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POLYGONS AND
QUADRILATERALS
16
1. From the adjoining diagram, find the value of x.
x
96°
Ans. In a quad. ABCD,
∠ DCB + ∠ BCE = 180°
Z
B
100°
75°
(Lincer pair angles)
∴
A
∠ DCB = 180° – ∠ BCE = 180° – 75 = 105°
⇒
x
The sum of interior angles of a quadrilateral = 360°
B
96°
x
+
100°
+
96°
+
105°
=
360°
∴
⇒ 301° + x = 360°
75°
100°
x = 360° – 301° = 59°
D
C E
2. If two of the angles of a quadrilateral are 76° and 138° and the other two angles
are equal find the size of equal angles.
Ans. Let the other two equal angles be x
Sum of interior angles of a quadrilateral = 360°.
76° + 138° + x + x = 360°
∴
214° + 2x = 360°
⇒
2x = 360° – 214°
⇒
2x = 146°
⇒
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146°
= 73°
2
Hence, the other two equal angles are 73 each.
3. A quadrilateral has 3 interior angles each equal to 85°. Find the size of the fourth
interior angle.
Ans. Let the fourth interior angle be x and the three interior angles are 85° each. (given)
Now, we know that
Sum of interior angles of a quadrilateral = 360°
85° + 85° + 85° + x = 360°
∴
255 + x = 360°
⇒
x = 360° – 255°.
⇒
Hence, the fourth interior angle = 105°.
⇒
ICSE Math Class VII
x=
1
Question Bank
4. From the adjoining diagram, find the value of x.
C
Ans.
Here ∠ ABC + 115° = 180° (Linear pair angles)
∴ ∠ ABC = 180° – 115 = 65°
D 210°
Since the sum of interior angles of a quadrilateral is 360°.
x + 210° + 48° + 65° = 360°
⇒
115°
48°
x + 323° = 360°
⇒
A
B
x = 360° – 323° = 37°
⇒
5. If one of the angles of quadrilateral is 105° and the remaining three angles are equal,
find the size of the equal angles.
Ans. One angle of quadrilateral = 150° and other three angles are equal i.e.x, x and x.
We know that,
Sum of angles of a quadrilateral is 360°
105° + x + x + x = 360°
⇒
105° + 3x = 360°
⇒
3x = 360° – 105° = 255°
⇒
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255°
= 85°
3
Hence, the size of the three equal angles is 85° each.
x=
⇒
6. If two of the angles of a quadrilateral are 76° 37′ and 57° 23′ and out of the remaining two angles one angle is 10°greater than the other, find these angles.
Ans. Two angles of a quadrilateral are 76° 37′ and 57° 23′ .
Let the other angles be x.
the one angle = 10° + x, two angles be x and x + 10°
We know that
sum of angles of a quadrilateral = 360°
∴
76° 37′ + 57° 23′ + x + 10° + x = 360°
133° 60′ + 10 + 2x = 360
⇒
144° + 2x = 360°
⇒
2x = 360° – 144° = 216°
⇒
216°
= 108°.
2
Hence, remaining two angles are x = 108° and 10 + x = 10° + 108° = 118°
7. Find the number of sides of a regular polygon if each of its interior angle is 165°
Ans. Given each interior angle = 165°, so each exterior angle = 180° – 165° = 15°.
∴
ICSE Math Class VII
x=
2
Question Bank
∴
8.
Ans.
∴
⇒
⇒
⇒
Number of sides of a polygon =
360°
= 24.
15°
Hence, the polygon has 24 sides.
The angles of a hexagon are (2x + 5)°, (3x – 5)°, (x + 40°),(2x + 20)°,
(2x + 25)° and (2x + 35)° . Find the value of x.
A hexagon has 6 sides
Sum of its interior angles = (2 × 6 – 4) right angles = (12 – 4) right angles
= 8 × 90 = 720°
(2x + 5)° + (3x – 5)° + (x + 40)° + (2x + 20)° + (2x + 25) ° + (2x + 35)° = 720°
2x + 5° + 3x – 5° + x + 40° + 2x + 20° + 2x + 25° + 2x + 35° = 720°
12x + 120° = 720°
12x = 720° – 120° = 600°
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600
∴ x=
. Hence, x = 50.
12
9. An octagon has 3 equal angles of 125° and five equal angles. Find the size of equal
angles.
Ans. An octagon has 8 sides.
