Quadrilaterals Test

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Quadrilaterals Test - 8

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Question 1
1 / -0

If the measures of the interior angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6, find the measure of its largest angle.

A
60^o

B
120^o

C
90^o

D
30^o

Solution

The interior angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6
3x + 4x + 5x + 6x = 360^o
18x = 360^o
x = 20^o
The largest angle 6x = 6 x 20° = 120^o
Question 2
1 / -0

In a quadrilateral, the measures of angles are in the ratio 1 : 4 : 3 : 2. The difference between the measures of the greatest and the smallest angles of the quadrilateral is

A
108°

B
54°

C
180°

D
360°

Solution

Suppose the angles are x, 4x, 3x and 2x.
Now, x + 4x + 3x + 2x = 10x = 360^o
x = 36^o
Thus, the smallest angle = 36° and the largest angle = 144°
Difference = 144^o - 36^o = 108°
Question 3
1 / -0

In the given circle, if AB and CD are diameters, then ACBD is a

A
square

B
trapezium

C
isosceles trapezium

D
rectangle

Solution

Since its given that AB & CD are diameters.
AB & CD must be equal to each other.
ABCD is a rectangle because diagonals of a rectangle are equal to each other.
Question 4
1 / -0

In a parallelogram ABCD, the bisectors of consecutive angles A and B intersect at P. The measure of APB is

A
90^o

B
60^o

C
120^o

D
180^o

E
None of these

Solution

The bisectors of consecutive angles A and B intersect at P.
∠A + ∠B = 180^o (Sum of adjacent angles of a parallelogram is 180^o)
∠A + B = 90^o.
∠PAB + ∠ABP = 90^o ...... (Equation 1)
In ABP, PAB + ABP + BPA = 180^o
BPA = 180^o – (PAB + ABP)
(PAB + ABP) = 180^o – BPA
From equation 1,
PAB + ABP = 90^o
180 – BPA = 90^o
BPA = 90^o
Question 5
1 / -0

In a quadrilateral ABCD, A + C = 180°, then B + D =

A
360°

B
100°

C
180°

D
80°

Solution

A + C = 180^o
A + B + C + D = 360^o (Sum of all the angles of a quadrilateral is 360⁰)
180^o + B + D = 360^o
B + D = 360^o - 180^o = 180^o
Question 6
1 / -0

If the lengths of two diagonals of a rhombus are 10 cm and 24 cm, then the length of each side of the rhombus is

A
13 cm

B
14 cm

C
cannot be determined

D
none of these

Solution

In ABCD
Diagonal AC = 10 cm
AO = 5 cm [ Diagonals bisect each other]
& OC = 5 cm
by, BO = 12 cm & OD = 12 cm [As BD = 24 cm]
Now, In AOB
(AB)² = (5)² + (12)²
= 25 + 144
= 169
AB = = 13
Side of rhombus = 13 cm
Question 7
1 / -0

In the given figure, TS = PQ, TS || PQ and the three sides TP, QR and RS are equal. If T = 102^o, then PQR is

A
138^o

B
162^o

C
88^o

D
120^o

E
None of these

Solution

TS = PQ, TS || PQ, TP = QR = RS
T = 102^o, T = PQS = 102^o (Opposite angles in a parallelogram are equal)
In QSR, QS = SR = RQ
We know that in an equilateral , each angle is 60^o.
So,SQR = 60^o, PQR = PQS + SQR = 102^o + 60^o = 162^o
Question 8
1 / -0

The angles of a quadrilateral are x°, (x – 20)°, (x + 50)° and 2x°. Find the measure of the greatest angle.

A
132°

B
180°

C
68°

D
None of these

Solution

The angles of a quadrilateral are
x^o, (x - 20)^o, (x + 50)^o & 2x^o
x^o + (x - 20)^o + (x + 50)^o + 2x^o = 360^o
5x^o + 30^o = 360^o
5x = 330^o
x = 66^o
The greatest angle is 2x^o = 132^o
Question 9
1 / -0

Two adjacent sides of a parallelogram are 6 cm and 7 cm. Its perimeter is

A
26 cm

B
36 cm

C
12 cm

D
none of these

Solution

Two adjacent sides of the parallelogram are 6 cm and 7 cm.
Perimeter = 2(6 cm + 7 cm) = 2 13 cm = 26 cm.
Question 10
1 / -0

