$\begin{array}{l} 8,24,26 \ \sin  ce, \ { \left( 8 \right) ^{ 2 } }+{ \left( { 24 } \right) ^{ 2 } }\ne { \left( { 26 } \right) ^{ 2 } } \end{array}$

Atp,

Given:- $ABC$ is a right angled triangle in which $AB = 5 \; cm$ and 

Given, $\angle ABC = 90$, M is a point of BC
In $\triangle ABM$, 
$AB^2 + BM^2 = AM^2$ (Pythagoras theorem)
$AB^2 = AM^2 - BM^2$ (I)

In $\triangle ABC$,
$AB^2 + BC^2 = AC^2$ (Pythagoras Theorem)
$AB^2 = AC^2 - BC^2$ (II)

Equating I and II,
$AM^2 - BM^2 = AC^2 - BC^2$
thus, $AM^2 + BC^2 = AC^2 + BM^2$

Consider the wire and pole system as a right angled triangle such that length of pole is the perpendicular, length of wire the hypotenuse and distance between the base of pole and the stack is the base, such that, H = 24 m, P = 18 m
Now, from Pythagoras theorem,
$H^2 = P^2 +B^2$
$24^2 = 18^2 + B^2$
$576 = 324 + B^2$
$B^2 = 252$
$B = 15.87$ m
Thus, distance between the base of the pole and the stack is 15.87 m

Let the altitude$=x:cm$
$\therefore$ The Base$=(x+4):cm$
and the Hypotenuse$=(x+8):cm$

Let the sides be $a=12$ and $b=18$

In $\triangle BAP$,
$BP^2 = AB^2+ AP^2$ (Pythagoras Theorem)

Since, Q is the mid-point of AB, then $AB = 2 AQ$
Hence, $BP^2 = (2 AQ)^2 + AP^2$
$BP^2 = 4 AQ^2 + AP^2$...(I)

In $triangle QAC$,
$QC^2 = AQ^2 + AC^2$ (Pythagoras theorem)
Since, P is the mid point of AC, $AC = 2 AP$
Hence, $QC^2 = AQ^2 + (2 AP)^2$ 
$QC^2 = AQ^2 + 4 AP^2$...(II)

By Adding I and II
$BP^2 + QC^2 = 4 AQ^2 + AP^2 + AQ^2 + 4 AP^2$
$BP^2 + QC^2 = 5 (AP^2 +AQ^2)$

In $\triangle QAP$,
$AQ^2 + AP^2 = PQ^2$
Hence, $BP^2 + QC^2 = 5 PQ^2$

In quadrilateral $ABCD$, 

Let the length of the shortest side be $x$ meters. Then,
Hypotenuse = $(2x +1)$ meters, Third side = $(x + 7)$ meters
(Hyppotenuse)$^2$ = sum of the square of the remaining two sides    ....[By pythagorous theroem]
$\Rightarrow (2x + 1)^2 = x^2 + (x + 7)^2$
$\Rightarrow 4x^2 + 4x + 1 = 2x^2 + 14x + 49$
$\Rightarrow 2x^2 - 10x - 48 = 0$
$\Rightarrow x^2 - 5x - 24 = 0$
$\Rightarrow x^2 - 8x + 3x - 24 = 0$
$\Rightarrow x(x - 8) + 3 (x - 8) = 0$
$\Rightarrow (x - 8) (x + 3) = 0$
$\Rightarrow x = 8, -3$
$\Rightarrow x = 8$             [$\because x = -3$ is not possible]
Hence, the lengths of the sides of the grassy land are $8$ m., $17$ m. and $15$ m.

Given: Hypotenuse of right triangle $= 25$ cm
Let the other sides be $x$ and $x + 5$
Thus, By Pythagoras Theorem:
$25^2 = x^2 + (x + 5)^2$
$625 = x^2 + x^2 + 25 + 10x$
$2x^2 + 10x - 600 = 0 $
$x^2 + 5x - 300 = 0 $
$x^2 + 20x - 15x - 300 = 0$
$(x + 20) (x - 15) = 0 $
$x = 15, -20$
Thus, the other two sides are $15$ cm and $20$ cm.

Let one side be $xcm$. Then the other side will be $(x+5)cm$. Therefore, from Pythagoras theorem
${x}^{2}+{(x+5)}^{2}={25}^{2}$
$\Rightarrow { x }^{ 2 }+{ x }^{ 2 }+10x+25=625$
$\Rightarrow { x }^{ 2 }+5x-300=0\quad \quad \Rightarrow { x }^{ 2 }+20x-15x-300=0$
$\Rightarrow x(x+20)-15(x+20)=0$
$\Rightarrow (x-15)(x+20)=0\quad \Rightarrow \quad x=15\quad or\quad x=-20$
Rejecting $x=-20$, we have length of one side $=15cm$ and that of the other side $=(15+5)cm=20cm$

We can observe,considering general numbers for right angled triangle,
$5^2 = 3^2 + 4^2$ (5, 3 are odd and 4 lies between 5, 3)
and  $13^2 = 5^2 + 12^2$ (5, 13 are odd and 12 lies between 5, 13)
By trial and error
$61^2 = 60^2 + 11^2$
Since (61, 11 are odd and 61 hypotenuse)
$61 -11 = 50$
Hence, option 'B' is correct.

