Question

Find the area of an isosceles triangle if its base is 16 cm and one of its equal sides is 17 cm long​

Answer 1

hey mate here's answer

let's see  answer

___________

perimeter of triangle = 10+10+16 = 36

semi perimeter = 36/2=18

area = √s(s-a)(s-b)(s-c)

√18(18-10)(18-10)(18-16)

√18×8×8×2

√2×3×3×4×2×4×2×2

2×2×3×4

48cm^2

Answer 2

Answer:

60 cm²

Step-by-step explanation:

Complete base = 16 cm

One of the equilateral side = 17 cm

We know, area of triangle = $\frac{1}{2}$ × base × height ( $\frac{1}{2}$ × b ×h)

Now, We need to find the height of triangle.

We also know that altitude of a triangle bisects the base. The isosceles trianle forms two right angled triangles. So, we take one of the halves of the triangle.

So, right-angular triangle's Base = $\frac{base}{2}$ = $\frac{16}{2}$ = 8 cm.

Now, using Pythagoerus theorum: H² = P² + B²

⇒ (17)² = P² + (8)²

⇒ 289 = P² + 64

⇒ P² = 289 - 64

⇒ P = $\sqrt{225}$

⇒ P = 15 cm (or Height 'H' = 15 cm)

Now, putting the values in area of triangle:

⇒ $\frac{1}{2}$ × 8 × 15

⇒ 4 × 15

⇒ 60 cm²