how to find area of an isosceles triangle?

If you know all three sides of a triangle, you can use Heron's formula

A^2 = s * (s - a) * (s - b) * (s - c)

s = (a + b + c) / 2
a , b , c are the sides of the triangle

s = (10 + 10 + 16) / 2 = 36/2 = 18

A^2 = 18 * (18 - 10) * (18 - 10) * (18 - 16)
A^2 = 18 * 8 * 8 * 2
A^2 = 36 * 8^2
A^2 = 6^2 * 8^2
A = 6 * 8
A = 48

22.
Method 1: Use Heron's formula to find the area of the triangle.
s = (10 + 10 + 16)/2 = 18

Area of the isosceles ΔABC
= √[18 × (18 - 10) × (18 - 10) × (18 - 16)] ft²
= 48 ft²

The answer: B) 48


Method 2: Use Pythagorean theorem
Draw the altitude BD from B to AC. BD ⊥ and bisect AC.

In ΔABD:
BD² + AD² = AB²
BD² + (16/2)² = (10)²
BD² = √(100 - 64)
BD = 6

Area of ΔABC
= (1/2) × 16 × 6 ft²
= 48 ft²

The answer: B) 48

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23.
Volume of the solid
= Weight / Density
= (360 lb) / (3 lb/ft³)
= 120 ft³

Width of the solid
= Volume / (Area of ΔABC)
= 120/48 ft
= 2.5 ft

Height using 1/2 triangle to make right triangle for Pythagorean is:
√(10² - 8²) = √100-64 = √36, so 6 ft high
2 * ½ (b*h) = 48 ft² area of triangle
360 lb (1ft³/ 3lb) /48 ft² = 2.5 ft wide

Area of ΔABC
= (1/2) × 16 × 6 ft^2
= 48 ft^2
Answer choice:
B) 48

Using Pythagoras' theorem the height of the isosceles triangle is 6 ft and its area is 0.5*16*6 = 48 square ft