If A(3, 5), B(– 5, – 4), C(7, 10) are the vertices of a parallelogram taken in order, then the co-ordinates of the fourth vertex are: -Maths 9th

If A(3, 5), B(– 5, – 4), C(7, 10) are the vertices of a parallelogram taken in order, then the co-ordinates of the fourth vertex are: -Maths 9th
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(c) RhombusCo-ordinates of P are \(\bigg(rac{-1-1}{2},rac{-1+4}{2}\bigg)\)i.e, \(\big(-1,rac{3}{2}\big)\)Co-ordinates of Q are \(\bigg(rac{-1+5}{2},rac{4+4}{2}\bigg)\)i.e, (2, 4)Co-ordinates of R are \(\bigg(rac{5+5}{2},rac{4-1}{2}\bigg)\)i.e, \(\big(5,rac{3}{2}\big)\)Co-ordinates of S are \(\bigg(rac{-1+5}{2},rac{-1-1}{2}\bigg)\)i.e, (2, -1)Now,⇒ PQ = QR = RS = SP ⇒ All sides are equalAlso, PR = \(\sqrt{(5+1)^2+\big(rac{3}{2}-rac{3}{2}\big)^2}\) = \(\sqrt{36}\) = 6SQ = \(\sqrt{(2-2)^2+(4+1)^2}\) = \(\sqrt{25}\) = 5⇒ PR ≠ SQ ⇒ Diagonals are not equal ⇒ PQRS is a rhombus.
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Last Answer : P ≡ (-3, 2), Q ≡ (-5, -5), R ≡ (2, -3), S ≡ (4, 4)∴ PQ = \(\sqrt{(-5+3)^2+(-5-2)^2}\) = \(\sqrt{4+49}\) = \(\sqrt{53}\)QR = \(\sqrt{(2+5)^2+(-3+5)^2}\) = \(\sqrt{49+4}\) = \ ... of rhombus = \(rac{1}{2}\) x (Product of length of diagonals) = \(rac{1}{2}\) x \(5\sqrt2\) x \(9\sqrt2\) = 45 sq. units.

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Description : In the adjoining figure, P and Q have co-ordinates (4, 6)and (0, 3) respectively. Find (i) the co-ordinates of R (ii) Area of quadrilateral OAPQ. -Maths 9th

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Description : If the co-ordinates of the mid-points of the sides of a triangle are (1, 1), (2, –3), (3, 4), find its incentre. -Maths 9th

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Description : What are the co-ordinates of the foot of the perpendicular from the point (2, 3) on the line x + y – 11 = 0 ? -Maths 9th

Last Answer : (d) 2x + 9y + 7 = 0PS being the median of ΔPQR, S is the mid-point of QR, i.e., Coordinates of S ≡ \(\bigg(rac{6+7}{2},rac{-1+3}{2}\bigg)\) = \(\bigg(rac{13}{2},1\bigg)\)Slope of line parallel to PS = Slope of PS= \(rac{1 ... y + 1) = \(rac{-2}{9}\)(x - 1), i.e., 9y + 9 = - 2x + 2 ⇒ 2x + 9y + 7 = 0.

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Description : The co-ordinates of P and Q are (–3, 4) and (2, 1) respectively. -Maths 9th

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Description : If vertices of a triangles are (1, k), (4, -3) and (-9, 7) and its area is 15 sq. units then find then the value of k. -Maths 9th

Last Answer : hope it helps if the vertices of a triangle are (1,k),(4,−3)(−9,7) area = 15 sq.units. find the value of k. Area of △ 21 [x1 (y2 −y3 )+x2 (y3 −y1 )+x3 (y1 −y2 )]=15 21 [1(−3−7)+ ... k+3)]=15 21 [(−10+28−4k−9k−27)]=15 −10+28−4k−9k−27=30 −10+28−13k−27=30 −13k=30+10+27−28 −13k=39 k=1339 k=−3 thank u

A vertex cover of an undirected graph G(V, E) is a subset V1 ⊆ V vertices such that (A) Each pair of vertices in V1 is connected by an edge (B) If (u, v) ∈ E then u ∈ V1 and v ∈ V1 (C) If (u, v) ∈ E then u ∈ V1 or v ∈ V1 (D) All pairs of vertices in V1 are not connected by an edge

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Last Answer : (b) 27√3 cm2.Let G be the centroid of ΔPQR. Then, PG = 6 cm.Now, \(rac{PG}{GS}\) = \(rac{2}{1}\) ⇒ GS = 3 cm∴PS = PG + GS = 9 cm (i)∴ If a is the length of a side of ΔPQR, then ... = 6√3 cm∴ Area of equilateral ΔPQR = \(rac{\sqrt3}{4}\) (a)2= \(rac{\sqrt3}{4}\) x (6√3)2 cm2 = 27√3 cm2.

