Question

the length of the hypotenuse of a right triangle exceeds the length of the bese by 2 cm and exceeds twice the length of the altitude by 1 cm. find the length o each side of the triangle.

Answer 1

b+2=c
b=c-2
.
2a+1=c
2a=c-1
a=(c-1)/2
.
a^2+b^2=c^2
((c-1)/2)^2+(c-2)^2=c^2

(c^2-2c+1)/4 + c^2-4c+4=c^2
c^2-2c+1+4(c^2-4c+4)=4c^2
c^2-2c+1+4c^2-16c+16=4c^2
c^2-18c+17=0
(c-17)(c-1)=0
c=17
b=15
a=8

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