Question

the area of a right triangle right angled at b is 24 cm if hyptneuse is 10 cm then determine the legth of the line segment which is perpendicular drawn from vertex b to the hyptneuse AC

Answer 1

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Answer 2

Let AB = x
BC = y

Let the perpendicular which is drawn from vertex B meets AC at D
Let BD = z

Now Let's find out the side of triangles


Given area = 24

=> 1/2 × xy = 24 (1/2 × base × height = area)

=> xy = 24 × 2

=> xy = 48 cm

Now x² + y² = 10² (Pythagoras Theorem)

Add 2xy on both sides

=> x² + y² + 2xy = 100 + 2xy

=> (x + y)² = 100 + 2(48)

=> (x + y)² = 100 + 96

=> (x + y)² = 196

=> x + y = √196

=> x + y = 14 cm

Now again, x² + y² = 10²

now subtract 2xy from both sides

=> x² + y² - 2xy = 100 - 2xy

=> (x - y)² = 100 - 96 (since 2xy = 96)

=. (x - y)² = 4

=> x - y = √4 = 2cm

Now x + y = 14
x - y = 2

Add the two equations,

=> x + y + x - y = 14 + 2

=> 2x = 16

=> x = 8 cm

Put the value of x in any equation and you will get the value of y as 6 cm

So the sides of triangle are 6 and 8cm

Now since BD is perpendicular to AC, ∆BDA and ∆BDC is right angled triangle.

Let AD = a

=> CD = 10 - a


So a² + z² = x²

=> a² + z² = 64 (x² = 8² = 64)

Again for ∆BDC,

(10 - a)² + z² = y²

=> 100 + a² - 20a + z² = 36

=> (a² + z²) - 20a = 36 - 100

=. 64 - 20a = -64

=> -20a = -64 - 64

=> -20a = -128

=> a = -128/-20

=> a = 6.4 cm

So a² + z² = x²

=> z² = x² - a²

=> z² = 64 - 40.96 {a² = (6.4)² = 40.96}

=> z² = 23.04

=> z = √(23.04)

=> z = 4.8 cm

=> BD = 4.8 cm



Answer :- 4.8 cm


Conclusion :-

The length of perpendicular drawn from the vertex = (8 × 6)/10

=> BD = (AB × BC)/AC

=> BD = product of perpendicular and base ÷ hypotenuse.