Question

An altitude of a triangle is five thirds the length of its corresponding base. If the altitude was

increased by 4cm and the base is decreased by 2cm, the area of triangle remains same. Find the

base and altitude of the triangle

Answer 1

Answer:

The base of the triangle is 12 cm and the altitude is 20 cm.

Step-by-step explanation:

Let the length of the base be $x$. Since the altitude of the triangle is five thirds the length of the base, the measure of altitude is  $\dfrac{5}{3}x$

We know that the area of a triangle  is,

$Area =\dfrac{1}{2}\times base \times altitude$

So the area of the given triangle is,

$Area =\dfrac{1}{2}\times x \times \frac{5}{3}x =\frac{5}{6}x^2$

Also, if we decrease the base length by 2, the new measurement will be $x-2$. If altitude is increased by 4 cm, the new altitude will be $\frac{5}{3}x+4$

Hence the area of the new triangle will be,

$Area= \dfrac{1}{2} (x-2)(\frac{5}{3}x+4)= \dfrac{1}{2} (x-2)(\frac{5x+12}{3})\\\\\ =\frac{1}{6} (x-2)(5x+12)=\frac{1}{6}(5x^2+12x-10x-24)= \frac{1}{6}(5x^2+2x-24)$

Since the area of both the triangles is the same, we can equate the two expressions for the area. Hence we get,

$\dfrac{5}{6}x^2= \dfrac{1}{6}(5x^2+2x-24) \\\\\implies 5x^2=5x^2+2x-24\\\\\implies 2x-24=0\\\\\implies 2x=24\\\\\implies x= \dfrac{24}{2}=12$

Thus the altitude is,

 $\dfrac{5}{3}x=\dfrac{5}{3}\times 12=20$

So the base of the triangle is 12 cm and the altitude is 20 cm.