Question

The length of the hypotenuse of a right-angled triangle exceeds the length of the base by 2 cm and exceeds twice the length of the altitude by 1 cm. Find the length of each side of the triangle.​

Answer 1

$\huge\underline\mathbb{SOLUTION:-}$

$\mathsf {Let,\:Base = x}$

$\mathsf {Altitude = y}$

$\mathsf {Hypotenuse = h}$

$\underline \mathsf{ATQ:-}$

$\mathsf {h = x + 2}$

$\mathsf {h = 2y + 1}$

$\implies \mathsf {x + 2 = 2y + 1}$

$\implies \mathsf {x + 2 - 1 = 2y}$

$\implies \mathsf {x - 1 = 2y}$

$\implies \mathsf {\frac{x - 1}{2} = y}$

$\mathsf {And\:x^2 + y^2 = h^2}$

$\implies \mathsf {x^2 + \big(\frac{x-1}{2}\big)^2 = (x + 2)^2}$

$\implies \mathsf {x^2 - 15x + x - 15 = 0}$

$\implies \mathsf {x^2 - 15x + x - 15 = 0}$

$\implies \mathsf {(x - 15) (x + 1) = 0}$

$\implies \mathsf {x = 15\:Or\:x = -1}$

$\underline\mathsf \blue {Base\:=\: 15\:Cm}$

$\underline\mathsf \blue {Altitude\:=\: \frac{x + 1}{2}\:=\:8\:Cm}$

$\underline\mathsf \blue {Hypotenuse \:=\:17\:Cm}$

Answer 2

Answer:

Let,Base=x

\mathsf {Altitude = y}Altitude=y

\mathsf {Hypotenuse = h}Hypotenuse=h

\underline \mathsf{ATQ:-}

ATQ:−

\mathsf {h = x + 2}h=x+2

\mathsf {h = 2y + 1}h=2y+1

\implies \mathsf {x + 2 = 2y + 1}⟹x+2=2y+1

\implies \mathsf {x + 2 - 1 = 2y}⟹x+2−1=2y

\implies \mathsf {x - 1 = 2y}⟹x−1=2y

\implies \mathsf {\frac{x - 1}{2} = y}⟹

2

x−1

=y

\mathsf {And\:x^2 + y^2 = h^2}Andx

2

+y

2

=h

2

\implies \mathsf {x^2 + \big(\frac{x-1}{2}\big)^2 = (x + 2)^2}⟹x

2

+(

2

x−1

)

2

=(x+2)

2

\implies \mathsf {x^2 - 15x + x - 15 = 0}⟹x

2

−15x+x−15=0

\implies \mathsf {x^2 - 15x + x - 15 = 0}⟹x

2

−15x+x−15=0

\implies \mathsf {(x - 15) (x + 1) = 0}⟹(x−15)(x+1)=0

\implies \mathsf {x = 15\:Or\:x = -1}⟹x=15Orx=−1

\underline\mathsf \blue {Base\:=\: 15\:Cm}

Base=15Cm

\underline\mathsf \blue {Altitude\:=\: \frac{x + 1}{2}\:=\:8\:Cm}

Altitude=

2

x+1

=8Cm

\underline\mathsf \blue {Hypotenuse \:=\:17\:Cm}

Hypotenuse=17Cm

Step-by-step explanation:

hope it will help you. ...............