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.
Answer
(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 = AC ........(1)
In triangle ABC,
P and Q are the mid-points of sides AB and BC respectively.
∴ PQ || AC and PQ = 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 = BD ........(3)
In triangle BCD,
Q and R are the mid-points of sides BC and CD respectively.
∴ QR || BD and QR = 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°.
Gautam had two sticks of equal length. He named them AB and CD. He then placed these two sticks in three different ways and joined their four vertices as shown below.
(i) What type of quadrilateral is in :
(a) Fig (i) ?
(b) Fig (ii) ?
(c) Fig (iii) ?
(ii) What is the sum of consecutive angles in Fig (iii) ?
(iii) If ∠BAD = 70° in Fig (iii), then what is the measure of ∠CDA ?
Answer
(i)
(a) In figure (i),
One pair of opposite sides are equal in length (AB = CD).
The diagonals are equal and they bisect each other.
If one pair of opposite sides of a quadrilateral are equal, and its diagonals are equal and bisect each other, the quadrilateral is a rectangle.
Hence, the quadrilateral in figure (i) is a rectangle.
(b) In figure (ii),
The diagonals of quadrilateral are equal and bisect each other. (AB = CD)
If the diagonals of a quadrilateral are equal and bisect each other, then the quadrilateral is a rectangle.
Hence, the quadrilateral in figure (ii) is a rectangle.
(c) In figure (iii),
Sides AB and CD are equal and parallel.
The diagonals of quadrilateral bisect each other.
Since, one pair of sides are equal and parallel and diagonals bisect each other, thus the quadrilateral is a parallelogram.
Hence, the quadrilateral in figure (iii) is a parallelogram.
(ii) In a parallelogram, sum of consecutive angles is equal to 180°.
Hence, the sum of consecutive angles in figure (iii) is 180°.
(iii) Sum of adjacent angles in a parallelogram is 180°.
Thus, in figure (iii),
⇒ ∠BAD + ∠CDA = 180°
⇒ 70° + ∠CDA = 180°
⇒ ∠CDA = 180° - 70° = 110°.
Hence, ∠CDA = 110°.