<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>
Answer:
Just put the values of x and y
x = 1
y = -2
2x – y = p
2(1) - (- 2) = p
p = 4
Answer:
32.08488741
Step-by-step explanation:
Answer:
1. The correct options are 2 and 4.
2. The measure of angle 6 is 60°.
Step-by-step explanation:
1.
It is given that a∥b , and c is not parallel to a or b. So, we can not establish any relationship among the angles on the line a,b and angles on the line c.
If a traversal line intersect two parallel line, then corresponding angles, alternate exterior angle and alternate interior angles are equal.


Therefore correct options are 2 and 4.
2.
It is given that AB∥CD and m∠3=120°.
Statements Reasons
m∥nm∠1=120° Given
∠5≅∠1 Corresponding angles are equal
m∠5=m∠1 Angle Congruence Postulate
m∠5=120° Substitution Property of Equality
m∠6+m∠5=180° Supplementary angle property
m∠6+120°=180° Substitution Property of Equality
m∠6=60° Subtraction Property of Equality
Therefore the measure of angle 6 is 60°.
Answer:
Step-by-step explanation:
First you need to find the area then divide that by 14