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D Divya Prakash Singh
The steps of constructions are: (i) Draw a line segment  = 10.3 cm (ii) Taking X and Y as centres and radius more than half of AB, draw two arcs which intersect each other at C and D. (iii) Join CD. Then CD is the required perpendicular bisector of . Now: (a) Take any point P on the bisector drawn. With the help of divider, we can check that =. (b) If M is the mid-point of , then ???????

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D Divya Prakash Singh
The steps of constructions are:  (a) Draw an angle of 40 degrees with the help of protractor, naming ∠ AOB. (b) Draw a line PQ. (c) Take any point M on PQ. (d) Place the compasses at O and draw an arc to cut the rays of ∠AOB at L and N. (e) Use the same compasses setting to draw an arc O as the centre, cutting MQ at X. (f) Set your compasses to length LN with the same radius. (g) Place the...

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D Divya Prakash Singh
The steps of constructions are: (a) Draw an angle 70 degrees with a protractor, i.e., ∠POQ = 70 degrees (b) Draw a ray AB. (c) Place the compasses at O and draw an arc to cut the rays of ∠POQ at L and M. (d) Use the same compasses, setting to draw an arc with A as the centre, cutting AB at X. (e) Set your compasses setting to the length LM with the same radius. (f) Place the compasses pointer...

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D Divya Prakash Singh
The steps of constructions are:  (a) Draw a line PQ and take a point O on it. (b) Taking O as the centre and convenient radius, mark an arc, which intersects PQ at A and B. (c) Taking A and B as centres and radius more than half of AB, draw two arcs intersecting each other at R. (d) Join OR. Thus, ∠QOR = ∠POQ = 90 . (e) Draw OD the bisector of ∠POR. Thus, ∠QOD is the required angle of 135. (f)...

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D Divya Prakash Singh
The steps of constructions are: 1. Draw a ray OA 2. Taking O as the centre and convenient radius, mark an arc, which intersects OA at X. 3. Taking X as a centre and the same radius, cut the previous arc at Y. Taking Y as the centre and the same radius, draw another arc intersecting the same arc at Z. 4. Taking Y and Z as centres and the same radius, draw two arcs intersecting each other at S....

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D Divya Prakash Singh
The steps of constructions are: 1. Draw a line PQ and take a point O on it. 2. Taking O as the centre and convenient radius, mark an arc, which intersects PQ at A and B. 3. Taking A and B as centres and radius more than half of AB, draw two arcs intersecting each other at R. Join OR. Thus, ∠QOR = ∠POR = 90°. 4. Draw OD the bisector of ∠POR. Thus, ∠QOD is required angle of 135°  

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D Divya Prakash Singh
The steps of constructions are: 1. Draw a ray OA 2. Taking O as the centre and convenient radius, mark an arc, which intersects OA at X. 3. Taking X as the centre and the same radius, cut the previous arc at Y. Taking Y as the centre and the same radius, draw another arc intersecting the same arc at Z. 4. Taking Y and Z as centres and the same radius, draw two arcs intersecting each other at...

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D Divya Prakash Singh
The steps of constructions are: 1. Draw a ray OA 2. Taking O as the centre and convenient radius, mark an arc, which intersects OA at P. 3. Taking P as the centre and same radius, cut previous arc at Q. Taking Q as the centre and the same radius cut the arc at S. Join OS. Thus, ∠AOS is the required angle of 120°.

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D Divya Prakash Singh
The steps of constructions are: 1. Draw a ray OA 2. Taking O as the centre and convenient radius, mark an arc, which intersects OA at X. 3. Taking X as the centre and the same radius, cut the previous arc at Y. Taking Y as the centre and the same radius, draw another arc intersecting the same arc at Z. 4. Taking Y and Z as centres and the same radius, draw two arcs intersecting each other at...

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D Divya Prakash Singh
The steps of constructions are: 1. Draw a ray OA. 2. Taking O as the centre and convenient radius, mark an arc, which intersects OA at P. 3. Taking P as the centre and the same radius, cut the previous arc at Q. Join OQ. Thus, ∠BOA is the required angle of 60°. 4. Put the pointer on P and mark an arc. 5. Put the pointer on Q and with the same radius, cut the previous arc at C. Thus, ∠COA is...

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D Divya Prakash Singh
The steps of constructions are: 1. Draw a ray OA 2. Taking O as the centre and convenient radius, mark an arc, which intersects OA at P. 3. Taking P as the centre and the same radius, cut the previous arc at Q. Join OQ. Thus,∠BOA is the required angle of 60°  

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D Divya Prakash Singh
The steps of constructions are: (a) Draw a ray OA. (b) At O, with the help of a protractor, construct ∠AOB = 153 degrees. (c) Draw OC as the bisector of ∠AOB. (d) Again, draw OD as bisector of ∠AOC. (e) Again, draw OE as bisector of,∠BOC. (f) Thus, OC, OD, and OE divide ∠AOB into four equal parts. 

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D Divya Prakash Singh
The steps of construction: (a) Draw a line PQ and take a point O on it. (b) Taking O as the centre and convenient radius, draw an arc that intersects PQ at A and B. (c) Taking A and B as centres and radius more than half of AB, draw two arcs which intersect each other at C. (d) Join OC. Thus, ∠COQ is the required right angle. (e) Taking B and E as centre and radius more than half of BE, draw...

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D Divya Prakash Singh
The steps of constructions are: 1. Draw a line  OA. 2. Using protractor and centre 'O draw an angle  AOB =147°. 3. Now taking 'O' as the centre and any radius draws an arc that intersects  'OA' and 'OB'  at p  and q. 4. Now take p and q as centres and radius more than half of PQ, draw arcs. 5. Both the arcs intersect at 'R' 6. Join 'OR' and produce it. 7. 'OR' is the required bisector of...

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D Divya Prakash Singh
Here, we will draw  using a protractor. We follow these steps: 1. Draw a ray OA. 2. Place the centre of the protractor on point O, and coincide line OA and Protractor line 3. Mark point B on 75 degrees. 4. Join OB Therefore  Now, we need to find its line of symmetry that is, we need to find its bisector. We follow these steps 1. Mark points C and D where the arc intersects OA and OB 2. Now,...

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D Divya Prakash Singh
The steps of constructions are: (i) Draw any angle with vertex O. (ii) Take a point A on one of its arms and B on another such that (iii) Draw perpendicular bisector of  and . (iv) Let them meet at P. Join PA and PB. With the help of divider, we obtained  that

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D Divya Prakash Singh
The steps of constructions are: (i) Draw the circle with O and radius 4 cm. (ii) Draw any two chords  and in this circle. (iii) Taking A and B as centres and radius more than half AB, draw two arcs which intersect each other at E and F. (iv) Join EF. Thus EF is the perpendicular bisector of chord . (v) Similarly draw GH the perpendicular bisector of chord . These two perpendicular bisectors...

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D Divya Prakash Singh
The steps of constructions are: (i) Draw a circle with centre C and radius 3.4 cm. (ii) Draw its diameter  (iii) Taking A and B as centres and radius more than half of it, draw two arcs which intersect each other at P and Q. (iv) Join PQ. Then PQ is the perpendicular bisector of  We observe that this perpendicular bisector of  passes through the centre C of the circle.

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D Divya Prakash Singh
The steps of constructions are: (i) Draw a circle with centre C and radius 3.4 cm. (ii) Draw any chord . (iii) Taking A and B as centres and radius more than half of , draw two arcs which cut each other at P and Q. (iv) Join PQ. Then PQ is the perpendicular bisector of . This perpendicular bisector of  passes through the centre C of the circle.
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