Question

angles is 7-8 find the angle of the trangle
17. The altitude of a triangle is three-fifth of the length of the corresponding base. If the
altitude is decreased by 4 cm and the corresponding base is increased by 10 cm, the area
of the triangle remains the same. Find the base and the altitude of the triangle.​

Answer 1

Answer:

Altitude = 12 cm

Corresponding Base = 20 cm

Step-by-step explanation:

Given that,

The altitude of a ∆ is 3/5 of the length of the corresponding base.

Let the base be x.

Therefore, altitude = 3x/5

Therefore, we have,

Area = ½ × x × 3x/5

Now, it's given that,

Altitude is decreased by 4 cm

Therefore, we have,

New altitude = (3x/5-4) cm

And,

Base is increased by 10 cm

Therefore, we have,

New base = (x+10) cm

Therefore, we have,

New area = ½ × (x+10) × (3x/5 -4)

But, its given that,

Area remains equal.

Therefore, we will get,

$= > \frac{1}{2} \times x \times \frac{3x}{5} = \frac{1}{2} \times (x + 10) \times ( \frac{3x}{5} - 4) \\ \\ = > x \times \frac{3x}{5} = (x + 10)( \frac{3x}{5} - 4) \\ \\ = > \frac{3}{5} {x}^{2} = \frac{3}{5} {x}^{2} - 4x + 6x - 40 \\ \\ = > 2x - 40 = 0 \\ \\ = > 2x = 40 \\ \\ = > x = 20$

Therefore, we have,

=> Base = 20 cm

=> Altitude = 20× 3/5 = 4×3 = 12 cm

Hence, the altitude and corresponding base is of 12 cm and 20 cm length respectively.