Question

In the right triangle, B is a point on AC such that AB + AD = BC + CD
a point on AC such that AB + AD = BC + CD. If AB = x, BC = h and CD = d, then find x
(in term of h and d).
In the right triangle. Bis a point on AC such that AB + AD = BC+ CD. IF AB = =, BC = h and CD = d, then find a
OR​

Answer 1

The value of x is $x =\frac{hd}{(2h + d)}$

Explanation:

Given that B is a point on AC in a right angled triangle such that $AB + AD = BC + CD$

Also, given that $AB=x$ , $BC=h$ and $CD=d$

We need to determine the value of x.

Let us substitute the values in $AB + AD = BC + CD$, we get,

$x + AD = h + d$

     $AD = h + d - x$

Now, we shall consider the right angled triangle ACD, we have,

$AD^2 = AC^2 + CD^2$

$AD^2 = (AB + BC)^2 + CD^2$

Substituting the values, we get,

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

Simplifying the terms, we have,

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

Subtracting the like terms, we get,

         $2 h d-2 x h-2 x d=2xh$

$2 h d-2 x h-2 x d-2xh=0$

          $2 h d-2 x d-4xh=0$

                  $-2 x d-4xh=-2 h d$

Taking out the common factor -2x, we get,

$-2 x( d+2h)=-2 h d$

Simplifying, we get,

$x( d+2h)= h d$

Dividing both sides by (d+2h), we get,

$x =\frac{hd}{(2h + d)}$

Therefore, the value of x is $x =\frac{hd}{(2h + d)}$

Learn more:

(1) In the right triangle, B is a point on AC such that AB + AD = BC + CD. If AB = x, BC = h and CD = d, then find x

(in term of h and d)

brainly.in/question/14466671

(2) In the right triangle, B is a point on AC such that AB+AD=BC+CD.If AB=x,BC=h and CD=d then find x(in terms of h and d)

brainly.in/question/14426177

Answer 2

The value of x is $x =\frac{hd}{(2h + d)}$

Explanation:

Given that B is a point on AC in a right angled triangle such that $AB + AD = BC + CD$

Also, given that $AB=x$ , $BC=h$ and $CD=d$

We need to determine the value of x.

Let us substitute the values in $AB + AD = BC + CD$, we get,

$x + AD = h + d$

     $AD = h + d - x$

Now, we shall consider the right angled triangle ACD, we have,

$AD^2 = AC^2 + CD^2$

$AD^2 = (AB + BC)^2 + CD^2$

Substituting the values, we get,

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

Simplifying the terms, we have,

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

Subtracting the like terms, we get,

         $2 h d-2 x h-2 x d=2xh$

$2 h d-2 x h-2 x d-2xh=0$

          $2 h d-2 x d-4xh=0$

                  $-2 x d-4xh=-2 h d$

Taking out the common factor -2x, we get,

$-2 x( d+2h)=-2 h d$

Simplifying, we get,

$x( d+2h)= h d$

Dividing both sides by (d+2h), we get,

$x =\frac{hd}{(2h + d)}$

Therefore, the value of x is $x =\frac{hd}{(2h + d)}$

Learn more:

(1) In the right triangle, B is a point on AC such that AB + AD = BC + CD. If AB = x, BC = h and CD = d, then find x

(in term of h and d)

brainly.in/question/14466671

(2) In the right triangle, B is a point on AC such that AB+AD=BC+CD.If AB=x,BC=h and CD=d then find x(in terms of h and d)

brainly.in/question/14426177