Step I: Take a point O on the plane paper and draw a circle of radius \[OA=6\text{ }cm\]. Also, draw a concentric circle of radius\[\text{OB}=8\text{ }cm\] |
Step II: Find the mid-point A of OB and draw a circle of radius BA = AO. Suppose this circle intersects the circle of radius 6 cm at P and O. |
Step III: Join BP and 8Q to get the desired tangents. |
A) Step I done clear
B) Step II done clear
C) Step I and Step II done clear
D) Step II and Step III done clear
View Solution play_arrowI. With centre O and radius = 6 cm, draw a circle. |
II. Taking a point b A on the circle and draw\[\angle AOB-{{120}^{o}}\]. |
III. Draw a perpendicular on OA at A. Draw another perpendicular on OB at B. |
IV. Let the two perpendiculars meet at C. |
A) Only Step I done clear
B) Only Step II done clear
C) Only Step III done clear
D) Only Step IV done clear
View Solution play_arrowStep I: Draw a circle of radius 4 cm and take a point P outside the circle and draw a secant PAB, intersecting the circle at A and B. |
Step II: Produce AP to C such that\[AP=CP\]. Draw a semicircle with CB as diameter. |
Step III: Draw PD 1 CB, intersecting the semicircle at D. With P as centre and PC as radius draw arcs to intersect the given circle at T and T. |
Step IV: Join PT and PT. Then, PT and PT are the required tangents. |
A) Only Step I done clear
B) Both Step I and Step II done clear
C) Only Step III done clear
D) Both Step II and Step IV done clear
View Solution play_arrowStep I: Join CA. Thus, \[\Delta ABC\] is obtained. |
Step II: Draw \[DE||BC,\]cutting AC produced at E. |
Step III: Extend AB to D such that \[AD=\frac{3}{2}AB=\left( \frac{3}{2}\times 4 \right)cm=6cm.\] |
Step IV: Draw a line segment AB = 4 cm. |
Step V: Draw a line \[GH||AB\]at a distance of 3 cm, intersecting BP at C. |
Step VI: Construct \[\angle ABP={{60}^{o}}~\] |
A) IV, VI, V, I, III, II done clear
B) IV, V, VI, I, Ill, II done clear
C) IV, V, I, III, II, VI done clear
D) V, IV, VI, III, I, II done clear
View Solution play_arrowStep I: Draw BC = 6 cm. |
Step II: At B construct \[\angle CBX={{60}^{o}}\] and at C construct |
Suppose BX and CY intersect at A. \[\Delta \,ABC\]so obtained is the given triangle. |
Step III: Construct an obtuse angle \[\angle CBZ\] at B on opposite side of vertex A of \[\Delta \,ABC.\] |
Step IV: Mark-off three (greater 3 of 2 in 3/2) points \[{{B}_{1}},{{B}_{2}},{{B}_{3}},\] on BZ such that\[B{{B}_{1}}={{B}_{1}}{{B}_{2}}={{B}_{2}}{{B}_{3}}.\]. |
Step V: Join \[{{B}_{2}}\] (the second point) to C and draw a line through \[{{B}_{3}}\] parallel to \[{{B}_{2}}C,\] intersecting the extended line segment BC at C'. |
Step VI: Draw a line through C' parallel to CA intersecting the extended line segment BA at A'. |
Triangle A'B'C so obtained is the required triangle such that \[\frac{A'B}{AB}=\frac{BC'}{BC}=\frac{A'C'}{AC}=\frac{3}{2}\] |
A) Step III done clear
B) Step IV done clear
C) Step V done clear
D) Step II done clear
View Solution play_arrowStep I: Draw \[AB=3.2\text{ }cm\] |
Step II: Construct an acute \[\angle BAX\]. |
Step III: On AX make \[3+5+1\] i.e. 9 equal parts and mark them as \[{{A}_{1}},{{A}_{2}},{{A}_{3}},{{A}_{4}},.........{{A}_{9}}\] |
Step IV: Join B to \[{{A}_{8}}\] From \[{{A}_{3}}\] draw \[{{A}_{3}}C\]parallel to \[{{A}_{8}}B\]. Point C divides AB internally in the ratio\[3:5\]. |
A) Thus, \[AC:CB=3:5.\] Step II done clear
B) Step III done clear
C) Step IV done clear
D) None of these done clear
View Solution play_arrowStep I: On BE, cut off 3 equal parts making \[{{B}_{1}},{{B}_{2}}\] and \[{{B}_{3}}\] |
Step II: Now, draw C'A' parallel to CA. Then, \[\Delta A'BC'\] is the required A whose sides are of the corresponding sides of the \[\Delta ABC\]. |
Step III: From point B draw an arc of \[2.3\text{ }cm\]and from point C draw an arc of \[2.9\text{ }cm\]cutting each other at point A. |
Step IV: Take\[BC=5\text{ }cm\]. |
Step V: Join \[{{B}_{3}}C\] and from \[{{B}_{2}}\] draw \[{{B}_{2}}C'\] parallel to \[{{B}_{3}}C,\] such that BC is 2/3 of BC. |
Step VI: On B make an acute \[\angle CBE\]downwards. |
Step VII: Join AB and AC. Then ABC is the required triangle. |
A) IV, III, VII, I, VI, V, II done clear
B) IV, V, I, VI, III, VII, II done clear
C) IV, III, VII, VI, I, V, II done clear
D) IV, VII. Ill, VI, V, I, II done clear
