Question

The altitude of a triangle is five- thirds the length of its corresponding base. If the altitude is increased by 4 cm and the base decreased by 2 cm, the area of the triangle would remain the same. Find the base and altitude of the triangle.​

Answer 1

Given :-

• The altitude of ∆ = 5/3 of its base
• The altitude + 4cm ,The base - 2 cm

To Find :-

• The base and altitude of ∆ = ?

Solution :-

To calculate the altitude and base of triangle at first we have to set up equation by applying formula. As given in the question that. The altitude of a triangle is five- thirds the length of its corresponding base. If the altitude is increased by 4 cm and the base decreased by 2 cm, the area of the triangle would remain the same. Let, assume the base of ∆ be x and altitude be 5x/3 .

Calculation begins :-

↠ Original area of ∆ = New area of ∆

↠ 1/2 × base × altitude = 1/2 × (base - 2) × (altitude + 4)

↠ 1/2 × X × 5x/3 = 1/2 × ( x - 2) × (5x/3 + 4)

↠ 5x²/3 = (x - 2) × (5x + 12/3)

↠ 5x²/3 = (x - 2)(5x + 12)/3

↠ 5x² = (x - 2)(5x + 12)

↠ 5x² = 5x²+ 2x - 24

↠ 5x² - 5x² = 2x - 24

↠ 2x = 0 + 24

↠ 2x = 24

↠ x = 12cm

Now calculate base and altitude :-

↠ The base of triangle (x) = 12 cm

↠ The altitude of triangle (5x/3) = 5(12)/3

↠ The altitude of triangle (5x/3) = 60/3

↠ The altitude of triangle (5x/3) = 20 cm

Hence,

• The base of triangle = 12 cm
• The altitude of triangle = 20 cm

Answer 2

Step-by-step explanation:

Given:-

The altitude of a triangle is five- thirds the length of its corresponding base. If the altitude is increased by 4 cm and the base decreased by 2 cm, the area of the triangle would remain the same.

To Find:-

The base and altitude of the triangle.

Solution:-

$\rm \: Let \: the \: lengt h \: of \: base \: of \: the \triangle \: b e \: x \: cm \\ \therefore \: \rm Altitude = \frac{5x}{3} \: cm \\ \rm \:Area \: of \triangle \: = \frac{1}{2} \bigg (base \times height \bigg) \\ \dashrightarrow \rm \: \frac{1}{2} \bigg(x \times \frac{5x}{3} \bigg) {cm}^{2}$

When the altitude is increased by 4 cm and the base is decreased by 2 cm, we have

$\rm \:New \: base = (x - 2)cm \\ \rm New \: altitude = \bigg( \frac{5x}{3} + 4 \bigg)cm \\ \rm \:Area \: of \: new \triangle \\ \looparrowright \: \frac{1}{2} \rm\bigg \{ \bigg( \frac{5x}{3} + 4 \bigg) \times (x - 2) \bigg \} \\ \rm \looparrowright \: \frac{1}{2} \bigg \{ \frac{5 {x}^{2} }{3} - \frac{10x}{3} + 4x - 8 \bigg \} \\ \rm \looparrowright \: \frac{5 {x}^{2} }{6} - \frac{5x}{3} + 2x - 4$

It is given that the area of the given triangle is same as the area of the new triangle. So,

$\therefore \: \rm \: \frac{5 {x}^{2} }{6} = \frac{5 {x}^{2} }{6} - \frac{5x}{3} + 2x - 4 \\ \rm \hookrightarrow \: \frac{5 {x}^{2} }{6} - \frac{5 {x}^{2} }{6} + \frac{5x}{3} - 2x = - 4 \\ \rm \hookrightarrow \: \frac{5x}{3} - 2x = - 4 \\ \rm \hookrightarrow \: 5x - 6x = - 12 \\ \rm \hookrightarrow \: - x = - 12 \\ \rm \hookrightarrow \: x = 12$

So,Base=12cm

Altitude=5x/3=5×12/3=20 cm