The orthocentre of a triangle whose vertices are (0, 0), (3, 0) and (0, 4) is -Maths 9th

The orthocentre of a triangle whose vertices are (0, 0), (3, 0) and (0, 4) is -Maths 9th
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(c) 60ºSince p is the length of perpendicular from origin on the straight line ax + by – p = 0.p = \(rac{|a.0+b.0-p|}{\sqrt{a^2+b^2}}\)⇒ 1 = \(\sqrt{a^2+b^2}\) ⇒ 1 = a2 + \(rac{3}{4}\)             \(\big(\because{b} = rac{\sqrt3}{2}\big)\)⇒ a2 = \(rac{1}{4}\) ⇒ a = \(rac{1}{2}.\)∴ Equation of the straight line is \(rac{1}{2}x\) + \(rac{\sqrt3}{2}y\) = p⇒ \(x\) cos 60° + y sin 60° = p Hence required angle is 60°, which is the angle between the perpendicular and the positive direction of x-axis.
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Description : A point within an equilateral triangle whose perimeter is 30 m is 2 m from one side and 3 m from another side. Find its distance from third side. -Maths 9th

Last Answer : answer:

In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram (ii) quadrilateral BEFC is a parallelogram (iii) AD || CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Description : In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram ( ... CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Last Answer : . Solution: (i) AB = DE and AB || DE (Given) Two opposite sides of a quadrilateral are equal and parallel to each other. Thus, quadrilateral ABED is a parallelogram (ii) Again BC = EF and BC || EF ... (Given) BC = EF (Given) AC = DF (Opposite sides of a parallelogram) , ΔABC ≅ ΔDEF [SSS congruency]

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that (i) ΔAPB ≅ ΔCQD (ii) AP = CQ -Maths 9th

Description : ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that (i) ΔAPB ≅ ΔCQD (ii) AP = CQ -Maths 9th

Last Answer : Q Solution: (i) In ΔAPB and ΔCQD, ∠ABP = ∠CDQ (Alternate interior angles) ∠APB = ∠CQD (= 90o as AP and CQ are perpendiculars) AB = CD (ABCD is a parallelogram) , ΔAPB ≅ ΔCQD [AAS congruency] (ii) As ΔAPB ≅ ΔCQD. , AP = CQ [CPCT]

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD . -Maths 9th

Description : ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD . -Maths 9th

Last Answer : In gm ABCD , AP and CQ are perpendicular from the vertices A and C on diagonal BD. Show that : (i) AAPB ≅ ACQD (ii) AP = CQ .