Sum of its interior angles = (2 × 8 – 4) right angles = (16 – 4) right angles
= 12 × 90° = 1080°.
Let each equal angle be x, so we have 3 × 125° + 5x = 1080°
375° + 5x = 1080°
⇒
5x = 1080° – 375° = 705°
⇒
705°
⇒ x = 141°
5
10. The sum of interior angles of a polygon is 1980°. How many sides this polygon has
?
Ans. Let n be the number of sides of polygon.
∴ Sum of interior angles of polygon = (2n – 4) right angles. = (2n – 4) 90°.
But (2n – 4) 90° = 1980°
(Given)
∴ x=
∴
⇒
1980°
= 22
90
2n = 22 + 4 = 26
2n – 4 =
26
= 13.
2
Hence, the polygon has 13 sides.
∴
ICSE Math Class VII
n=
3
Question Bank
11. An exterior angle of a regular polygone is one-third of its interior angle. Find the
number of sides of the polygon.
Ans. Let exterior angle be x°, then interior angle = (180 – x)°
1
1
According to given, x = (180 – x ) ⇒ x = 60 – x
3
3
⇒ x+
1
3x + x
x = 60 ⇒
= 60
3
3
⇒ 4x = 60 × 3 ⇒ x =
60 × 3
= 45°
41
Z
B
∴ Exterior angle of a polygon = 45°
Then, number of sides of polygon
=
360°
= 8.
45°
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Hence, the polygone has 8 sides.
12. The measures of the angles of a quadrilateral are in the ratio 2 : 4 : 5 : 7. Find the
measures of each of its angles.
Ans. Let the measure of angles of a quadrilateral are 2x, 4x, 5x and 7x
2x + 4x + 5x + 7x = 630°
∴
18x = 360°
⇒
360°
= 20°
18
Hence, the measure of angle of a quadrilateral are :
2(20)°, 4 (20)°, 5 (20)° and 7 (20)° i.e. 40°, 80°, 100° and 140°.
A quadrilateral has three acute angles, each measuring 75°. What is the measure of
its fourth angle ?
Let the measure of the fourth angle be x°
x + 75° + 75° + 75° = 360°
(by property)
∴
x + 225° = 360°
⇒
x = 360° – 225° = 135°
⇒ ∴
Hence, the measure of the fourth angle = 135°.
One angle of a parallelogram measures 80°. Find the measure of each of its remaining angles ?
We know that in parallelogram opposite angles are equal. Let the adjacent angle of
80° be x
x + 80° + x + 80° = 360°
(by property)
∴
2x + 160° = 360°
⇒
⇒
13.
Ans.
14.
Ans.
ICSE Math Class VII
x=
4
Question Bank
2x = 360° – 160° = 200°
⇒
200°
= 100°
2
∴ The other three angles of the parallelogram are 100°, 80°, 100°.
15. Two adjacent angles of a parallelogram are in the ratio 1 : 3. Find the measure of
each of its angles.
Ans. Let the adjacent angles are x and 3x.
x + 3x + x + 3x = 360°
(by property)
∴
8x = 360°
⇒
x=
360°
= 45°
8
Hence, two adjacent angles are 45° and 3 × 45° i.e. 45° and 135°.
The lengths of two adjacent sides of a parallelogram are 7 cm and 5 cm respectively. Find the perimeter of the parallelogram.
Lengths of two adjacent sides of a parallelogram are 7 cm and 5cm.
∴ The other two sides of a parallelogram are also 7 cm an 5 cm.
∴ Perimeter of the parallelogram = 2 (sum of adjacent sides) = 2 (7 + 5) cm2 =
2 (12) cm2 = 24 cm2
Two sides of a parallelogram are in the ratio 5 : 3 and its perimeter is 48 cm. Find
the length of each of its sides.
Let the two sides of a parallelogram are 5x and 3x
∴ Perimeter of parallelogram = 2 (sum of adjacent sides)
48 = 2 (5x + 3x)
⇒
48 = 2 (8x)
⇒
48 = 16x
⇒
16x = 48
⇒
x=
⇒
16.
Ans.
17.
Ans.
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48
=3
16
Hence, the two adjacent sides of a parallelogram are 5 × 3 cm and 3 × 3 cm i.e 15
cm and 9 cm and other two sides are also 15 cm and 9 cm.
18. Five of the angles of a hexagon are each 115°. Calculate the measure of the sixth
angle.
