Question

the length of hypotenuse of right triangle exceeds the length of its base by 2cm and exceeds twice the length of altitude by 1 cm find the length of each side of triangle

Answer 1

let the hypotenuse of right angled triangle be x. Then the base of right angled triangle =(x-2).And it is given that the hypotenuse of triangle exceeds twice the length of altitude by 1cm => x= 2(length of altitude)+1=> (x-1)/2 =length of altitude. In a right angled triangle, Hypotenuse square= Altitude square+ base square =>
${x}^{2} = {(x - 2)}^{2} + { \frac{(x - 1)}{4} }^{2}$
multiply by 4, we get
$4 {x}^{2} = 4 {(x - 2)}^{2} + {(x - 1)}^{2}$
=>
$4 {x}^{2} = 4( {x}^{2} - 4x + 4) + {x}^{2} - 2x + 1$
=>
$4 {x}^{2} = 4 {x}^{2} - 16x + 16 + {x}^{2} - 2x + 1$
=>
${x}^{2} -18x + 17 = 0$
=>
${x}^{2} - 17x - x + 17 = 0$
=>
$x(x - 17) - 1(x - 17) = 0$
=>
$(x - 1)(x - 17) = 0$
=> x=1 or 17 but 1 is not 1 So length of hypotenuse is 17 cm , length of base is 15 cm and length of altitude is 8 cm .Hope it helps you...

Answer 2

Let base = x cm

Hypotenuse = (x+2)cm

Given,

Hypotenuse = 2 × altitude + 1 cm

$\longrightarrow$ $\sf{x+2=2×altitude+1cm}$

$\longrightarrow$ $\sf{altitude=\frac{1}{2}(x+1)cm}$

By Pythagoras theorem,

$\large\sf\purple{{hypotenuse}^{2} = {base}^{2} + {altitude}^{2}}$

$\longrightarrow$$\sf\orange{ ({x + 2)}^{2} = {x}^{2} + \frac{1}{4} {(x + 1)}^{2}}$

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

$\longrightarrow$$\sf\orange{{x}^{2} - 14x - 15 = 0}$

$\longrightarrow$$\sf\orange{(x - 15)(x + 1) = 0}$

$\longrightarrow$$\sf\orange{x = 15 \: or \: x = - 1}$

As length cannot be negative, x ≠ -1, so x = 15

Hence,

base = 15cm

Hypotenuse = 15+2 = 17cm

Altitude = $\sf{\frac{1}{2}(15+1)=8cm}$