC ~?A’BC’ AB AC BC A'B A'C' BC' ? ? ? 5 7 BB BC BC' BB ? 5 7 ? BC' 7 BC 5 ? A'B A'C' BC' 7 AB AC BC 5 ? ? ? ? 5) Construct an isosceles triangle whose base is 5 cm and altitude is 4 cm. Then another triangle whose sides are 1 1 2 times the corresponding sides of the isosceles triangle. Solution: CONSTRUCTION Steps Of Construction 1) Draw BC =5cm. Draw perpendicular bisector of BC and name the mid-point of BC as O 2) With O as centre and radius 4 cm draw an arc intersecting the perpendicular bisector of BC. Name the point of intersection as A. 3) Join AB and AC. This is the triangle ABC. 4) Now draw a ray BX making an acute angle with BC. 5) Locate 3 equal points B 1 , B 2 and B 3 such that BB 1 =B 1 B 2 = B 2 B 3 6) Join B 2 C and through B 3 draw a line B 3 C’ parallel to B 2 C intersecting BC extended at C’. 7) Through C’ draw a line A’C’ parallel to AC intersecting BA at A’. Thus ?A’BC’ is the required triangle 6) Draw a triangle ABC with sides BC=9 cm and AB=6 cm and ABC ? = 55 ? . Then Construct a triangle whose sides are 3 4 of the corresponding sides of the triangle ABC. Solution: Steps Of Construction 1) Draw BC= 9cm. 2) At the point B draw a line BA =4cm making an angle of 55 ?. Join A to C to get ?ABC 3) Draw a ray BX making acute angle with the line segment BC and divide it into 4 equal parts BB 1 , BB 2 , BB 3 & BB 4 .Read More

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