Question

AN ALTITUDE OF A TRIANGLE IS 5/3 THE LENGTH OF ITS CORRESPONDING BASE. IF THE ALTITUDE IS INCREASED BY 4CM AND THE BASE DECREASEDBY2CM,THE AREA OF THE TRIANGLE REMAIN THE SAME .FIND THE BASE AND THE ALTITUDE OF THE TRIANGLE

Answer 1

AN ALTITUDE OF A TRIANGLE IS 5/3 THE LENGTH OF ITS CORRESPONDING BASE. IF

THE ALTITUDE IS INCREASED BY 4CM AND THE BASE DECREASED BY 2CM, THE AREA OF THE

TRIANGLE REMAIN THE SAME. FIND THE BASE AND THE ALTITUDE OF THE TRIANGLE.

**Solution:**

Let an altitude will be $h$ and it's corresponding base will be $a$. According to condition $h = (5/3)a$, and the area of triangle will be $S = (1/2)^* (5/3)a^*a$. If the altitude is increased by 4 cm and the base decreased by 2 cm, the area will be $S = (1/2)^* ((5/3)a + 4)^* (a - 2)$.

According to condition, the area of triangle remain the same, so let's make the equation

$$(1/2)^* (5/3)a^*a = (1/2)^* ((5/3)a + 4)^* (a - 2)$$ $$(5/3)a^*a = ((5/3)a + 4)^* (a - 2)$$ $$12a^2 = 34$$ $$a^2 = \frac{17}{6} \Rightarrow a = \sqrt{17/6} \Rightarrow h = 5/3\sqrt{17/6}$$

Answer: $a = \sqrt{17/6}, h = 5/3\sqrt{17/6}$