Advertisements
Advertisements
Question
The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rhombus, if ______.
Options
PQRS is a rhombus
PQRS is a parallelogram
diagonals of PQRS are perpendicular
diagonals of PQRS are equal
Advertisements
Solution
The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rhombus, if diagonals of PQRS are equal.
Explanation:
Given, the quadrilateral ABCD is a rhombus.
So, sides AB, BC, CD and AD are equal.
Now, in ΔPQS, we have
D and C are the mid-points of PQ and PS.
So, `DC = 1/2 QS` [By mid-point theorem] ...(i)
Similarly, in ΔPSR, `BC = 1/2 PR` [By mid-point theorem] ...(ii)
As BC = DC ...[Since, ABCD is a rhombus]
∴ `1/2 QS = 1/2 PR` ...[From equations (i) and (ii)]
⇒ QS = PR
Hence, diagonals of PQRS are equal.
APPEARS IN
RELATED QUESTIONS
ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
- D is the mid-point of AC
- MD ⊥ AC
- CM = MA = `1/2AB`
ABCD is a square E, F, G and H are points on AB, BC, CD and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is a square.
In the given figure, ΔABC is an equilateral traingle. Points F, D and E are midpoints of side AB, side BC, side AC respectively. Show that ΔFED is an equilateral traingle.

D and F are midpoints of sides AB and AC of a triangle ABC. A line through F and parallel to AB meets BC at point E.
- Prove that BDFE is a parallelogram
- Find AB, if EF = 4.8 cm.
D, E, and F are the mid-points of the sides AB, BC, and CA respectively of ΔABC. AE meets DF at O. P and Q are the mid-points of OB and OC respectively. Prove that DPQF is a parallelogram.
In triangle ABC, D and E are points on side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC which meet side AC at points F and G respectively. Through F and G, lines are drawn parallel to AB which meets side BC at points M and N respectively. Prove that: BM = MN = NC.
In parallelogram PQRS, L is mid-point of side SR and SN is drawn parallel to LQ which meets RQ produced at N and cuts side PQ at M. Prove that M is the mid-point of PQ.
In the given figure, ABCD is a trapezium. P and Q are the midpoints of non-parallel side AD and BC respectively. Find: DC, if AB = 20 cm and PQ = 14 cm
In ΔABC, the medians BE and CD are produced to the points P and Q respectively such that BE = EP and CD = DQ. Prove that: Q A and P are collinear.
In ΔABC, D and E are the midpoints of the sides AB and BC respectively. F is any point on the side AC. Also, EF is parallel to AB. Prove that BFED is a parallelogram.
Remark: Figure is incorrect in Question
