Bissoy.com
Doctors Medicines MCQs Q&A

Given are the steps to construct an equilateral triangle when the altitude of a triangle is given. Using, T-square, set-squares only. Arrange the steps. i. Join R, Q; T, Q. Q, R, T is the required triangle ii. With a T-square, draw a line AB of any length iii. From a point P on AB draw a perpendicular PQ of given altitude length iv. With 30o-60o set-squares, draw a line making 30o with PQ at Q on both sides cutting at R, T

Given are the steps to construct an equilateral triangle when the altitude of a triangle is given. Using, T-square, set-squares only. Arrange the steps. i. Join R, Q; T, Q. Q, R, T is the required triangle ii. With a T-square, draw a line AB of any length iii. From a point P on AB draw a perpendicular PQ of given altitude length iv. With 30o-60o set-squares, draw a line making 30o with PQ at Q on both sides cutting at R, T Correct Answer ii, iii, iv, i

Here gives the simple procedure since T-square and 30o-60o set-squares. The interior angles are 60o, 60o, 60o (180o /3 = 60o). Altitude divides the sides of equilateral triangle equally. Set- squares are used for purpose of 30o.

Related Questions

Given are the steps to construct an equilateral triangle, when the length of side is given. Using, T-square, set-squares only. Arrange the steps. i. The both 2 lines meet at C. ABC is required triangle ii. With a T-square, draw a line AB with given length iii. With 30o-60o set-squares, draw a line making 60o with AB at A iv. With 30o-60o set-squares, draw a line making 60o with AB at B
Given are the steps to construct an equilateral triangle, with help of a compass, when the length of altitude is given. Arrange the steps. i. Draw a line AB of any length. At any point P on AB, draw a perpendicular PQ equal to altitude length given ii. Draw bisectors of CE and CF to intersect AB at R and T respectively.QRT is required triangle iii. With center Q and any radius, draw an arc intersecting PQ at C iv. With center C and the same radius, draw arcs cutting the 1st arc at E and F
Given are the steps to construct regular polygon of any number of sides. Arrange the steps. i. Draw the perpendicular bisector of AB to cut the line AP in 4 and the arc AP in 6 ii. The midpoint of 4 and 6 gives 5 and extension of that line along the equidistant points 7, 8, etc gives the centers for different polygons with that number of sides and the radius is AN (N is from 4, 5, 6, 7, so on to N) iii. Join A and P. With center B and radius AB, draw the quadrant AP iv. Draw a line AB of given length. At B, draw a line BP perpendicular and equal to AB
Given are the steps to construct an equilateral triangle, with help of a compass, when the length of a side is given. Arrange the steps. i. Draw a line AB with given length ii. Draw lines joining C with A and B iii. ABC is required equilateral triangle iv. With centers A and B and radius equal to AB, draw arcs cutting each other at C
Given are the steps to draw a perpendicular to a line at a point within the line when the point is near the centre of a line. Arrange the steps. Let AB be the line and P be the point in it i. P as centre, take convenient radius R1 and draw arcs on the two sides of P on the line at C, D. ii. Join E and P iii. The line EP is perpendicular to AB iv. Then from C, D as centre, take R2 radius (greater than R1), draw arcs which cut at E.
Steps are given to determine the centre of curvature at a given point on a conic. Arrange the steps. Let P be the given point on the conic and F is the focus. Join P with F. Draw a line NR perpendicular to PN and cutting PF or PF-extended at R. Draw a line RO perpendicular to PR and cutting PN-extended at O which is centre of curvature. At P, draw a normal PN, cutting the axis at N.
Given are the steps to draw a perpendicular to a line from a point outside the line, when the point is nearer the centre of line. Arrange the steps. Let AB be the line and P be the point outside the line i. Take P as centre and take some convenient radius draw arcs which cut AB at C, D. ii. Join E, F and extend it, which is perpendicular to AB. iii. From C, D with radius R1 (more than half of CD), draw arcs which cut each other at E. iv. Again from C, D with radius R2 (more than R1), draw arcs which cut each other at F.
Given are the steps to draw a perpendicular to a line at a point within the line when the point is near an end of the line. Arrange the steps. Let AB be the line and P be the point in it. i. Join the D and P. ii. With any point O draw an arc (more than a semicircle) with a radius of OP, cuts AB at C. iii. Join the C and O and extend till it cuts the large arc at D. iv. DP gives the perpendicular to AB.
Steps are given to find the normal and tangent for a cycloid. Arrange the steps if C is the centre for generating circle and PA is the directing line. N is the point on cycloid. i. Through M, draw a line MO perpendicular to the directing line PA and cutting at O. ii. With centre N and radius equal to radius of generating circle, draw an arc cutting locus of C at M. iii. Draw a perpendicular to ON at N which is tangent. iv. Draw a line joining O and N which is normal.
Given are the steps to draw a perpendicular to a line from a point outside the line, when the point is near the centre of line. Arrange the steps. Let AB be the line and P be the point outside the line i. The line EP is perpendicular to AB ii. From P take convenient radius and draw arcs which cut AB at two places, say C, D. iii. Join E and P. iv. Now from centers C, D draw arc with radius (more than half of CD), which cut each other at E.