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This mock test of Olympiad Test: Understanding Quadrilaterals for Class 8 helps you for every Class 8 entrance exam.
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QUESTION: 1

ABCD is a quadrilateral. If AC and BD bisect each other, what is ABCD?

Solution:

QUESTION: 2

ABCD is a parallelogram. The angle bisectors of ∠A and ∠D meet at O. What is the measure of ∠AOD?

Solution:

QUESTION: 3

The diagonal of a rectangle is 10 cm and its breadth is6 cm. What is its length?

Solution:

Apply Pythagoras Theorem,

(Diagonal)^{2 }= (Breadth)^{2 }+ (Length)^{2}

100 = 36+(Length)^{2}

Length=8cm

QUESTION: 4

ABCD and MNOP are quadrilaterals as shown in the figure.

Which of the following is correct?

Solution:

QUESTION: 5

What do you call a parallelogram which has equal diagonals?

Solution:

QUESTION: 6

In a square ABCD, the diagonals bisect at O. What type of a triangle is AOB?

Solution:

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

QUESTION: 7

The perimeter of a parallelogram is180 cm. If one side exceeds the other by 10 cm, what are the sides of the parallelogram?

Solution:

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

QUESTION: 8

In the quadrilateral ABCD, the diagonals AC and BD are equal and perpendicular to each other. What type of a quadrilateral is ABCD?

Solution:

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

QUESTION: 9

ABCD is a parallelogram as shown in the figure. If AB = 2AD and P is the mid-point of AB, what is the measure of ∠CPD ?

Solution:

As shown in the figure, since P is the midpoint of AB and AB = 2AD,we have AB = 2AP = 2AD or AP = AD. i.e.,

triangle ADP is an isosceles triangle. If ∠ADP = x^{o} and ∠APD = x^{o}, then ∠A = 180^{o}−2x^{o}.

Since ∠B is adjacent to ∠A, in ABCD ∠B = 180^{o}−(180^{o}−2x) = 2x. In ΔCBP,x^{o}+x^{o}+2x^{o }= 180^{o}(Angle sum property)

⇒ 4x^{o} = 180^{o }⇒ x^{o }= 45^{o} ∴∠CPD = 180^{o}−2x^{o} = 180^{o}−2×45^{o }= 90^{o}

QUESTION: 10

In a parallelogram ABCD, if AB = 2x+5, CD = y+1, AD= y+5 and BC = 3x−4,what is the ratio of AB and BC?

Solution:

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

QUESTION: 11

If ABCD is an isosceles trapezium, what is the measure of ∠C?

Solution:

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

QUESTION: 12

A diagonal of a rectangle is inclined to one side of the rectangle at 25^{o}. What is the measure of the acute angle between the diagonals?

Solution:

Since ∠CAB = 25^{o} ∠CAB = 65^{o}

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

∴ ∠OBC = 65^{o}

⇒ ∠BOC = 50^{o}

QUESTION: 13

If angles P, Q, R and S of the quadrilateral PQRS, taken in order, are in the ratio 3:7:6:4, what is PQRS?

Solution:

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.

QUESTION: 14

If AB and CD are diameters, what is ACBD?

Solution:

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.

QUESTION: 15

If two adjacent angles of a parallelogram are in the ratio 3:2, what are their measures?

Solution:

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}.

QUESTION: 16

ABC and DEF are straight lines.

Find the value of 'x',

Solution:

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}

QUESTION: 17

ABCD is a rectangle. Its diagonals meet at O.

Find x; if OA = 2x+4 and OD = 3x+1.

Solution:

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.

QUESTION: 18

ABCD is a rhombus.

Find the respective values of x, y and z.

Solution:

x = OB = OD (Diagonals bisect) = 5

y = OA = OC (Diagonals bisect) = 12

z = side of the rhombus = 13 (All sides are equal).

QUESTION: 19

In the figure, ABCD is a parallelogram.

Find the respective values of x, y and z.

Solution:

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)

QUESTION: 20

In the figure, ABCD is a rhombus and ABDE is a parallelogram.

Given that EDC is a straight line and ∠AED = 36^{o} find ∠BAD.

Solution:

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}

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