In the right angled triangle, AB+AD=BC+CD. If AB=x, BC=h and CD=d, then find x in terms of h and d.

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A
$x = \frac{h d}{2 h + d}$
B
$x = \frac{h}{d + 2 h}$
C
$x = \frac{h}{d + 2}$
D
$x = \frac{4 h}{d + 2}$

Solution:

It is given that in the right angled triangle,

AB + AD = BC + CD

AB = x

BC = h

and

CD = d

The Pythagoras theorem states that the square of the hypotenuse side is equal to the sum of squares of the other two sides.
It is given that

AB + AD = BC + CD

\Rightarrow AD = BC + CD - AB

\Rightarrow AD = h + d - x

……….(1)
Apply Pythagoras theorem in

ΔACD

{AD}^{2} = {AC}^{2} + {DC}^{2}

\Rightarrow {(h + d - x)}^{2} = {(x + h)}^{2} + d^{2}

[From equation (1)]

\Rightarrow {(h + d - x)}^{2} - {(x + h)}^{2} = d^{2}

Use the identity that

a^{2} - b^{2} = (a + b) (a - b)

.
Thus,

(h + d - x - x - h) (h + d - x + x + h) = d^{2}

\Rightarrow (d - 2 x) (2 h + d) = d^{2}

\Rightarrow 2 h d + d^{2} - 4 x h - 2 xd = d^{2}

\Rightarrow 2 (h d - 2 x h - xd) = 0

\Rightarrow 2 h d = 4 h x + 2 xd

\Rightarrow 2 h d = 2 (2 h + d)

\Rightarrow x = \frac{h d}{2 h + d}

Therefore, the value of x in terms of h and d is

x = \frac{h d}{2 h + d}

.
Hence, option 1 is correct.

In the right angled triangle, $AB + AD = BC + CD$ . If $AB = x$ , $BC = h$ and $CD = d$ , then find x in terms of h and d.

Solution:

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