View Solution play_arrowStep I: Join OA and bisect it. Let P is the mid-point of OA. |
Step II: Join AB and AC. AB and AC are the required tangents. Length of tangents\[=11\text{ }cm\]. |
Step III: With O as centre, draw a circle of radius 5 cm. |
Step IV: Taking P as centre and PO as radius, draw a circle intersecting the given circle at the points B and C. |
Step V: Take a point A at a distance of \[12\text{ }cm\]from O. |
A) III, V, I, IV, II done clear
B) III, V, IV, I, II done clear
C) II, V, IV, III, I done clear
D) ll. IV, II, I, III done clear
View Solution play_arrowStep I: Draw a circle with O as centre and radius 2 cm. |
Step II: Mark a point P outside the circle such that\[OP=2.25\text{ }cm\]. |
Step III: Join \[OP=4.5\text{ }cm\]and bisect it at M. |
Step IV: Draw a circle with M as centre and radius equal to MP to intersect the given circle at the points T and T'. |
Step V: Joint PT and PT'. Then. PT and PT are the required tangents. |
A) Step V done clear
B) Step IV done clear
C) Step II done clear
D) None of these done clear
View Solution play_arrowStep I: Draw a circle with centre O and radius\[2.5\text{ }cm\]. |
Step II: Draw a radius OA of this circle and produce it to B. |
Step III: Construct an angle \[\angle AOP\] equal to the complement of \[{{30}^{o}}\] i.e. equal to\[{{150}^{o}}\]. |
Step IV: Draw perpendicular to OP at P which intersects OA produced at Q. |
A) Both I and III done clear
B) Only III done clear
C) Both III and IV done clear
D) Only I done clear
View Solution play_arrowStep I: Draw any diameter AOB of this circle. |
Step II: Draw AM 1 AB and\[CN\bot OC\]. |
Let AM and CN intersect each other at P. |
Then PA and PC are the desired tangents to the given circle, inclined at an angle of \[{{60}^{o}}\]. |
Step III: Draw a circle with O as centre and radius 3 cm. |
Step IV: Construct \[\angle BOC={{60}^{o}}\]such that radius OC meets the circle at C. |
A) III, I, IV, II done clear
B) III, II, IV, I done clear
C) II, I, IV, III done clear
D) IV, II, III, I done clear
View Solution play_arrowStep I: Bisect the line segment OP and let the point of bisection be M. |
Step II: Taking M as centre and OM as radius, draw a circle. Let it intersect the given circle at the point Q and R. |
Step III: Draw a circle of radius 3 cm. |
Step IV: Join PQ and PR. |
Step V: Take an external point P which is 10 cm away from its centre. Join OP. |
A) III. V. I, II, IV done clear
B) III, I. V, IV, II done clear
C) III, V, I. IV, II done clear
D) III, V, II, I, IV done clear
View Solution play_arrowStep I: Draw \[\Delta ABC\] and perpendicular BD from B on AC. |
Step II: Draw a circle with BC as a diameter. |
This circle will pass through D. |
Step III: Let O be the mid-point of BC. |
Join AO. |
Step IV: Draw a circle with AO as diameter. This circle cuts the circle drawn in step II at S and P. Join /AO, AP and AB are desired tangents drawn from A to the circle passing through B, C and D. |
A) Only Step I done clear
B) Only Step II done clear
C) Only Step III done clear
D) Only Step IV done clear
View Solution play_arrowStep I: Draw a ray BY parallel to AX by making \[\angle ABY\]equal to \[\angle BAX\]. |
Step II: Join \[{{A}_{3}}{{B}_{4}}\]. Suppose it intersects AB at a point P. Then, P is the point dividing AB internally in the ratio \[3:4\]. |
Step III: Draw the line segment AB of length 8 cm. |
Step IV: Mark of three point \[{{A}_{1}},{{A}_{2}},{{A}_{4}}\]on AX and 4 points \[{{B}_{1}},{{B}_{2}},{{B}_{3}},{{B}_{4}}\] on BY such that \[A{{A}_{1}}={{A}_{1}}{{A}_{2}}={{A}_{2}}{{A}_{3}}=B{{B}_{1}}={{B}_{1}}{{B}_{2}}\]\[={{B}_{2}}{{B}_{3}}={{B}_{3}}{{B}_{4}}\]. |
Step V: Draw any ray AX making an acute angle\[~\angle BAX\] with AB. |
A) III, V, I, II. IV done clear
B) III, IV. I, V, II done clear
C) III, I, V, IV, II done clear
D) III, V, I, IV, II done clear
View Solution play_arrowStep I: Draw a circle of radius 5 cm. |
Step II: Mark a point P on it. |
Step III: Draw any chord PQ. |
Step IV: Take a point R in the minor arc QP. |
Step V: Join PR and RQ. |
Step VI: Make\[~\angle QPT=\angle PRQ\]. |
Step VII: Produce TP to T. Then, PT is the required tangent at P. |
A) Step II done clear
B) Step IV done clear
C) Step VI done clear
D) None of these done clear
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