Question

(11) The base of a triangle is two-fifths its
altitude. If the base is increased by
2 cm and the altitude by 1.5 cm, the area
increases by 34 cm. Find the base and
altitude of the triangle.​

Answer 1

Answer:

the correct answer is Altitude=15cm and Base=34cm

Answer 2

The base of a triangle (b) = 10 cm and

The altitude of a triangle (h) = 25 cm

Step-by-step explanation:

Let the base of a triangle (b) = $\dfrac{2}{5} x$ and

The altitude of a triangle (h) = x

To find, the base and  altitude of the triangle = ?

We know that,

The area of the triangle = $\dfrac{1}{2}$ × Base(b) × Height(h)

The area of triangle = $\dfrac{1}{2}$ × $\dfrac{2}{5} x$  × x = $\dfrac{x^2}{5}$               ........... (1)

The base of a new triangle (b) = $(\dfrac{2}{5} x+2)cm$ and

The altitude of a new triangle (h) = (x + 1.5 cm)

The area of the new triangle = $\dfrac{1}{2}$ × $(\dfrac{2}{5} x+2)cm$  × (x + 1.5 cm)  

= $\dfrac{1}{2} \times (\dfrac{2x+10}{5 })(x+1.5)$

= $\dfrac{1}{10} \times(2x^2+3x+10x+15)$

= $\dfrac{1}{10} \times(2x^2+13x+15)$                             ................. (2)

According to question,

$\dfrac{x^2}{5}+34$ = $\dfrac{1}{10} \times(2x^2+13x+15)$    

⇒ $\dfrac{x^2+170}{5}$  = $\dfrac{1}{10} \times(2x^2+13x+15)$  

⇒  $2x^2+340=2x^2+13x+15$

⇒ $340=13x+15$

⇒ 13x = 340 - 15 = 325

⇒ x = $\dfrac{325}{13} =25$

∴ The base of a triangle (b) = $\dfrac{2}{5} \times 25$ = 10 cm and

The altitude of a triangle (h) = 25 cm