Question

the altitude of a traingle is two third the length of its corresponding base . if the altitude increased by 4 cm and the base decreased by 2cm , the area of the traingle remain the same find the base and altitude of the traingle.​

Answer 1

Let x be the base of the triangle.

The altitude of a traingle is two third the length of its corresponding base.

→ Altitude (height) = $\sf{\frac{2M}{3}}$

Now,

Area of triangle = 1/2 × base × height

=> $\sf{\frac{1}{2}\:\times\:M\:\times\:\frac{2M}{3}}$

=> $\sf{\frac{M^2}{3}}$

Now, if the altitude increased by 4 cm and the base decreased by 2cm, the area of the traingle remain the same.

New altitude = $\sf{\frac{2M}{3}\:+\:4}$

New base = $\sf{M\:-\:2}$

According to question,

=> $\sf{\frac{M^2}{3}\:=\:\frac{1}{2}\:\times\:(M\:-\:2)\:(\frac{2M}{3}\:+\:4)}$

=> $\sf{\frac{M^2}{3}\:=\:\frac{M\:-\:2}{2}\:\times\:(\frac{2M\:+\:12}{3})}$

=> $\sf{2M^2\:=\:M\:-\:2\:(2M\:+\:12)}$

=> $\sf{2M^2\:=\:2M^2\:+\:12M\:-\:4M\:-\:24}$

=> $\sf{0\:=\:+8M\:-\:24}$

=> $\sf{8M\:=\:24}$

=> $\sf{M\:=\:3}$

So,

Altitude = $\sf{\frac{2(3)}{3}}$

=> $\sf{2}$

•°• Base of triangle = 3 cm and altitude = 2 cm

Answer 2

Given that :-

• Altitude is 2/3 of base of triangle.
• If base is deceased by 2 cm and altitude is increased by 4 cm then area remains same .

Solution :-

As we know that area of a triangle is

$\huge\frac{1}{2}$×base×height .

Let base of triangle = b cm

Altitude (height) = $\frac{2}{3}$b cm

Area before increase or decrease in height and base :-

$= > \frac{1}{2} \times b \times \frac{2}{3} b$

$area \: = > \frac{ {b}^{2} }{3} {cm}^{2}$

Now after given changes height and base will become :-

Base = b -2 cm

Height = $\frac{2}{3}$b + 4 cm

$area \: = \frac{1}{2} ( \frac{2}{3} b + 4)(b - 2)$

$= > \frac{1}{2} \times 2( \frac{1}{3} b + 2)(b - 2)$

$= > \frac{ {b}^{2} }{3} + 2b - \frac{2b}{3} - 4$

As given that area will remain same,So

$\frac{ {b}^{2} }{3} = \frac{ {b}^{2} }{3} + 2b - \frac{2b}{3} - 4$

$= > 4 = 2b - \frac{2b}{3}$

$= > 4 = \frac{6b - 2b}{3}$

$= > 4 = \frac{4b}{3}$

12 = 4b

base. = 3 cm

$\frac{2b}{3} = altitude$

Altitude = 2 cm