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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.

Given: ABC is an equilateral triangle. O is any point in the interior of the triangle. Perpendiculars OD, OE and OF are drawn from 0 on the sides BC, AB and AC respectively of ΔABC.
To Prove: OD + OE + OF = constant
Construction: Join O to A, B and C. Draw AH
⊥ BC.
Proof: ∵ ΔABC is an equilateral triangle AB = BC = CA

$ar left parenthesis increment AOB right parenthesis equals fraction numerator left parenthesis AB right parenthesis left parenthesis OE right parenthesis over denominator 2 end fraction equals fraction numerator left parenthesis BC right parenthesis left parenthesis OE right parenthesis over denominator 2 end fraction space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis space space space space space space space space space space space space space space space space vertical line space because space AB space equals space BC$

Given: ABC is an equilateral triangle. O is any point in the interior

$ar left parenthesis increment BOC right parenthesis equals fraction numerator left parenthesis BC right parenthesis left parenthesis OD right parenthesis over denominator 2 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis ar space left parenthesis increment COA right parenthesis equals fraction numerator left parenthesis CA right parenthesis left parenthesis OF right parenthesis over denominator 2 end fraction equals fraction numerator left parenthesis BC right parenthesis left parenthesis OF right parenthesis over denominator 2 end fraction space space space space space space space... left parenthesis 3 right parenthesis space space space space space space vertical line space because space CA space equals space BC$

Adding (1), (2) and (3), we get

$space space space space space space space ar space left parenthesis increment AOB right parenthesis plus ar left parenthesis increment BOC right parenthesis plus ar left parenthesis increment COA right parenthesis rightwards double arrow space space space ar space left parenthesis increment ABC right parenthesis equals fraction numerator left parenthesis BC right parenthesis left parenthesis OE right parenthesis over denominator 2 end fraction plus fraction numerator left parenthesis BC right parenthesis left parenthesis OD right parenthesis over denominator 2 end fraction plus fraction numerator left parenthesis BC right parenthesis left parenthesis OF right parenthesis over denominator 2 end fraction rightwards double arrow space space space 1 half left parenthesis BC right parenthesis left parenthesis AH right parenthesis equals 1 half BC left parenthesis OE plus OD plus OF right parenthesis rightwards double arrow space space space space AH equals OD plus OE plus OF$

⇐ OD + OE + OF = AH which is constant for a given triangle.

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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) $equals 1 third space ar left parenthesis increment ABC right parenthesis$

293 Views

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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