Introduction Simple Construction:
Basic concepts on simple construction:
A chord in the perpendicular bisector passes through centre of a circle.
The tangent to the circle is perpendicular to the line joining the centre and the point of contact.
Tangents drawn from an external point to the circle are equal.
The same segment of aggles are equal.
The angle which tangent makes with a chord is equal to the angle in the alternate segment.
Problem based on construction division of line segment:
Draw a line segment of length 5.6cm and divide it in the ratio 3:5.
Solution:
Steps of simple construction:
Step 1: Draw a line segment AB = 5.6cm.
Step 2: Draw any ray AX making an acute angle BAX with AB.
Step 3: On ray AX, starting from A, mark 3 + 5 = 8 equal arcs.
AA1, A1A2, A2A3, A3A4, A4A5, A5A6, A6A7, A7A8
Step 4: Join A8B
Step 5: From A5, draw A5p || A8 at P.
Thus P divides AB in the ratio 3:5. On measuring two parts. AP = 1.9cm and PB = 3.7cm (approx)
Justification:
In ABA8, pA5||BA8
Therefore, By Basic proportionality theorem
`(AP) / (PB) ` = `(A A5 )/( A5A8 )` = `3 / 5`
`(AP )/(PB) ` = `3 / 5` So, AP : PB = 3: 5
Understanding Lines Parallel is always challenging for me but thanks to all math help websites to help me out.
Problems Based on simple Construction of a Triangle:
Make a Construction of a triangle of sides 4 cm , 5 cm and 6 cm and then a triangle similar to it whose sides are 2 / 3 of the corresponding sides of the first triangle.
Solution:
Steps of simple Construction:
Step 1: Draw a line segment BC = 6cm.
Step 2: Draw an arc with B as centre and radius equal to 5 cm.
Step 3: Draw an arc, with C as centre and radius equal to 4 cm intersecting the previous drawn arc at A.
Step 4: Join AB and AC, then triangle ABC is the required triangle.
Step 5: Below BC, make an acute angle CBX.
Step 6: Along BX, mark off three points: B1, B2, B3 such that BB1 = B1B2 = B2B3
Step 7: Join B3C.
Step 8: From B2, Draw B2D||B3C, meeting BC at d.
Step 9: From D, draw ED || AC, meeting BA at E. Then we have triangle EDB which is the required triangle.
Justification:
Since DE || CA
Therefore, Triangle ABC similarly triangle EBD
And `(EB) / (AB)` = `(BD) /( BC )` = `(DE) / (CA)` = `(2 / 3)` .
Basic concepts on simple construction:
A chord in the perpendicular bisector passes through centre of a circle.
The tangent to the circle is perpendicular to the line joining the centre and the point of contact.
Tangents drawn from an external point to the circle are equal.
The same segment of aggles are equal.
The angle which tangent makes with a chord is equal to the angle in the alternate segment.
Problem based on construction division of line segment:
Draw a line segment of length 5.6cm and divide it in the ratio 3:5.
Solution:
Steps of simple construction:
Step 1: Draw a line segment AB = 5.6cm.
Step 2: Draw any ray AX making an acute angle BAX with AB.
Step 3: On ray AX, starting from A, mark 3 + 5 = 8 equal arcs.
AA1, A1A2, A2A3, A3A4, A4A5, A5A6, A6A7, A7A8
Step 4: Join A8B
Step 5: From A5, draw A5p || A8 at P.
Thus P divides AB in the ratio 3:5. On measuring two parts. AP = 1.9cm and PB = 3.7cm (approx)
Justification:
In ABA8, pA5||BA8
Therefore, By Basic proportionality theorem
`(AP) / (PB) ` = `(A A5 )/( A5A8 )` = `3 / 5`
`(AP )/(PB) ` = `3 / 5` So, AP : PB = 3: 5
Understanding Lines Parallel is always challenging for me but thanks to all math help websites to help me out.
Problems Based on simple Construction of a Triangle:
Make a Construction of a triangle of sides 4 cm , 5 cm and 6 cm and then a triangle similar to it whose sides are 2 / 3 of the corresponding sides of the first triangle.
Solution:
Steps of simple Construction:
Step 1: Draw a line segment BC = 6cm.
Step 2: Draw an arc with B as centre and radius equal to 5 cm.
Step 3: Draw an arc, with C as centre and radius equal to 4 cm intersecting the previous drawn arc at A.
Step 4: Join AB and AC, then triangle ABC is the required triangle.
Step 5: Below BC, make an acute angle CBX.
Step 6: Along BX, mark off three points: B1, B2, B3 such that BB1 = B1B2 = B2B3
Step 7: Join B3C.
Step 8: From B2, Draw B2D||B3C, meeting BC at d.
Step 9: From D, draw ED || AC, meeting BA at E. Then we have triangle EDB which is the required triangle.
Justification:
Since DE || CA
Therefore, Triangle ABC similarly triangle EBD
And `(EB) / (AB)` = `(BD) /( BC )` = `(DE) / (CA)` = `(2 / 3)` .
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