CONSTRUCTIONS EXERCISE 11.1
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1 CONSTRUCTIONS EXERCISE 11.1 In each of the following, give the justification of the construction also: Q1. Draw a line segment of length 7.6cm and divides it in the 5:8. Measure the two parts. Step-I : Draw line segment =7.6cm and draw an acute angle with and produce the ray from A to X. Step-II : Locate 13=(5+8) points A 1, A 2, A 3 A 13 such that AA 1 = A 1 A 2 = A 2 A 3 = A 12 A 13 Step-III : Join BA 13. Step-IV: Through the pointa 5, draw a line parallel to BA 13 (by making an angle equal to AA 13 B) at A 5 intersecting at the point C. Then AC:CB= 5:8 AC=2.9 cm, BC= 4.7cm respectively. Since A 5 c is parallel to A 13 B By using basic proportionality a theorem AA 5 A 5 A 13 BC AA 5 A 5 A 13 = 5 8 Using equation and equation, we get
2 AA 5 = 5 A 5 A 13 BC 8 This justify the C divides in the ratio of 5:8. Q2. Construct a triangle sides 4cm, 5cm, 6cm and then a triangle similar it to whose sides are 3 the corresponding sides of the first triangle. 2 Step-I : Draw line segment =4, BC=5cm, AC=6cm and draw an acute angle with and produce the ray from A to X which is opposite to vertex C Step-II : Locate 3=(the greater of 2 and 3 in 3 2 ) points A 1, A 2, A 3 such that AA 1 = A 1 A 2 = A 2 A 3 Step-III : Join BA 3. Step-IV : Through the pointa 2, draw a line parallel to BA 3 to intersecting at the point B. C is the required triangle. To Proof: = 2 3 BC = 2 3 Bc Ac = 2 3 Ac B C ll BC C = C (Corresponding In C and A B C C = C B AC = BAC (Corresponding (Common C ~ A B C (AA similarity criterion) = B C BC AC
3 In AA 2 B and AA 3 B A 2 = A 3 (Corresponding AA 2 B = AA 3 B (Common AA 2 B ~ AA 3 B (AA similarity criterion) = AA 2 AA 3 = 2 3 By equation and equation = B C BC AC = 2 3 = 2 3 BC = 2 3 Bc Ac = 2 3 Ac Q3. Construct a triangle with sides 5cm, 6cm, 7cm and then another triangle whose sides are 7 of the corresponding sides of the first triangle. 5 Step-I : Draw line segment =5, BC=6cm, AC=7cm and draw an acute angle with and produce the ray from A to X which is opposite to vertex C Step-II : Locate 7=(the 7greater of between 7 and 5 in 7 5 ) points A 1, A 2, A 3,. A 7 such that AA 1 = A 1 A 2 = A 2 A 3 =..A 6 A 7 Step-III : Join BA 5. Step-IV :Through the pointa 7, draw a line parallel to BA 5 to intersecting extended line at the point B. Step-V : Draw a line through B parallel to BC intersecting the extended line AC at C. C is required triangle.
4 To Proof : = 7 5 B C = 7 5 BC AC = 7 5 AC B C ll BC C = C (Corresponding In C and A B C C = C BAC = B AC (Corresponding (Common C ~ C (AA similarity criterion) B C AC In AA 5 B and AA 7 B A 5 = A 7 (Common AA 5 B = AA 7 B (Corresponding AA 2 B ~ AA 3 B (AA similarity criterion) = AA 5 AA 7 = 5 7 By equation and equation = 5 B C AC 7 = 5 7 BC = 5 7 Bc Ac = 5 7 Ac Q4. Construct an isosceles triangle whose base is 8cm and the altitude 4cm and then another triangle whose sides are1 1 times the corresponding sides of 2 the isosceles triangle.
