Find the area of a parallelogram given in the figure. Also, find the length of the altitude from vertex A on the side DC. -Maths 9th

Find the area of a parallelogram given in the figure. Also, find the length of the altitude from vertex A on the side DC. -Maths 9th
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1 Answer

Answer :

=3 x 3 x 5 x 2 cm2 Area of parallelogram ABCD = 2 x 90 = 180 cm2 …(ii) Let altitude of a parallelogram be h. Also, area of parallelogram ABCD =Base x Altitude ⇒  180 = DC x h [from Eq. (ii)] ⇒  180 = 12 x h ∴ h = 180/12= 15 cm Hence, the area of parallelogram is 180 cm2 and the length of altitude is 15 cm.
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Related questions

Find the area of a parallelogram given in the figure. Also, find the length of the altitude from vertex A on the side DC. -Maths 9th

Description : Find the area of a parallelogram given in the figure. Also, find the length of the altitude from vertex A on the side DC. -Maths 9th

Last Answer : Weknowthatthediagonalofaparallelogram(∥gm)dividesitintotwocongruenttriangles.SoAreaof∥gmABCD=2 Areaof△BCD.AccordingtoHeron′sformulathearea(A)oftrianglewithsidesa,b&cisgivenasA=2[s(s−a)(s−b) ... 90=180Areaof∥gm=base heightHeightofaltitudefromvertexAonsideCDoftheof∥gm=baseCDareaof∥gmABCD =12180 =15cm

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Last Answer : In ||gm ABCD, AC is the diagonal ∴ ar(△ABC) = ar(△ADC) = 1/2 ar ||gm ABCD) In△ADC, AL is the median ∴ ar(△ADL) = ar(△ACL)= 1/2 ar(△ADC) = 1/4 ar (||gm ABCD) Now, ar(quad.ABCL) = ar(△ABC) + ar(△ACL) = 3/4 ar ... ar(||gm ABCD) = 96 cm2 ∴ ar(△ADC) = 1/2 ar(||gm ABCD) = 1/2 96 = 48 cm2

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The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60°. -Maths 9th

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Last Answer : (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 ... \sqrt{(2-2)^2+(4+1)^2}\) = \(\sqrt{25}\) = 5⇒ PR ≠ SQ ⇒ Diagonals are not equal ⇒ PQRS is a rhombus.

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Description : Write the coordinates of the vertices of a rectangle whose length and breadth are 6 and 3 units respectively, one vertex at the origin, the longer side lies on the y-axis and one of the vertices lies in the second quadrant. -Maths 9th

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In the figure, the area of parallelogram ABCD is -Maths 9th

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PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram . -Maths 9th

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Last Answer : This answer was deleted by our moderators...

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Description : Write the coordinates of the vertices of a rectangle whose lenght and breadth are 7 and 4 units respectively,one vertex atthe the origin,the longer side lies on the x-axis and one of the vertices lies in the third quadrant. -Maths 9th

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O is any point on the diagonal PR of a parallelogram PQRS (figure). -Maths 9th

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Last Answer : According to question prove that ar(ΔPSO) = ar(ΔPQO).

In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE = AB. Prove that ED bisects BC -Maths 9th

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Ankush prepare a poster in the form of parallelogram , as in figure. -Maths 9th

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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

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Prove that the figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a parallelogram. -Maths 9th

Description : Prove that the figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a parallelogram. -Maths 9th

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If the distance from the vertex to the centroid of an equilateral triangle is 6 cm, then what is the area of the triangle? -Maths 9th

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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.

The area of a triangle is 5. Two of its vertices are (2, 1) and (3, –2). The third vertex is (x, y) -Maths 9th

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Last Answer : Let A(x1, y1) = (3, 4), B(x2, y2) ≡ (0, 5), C(x3, y3) ≡ (2, -1)and D(x4, y4) ≡ (3, -2) be the vertices of quadrilateral ABCD.Area of quad. ABCD = \(rac{1}{2}\) |{(x1 y2 - x2 y1) + (x2y3 - x3y2) + (x3y4 - x4y3) ... ) + (12 + 6)}|= \(rac{1}{2}\) |{15 - 11 + 0 + 18}| = \(rac{1}{2}\)x 22 = 11 sq. units.