Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment. -Maths 9th

Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment. -Maths 9th
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Steps of Construction  (i) Draw a line segment AB = 5.8 cm. (ii) Taking A as centre and radius more than 1/2AB, draw two arcs, one on either side of AB.  (iii) Taking B as centre and the same radius draw two arcs, cutting the previously drawn arcs at points C and D respectively.  (iv) Join CD, intersecting AB at point P. Then, line CPD is the required perpendicular bisector of AB.
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Description : If a point on the perpendicular bisector of a segment, then it is?

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If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

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Description : A straight line passes through the points (5, 0) and (0, 3). The length of the perpendicular from the point (4, 4) on the line is: -Maths 9th

Last Answer : (b) \(rac{\sqrt{17}}{2}\)Equation of the line through the points (5, 0) and (0, 3) y - 0 = \(rac{3-0}{0-5}\) (x - 5)⇒ y = \(rac{-3}{5}\)(x - 5)⇒ 5y + 3x - 15 = 0 ∴ Distance of perpendicular from ... (rac{|20+12-15|}{\sqrt{25+9}{}}\) = \(rac{17}{\sqrt{34}}\) units. = \(rac{\sqrt{17}}{2}\) units.

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ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

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Last Answer : Solution: (i) In ΔACB, M is the midpoint of AB and MD || BC , D is the midpoint of AC (Converse of mid point theorem) (ii) ∠ACB = ∠ADM (Corresponding angles) also, ∠ACB = 90° , ∠ADM = 90° and MD ⊥ AC (iii ... SAS congruency] AM = CM [CPCT] also, AM = ½ AB (M is midpoint of AB) Hence, CM = MA = ½ AB