Prashant had a plot of land in the shape of a quadrilateral. He constructed his house in the middle by joining the mid-points of the four sides of the land and used the remaining four portions at the four ends for different purposes, like a small garden, swimming pool, etc.

(i) What type of a quadrilateral is PQRS ?

(ii) What are the lengths of adjacent sides of the quadrilateral PQRS, if their ratio is 1 : 2 and the perimeter of the quadrilateral is 180 m ?

(iii) In quadrilateral PQRS, if ∠PSQ = 30° and ∠QRS = 110°, find ∠SQP.

(i) Join AC and BD.

By mid-point theorem,

In a triangle, the line segment joining the midpoints of any two sides is parallel to the third side and is half of the length of the third side.

In triangle ADC,

S and R are the mid-points of sides AD and DC respectively.

∴ SR || AC and SR = $\dfrac{1}{2}$ AC ……..(1)

In triangle ABC,

P and Q are the mid-points of sides AB and BC respectively.

∴ PQ || AC and PQ = $\dfrac{1}{2}$ AC ……..(2)

From equation (1) and (2), we get :

PQ || SR and PQ = SR.

In triangle ABD,

P and S are the mid-points of sides AB and AD respectively.

∴ PS || BD and PS = $\dfrac{1}{2}$ BD ……..(3)

In triangle BCD,

Q and R are the mid-points of sides BC and CD respectively.

∴ QR || BD and QR = $\dfrac{1}{2}$ BD ……..(4)

From equation (3) and (4), we get :

PS || QR and PS = QR.

Since, opposite sides of quadrilateral PQRS are equal and parallel.

Thus, PQRS is a parallelogram.

Hence, PQRS is a parallelogram.

(ii) Given,

Length of adjacent sides of the quadrilateral PQRS are in the ratio 1 : 2.

Let PQ = x and PS = 2x.

Thus, PQ = SR = x and PS = QR = 2x.

Perimeter of quadrilateral PQRS = PQ + QR + SR + PS

= x + 2x + x + 2x = 6x

Given,

Perimeter = 180 m

⇒ 6x = 180

⇒ x = 30

Therefore, PQ = SR = 30 m and PS = QR = 60 m.

Hence, the lengths of adjacent sides of the quadrilateral PQRS are 30 m and 60 m.

(iii) In a //gm,

Opposite angles are equal.

Thus, in //gm PQRS,

∠QPS = ∠QRS = 110°

In triangle QPS,

By angle sum property of triangle,

⇒ ∠PSQ + ∠QPS + ∠SQP = 180°

⇒ 30° + 110° + ∠SQP = 180°

⇒ 140° + ∠SQP = 180°

⇒ ∠SQP = 180° - 140° = 40°.

Hence, ∠SQP = 40°.