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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).

In right triangle BAC,
AB²+ AC² = BC²| By Pythagoras Theorem
⇒ 6² + AC² = 10² ⇒ AC = 8 cm

$therefore space space space ar left parenthesis increment ABC right parenthesis equals 1 half cross times space Base space cross times thin space Corresponding space altitude space space space space space space space space space space space equals space 1 half cross times AB cross times AC space space space space space space space space space space space equals space 1 half cross times 6 cross times 8 space space space space space space space space space space space space equals space 24 space cm squared space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis$

∵ AD is a median of ΔABC ∴ ar(ΔABD) = ar(ΔACD)
∵ A median of a triangle divides it into two triangles of equal areas

$equals space 1 half space ar left parenthesis ABC right parenthesis equals space 1 half straight x space 24 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space vertical line space From space left parenthesis 1 right parenthesis equals space space 12 space cm squared space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis space space space$

∵ BA = AE ∴ A is the mid-point of BE ∴ DA is a median of ΔBDE

$therefore space space ar space left parenthesis increment ABD right parenthesis equals 1 half ar left parenthesis increment BED right parenthesis$
∵ A median of a triangle divides it into two triangles of equal areas

$rightwards double arrow space space space space space ar left parenthesis increment ABD right parenthesis equals 1 half ar left parenthesis increment BED right parenthesis rightwards double arrow space space space space space 12 equals 1 half ar left parenthesis increment BED right parenthesis space space space space space space space vertical line space From space left parenthesis 3 right parenthesis rightwards double arrow space space space space space ar left parenthesis increment BED right parenthesis equals 24 space cm squared$

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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.

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