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Areas of Parallelograms and Triangles

Short Answer Type

51. Given two points A and B and a positive real number k. Find the locus of a point P such that ar(ΔPAB) = k.

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52. Show that the line segment joining the mid-points of a pair of opposite sides of a parallelogram divides it into two equal parallelograms.

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53. In an equilateral triangle, O is any point is the interior of the triangle and perpendiculars are drawn from O to the sides. Prove that the sum of these perpendicular line segments is constant.

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54. ΔABC and ΔABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that ar (ΔABC) = ar (ΔABD).

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55. ΔABC and ΔABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that ar (ΔABC) = ar (ΔABD).

ar left parenthesis quad. space ABCD right parenthesis equals 1 half BD. left parenthesis AM plus CN right parenthesis.

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Long Answer Type

56.

ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that

(i) ar(ΔADO) = ar(ΔCDO)
(ii) ar(ΔABP) = ar(ΔCBP)

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Short Answer Type

57. In the given figure, ΔABC is right angled at B, and BD is its median. E is the mid-point of BD. If AB = 6 cm, AC = 10 cm, calculate area of ΔBEC.

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Long Answer Type

58. In the given figure, ΔABC is right angled at A and AD is its median. BA is produced to E such that BA = AE. ED is joined. If AB = 6 cm, BC = 10 cm, find ar(ΔBED).

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59.

The medians of ΔABC intersect at G. Prove that

ar(ΔAGB) = ar(ΔAGC) = ar(ΔBGC) $equals 1 third space ar left parenthesis increment ABC right parenthesis$

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Short Answer Type

60. Triangles ABC and DBC are on the same base BC with vertices A and D on opposite sides of BC such that ar (ΔABC) = ar(ΔDBC). Show that BC bisects AD.

Given: Triangles ABC and DBC are on the same base BC with vertices A and D on opposite sides of BC such that ar(ΔABC) = ar(ΔDBC).
To Prove: BC bisects AD.

Given: Triangles ABC and DBC are on the same base BC with vertices A

Proof: ar(ΔABC) = ar(ΔDBC) | Given

$rightwards double arrow space 1 half cross times BC cross times AM equals space 1 half cross times BC cross times DN$

| Area of a triangle = $equals 1 half$ x Base x Corresponding altitude
$rightwards double arrow$ AM = DN ...(1)

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