Respuesta :
Kite has two pairs of adjacant sides equal. So side QR is either equal to QT or to RS. If side QR equals QT. So then QT = 17
Diagonals in kite cross each other at right angle. So in right trinagle QPT, QT is hypotenuse. Now if QT = 17, and we are given PT = 25. We cannot have hypotenuse smaller than other side. So QR cannot be equal to side QT. So QR has to be equal to other adjacent side RS. So we will have kite QRST as shown in figure.
So RS = 17. If side QR and RS are equal so diagonal RT will bisect(divide in half) diagonal QS then.
AS QS = 30 (given) so PS = PQ = [tex] \frac{30}{2} = 15 [/tex]
Now in right triangle RPS we have
RS = 17
PS = 15
We have to find other side RP.
For that we will use pythagorean formula
[tex] c^{2} = a^{2} + b^{2} [/tex] where c is the hypotenuse side and a and b are the other two sides of triangle
So in right triangle RPS we will have hypotenuse as RS (opposite ot right angle) so we will have
[tex] RS^{2} = PS^{2} + RP^{2} [/tex]
Now plug values in equation. Plug 17 in RS place, 15 in PS place and RP is known so let that be x
[tex] 17^{2} = x^{2} + 15^{2} [/tex]
[tex] 289 = x^{2} + 225 [/tex]
now solve for x as shown
[tex] 289 - 225 = x^{2} + 225 - 225 [/tex]
[tex] 64 = x^{2} [/tex]
[tex] \sqrt{64} = \sqrt{x^2} [/tex]
8 = x
So RP = 8
Also we are given PT = 25
So diagonal RT= RP + PT = 8 + 25 = 33m
Area of kite is given by formula
[tex] \frac{pq}{2} [/tex] where p and q are the diagonals in kite
Here we have digonal RT= 33m and diagonal QS = 30m
So area = [tex] \frac{33 \times 30}{2} = \frac{990}{2} =495 m^{2} [/tex]
so final answer for area of kite is 495 [tex] m^{2} [/tex]
Answer:
In the given kite QRST,
PQ = PS = 15 meter
QR = 17 meter
We have to find the area of the kite.
Since kite is in the form of a rhombus.
and rhombus is = (Diagonal QS) × (Diagonal RT)
In Δ QRP,
17² = 15² + RP²
Step-by-step explanation:
So RT = RP + PT = 8 + 25 = 33 meter.
Now area of kite = (30) (33) = 495 meter²
