(20點) F.2 Maths Question (Pythagors' Theorem)

3.
the length of the sqaure ABCD is √(576)
the length of the sqaure DEFG is √(324)
by Pythagors' Theorem
AG²
=AD²+DG²
=576+324
=900
so AG=√(900)=30

4.
by Pythagors' Theorem
EC²
=ED²+CD²
=3²2+4²
=5²
so EC=5
Area of ABCE = 5²= 25
Area of ECD = 3x4 = 12
Area of ABCDE = 25+12=37

5.
by Pythagors' Theorem
RP²+PQ²=RQ²=14²
RP²+9²=14²
RP²=115
AREA OF PQR
=RPxPQ
=√(115)*9
=96.5

6.
by Pythagors' Theorem
CB²=DA²+(12-6)²
100=DA²+36
DA²=64
DA=h=8

the length of the sqaure ABCD is √(576)
the length of the sqaure DEFG is √(324)
by Pythagors' Theorem
AG²
=AD²+DG²
=576+324
=900
so AG=√(900)=30

EC²
=ED²+CD²
=3²2+4²
=5²
so EC=5
Area of ABCE = 5²= 25
Area of ECD = 3x4 = 12
Area of ABCDE = 25+12=37

RP²+PQ²=RQ²=14²
RP²+9²=14²
RP²=115
AREA OF PQR
=RPxPQ
=√(115)*9
=96.5

CB²=DA²+(12-6)²
100=DA²+36
DA²=64
DA=h=8

area ABCD = 567
=> AD = square root of 576 = 24

area DEFG = 324
=> DG = square root of 324 = 18

therefore,
square of AG = square of AD + square of DG = 576 + 324 = 900
AG = square root of 900 = 30