Can someone answer this for me?

Work Shown:

Break the figure up as shown by the dashed lines. Think of cutting along those lines to form 3 separate figures. The trapezoid on the left has area

A = h*(b1+b2)/2

A = 5*(11+12)/2

A = 57.5

The rectangle in the middle has area

A = L*W

A = 12*15

A = 180

The trapezoid on the right has area

A = h*(b1+b2)/2

A = 10*(12+9)/2

A = 105

Overall, the entire figure has area of 57.5+180+105 = 342.5 square meters

1 can costs £6 and it covers 15 square meters. We need to cover 342.5 square meters

342.5/15 = 22.8333333333333 approximately

Round up to the nearest whole number to get 23

We need 23 cans to be able to paint 342.5 square meters or more

Note that 22 cans won't be enough because 22*15 = 330 is short of 342.5

While 23 cans is more than enough since 23*15 = 345. It's better to have leftovers than come up short.

Since each can costs £6 and we need 23 of them, the total cost would then come to 6*23 = 138 pounds

The figure consists of two trapeziums and a rectangle

Area of figure = Area of rectangle + Area of the two trapeziums

Area of rectangle = length × width

= 12m × 15m = 180m²

Area of first trapezium = 1/2(a+b) × height

= 1/2(12+11)×5

= 57.5m²

Area of second trapezium = 1/2(a+b) × height

= 1/2(9+12)×10

= 105m²

Area of figure = 180 + 57.5 + 105 = 342.5m²

If 15m² = 6£

342.5 = 342.5/15 ×6

= £ 137

Hope this helps.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

step-by-step explanation:

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