Question

Pythagoras ' Theorem - Solving Complex Problems​

Answer 1

see explanation

Explanation:

using Pythagoras' identity in right triangle MOP to find OP

OP² + MP² = MO²

OP² + 7² = 25²

OP² + 49 = 625 ( subtract 49 from both sides )

OP² = 576 ( take square root of both sides )

OP =
`√(576)` = 24

Using Pythagoras' identity in right triangle NOP

PN² + OP² = NO²

PN² + 576 = 30²

PN² + 576 = 900 ( subtract 576 from both sides )

PN² = 324 ( take square root of both sides )

PN =
`√(324)` = 18 cm

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Consider a vertical line from C to AB parallel to AD

then using Pythagoras' identity in the formed right triangle

BC² = AD² + (10 - 7)²

BC² = 4² + 3² = 16 + 9 = 25 ( take square root of both sides )

BC =
`√(25)` = 5

then perimeter (P) of ABCD is

P = AB + BC + CD + AD = 10 + 5 + 7 + 4 = 26 cm

The perimeter of the square is therefore 26 cm

with side s = 26 ÷ 4 = 6.5 cm

the area (A) of the square is calculated as

A = s² = 6.5² = 42.25 cm²

Answer 2

1.you don't need to calculate 25²,7²,30² and 16² because you have a formula a²-b²=(a+b)(a-b)

OP²=MO²-MP²

OP²=25²-7²

OP²=(25+7)(25-7)

=32(8)

=256

OP=√256

OP=16 cm

NP²=NO²-OP²

NP²=30²-16²

=(30+16)(30-16)

=46(14)

=144

NP=√144

=12 cm

2.draw a straight line from the angle C and name that line with CE

CE=AD and AE=CD (because AECD is a rechtangular)

So EB=10-7 =3 cm

BC²=EC²+EB²

=4²+3²

=16+9

=25

BC=√25

=5 cm

for area of the square? i don't understand this question...