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तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएस.एस.एल.सी. (इंग्रजी माध्यम) इयत्ता १०

Let P(11, 7), Q(13.5, 4) and R(9.5, 4) be the midpoints of the sides AB, BC and AC respectively of ∆ABC . Find the coordinates of the vertices A, B and C. Hence find the area of ∆ABC and compare this - Mathematics

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प्रश्न

Let P(11, 7), Q(13.5, 4) and R(9.5, 4) be the midpoints of the sides AB, BC and AC respectively of ∆ABC. Find the coordinates of the vertices A, B and C. Hence find the area of ∆ABC and compare this with area of ∆PQR.

बेरीज

उत्तर

Let the vertices of the ∆ABC be A(x1, y1), B(x2, y2), C(x3, y3)

Mid point of AB = (x1+x22,y1+y22)

(11, 7) = (x1+x22,y1+y22)

x1+x22 = 11

x1 + x2 = 22   ...(1)

y1+y22 = 7

y1 + y2 = 14  ...(2)

Mid point of BC = (x2+x32,y2+y32) 

⇒ (13.5, 4) = (x2+x32,y2+y32)

x2+x32 = 13.5

x2 + x3 = 27   ...(3)

y2+y32 = 4

y2 + y3 = 8   ...(4)

Mid point of AC = (x1+x32,y1+y32)

(9.5, 4) = (x1+x22,y1+y32)

x1+x32 = 9.5

x1 + x3 = 19   ...(5)

y1+y32 = 4

y1 + y3 = 8   ...(6)

Add (1), (3) and (5)

2x1 + 2x2 + 2x3 = 22 + 27 + 19

2(x1 + x2 + x3) = 68

x1 + x2 + x3 = 34

From (1) ⇒ x1 + x2 = 22

x3 = 34 – 22 = 12

From (3) ⇒ x2 + x3 = 27

x1 = 34 – 27 = 7

From (5) ⇒ x1 + x3 = 19

 x2 = 34 – 19 = 15

Add (2), (4) and (6)

2y1 + 2y2 + 2y3 = 14 + 8 + 8

2(y1 + y2 + y3) = 30

y1 + y2 + y3 = 15

From (2) ⇒ y1 + y2 = 14

y3 = 15 – 14 = 1

From (4) ⇒ y2 + y3 = 18

y1 = 15 – 8 = 7

From (6) ⇒ y1 + y3 = 8

y2 = 15 – 8 = 7

The vertices of a ΔABC are A(7, 7), B(15, 7) and C(12, 1)

Area of ΔABC = 12[(x1y2+x2y3+x3y1)-(x2y1+x3y2+x1y3)]

= 12[(7+84+105)-(84+15+49)]

= 12[196-148]

= 12×48

= 24 sq. units

Area of ΔPRQ = 12[(44+8+94.5)-(66.5+54+44)]

= 12[176.5-164.5]

= 12×12

= 6 sq. units

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पाठ 5: Coordinate Geometry - Exercise 5.1 [पृष्ठ २१२]

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सामाचीर कलवी Mathematics [English] Class 10 SSLC TN Board
पाठ 5 Coordinate Geometry
Exercise 5.1 | Q 8 | पृष्ठ २१२

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