5 Step-I : Draw line segment =8cm, And draw the same line segment from point A and point B on the both side of the line segment and mark the both intersecting point O and O and OO intersect at D. Step-II : from D cut an arc 4cm on the extended line segment of OO at point C. An isosceles triangle C is formed, having CD(altitude) as 4cm and as 8 cm., A 2, A 3,. A 7 such that AA 1 = A 1 A 2 = A 2 A 3 =..A 6 A 7 Step-III : Draw a ray AX an acute angle with line segment on the opposite side on opposite side on vertex C. Step-IV :Locate 3 points (as 3 is greater between 3 and 2) A 1, A 2, A 3 on AX such that AA 1 =A 1 A 2 =A 2 A 3. Step-V : Join BA2 and draw a line through A3 parallel to BA2 to intersect extended line segment at point B. Step-VI : Draw a line through B parallel to BC intersecting the extended line segment AC at C. C is the required triangle. To Proof : = 3 2 B C = 3 2 BC AC = 3 2 AC In C and A B C C = C BAC = B AC (Corresponding (Common C ~ C (AA similarity criterion) B C AC In AA 2 B and AA 3 B A 2 = A 3 (Common
6 AA 2 B = AA 3 B (Corresponding AA 2 B ~ AA 3 B (AA similarity criterion) = AA 2 AA 3 = 2 3 By equation and equation = 2 B C AC 3 = 3 2 BC = 3 2 Bc Ac = 3 2 Ac Q5. Draw a triangle C with sides BC= 6cm, =5cm, and C=60. Then, construct a triangle whose sides are 3 times the corresponding sides of 4 the triangle C. StepI: Draw a C with BC=6cm, =5cm, C=60. StepII: Draw a ray BX making an actue angle with BC on the opposite side of vertex A. StepIII: Locate 4 points (as 4 is greater between 3 and 4),B 1, B 2, B 3, B 4 On the line segment BX. StepIV: Join B 4 C and draw a line through B 3, parallel to B 4 C intersecting BC at C. StepV: Draw a line through C parallel to AC intersecting at A. A B C is required triangle. To Proof :A B = 3 4 BC = 3 4 BC A C = 3 4 AC
7 In A B C and C A C B = ACB (Corresponding A BC = C (Common A B C ~ C (AA similarity criterion) A B = A C BC AC In B 3 BC and BB 4 C B 3 BC = B 4 BC (Common BB 3 C = BB 4 C (Corresponding B 3 BC ~ BB 4 C (AA similarity criterion) BC BC = BB 3 BB 4 BC BC = 3 4 By equation and equation A B = 3 BC = 3 BC A C = 3 AC Q6.Draw a triangle C with sides BC= 7cm, B=45, A=105, Then construct a triangle whose sides are 4 times the corresponding sides of the 3 triangle C. A+ B+ c=180
8 c=180 C= 30 Step I: Draw a triangle C with side of BC=7cm, B= 45, C= 30. Step II: Draw a ray BX making an actue angle with BC on the opposite side of vertex A. Step III: Locate 4 points (as 4 is greater between 4 and 3) B 1, B 2, B 3, B 4 on BX. Step IV: Join B 3 C, Draw a line through B4 parallel to B 3 C intersecting extended BC at C. Step V: By C, draw a line parallel to AC intersecting extended line segment at C. A B C is the required triangle. To Proof :A B = 4 3 BC = 4 3 BC A C = 4 3 AC In A B C and C A C B = ACB A BC = C (Corresponding (Common A B C ~ C (AA similarity criterion) A B BC A C, In BB 3 C and BB 4 C B 3 BC = B 4 BC (Common
9 BB 3 C = BB 4 C^ (Corresponding BB 3 C ~ BB 4 C (AA similarity criterion) BC BC = BB 3 BB 4 BC BC = 3 4 By equation and equation A B BC A C = 3 4 A B = 4 3 BC = 4 3 BC A C = 4 3 AC Q7.Draw a right triangle in which the sides (other than hypotenuse) are of length 4cm and 3cm. Then construct another triangle whose sides are 5 3 times the corresponding sides of the triangle C. Step I: Draw a line segment =4cm, Draw a ray PA making 90 with it. Step II: Draw an arc of 3cm on the ray PA at C. Join BC triangle C is required. Step III: Draw a ray AX making an actue angle with, opposite to C. Step IV: Locate 5 points (as 5 is the greater between 5 and 3)A 1, A 2, A 3, A 4, A 5. Step V: Join A 3 B.Draw a line through A parallel toa 3 B intersecting extended line segment at B. Step VI: By B draw a line parallel BC intersecting extended line segment AC at C. Triangle C is the triangle. By constructio5 To Proof : = 5 3 B C = 5 3 BC AC = 4 3 AC
10 In C and C C = C B AC = BAC (Corresponding (Common C ~ C (AA similarity criterion) A B B C AC, In AA 3 B and AA 5 B A 3 = A 5 (Common AA 3 B = AA 5 B (Corresponding AA 3 B ~ AA 5 B (AA similarity criterion) = AA 3 AA 5 = 3 5 By equation and equation B C AC = 3 5 A B = 5 3 BC = 5 3 BC AC = 5 3 AC
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