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Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

Description : Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

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A point E is taken on the side BC of a parallelogram ABCD. -Maths 9th

Description : A point E is taken on the side BC of a parallelogram ABCD. -Maths 9th

Last Answer : Given ABCD is a parallelogram and E is a point on BC. AE and DC are produced to meet at F. AB||CD anti BC||AD ,..(i)

Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ . -Maths 9th

Description : Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ . -Maths 9th

Last Answer : Join AQ and PC . Since ABCD is a parallelogram . ⇒ AB | | DC ⇒ AP | | QC ∵ AP and QC are parts of AB and DC respectively] Also, AP = CQ [given] Thus, APCQ is a parallelogram . We know that diagonals of a parallelogram bisect each other . Hence AC and PQ bisect each other .

Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

Description : Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

Last Answer : According to question parallelogram ABCD such that AP = CQ.

In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. -Maths 9th

Description : In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. -Maths 9th

Last Answer : Given In a parallelogram PSDA, points 0 and R are on PS such that PQ = QR = RS and PA || QB || RC. To prove ar (PQE) = ar (CFD) Proof In parallelogram PABQ, and PA||QB [given] So, ... = ΔDCF [by ASA congruence rule] ∴ ar (ΔPQE) = ar (ΔCFD) [since,congruent figures have equal area] Hence proved.

A point E is taken on the side BC of a parallelogram ABCD. -Maths 9th

Description : A point E is taken on the side BC of a parallelogram ABCD. -Maths 9th

Last Answer : Given ABCD is a parallelogram and E is a point on BC. AE and DC are produced to meet at F. AB||CD anti BC||AD ,..(i)

In Fig. 8.40, points M and N are taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AM = CN. Show that AC and MN bisect each other. -Maths 9th

Description : In Fig. 8.40, points M and N are taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AM = CN. Show that AC and MN bisect each other. -Maths 9th

Last Answer : Solution :-

Name the quadrilateral ABCD, the coordinates of whose vertices are A(3, 5), B(1, 1), C(5, 3) and D(7, 7). -Maths 9th

Description : Name the quadrilateral ABCD, the coordinates of whose vertices are A(3, 5), B(1, 1), C(5, 3) and D(7, 7). -Maths 9th

Last Answer : (b) Equilateral, \(2\sqrt3a^2\) sq. unitsLet A(a, a), B(-a, -a) and C \((-\sqrt3a,\sqrt3a)\) be the vertices of ΔABC. Then,AB = \(\sqrt{(a+a)^2+(a+a)^2}\) = \(\sqrt{4a^2+4a^2}\) = \(\sqrt{8a^2}\) = \( ... 4}\) (side)2 = \(rac{\sqrt3}{4}\) x (\(2\sqrt2a\))2= \(rac{\sqrt3}{4}\) x 8a2 = \(2\sqrt3a^2\).

The vertices A(4, 5), B(7, 6), C(4, 3) and D(1, 2) from the quadrilateral ABCD whose special name is -Maths 9th

Description : The vertices A(4, 5), B(7, 6), C(4, 3) and D(1, 2) from the quadrilateral ABCD whose special name is -Maths 9th

Last Answer : (a) (5, 2)Let A(8, 6) , B(8, -2) and C(2, -2) be the vertices of the given triangle and P(x, y) be the circum-centre of this triangle. Then, PA2 = PB2 = PC2 Now, PA2 = PB2 ⇒ (x - 8)2 + (y - 6)2 = (x ... 4 = x2 - 4x + 4 + y2 + 4y + 4 ⇒ 12x = 60 ⇒ x = 5. ∴ Co-ordinates of the circumcentre are (5, 2).

The point A(0, 0), B(1, 7) and C(5, 1) are the vertices of a triangle. Find the length of the perpendicular from -Maths 9th

Description : The point A(0, 0), B(1, 7) and C(5, 1) are the vertices of a triangle. Find the length of the perpendicular from -Maths 9th

Last Answer : (b) 3x - y = 0 Given lines are 3x - y - 3 = 0 and 3x - y + 5 = 0. Line parallel to the given lines can be written as 3x - y + c = 0 ...(i) Let us taken a point, say ... 5c + 15 = - 3c + 15 ⇒ 8c = 0 ⇒ c = 0. Substituting c = 0 in (i), the required equation is 3x - y = 0.