Solution: Steps to construct: Step 1: Draw a regular hexagon of sides 4cm. Step 2: Draw the angle bisector of A and B. which intersects each other at point O. Step 3: Draw OL perpendicular to AB....
Solution: Steps to construct: Step 1: Consider a point O on a line, with center O, and radius 3cm, draw a circle. Step 2: Extend its diameters on both sides and cut off OP = OQ = 7cm. Step 3: Mark...
Solution: Steps to construct: Step 1: Draw a line segment AB = 6cm. Step 2: Draw its perpendicular bisector bisecting it at point O. Step 3: With center O and radius OB, draw a circle. Step 4:...
Solution: Steps to construct: Step 1: Draw a circle with center O and radius 4cm and mark that point as A. Step 2: Take a point P such that OP = 7cm. Step 3: Bisect OB at M. Step 4: With center M...
(B) inthe figure given below, AB is the diameter of the semi-circle ABCDE with centre O. If AE = ED and ∠BCD = 140°, find ∠AED and ∠EBD. Also Prove that OE is parallel to BD. Solution: (a) Join DB,...
(a) In figure (i) given below, triangle ABC is circumscribed, find x. (b) In figure (ii) given below, quadrilateral ABCD is circumscribed, find x. Solution: (a) From A, AP and AQ are the tangents...
Solution: Two concentric circles with center O OP and OB are the radii of the circles respectively, then OP = 5 cm, OB = 13 cm. Ab is the chord of outer circle which touches the inner circle at P....
Solution: Radius of the circle = 6 cm and length of tangent = 8 cm Let OP be the distance i.e. OA = 6 cm, AP = 8 cm . OA is the radius OA ⊥ AP Now In right ∆OAP, OP2 = OA2 + AP2 (By Pythagoras...
Solution: In a circle with center O and radius 3cm and p is at a distance of 5cm. That is OT = 3 cm, OP = 5 cm OT is the radius of the circle OT ⊥ PT Now in right ∆ OTP, by Pythagoras axiom, OP2 =...
Solution:- From the dimensions given in the question, Consider the ∆PQR So, PQ/PA = 12.5/5 = 2.5/1 PR/PB = (PB + BR)/PB = (4 + 6)/4 = 10/4 = 2.5 By comparing both the results, 2.5 = 2.5 Therefore,...
Solution:- From the question it is given that, ABC is a triangle in which AB = AC. P is a point on the side BC such that PM ⊥ AB and PN ⊥ AC. We have to prove that, BM x NP = CN x MP Consider the...
Solution: Given line: 3x – 4y + 12 = 0 The slope of the line is given by 3x – 4y + 12 = 0 ⇒ 4y = 3x + 12 y = (3/4) x + 3 Thus, slope (m1) = ¾ Now, let the slope of the line perpendicular to the...