Question

The altitude of a triangle is two third the length of its corresponding base. If the altitude increased by 4 cm and the Base decreased by 2 cm the area of a triangle remains of the same .Find the Base and altitude of the traingle

Answer 1

let the base=xcm
altitude=2/3xcm
area of triangle=1/2×base×height
=1/2×x×2/3x
=1/2×2/3x2cm2(1)
area=1/2(x-2)(2x+12/3)_(2)
1÷2×2÷3x2=1÷2(x-2)(2x+12÷3)
2x2=1÷2(x-2)(2x+12÷3)×2×3
2x2=(x-2)(2x+12)
2x2=2x2+12x-4x-24
2x2=2x2+8x-24
2x2-2x2-8x=-24
-8x=-24
x=24÷8
=3
so,base=3cm
and,altitude=2/3x
=2/3×3
=2cm.
I think it will help you

Answer 2

Answer:

Base is 3 cm and altitude is 2 cm

Step-by-step explanation:

Let the base be x

We are given that The altitude of a triangle is two third the length of its corresponding base.

So, Altitude = $\frac{2}{3}x$

Area of triangle = $\frac{1}{2} \times Base \times Height$

Area of triangle = $\frac{1}{2} \times x \times \frac{2}{3}x$

Now we are given that the altitude increased by 4 cm and the Base decreased by 2 cm

Length of altitude = $\frac{2}{3}x+4$

Length of base = x-2

Area of triangle = $\frac{1}{2} \times Base \times Height$

Area of triangle = $\frac{1}{2} \times (x-2) \times (\frac{2}{3}x+4)$

Area of triangle = $\frac{1}{2} \times (x-2) \times (\frac{2}{3}x+4)$

Now we are given that  the area of a triangle remains of the same

So,  $\frac{1}{2} \times (x-2) \times (\frac{2}{3}x+4)= \frac{1}{2} \times x \times \frac{2}{3}x$

x=3

So, base = 3 cm

Altitude = $\frac{2}{3}x=\frac{2}{3} \times 3 = 2 cm$

Hence Base is 3 cm and altitude is 2 cm