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Area and perimeter of geometric functions

Area and perimeter of geometric functions-example-1
User Sharada
by
8.3k points

2 Answers

4 votes

Answer: OPTION C

Explanation:

Calculate the height AD with Pythagorean Theorem:


AD=\sqrt{20^(2)-16^(2)}=12

Calculate the base CD with Pythagorean Theorem:


CD=\sqrt{15^(2)-12^(2)}=9

The perimeter is the sum of all the sides, therefore the perimeter is:


P=9+15+16+20=60

User Wizard Of Ogz
by
9.2k points
4 votes

Answer:

C. 60

Explanation:

From triangle ADB,


|AD|^2+|BD|^2=|AB|^2


|AD|^2+16^2=20^2


|AD|^2=20^2-16^2


|AD|^2=400-256


|AD|^2=144


|AD|=√(144)


|AD|=12

From triangle ADC,


|CD|^2+12^2=15^2


|CD|^2+144=225


|CD|^2=225-144


|CD|^2=81


|CD|=9

The perimeter is the distance around the figure.

The perimeter


=15+20+16+9


=60

User Nicholas Humphrey
by
7.6k points

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