Question

find the area of an isosceles triangle having base as x cm and one side 7cm... I Will mark as BRAINLIEST answer... wrong answers will be reported ​

Answer 1

Step-by-step explanation:

By using Pythagoras theorem,

Half the length of base is (x/2)cm.

Let the height of the triangle be 'h' cm.

So,

${h}^{2} + \frac{ {x}^{2} }{4} = 49 \\ {h}^{2} = 49 - \frac{ {x}^{2} }{4} \\ h = \frac{ \sqrt{196 - {x}^{2} } }{2}$

So, the area of triangle will be

$\frac{1}{2} hx \\ = \frac{x \sqrt{196 - {x}^{2} } }{4}$

This is the required area.

Thank you

Answer 2

Answer:

Area = (x/2) * √49 - x²/4 cm²

Step-by-step explanation:

Given, Base = x cm and Side = 7 cm.

In ΔADB,

AD² + BD² = AB²

⇒ AD² = AB² - BD²

⇒ AD² = (7)² - (x/2)²

⇒ AD² = (49 - x²/4)²

⇒ AD = √49 - x²/4

Now,

Area = (1/2) * Base * Height

⇒ (1/2) * BC * AD

⇒ (1/2) * x * √49 - x²/4

⇒ (x/2) * √49 - x²/4

Therefore,

Area = (x/2) * √49 - x²/4 cm²

Hope it helps!