Problem 1: ConstopicIder a quadrilateral ABCD. It is given that the perimeter of quadrilateral ABCD is 40 m. The stopicIde AB = x + 5 m, the stopicIde BC = x + 4m, and the stopicIde CD = 2x – 3m, and the stopicIde DA = 4x – 8 m. You will have to find the quadrilateral’s shortest stopicIde length. 

Solution: 

Now, the Perimeter of a quadrilateral = Sum of all stopicIdes.

Therefore, AB+BC+CD+DA = 40m

=> (x+5)+(x+4)+(2x-3)+(4x-8) = 40m

=>  8x+5+4-3-8=40

=>  8x-2=40

=>  8x=40+2

=>  8x=42

=>  x=42/8=5.25m

stopicIde AB = x + 5 m, the stopicIde BC = x + 4m, and the stopicIde CD = 2x – 3m, and the stopicIde DA = 4x – 8m.

That gives, AB = 5.25 + 5 = 10.25 m, 

BC = 5.25 + 4 = 9.25 m, 

CD = 2(5.25)-3 = 10.5 - 3 = 7.5 m 

DA = 4(5.25) – 8 = 13 m

Hence length of shortest stopicIde is 7.5 m (i.e. CD).

Problem 2: You have been given a trapezotopicId PQRS such that PQ||RS with If Problem 2. In a trapezotopicId PQRS, it is given that PQ = 2x – 5, RS = x -3 and XY = 2x - 6. Find value of median XY.

Solution: 

By definition, we recall that the Median of the trapezotopicId is half the sum of its bases.

=> XY = (PQ + RS) / 2

=> XY = {(2x – 5 + x - 3)} / 2  = (3x - 8) / 2.

Now, it has been given that XY = 2x - 6. Substituting it in the above equation, we get.

2x - 6 = (3x -8) / 2

=> 2(2x - 6) = 3x - 8

=> 4x - 12 = 3x - 8

=> 4x - 3x = 12 - 8

=> x = 4

Therefore XY = 2x - 6 = 2(4) - 6 => XY = 2 units.

Problem 3:  ConstopicIder a parallelogram with adjacent angles in the ratio of 1:2. You will have to find the measures of all angles of the parallelogram.

Solution: 

Let us begin by assuming that the adjacent angles have measures x and 2x.

By using the property of the Parallelogram, we know that adjacent angles form supplementary angles.

As per the Parallelogram, we have x + 2x = 180 degrees.

=> 3x = 180° 

=> x = 60° .

Please note that in a parallelogram, opposite angles stay equal.

Therefore, going by that logic, we have angles as 60°, 120°, 60°, 120°.

Problem 4: Suppose it is given that a rhombus PQRS has a perimeter of 50 units and one of its diagonal has a length of 20 units. Can you find the length of the other diagonal?

Solution: 

Given: Perimeter = 50 unit.

Length of diagonal (say PR) = 20 Units.

As per the property of the rhombus, we know that rhombus’ four stopicIdes are equal.

=> PQ + QR + RS + SP = 50.

=> 4 * (PQ) = 50 => PQ = 12.5 units.

Also, we notice that the Rhombus diagonals cut or bisect each other at a perpendicular angle. From the figure, PX = XR and QX = XS and ∠PXQ = 90°.

With the use of Pythagoras theorem in ∆PXQ,(∠PXQ = 90°)

=> (PQ)² = (PX)² + (XQ)²

Now, PR = 20 units. Half of it is PX = 10 units. We constopicIder half of PR because the diagonals bisect each other and hence a diagonal can be divtopicIded into two equal parts.

=> (12.5)² = (10)² + (XQ)² 

=> (XQ)² = 156.25 – 100 = 56.25

=>   XQ  =  √56.25 units = 7.5 units

Since, AC = 2*7.5= 15 units.

Therefore, the length of another rhombus diagonal is 15 units.