Question

the length (in metres) of the sides of a right angled triangle 2x-1,4x and 4x+1, x>0 find value of x and the area of triangle .​

Answer 1

Answer:

$\sf{The \ value \ of \ x \ is \ 6 \ m \ and \ the}$

$\sf{area \ of \ the \ triangle \ is \ 132 \ m^{2}}$

Given:

$\sf{Sides \ of \ right \ angled \ triangle \ ate}$

$\sf{2x-1, \ 4x \ and \ 4x+1.}$

To find:

$\sf{The \ value \ of \ x \ and \ the \ area \ of \ the}$

$\sf{triangle.}$

Solution:

$\sf{By \ Pythagoras \ theorem}$

$\sf{(4x+1)^{2}=(2x-1)^{2}+(4x)^{2}}$

$\sf{\therefore{16x^{2}+8x+1=(2x^{2}-4x+1)+16x^{2}}}$

$\sf{\therefore{8x+1=2x^{2}-4x+1}}$

$\sf{\therefore{2x^{2}-12x=0}}$

$\sf{\therefore{x(2x-12)=0}}$

$\sf{\therefore{x=0 \ or \ 2x-12=0}}$

$\sf{But, \ x \ is \ bigger \ than \ zero.}$

$\sf{\therefore{x\neq \ 0}}$

$\sf{\therefore{2x-12=0}}$

$\sf{\therefore{2x=12}}$

$\sf{\therefore{x=\dfrac{12}{2}}}$

$\boxed{\sf{\therefore{x=6}}}$

$\sf{Area \ of \ right \ angled \ triangle,}$

$\sf{Here, \ base=2x-1}$

$\sf{Base=2(6)-1}$

$\sf{\therefore{Base(b)=12-1=11 \ m}}$

$\sf{Height=4x}$

$\sf{\therefore{Height (h)=4(6)=24 \ m}}$

$\boxed{\sf{Area \ of \ triangle=\dfrac{1}{2}\times \ h\times \ b}}$

$\sf{\therefore{A(\triangle)=\dfrac{1}{2}\times24\times11}}$

$\sf{\therefore{A(\triangle)=12\times11}}$

$\sf{\therefore{A(\triangle)=132 \ m^{2}}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ x \ is \ 6 \ m \ and \ the}}}$

$\sf\purple{\tt{area \ of \ the \ triangle \ is \ 132 \ m^{2}}}$