Question

in right triangle ACD,AB+AD is equal to BC+CD. if AB is equal to X,BC is equal to h and CD is equal to d then x is equal to​

Answer 1

The value of $x$ is $\frac{d^{2}+hd}{2h+d}$

Step-by-step explanation:

Given,

In $\triangle ACD$ is right-angle triangle where $AB+AD=BC+CD$,$AB=x,BC=h\ and\ CD=d$

From figure,

Lets take, $AD=y$

$x+y=h+d$

⇒$y=h+d-x$__1

And,

From Pythagoras Theorem,

$(x+h)^{2}=y^{2}+d^{2}$__2

Plug equation-1 value in equation-2,

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

⇒$x^{2}+h^{2}+2xh=h^{2}+d^{2}+x^{2}+2hd-2dx-2hx+d^{2}$

⇒$2xh=d^{2}+2hd-2dx-2hx+d^{2}$

⇒$4xh+2xd=2d^{2}+2hd$

⇒$2x(2h+d)=2(d^{2}+hd)$

⇒$x=\frac{2(d^{2}+hd)}{2(2h+d)}$

∴$x=\frac{d^{2}+hd}{2h+d}$

So, The value of $x$ is $\frac{d^{2}+hd}{2h+d}$

Answer 2

Answer:

hd

2h+d

Step-by-step explanation:

please refer the above attached photo for the answer. Remember, THIS TYPE OF QUESTIONS MAY CANE FOR EXAMS!!!

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