Question

in a rhombus pqrs, if pq=3x-7 and qr=x+3,find ps

Answer 1

We know that in rhombus all sides are equal
So,
PQ=QR
3x-7=x+3
2x=10
x=5
putting value of x in QR=x+3
5+3=8
So,PQ=PS=QR=RS=8

Answer 2

Answer:

PS = 8

Step-by-step explanation:

We know that

In rhombus all sides are equal.

So,

PQ = QR

$3x - 7 = x + 3$

$3x - x = 3 + 7$

$2x = 10$

$x = \frac{10}{2}$

$x = 5$

$Put\ the\ value\ of\ x\ in\ PQ = 3x -7$

$3\times5\ -\ 7$

= 15 - 7

= 8

As we know that

PQ = QR = RS = PS

Hence PS will be 8.