Olympiad Test: Understanding Quadrilaterals - Class 8 MCQ

Explanation:

• When AC and BD bisect each other, it means that they intersect at their midpoints.


• If a quadrilateral has diagonals that bisect each other, it must be a parallelogram.


• All squares are parallelograms because they have opposite sides that are parallel, so a square is a special type of parallelogram.


• Similarly, all rectangles are parallelograms because they also have opposite sides that are parallel, so a rectangle is a special type of parallelogram.


• Therefore, if AC and BD bisect each other, the quadrilateral ABCD can be a square, a parallelogram, or a rectangle, making the correct answer D: All of the above.

 

Apply Pythagoras Theorem,
(Diagonal)² = (Breadth)² + (Length)²
100 = 36+(Length)²
Length=8cm

p+q+r+s = 360 [ sum of interior angles of quadrilateral ]
w+x+y+z = 360 [ sum of interior angles of quadrilateral ]

A parallelogram which has equal diagonals are called rectangle. Trapezium, rhombus and kite does not contain equal diagonal.
Proof for rectangle contains equal diagonals are given below :-

⇒ In given figure ABCD is a rectangle.

⇒ In △ABC and △DCB

⇒ BC = CB [Common]

⇒ AB = DC [Opposite sides of rectangle]

⇒ ∠ABC = ∠DCA = 90∘ [Each angle of rectangle is 90∘]

⇒ △ABC ≅ △DCB [By SAS property]

∴ AC = BD [By CPCT]

Since diagonals of a square are equal and bisect at right angles, triangle AOB is an isosceles right angled triangle.

Let one side of the parallelogram be 'x' cm. Then adjacent side is (x+10)cm. 

∴ Perimeter = x+(x+10)+x+(x+10) = 180

(Given) ⇒ 4x+20 = 180 or x = 40cm

∴ x+10 = 50cm

A quadrilateral in which the diagonals are equal and perpendicular is called a square.

We know that in a parallelogram opposite sides are equal.       

∴ AB = CD or  2x+5=y+1 and 

AD = BC or y+5=3x−4

2x−y =−4 ....(i) 
y−3x = −9 .....(ii)

Adding (i) and (ii),

we get −x =−13 or x = 13

and y = 30.

Substituting, we have 

AB = 31 cm and BC = 35 cm

∴The required ratio = 31:35

From definition, we know that in an isosceles trapezium the non-parallel sides are equal or AD = BC in the  figure.  Drop perpendiculars AE and BF to CD. Triangles AED and BFC are congruent by R.H.S congruency. Hence, ∠D = ∠C

Since ∠CAB = 25^o ∠CAB = 65^o

Let diagonals meet at O. ΔOCB is an isosceles triangle. 

∴ ∠OBC = 65^o

⇒ ∠BOC = 50^o

Let the angles be 3x, 7x , 6x and 4x.  
∴ 3x+7x+6x+4x = 360^o or 20x = 360^o or x = 18^o.
The angles are 54^o,126^o,108^o and 72^o.
We see that adjacent angles are supplementary but opposite angles-are not equal. Clearly, it is a trapezium.

Since the angle in a semicircle is a right angle, clearly ∠A = ∠C = ∠B = ∠D = 90^o
The diagonals (diameters) are equal but they are not intersecting (bisecting) at right angles. Hence, it is not a square and can be only a rectangle.

Let the angles be 3x and 2x. We have, 3x+2x = 180^o ⇒ 5x = 180^o ⇒ x = 36^o
∴ The angles are 36^o×3 and 36^o×2 = 108^o and 72^o.

In the given figure. ∠ABF+∠FBC = 180^o. 

70^o+∠FBC = 180^o ⇒ ∠FBC = 180^o−70^o = 110^o

 Now,  ∠DEC+∠CEF = 180^o

 ∠CEF = 180^o−60^o = 120^o

 Now, ∠FBC+∠BCE+∠CEF+∠BFE = 36^o

 290^o+x = 360^o ⇒ x = 70^o

OD is half of the diagonal BD and OA  is half of the diagonal AC. Diagonals are equal. So, their halves are also equal. Therefore, 3x + 1 = 2x + 4
 ⇒ x = 3.

x = OB = OD (Diagonals bisect) = 5 
y = OA = OC (Diagonals bisect) = 12
z = side of the rhombus = 13 (All sides are equal).

C is opposite to A. So, 
x = 100^o (Opposite angles property.) 
y = 100^o (Measure of angle corresponding to ∠x.) 
z = 80^o (Since ∠y,∠z is a linear pair)

BDC = AED = 36^o (Corresponding  s, AE  BD.) ABD= BDC = 36^o (Alternate  s, AB DC) ADB = ABD = 36^o (Base angles of isosceles, since AB = DC) BAD = 180^o−ABD−ADB (Angle sum of a triangle.) = 180^o−36^o−36^o = 108^o