The diagonals of a rectangle PQRS meet at O. If QOR = 46⁰, then POS is

A
67⁰

B
46⁰

C
44⁰

D
60⁰

Solution

QOR = 46^o
QOR = POS = 46^o
(Vertically opposite angles are equal)
Question 11
1 / -0

Two adjacent angles of a parallelogram are in the ratio 4 : 5. The angles are

A
180°, 180°

B
36°, 144°

C
80°, 100°

D
none of these

Solution

The adjacent angles of the parallelogram are in the ratio 4 : 5
4x + 5x = 180^o (Sum of adjacent angles of parallelogram)
9x = 180^o
x = 20^o
Angles are 4x = 80^o and 5x = 100^o
Question 12
1 / -0

In a rhombus ABCD, A = 60° and AB = 8 cm. Find the diagonal BD.

A
cm

B
8 cm

C
12 cm

D
13 cm

Solution

As diagonals of a rhombus bisect each other at 90^o
BO = OD
Now, In AOB
Sin 30^o =
8 = OB
OB = 4 cm
BD = OB + OD
= 4 + 4 = 8 cm
Question 13
1 / -0

The sides of a rectangle are in the ratio 3 : 4. If its perimeter is 140 cm, then its largest side is

A
15 cm

B
20 cm

C
40 cm

D
30 cm

Solution

A + B = 180^o (Sum Adjacent angles of parallelogram are equal)
1 = 2 4 = 3 (A + B) = (180^o) = 90^o
2 = 3 = 90^o ARB + 2 + 3 = 180^o
ARB + 90^o = 180^o ARB = 90^o
2 ARB = 180^o = A + B = A + D = B + C
Question 14
1 / -0

In the adjoining figure, AR and BR are angle bisectors of A and B of the parallelogram ABCD which meet at point R as shown. Then 2ARB is equal to

A
A + B

B
A + D

C
B + C

D
all of the above
Question 15
1 / -0

In a parallelogram ABCD, diagonals AC and BD intersect at O. If AO = 7 cm, then AC =

A
5 cm

B
20 cm

C
14 cm

D
none of these

Solution

In a parallelogram ABCD, diagonals AC and BD intersect at O. If AO = 7 cm, then AC = 14 cm because the diagonals of parallelogram bisect each other.
Question 16
1 / -0

In the given figure, the area of the quadrilateral ABCD is

A
21 m²

B
24 m²

C
42 m²

D
18 m²

Solution

Area of quadrilateral = (Area of 2 triangles)
= (Area of ΔABC + Area of ΔACD)
=
= 9 + 12 = 21
Area of quadrilateral ABCD = 21 m²
Question 17
1 / -0

In a parallelogram, the sum of the angle bisectors of two adjacent angles is

A
30°

B
45°

C
60°

D
90°

Solution

ABCD is a parallelogram.
We know that: A + B = 180° (Adjacent sides are supplementary, i.e. sum of the angles is 180°)
Therefore, = 90°
Question 18
1 / -0

In a rhombus ABCD, if AB = AC then BCD =

A
60°

B
120°

C
72°

D
108°
Question 19
1 / -0

In the given figure, AR = 7 cm, AP = 5 cm, PB = 4 cm and RC = 2 cm. The area of the quadrilateral APBC is

A
20 cm²

B
13 cm²

C
23cm²

D
28 cm²

Solution

P = Q = R = A =
Given, AR = 7 cm, AP = 5 cm,

∴ APQR is a rectangle.
So, BQ = 3 cm and CQ = 3 cm
The area of the quadrilateral APBC = Area of rectangle APQR - ( Area of triangle ARC + Area of triangle BQC)
= 5 × 7 - = 35 - (23/2) = 23
Question 20
1 / -0

In a parallelogram ABCD D = 50°, then the measurement of A =

A
130°

B
65°

C
90°

D
75°

Solution

Since opposite angles of a parallelogram are equal to each other
B = D = 50^o
Let A = C = x
x + x + 50^o + 50^o = 360^o [sum of all angles of a quadrilateral is 360^o]
2x + 100 = 360
2x = 260
x = 130
A = C = 130^o