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Ch11 ex11.1 q 4,5,7, |
| Answer» \xa0Construction Procedure:1. Draw a line segment BC with the measure of 8 cm.2. Now draw the perpendicular bisector of the line segment BC and intersect at the point D3. Take the point D as centre and draw an arc with the radius of 4 cm which intersect the perpendicular bisector at the point A4. Now join the lines AB and AC and the triangle is the required triangle.5. Draw a ray BX which makes an acute angle with the line BC on the side opposite to the vertex A.6. Locate the 3 points B1, B2\xa0and B3\xa0on the ray BX such that BB1\xa0= B1B2\xa0= B2B37. Join the points B2C and draw a line from B3\xa0which is parallel to the line B2C where it intersects the extended line segment BC at point C’.8. Now, draw a line from C’ the extended line segment AC at A’ which is parallel to the line AC and it intersects to make a triangle.9. Therefore, ΔA’BC’ is the required triangle.Justification:The construction of the given problem can be justified by proving thatA’B\xa0= (3/2)ABBC’ = (3/2)BCA’C’= (3/2)ACFrom the construction, we get A’C’ || AC∴ ∠ A’C’B = ∠ACB (Corresponding angles)In ΔA’BC’ and ΔABC,∠B = ∠B (common)∠A’BC’ = ∠ACB∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)Therefore, A’B/AB = BC’/BC= A’C’/ACSince the corresponding sides of the similar triangle are in the same ratio, it becomesA’B/AB = BC’/BC= A’C’/AC = 3/2Hence, justified.5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.Construction Procedure:1. Draw a ΔABC with base side BC = 6 cm, and AB = 5 cm and ∠ABC = 60°.2. Draw a ray BX which makes an acute angle with BC on the opposite side of vertex A.3. Locate 4 points (as 4 is greater in 3 and 4), such as B1, B2, B3, B4, on line segment BX.4. Join the points B4C and also draw a line through B3, parallel to B4C intersecting the line segment BC at C’.5. Draw a line through C’ parallel to the line AC which intersects the line AB at A’.6. Therefore, ΔA’BC’ is the required triangle.Justification:The construction of the given problem can be justified by proving thatSince the scale factor is 3/4 , we need to proveA’B\xa0= (3/4)ABBC’ = (3/4)BCA’C’= (3/4)ACFrom the construction, we get A’C’ || ACIn ΔA’BC’ and ΔABC,∴ ∠ A’C’B = ∠ACB (Corresponding angles)∠B = ∠B (common)∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)Since the corresponding sides of the similar triangle are in the same ratio, it becomesTherefore, A’B/AB = BC’/BC= A’C’/ACSo, it becomes A’B/AB = BC’/BC= A’C’/AC = 3/4Hence, justified.6. Draw a triangle ABC with side BC = 7 cm, ∠ B = 45°, ∠ A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of ∆ ABC.To find ∠C:Given:∠B = 45°, ∠A = 105°We know that,Sum of all interior angles in a triangle is 180°.∠A+∠B +∠C = 180°105°+45°+∠C = 180°∠C = 180° − 150°∠C = 30°So, from the property of triangle, we get ∠C = 30°Construction Procedure:The required triangle can be drawn as follows.1. Draw a ΔABC with side measures of base BC = 7 cm, ∠B = 45°, and ∠C = 30°.2. Draw a ray BX makes an acute angle with BC on the opposite side of vertex A.3. Locate 4 points (as 4 is greater in 4 and 3), such as B1, B2, B3, B4, on the ray BX.4. Join the points B3C.5. Draw a line through B4 parallel to B3C which intersects the extended line BC at C’.6. Through C’, draw a line parallel to the line AC that intersects the extended line segment at C’.7. Therefore, ΔA’BC’ is the required triangle.Justification:The construction of the given problem can be justified by proving thatSince the scale factor is 4/3, we need to proveA’B\xa0= (4/3)ABBC’ = (4/3)BCA’C’= (4/3)ACFrom the construction, we get A’C’ || ACIn ΔA’BC’ and ΔABC,∴ ∠A’C’B = ∠ACB (Corresponding angles)∠B = ∠B (common)∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)Since the corresponding sides of the similar triangle are in the same ratio, it becomesTherefore, A’B/AB = BC’/BC= A’C’/ACSo, it becomes A’B/AB = BC’/BC= A’C’/AC = 4/3Hence, justified.7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.Given:The sides other than hypotenuse are of lengths 4cm and 3cm. It defines that the sides are perpendicular to each otherConstruction Procedure:The required triangle can be drawn as follows.1. Draw a line segment BC =3 cm.2. Now measure and draw ∠= 90°3. Take B as centre and draw an arc with the radius of 4 cm and intersects the ray at the point B.4. Now, join the lines AC and the triangle ABC is the required triangle.5. Draw a ray BX makes an acute angle with BC on the opposite side of vertex A.6. Locate 5 such as B1, B2, B3, B4, on the ray BX such that such that BB1\xa0= B1B2\xa0= B2B3= B3B4\xa0= B4B57. Join the points B3C.8. Draw a line through B5 parallel to B3C which intersects the extended line BC at C’.9. Through C’, draw a line parallel to the line AC that intersects the extended line AB at A’.10. Therefore, ΔA’BC’ is the required triangle.Justification:The construction of the given problem can be justified by proving thatSince the scale factor is 5/3, we need to proveA’B\xa0= (5/3)ABBC’ = (5/3)BCA’C’= (5/3)ACFrom the construction, we get A’C’ || ACIn ΔA’BC’ and ΔABC,∴ ∠ A’C’B = ∠ACB (Corresponding angles)∠B = ∠B (common)∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)Since the corresponding sides of the similar triangle are in the same ratio, it becomesTherefore, A’B/AB = BC’/BC= A’C’/ACSo, it becomes A’B/AB = BC’/BC= A’C’/AC = 5/3Hence, justified. | |