I need help with this problem please.

Mathematics

2 answers:

erastovalidia [21]3 years ago

6 0

Answer:

1725 sq.ft

Step-by-step explanation:

Given

Parallel sides of the trapeziod yard = 50ft , 65ft

Height of the trapezoid = 30ft

Therefore, area of the trapezoid yard

$= \frac{b1 + b2}{2} \times h$

$= \frac{50ft + 65ft}{2} \times 30ft$

$= \frac{115ft}{2} \times 30ft$

= 57.5 ft × 30 ft

$= 1725 {ft}^{2} (ans)$

Vera_Pavlovna [14]3 years ago

5 0

Answer:

1725ft^2

Step-by-step explanation:

Area of trapezoid:

$\frac{b_1+b_2}{2}*h$

Substitute the values and solve:

$\frac{50+65}{2}*30=1725$ ft^2

You might be interested in

a rectangular section of land has a length of 5•10^3 meters and a width of 6•10^2 meters. what is the area of land in square met

monitta

A = lw
A = (5 × 10^3)(6 × 10^2)
= 5000 × 600
= 300000
= 3 × 10^6 m^2

4 0

3 years ago

Can someone help me with this?

Marianna [84]

Look at the rectangle. The bottom side measures 8 miles, so the top side also measures 8 miles.

Now look at the upper right triangle. For the angle of 32 deg, the top side of the triangle is the adjacent leg of the triangle. The air distance, marked x, is the hypotenuse of the triangle.

The trig ratio that relates the adjacent leg and the hypotenuse is the cosine.

$\cos \theta = \dfrac{adj}{hyp}$

$\cos 32^\circ = \dfrac{8~miles}{x}$

$x = \dfrac{8~miles}{\cos 32^\circ}$

$x = 9.4 ~miles$