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Quadrilaterals

Short Answer Type

31. In the given figure, PQRS is a parallelogram. B and D are mid-points of the sides QR and PS respectively. Show that ABCD is a parallelogram.

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32. In the given figure, PQRS is a parallelogram and ∠SPQ = 60°. If the bisectors of ∠P and ∠Q meet at A on RS, prove that A is the mid-point of RS.

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33.

The angles A, B, C and D of a quadrilateral have measures in the ratio 2 : 4 : 5 : 7. Find the measures of these angles. What type of quadrilateral is it? Give reasons.

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34. If the diagonals of a quadrilateral bisect each other, prove that it is a parallelogram.

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35. In a parallelogram PQRS, if ∠QRS = 2x, ∠PQS = 4x and ∠PSQ = 4x, find the angles of the parallelogram.

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36.

In the figure, ABCD is a parallelogram in which AB is produced to E so that AB = BE

(a) Prove that ED bisects BC
(b) If AD = 10 cm, find OB.

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Long Answer Type

37. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show that:

left parenthesis straight I right parenthesis space space SR parallel to space AC space and space SR space equals 1 half space AC

(ii) PQ = SR
(iii) PQRS is a parallelogram.

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Short Answer Type

38. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

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39. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

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40. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see figure). Show that F is the midpoint of BC.

Given: ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F.
To Prove: F is the mid-point of BC.
Proof: Let DB intersect EF at G.
In ∆DAB,
∵ E is the mid-point of DA and EG || AB
∴ G is the mid-point of D
| By converse of mid-point theorem
Again, in ∆BDC,
∵ G is the mid-point of BD and GF || AB || DC
∴ F is the mid-point of BC.
| By converse of mid-point theorem

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