Question

4. The lengths of the sides of a right triangle are (2x - 1) m, (4x) m and (4x + 1) m, where x > 0. Find
(i) the value of x,
(ii) the area of the triangle.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given three sides of a right angled triangle as

First side = 2x - 1 m

Second side = 4x m

Third side = 4x + 1 m

Now,

4x + 1 > 4x > 2x - 1

So, it implies,

Longer side of a right angle triangle is 4x + 1.

Now,

We know that

Pythagoras Theorem,

$\begin{gathered}{\boxed{\sf{\pink{(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2}}}}\end{gathered}$

So, on substituting the values, we get

$\rm :\longmapsto\: {(4x + 1)}^{2} = {(4x)}^{2} + {(2x - 1)}^{2}$

$\rm :\longmapsto\: {16x}^{2} + 1 + 8x = {16x}^{2} + {4x}^{2} + 1 - 4x$

$\rm :\longmapsto\:8x = {4x}^{2} - 4x$

$\rm :\longmapsto\:{4x}^{2} - 4x - 8x = 0$

$\rm :\longmapsto\:{4x}^{2} -12x = 0$

$\rm :\longmapsto\:4x(x - 3) = 0$

$\bf\implies \:x = 3$

Hence, sides of triangle are

First side = 2x - 1 = 6 - 1 = 5 m

Second side = 4x = 12 m

Third side = 4x + 1 = 13 m

Now, we know area of triangle is given by

$\rm :\longmapsto\:Area_{\triangle} = \dfrac{1}{2} \times base × height$

$\rm \: = \: \dfrac{1}{2} \times 5 \times 12$

$\rm \: = \: 5 \times 6$

$\rm \: = \: 30 \: {m}^{2}$

Additional Information :-

$\purple{ \boxed{ \bf{Area_{rectangle} = length \times breadth}}}$

$\purple{ \boxed{ \bf{Area_{square} = {side}^{2}}}}$

$\purple{ \boxed{ \bf{Area_{square} = \dfrac{1}{2} \times {(diagonal)}^{2}}}}$

$\purple{ \boxed{ \bf{Area_{circle} = \pi \: {r}^{2} }}}$