<em>Question:</em>
The area of the kite is 48 cm². What are the lengths of the diagonals PR and QS?
________
<em>Solution:</em>
You can split the kite into two isosceles triangles: PSR and PQR.
Assume that both diagonals intersect each other at the point O.
• Area of the triangle PSR:
m(PR) · m(OS)
A₁ = ————————
2
(x + x) · x
A₁ = ——————
2
2x · x
A₁ = ————
2
A₁ = x² (i)
• Area of the triangle PQR:
m(PR) · m(PQ)
A₂ = ————————
2
(x + x) · 2x
A₂ = ——————
2
2x · 2x
A₂ = ————
2
4x²
A₂ = ———
2
A₂ = 2x² (ii)
So the total area of the kite is
A = A₁ + A₂ = 48
Then,
x² + 2x² = 48
3x² = 48
48
x² = ———
3
x² = 16
x = √16
x = 4 cm
• Length of the diagonal PR:
m(PR) = x + x
m(PR) = 2x
m(PR) = 2 · 4
m(PR) = 8 cm
<span>• </span>Length of the diagonal SQ:
m(SQ) = x + 2x
m(SQ) = 3x
m(SQ) = 3 · 4
m(SQ) = 12 cm
I hope this helps. =)
Tags: <em>polygon area triangle plane geometry</em>
A.) 7^4x = 10
log base 10 (7^4X) = log base 10 (10)
4x log base 10 (7) = 1
4x (0.8451) = 1
3.3804x = 1
x = 0.2958
b.) ln(2) + ln(4x-1) = 5
ln (2 * 4x-1) = 5
ln (8x-2) = 5
log base (3) (8x-2) = 5
e^5 = 8x-2
e^5+2 = 8x
x = 18.8016
The team go skirt pop pop pop pop pop pop...
Answer:
-7
Step-by-step explanation:
it is possible because x is just a letter representing the unknown nos
Maybe it's no. 1 bcz WX=YZ.