Consider the two trees as perpendiculars on the ground forming a right triangle, where the distance between their bases is the base $(B)$, the distance between their tops is the Hypotenuse $(H)$ and difference between their lengths is the perpendicular $(P)$.
Hypotenuse, $H = 17$ m
Perpendicular, $P=$ distance between their lengths $= 28 - 20 = 8$ m
Thus, By Pythagoras Theorem,
$H^2 = P^2 + B^2$
$17^2 = 8^2 + B^2$
$B^2 = 289 - 64$
$B = 15m$.

$0.35 \ m = 0.35 \times 100 \ cm = 35 \ cm.$
We have,
$(hypotenuse)^2\, =\, (side)^2\, +\, (side)^2$
$=\, (12)^2\, +\, (35)^2$
$= 144 + 1225$
$= 1369$
Hypotenuse $=\sqrt{1369} = 37 \ cm.$

$In\triangle ABC $

As the sides are of a right angled triangle, we have
$ {Hypotenuse}^{2} = {Side1}^{2} + {Side2}^{2} $ where hypotenuse is the largest side.
$ {(3x+4)}^{2}  = {(3x+3)}^{2} + {x}^{2} $
$ => 9{x}^{2} + 16 + 24x = 9{x}^{2} + 9 + 18x + {x}^{2}  $
$ => {x}^{2} - 6x - 7 = 0 $
$ => {x}^{2} - 7x + x - 7 = 0 $
$ => x(x-7) + (x-7) = 0 $
$ => (x-7)(x+1) = 0 $
$ => x = 7, -1 $
As the side cannot be negative, $ x = 7 $.

The field is in the shape of a right angled triangle.

As we know $2m,$ $m^{2}+1, m^{2}-1$ form a Pythagorean triplet for any number $m > 1$.
Let us assume $2m = 10$
$\Rightarrow m = 5$
Therefore, $m^{2}+1$ $=$ $5^{2}+1$
$\Rightarrow 25 + 1 = 26$
$m^{2}-1$ = $5^{2}-1$
and $\Rightarrow 25 - 1 = 24$
Hence, the other two members are $24$ and $26$.

As we know $2m,$ $m^{2}+1, m^{2}-1$ form a Pythagorean triplet for any number $m > 1$.
Let us assume $2m = 8$
$\Rightarrow m = 4$
Therefore, $m^{2}+1$ $=$ $4^{2}+1$
$\Rightarrow 16 + 1 = 17$
and $m^{2}-1$ $=$ $4^{2}-1$
$\Rightarrow 16 - 1 = 15$
Hence, the other two members are $15$ and $17$.

As we know $2m$, $m^{2}+1, m^{2}-1$ form a Pythagorean triplet for any number $m > 1$.

For a set of numbers to be sides of a right angled triangle, the square of the longest side must equal the sum of the squares of the other two sides.

The legs of a right angled triangle are in the ratio $3:1$ and they are whole numbers.

It will form a right angled triangle where one side is $12$ miles and the other sides is $9$ miles. 

While travelling to $10$ miles East and then $24$ miles South the boat is moving on the path to form a right angled triangle.
The distance $c$ then would be the hypotenuses of the triangle, which can be calculated using Pythagorean Theorem which is ${ a }^{ 2 }{ +b }^{ 2 }={ c }^{ 2 }$.    
As we have the measurements of the two sides making the right angle: 
$\Rightarrow { 10 }^{ 2 }{ +24 }^{ 2 }={ c }^{ 2 }$
$\Rightarrow 100 + 576 =$ ${ c }^{ 2 }$
$\Rightarrow { c }^{ 2 }$ $= 676$
$\Rightarrow { c }$ $= 26$ miles
Hence, the option C is the right answer.
Ans: C

The resultant distance she traveled towards south is equal to $30-10 = 20$ miles

In $\triangle ABC$ and $\triangle RST$,

hypt. $AC=$ hypt. $RT$

$\therefore AC^2 = RT^2$

$\therefore { AB }^{ 2 }+{ BC }^{ 2 }={ RS }^{ 2 }+{ ST }^{ 2 }$.

Let $a=25 \ m, \ b=39 \ m, \ c=56 \ m$

Given, $\angle ABC=90^{\circ},$

Suppose a pilgrim starts from point $A$.

$\angle PRQ = 90^{\circ}\Rightarrow PR = \sqrt {(PQ)^{2} - (RQ)^{2}}$

$BP=PQ=QC=x(let)$

Let $\triangle ABC$ be right angled at B.

Then, by Pythagoras theorem,

$AC^2 = AB^2+BC^2$

But $AC^2 = 2 \times AB . BC$     .....Given

Hence, $AB^2 + BC^2 = 2 \times AB. BC$

$AB^2 + BC^2 - 2 \times AB. BC = 0$

$(AB - BC)^2 = 0$

So, $AB = BC$

Therefore, $\triangle ABC$ is right isosceles triangle.

So, $\angle A = \angle C = 45^o$.

Let $\theta =\csc ^{ -1 }{ \left( \dfrac { x+4 }{ 5 }  \right)  }$ implies that $\csc { (\theta )=\dfrac { x+4 }{ 5 }  }$ and therefore, $\sin { (\theta ) } =\dfrac { 5 }{ x+4 }$.

The altitude to the base of an isosceles triangle also bisects the vertex angle, so $m\angle ADE=30$.