ada yg bisa bantu in sayang

Mapel : Matematika
Bab : Bangun Datar

Layang-Layang

Titik tengah layang-layang saya beri nama O

AB = 20 cm
BC = 13 cm
BD = 24 cm

OB = 24 cm : 2 = 12 cm

OC = √BC² - OB²
OC = √13² - 12²
OC = √169 - 144
OC = √25
OC = 5 cm

OA = √AB² - OB²
OA = √20² - 12²
OA = √400 - 144
OA = √256
OA = 16 cm

Panjang diagonal yang lainnya
AC = OC + OA
AC = 5 cm + 16 cm
AC = 21 cm

Luas layang-layang
= 1/2 x d1 x d2
= 1/2 x BD x AC
= 1/2 x 24 cm x 21 cm
= 252 cm²

[tex]AB = 20 \: cm \\ BC = 13 \: cm \\ BD = 24 \: cm \\ BO = 12 \: cm \\ AO = \sqrt {{20^2} - {12^2} } \: = \sqrt{400 - 144} = \sqrt{256 } = 16 \: cm \\
CO =\sqrt {{13^2} - {12^2} } \: = \sqrt{400 - 144} = \sqrt{25 } = 5 \: cm \\ AC \: = AO + CO \: = 16 + 5 = 21 \\ luas \: = \frac{1}{2} \times AC \times BD = \frac{1}{2} \times 21 \times 24 = 252 \: {cm}^{2}
[/tex]