Question

. Kite QRST has a short diagonal of QS and a long diagonal of RT. The diagonals intersect at point P. Side QR = 17m, diagonal QS = 30m, and segment PT = 25m. Find the area of kite QRST.

8m2
33m2
495m2
990m2

Answer 1

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 =

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 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. Now plug values in equation. Plug 17 in RS place, 15 in PS place and RP is known so let that be x

now solve for x as shown

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

where p and q are the diagonals in kite

Here we have digonal RT= 33m and diagonal QS = 30m

So area =

so final answer for area of kite is 495

Explanation:

Answer 2

495 meter²

Explanation:

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 =
`(1)/(2)` (Diagonal QS) × (Diagonal RT)

In Δ QRP,

17² = 15² + RP²

`\sqrt{17^(2)-15^(2)} = RP`

`RP=√(289-225)`

= √64 = 8 meters.

So RT = RP + PT = 8 + 25 = 33 meter.

Now area of kite =
`(1)/(2)` (30) (33) = 495 meter²