Geometry
The perimeter of kite is 172.11 cm
Explanation:
ΔABD and ΔBCD are isosceles triangles
Join the two points A and C and the let the point be O where this lines cuts BD
Given
CO = 2 × AO ( CO is height of ΔBCD, AO is height of ΔA
BD )
and
BD = 1.5 × CO
Area of Kite = Area of ΔBCD + Area of ΔABD
If the height of ΔABD = h then height of ΔBCD =2×h
DB = 1.5 × 2×h = 3×h
Area of ΔABD =
×
×
![2h](https://img.qammunity.org/2021/formulas/mathematics/high-school/h7cvngtsqzvgfu21yih5hdah2q85jply9z.png)
Area of ΔBCD =
×
×
![h](https://img.qammunity.org/2021/formulas/mathematics/middle-school/l8r1z2p4kda68gan99c9x827adhuu5hbu7.png)
Area of Kite = 1800 = Area of ΔABD + Area of ΔBCD
1800 =
×
+
×
=
×
![h^(2)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/vsrxgg6exfxievooo0n6149fyfmuqqkcj3.png)
⇒
![h^(2) = 400\\](https://img.qammunity.org/2021/formulas/mathematics/middle-school/3fqr39yyhmzvkmnu7o9zck63d5oci4rrwe.png)
⇒ h = 20
![AB^(2) = (20)^(2) + (30 )^(2)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/1g8zqc5umjbmop8m7a36jv47zmr45szppa.png)
AB =
cm
![BC^(2) = 40^(2) + 30^(2)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/sj5yadys5f3j5jaldgefzodnw824g8rhsg.png)
BC = 50 cm
So the perimeter of kite = AB + BC +CD +AD
⇒
![10√(13) + 50 + 50 + 10√(13)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/s344kcb3iav6qtjfmr8tnbnyigk0dko00d.png)
⇒100 + 20
![√(13)](https://img.qammunity.org/2021/formulas/mathematics/college/8v0laomqr8x2mvoubh414vl5rnw9efjmph.png)
⇒172.11 cm
Hence the perimeter of kite is 172.11 cm