Question

What is the area of the figure?a 28 ydb 40 ydc 52 ydd 64 yd

Answer 1

Answer:

b. 40 yd²

Step-by-step explanation:

We have the square and the right triangle.

The formula of an area of a square:

$A_{\boxed{}}=a^2$

a - length of side

The formula of an area of a right triangle:

$A_\triangle=\dfrac{ab}{2}$

a,b - legs

We have:

SQUARE:

a = 4 yd

$A_{\boxed{}}=4^2=16\ yd^2$

RIGHT TRIANGLE:

a = 6 yd, b = 4 yd + 4 yd = 8 yd

$A_\triangle=\dfrac{(6)(8)}{2}=(3)(8)=24\ yd^2$

The area of the figure:

$A=A_{\boxed{}}+A_\triangle\\\\A=16\ yd^2+24\ yd^2=40\ yd^2$

Answer 2

Answer:

b

Step-by-step explanation:

The figure is composed of a square and a right triangle

area of a square = s² ( s is the length of a side )

area of triangle = $\frac{1}{2}$ bh ( b is the base and h the height )

Total area = 4² + ( 0.5 × 6 × 8 ) = 16 + 24 = 40 yd² → b