
10th_hotsquestionanswersmathunit10constructions  
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Q. Construct a tangent to a circle of radius 3 cm at a point P on it without using the centre of the circle.
Steps of construction: 1. Draw a circle of radius 3 cm 2. Draw a cord PQ 4.Take any point R on the major arc PQ and join PR and QR 5. Make < PRQ = < QPT 6. PT is required tangent Q. Construction of a tangent to a circle of radius 3 cm (using the centre). We remember that: In a circle, the radius drawn at the point of contact is perpendicular to the tangent at that point. Steps of construction: 1. Draw a circle of radius 3 cm 2. Take a point p on the boundary and join it to Centre O of circle 4.Take any point R on the major arc PQ and join PR and QR 5. Make < PRQ = < QPT 6. PT is required tangent Q. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠ B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle. Steps of constructions: 1. Draw BC = 8 cm 2. Make <CBX = 90 degree 3. Cut off AB = 6 cm along AX 4. Join AC. ABC is required triangle 5. Draw BD perpendicular to AC from B 6. Draw perpendicular bisector of BC and CD let meet BC at E 7. Taking E as a centre and CE as a radius draw circle passing through B,D,C 8. Join AE and bisect it. Let F be the midpoint of AE. 9. Taking F as centre and FE as its radius, draw a circle which will intersect the circle at point B and G. Join AG. 10. AB and AG are the required tangents. Q.Construction a tangent to a circle of radius 3 cm from a point out side the circle without using its centre.
Steps of constructions: 1. Dra a circle of radius 3 cm 2. Draw a secant PAB to the circle. 3. Draw bisector of PB let it be at M. 4. Draw a semicircle taking M as a centre and PM as a radius 5. Through A is drawn perpendicular to AB which intersect the semicircle at C. 6.Taking P as centre and PC as radius, arcs are drawn to intersect the given circle at Q and R. 7. Join PQ and PR which is the required tangent. Q. Draw a circle of radius 3 cm. From an external point 7 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Given: Radius of the circle = 3 cm. OP = 7 cm. Construction Steps: (i) With O as the centre draw a circle of radius 3 cm. (ii) Mark a point P at a distance of 7 cm from O and join OP. (iii) Draw the perpendicular bisector of OP. Let it meet OP at M. (iv) With M as centre and MO as radius, draw another circle. (v) Let the two circles intersect at T and Tl. (vi) Join PT and PT'. They are the required tangents. Length of the tangent, PT = 6.3 cm Q. Draw a line segment of length 7 cm and divide it in the ratio 2 : 3.
Step of constructions : 1. Draw AB = 7 cm 2. Draw ray AX making a suitable acute angle with AB. 3. Cut 2 + 3 = 5 equal segments AA1, A1A2, A2A3, A3A4 and A4A5 on AX. 4. Join A5 with B. 5. Through A2 ,Draw A2P parallel to A5B by making corresponding angles AA2P and AA5B equal. 6. The line through A2 and parallel to A5B will meet the given line segment at point P. Then P is the required point which divides AB in the ratio 2 : 3 i.e. AP : PB = 2 : 3 Q. Construct a triangle similar to a given triangle with sides 7 cm, 9 cm and 10 cm and whose sides are 5/7 th of the corresponding sides of the given triangle.
Steps of Constructions : 1. With the given measurements construct the triangle ABC in which AB = 7 cm, BC = 9 cm andAC = 10 cm 2. Draw a ray AP, making any suitable angle with AB and on opposite side of vertex C 3. Starting from A, cut off seven equal linesegments AX1, X1X2, X2X3, X3X4, X4X5, X5X6 and X6X7 on AP. 4. Join BX1 and draw a line X5B' parallel to X7B which meets AB at B'. 5. Through B', draw B'C'  BC which meets AC at point C'. The DAB'C', so obtained, is similar to the given D ABC and each side of DAB'C' is 5/7 times the corresponding side of tri. ABC. Q. 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. Step of Constructions : 1. Draw BC = 4 cm 2. At B, draw a ray BP making angle 90° with BC i.e. <PBC = 90° 3. From BP, cut BA = 3 cm 4. Join A and C to get the given DABC 5. Through vertex B, draw ray BX making any suitable angle with BC. 6. On BX cut 5 equal line segments BB1 = B1B2 = B2B3 = B3B4 = B4B5. 7. Join B3 to C. 8. Through B5, draw a line parallel to B3C to meet BC produced at point C'. 9. Through C', draw a line parallel to side CA to meet BA produced to A'.D A'BC' is the required triangle New way to construct similar triangle Smaller in size than given triangle View Video
New way to construct similar triangle Smaller in size than given triangle View Video New way to construct similar triangle Larger in size than given triangle View Video 

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