Answer
Verified
497.4k+ views
Hint: First of all, draw a line AB = 10 cm and mark point P such that AP = 4 cm. Now mark two points on both sides of P on line AB equidistant from P. Now with the centers at these points, draw an arc above line AB and proceed to get the required perpendicular.
Complete step-by-step answer:
Here, we have to draw a line segment AB of length 10 cm. Then, we have to mark a point P on AB such that AP = 4 cm. Finally, we have to draw a line through P which would be perpendicular to AB.
(i) Let us first draw the line AB = 10 cm with the help of a ruler as follows:
(ii) Now, with A as the center and 4 cm as the radius, we will draw an arc on AB as
(iii) We will name this point as P such that AP = 4 cm.
(iv) Now, we will take any radius, say 3 cm. With P as center and radius = 3 cm. We will draw an arc on the left side of P on line AB as,
(v) We will name this as point C.
(vi) Now with the same radius that is 3 cm and with P as the center, we will again draw an arc on the right side of P on the line AB as,
(vii) We will name this as point D.
(viii) Now with D as center and radius greater than DP, we will draw an arc above line AB as,
(ix) Now with C as the center and with the same radius as that of the previous step, we will draw an arc cutting the previous arc and we will name this point of intersection of two arcs as O.
(x) Now we will join OP and extend it.
Hence, we get OP as the required perpendicular on line AB through P.
Note: Some students make this mistake of constructing the perpendicular bisector of AB but they must keep in mind that they have to construct perpendicular passing through P. Also, students can cross-check their construction by measuring if angle \[\angle OPD\] is \[{{90}^{o}}\] or not.
Complete step-by-step answer:
Here, we have to draw a line segment AB of length 10 cm. Then, we have to mark a point P on AB such that AP = 4 cm. Finally, we have to draw a line through P which would be perpendicular to AB.
(i) Let us first draw the line AB = 10 cm with the help of a ruler as follows:
(ii) Now, with A as the center and 4 cm as the radius, we will draw an arc on AB as
(iii) We will name this point as P such that AP = 4 cm.
(iv) Now, we will take any radius, say 3 cm. With P as center and radius = 3 cm. We will draw an arc on the left side of P on line AB as,
(v) We will name this as point C.
(vi) Now with the same radius that is 3 cm and with P as the center, we will again draw an arc on the right side of P on the line AB as,
(vii) We will name this as point D.
(viii) Now with D as center and radius greater than DP, we will draw an arc above line AB as,
(ix) Now with C as the center and with the same radius as that of the previous step, we will draw an arc cutting the previous arc and we will name this point of intersection of two arcs as O.
(x) Now we will join OP and extend it.
Hence, we get OP as the required perpendicular on line AB through P.
Note: Some students make this mistake of constructing the perpendicular bisector of AB but they must keep in mind that they have to construct perpendicular passing through P. Also, students can cross-check their construction by measuring if angle \[\angle OPD\] is \[{{90}^{o}}\] or not.
Recently Updated Pages
Two spheres of masses m and M are situated in air and class 9 physics CBSE
Glycerol can be separated from spentlye in soap industry class 9 chemistry CBSE
Glycerol can be separated from spentlye in soap industry class 9 chemistry CBSE
Glycerol can be separated from spentlye in soap industry class 9 chemistry CBSE
Glycerol can be separated from spentlye in soap industry class 9 chemistry CBSE
Glycerol can be separated from spentlye in soap industry class 9 chemistry CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
Who was the leader of the Bolshevik Party A Leon Trotsky class 9 social science CBSE
What is pollution? How many types of pollution? Define it
Voters list is known as A Ticket B Nomination form class 9 social science CBSE
The president of the constituent assembly was A Dr class 9 social science CBSE