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In the above figure, OS = OQ and PR = 2OR = 2OR, and also, OR = OS, then, PQRS is not a
∵ O is the bisector of both the diagonals.
∴ PQRS is a parallelogram.
∵ Diagonals of the parallelogram are equal.
∴ PQRS may be a square or rectangle.
In the adjoining figure, AD  BC and AB and DC are not parallel, then ∠B =
∵ AB is parallel to BC, and, AB is transversal.
∴ ∠A + ∠B = 180°
⇒ ∠B = 180°  ∠A = 180°  110° = 70°
In ∆LAM,
∠ALM + 90° + ∠M = 180°
⇒∠M = 65°
∴x=∠M= 60° (∵KLMN is a parallelogram).
In ∆BLK,
∠BLK + 65° + 90° = 180°
⇒ ∠M = 180°  90° = 90°
⇒∠BLK = 25°
∵ ∠L = 180°  65° = y + 25° + 25°
⇒y = 65°
ABCD is a parallelogram.
The measure of ∠ADO is :
∠OBC = ∠ODA = 35°
(Alternate Interior ∠S).
The quadrilateral formed by joining the midpoints of a given quadrilateral will be (surely) :
ABCD is a parallelogram and P,Q,R,S are the midpoints of AB, BC, CD, DA respectively. Consider, AC as a diagonal of ABCD.
Now, According to midpoint theorem,
PQ =1/2 AC and PQ is parallel to AC,
and SR = 1/2 AC and PQ is parallel to AC.
∴ PQ = RS and PQ  RS  AC.
∴ PQRS will be a parallelogram.
ABCD is a rhombus and ABEF is a square find ‘a’.
∵ AB is a common side in both square and rhombus.
∴ AB = BC = CD = DA = FA = FE = EB
In ∆EBC,
∠BEC + ∠C + ∠B = 180° [∠C = ∠BEC]
[∵ EB = BC]
[∠B = 90° + 60° = 150°]
∵ AE is the diagonal of square. ∴
∴∠AEF = 45°
Now,
∵∠ E is the angle of a square,
∴∠E = 90°
⇒ 45° + 15° + a = 90°
⇒ a = 30°
ABCD is a rectangle , with, ED = DC, ∠BOC = 120°, ∠CED = 30°.
Find ‘x’ from the adjoining figure.
From ∆EDC,
∠EDC = 180°  2 × 30° = 120°
∠EDG = ∠EDC  ∠D = 120°  90° = 30°
Also,
The adjacent sides of a rectangle are in the ratio 5: 13 and its area is 135 cm^{2}. The perimeter of the rectangle is :
5K × 13 K = 195
⇒ K = 3
∴ Sides of the rectangle are 15 cm, 39 cm
∴ Perimeter = 2 (15 + 39)
= 2 × 54
= 108 cm.
∵ AE = EB.
∴ ∠BAE = ∠ABE = x.
∵ ΔDO = OB, and , AO = OC
In ΔABE,
∠DAE + ∠ABE = ∠AEF
⇒ x + x = 50°
⇒ 2x = 50°
⇒ x = 25°
ABCD is a square, P,Q,R,S are the midpoints of AB, BC, CD and DA respectively. If the perimeter of ABCD is cm, then perimeter of PQRS is :
In ΔAPS,
∴ Perimeter (PQRS) = 4 × 4 cm
= 16 cm
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