Question

Find the base of an isosceles triangle whose area is 4√15 sq. cm. and the length of one of the equal side is 8 cm.​

Answer 1

Answer:

Step-by-step explanation:Given: Area = 4√15 and length of one side = 8

Let b be the base and h be the height.

Let the base be 2b.

Area of the triangle = 1/2 × 2b × h

4√15 = b × h

h = 4√15/b

Now, use pythagoras theorem,

b² + h² = 8²

b² + (4√15/b)² = 8²

b⁴ - 64b² + 240 =0

b⁴ - 60b² - 4b² + 240 =0

(b² - 60)(b² - 4) = 0

(b - 2√15)(b + 2√15)(b - 2)(b + 2) = 0

Negative values should be neglected.

When b is 2√15 then h is 2.

when b is 2 then h is 2√15.

Hence, the base could be 4√15 then the height will be 2 and the base could be 4 then the height will be 2√15.

Answer 2

Answer: The base could be 4√15 then the height will be 2 and the base could be 4 then the height will be 2√15 when area of an isosceles triangle is 4√15 sq. cm. and the length of  equal side is 8 cm.

Step-by-step explanation:
It is given that,
Area of an isosceles triangle = 4√15cm²
length of one of the equal side = 8cm

Let 2b be the base of an isosceles triangle and h be the height of the same.

Area of the triangle = $\frac{1}{2}$ ×$b$ × $h$ =$\frac{1}{2}$×$2b$×$h$

Thus, it can be written as 4√15 = b × h

h = 4√15/b

Now, using the Pythagoras theorem,

⇒b² + h² = 8²

⇒b² + (4√15/b)² = 8²

⇒b⁴ - 64b² + 240 =0

⇒b⁴ - 60b² - 4b² + 240 =0

⇒(b² - 60)(b² - 4) = 0

⇒(b - 2√15)(b + 2√15)(b - 2)(b + 2) = 0

Negative values should be neglected it couldn't be taken

When b value is 2√15 then h is 2.

when b value is 2 then h is 2√15.

Hence, the base could be 4√15 then the height will be 2 and the base could be 4 then the height will be 2√15 when area of an isosceles triangle is 4√15 sq. cm. and the length of equal side is 8 cm.

To find similar questions, click here:

https://brainly.in/question/1141167

https://brainly.in/question/33774149
#SPJ2