Answer:
Perimeter of i - 22CM
Area of i - 13CM^2
Perimeter of ii -38CM
Area of ii -66CM^2
Perimeter of iii -30CM
Area of iii- 42CM^2
Perimeter of iv - 50CM
Area of iv- 126CM^2
Step-by-step explanation:
SHAPE I:
PERIMETER = S + S + S + S + S +S
= 7 + 1 + 5 +3 +4 +2
= 22CM
AREA = Part a - 4 * 2 = 8cm^2 part B - 5 *1 = 5cm^2
Total = 8 + 5 = 13cm^2
SHAPE II:
PERIMETER = S + S + S + S + S +S
= 4 + 4 +5 + 6 + 9 + 10
= 38 CM
AREA = Part a - 5 * 10 = 50 cm^2 part B - 4 *4 =16 cm^2
Total = 50+16 =66 cm^2
SHAPE III:
PERIMETER = S + S + S + S + S +S
= 9 + 2 + 3 + 4 + 6 + 6
= 30CM
AREA = Part a - 6 * 6 = 42cm^2 part B - 3 * 2= 6cm^2
Total = 36 + 6 =42 cm^2
SHAPE IV:
PERIMETER = S + S + S + S + S +S
= 9 + 10 + 4 + 6 + 6 + 15
= 50 CM
AREA = Part a -15 * 6 = 90 cm^2 part B - 9 *4 = 36cm^2
Total = 90 + 36 = 126cm^2
HOPE THIS HELPED
The answer is A. 150x + 50.
So to find the answer you need to use SOHCAHTOA
Sine= Opposite/ Adjacent
Cosine= Adjacent/Hypotenuse
Tan = opposite/adjacent
angle ADC is 60 so to find the length of line ABC we would use the angle and the line DC.
We would use Tan with opposite over adjacent
Tan 60 = opp. / 5 root 3
opp. = 15
so that is the length of the entire thing, all you need to do now is find the length of the single line of BC which we can use the same equation with just Tan 30 now.
Tan 30 = opp. / 5 root 3
opp. = 5
next you need to take the entire ABC length and subtract the length of BC to get AB
15-5 = 10
so the answer would be C

Let's solve ~


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Answer:
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Step-by-step explanation: