Question

Find the altitude of the triangle with vertices P(8,20), Q (2,3) and R (6,6) drawn from P to the base QR​

Answer 1

Answer is 6.4 units = 32/5 units.

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The idea is to equate the area of triangle found by 2 different formulae.

P= (x1, y1)= (8,20). Q= (x2, y2)= (2,3). R= (x3, y3)= (6,6).

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Since, area of triangle in co-ordinate geometry is given by,

Area=   (1/2) [x1 (y2- y3 )+ x2 (y3-y1 )+ x3(y1-y2)]

Area= 16 sq. units.

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But, we know that, area of any triangle can be given by,

Area= Base x Altitude /2

Hence, Area= QR x Altitude/2.

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Now, the length of QR can be calculated by Euclidean distance:

QR= squareroot[ (x2-x3)^2 + (y2-y3)^2 ]

QR= squareroot[ 4x4 + 3x3 ]

QR= 5

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Hence, Area= 16= 5 x Altitude/2

Thus, Altitude= 32/5 = 6.4 units.