Question

The hypotenuse of a right triangle exceeds the base by 2 and exceed twice the length of the altitude by 1,find the dimensions of the triangle

Answer 1

Answer:h=17, p=8 & b=15

Step-by-step explanation:

Here, h=b+2 and h=2p+1

Therefore b=(h-2) & p=(h-1)/2

Now for right triangle,

h^2=p^2+b^2

Or, h^2={(h-1)/2}^2+(h-2)^2

=(h^2-2h+1)/4+(h^2-4h+4)

={h^2-2h+1+4h^2-16h+16}/4

=(5h^2-18h+17)/4

4h^2 = 5h^2-18h+17

Or,h^2-18h+17=0

Or, h^2-17h-h+17=0

Or, h(h-17)-1(h-17)=0

Or, (h-17) (h-1)=0

Therefore h=17, 1

When h=17, b=15 and p=8

When h=1, b=-1 ,& p=0

Therefore h=1 is not possible because length cannot be zero or negative

Hence h=17 is required dimension and accordingly p=8 and b=15.

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