Ans. In a hexagon, n = 6
Sum of its interior angles = (2x – 4) right angles = (2 × 6 – 4) × 90°
= (12 – 4) × 90° = 8 × 90° = 720°
Sum of its 5 angles = 5 × 115° = 575°. Hence, sixth angles = 720° – 575° = 145°
⇒
ICSE Math Class VII
x=
5
Question Bank
19. The angles of a heptagon are
(x + 3)°, (2x + 5)°, (x + 8)°, (3x + 1)°,
(5x – 6)°, (2x + 9)° and (x – 5)° . Calculate x .
Ans. In a heptagon, n = 7
Sum of its interior angles of heptagon
= (2n – 4) right angles = (2 × 7 – 4) × 90° = (14 – 4) × 90°
= 10 × 90° = 900°
But sum of angles of heptagon
(x + 3)° + (2x + 5)° + (x + 8)° + (3x + 1)° + (5x – 6)° + (2x + 9)° + (x – 5)° = 900°
⇒ = x + 3 + 2x + 5 + x + 8 + 3x + 1 + 5x – 6 + 2x + 9 + x – 5 = 900°
⇒ 15x + 15 = 900° ⇒ 15x = 900 – 15 = 885
Z
B
885
= 59° . Hence, x = 59°
15
20. An octagon has three angles each of measure 115°. If all the remaining angles have
equal measure, find the measure of each of these remaining angles.
Ans. In an octagon the number of sides, n = 8
Sum of angles of an octagon = (2n – 4) right angles
= (2 × 8 –4) × 90° = (16 – 4) × 90° = 12 × 90° = 1080°
Sum of three angles of octagon are = 115° × 3 = 345°
and sum of remaining five angles of octagon are = 1080° – 345° = 735°
⇒
x=
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735°
= 147°
5
21. The sum of the interior angles of a polygon is 2160°. How many sides does this
polygon have ?
Ans. Sum of the interior angles of a polygon = 2160°
Let number of sides of the polygon = be n.
Sum of angles of a polygon = (2n – 4) right angle = 2160°
(2n – 4) × 90° = 2160°
⇒
Hence, measure of each angle =
2n – 4 =
⇒
⇒
2160
= 24°
90
2n = 24 + 4 = 28 ⇒ n =
28
= 14
2
Number of sides = 14.
∴
22. The angles of a pentagon are x°, (x – 10)°, (x + 20)°, (2x – 44)° and (2x – 70)°.
Calculate x.
ICSE Math Class VII
6
Question Bank
Ans. We know that Sum of interior angles of a pentagon = 540°
∴ x° + (x – 10)° + (x + 20)° + (2x – 44)° + (2x – 70)° = 540°
⇒ x + x + x + 2x + 2x – 10° + 20° – 44° – 70° = 540°
7x + 20° – 124° = 540°
⇒
7x – 104° = 540° ⇒ 7x = 540° + 104° = 644°
⇒
Hence, x =
644
= 92°
7
23. One angle of a heptagon is 162° and each other angle is x°. Find the value of x.
Ans. The number of sides of heptagon is n = 7
Sum of interior angles of a heptagon
= (2 × 7 – 4) right angles = (14 – 4) right angles = 10 × 90° = 900°
Then,
162° + 6x = 900°
6x = 900° – 162° = 738°
⇒
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x=
∴
Z
B
738°
= 123°
6
24. The sum of interior angles of a polygon is 1260°. How many sides this polygon has
?
Ans. Let n be the number of sides of the polygon, Sum of interior angles of the polygon
= (2n – 4) right angles = (2n – 4) × 90°
As per conditions (2n – 4) × 90 = 1260°
⇒
⇒
⇒
2n – 4 =
1260
90
2n – 4 = 14
2n = 14 + 4 = 18
18
= 9.
2
Hence, the polygon has 9 sides.
25. The perimeter of a parallelogram is 88 cm and one of its adjacent sides is longer
than the other by 10 cm. Find the length of each of its sides.
Ans. Let one side of the parallelogram be x cm and its adjacent side = x + 10 cm
∴ Perimeter of the parallelogram = 88 cm
2 (sum of adjacent sides) = 88 cm
2 (x + x + 10) = 88 ⇒ 4x + 20 = 88
⇒
4x = 88 – 20 = 68
⇒
∴
ICSE Math Class VII
n=
7
Question Bank
68
= 17
4
Thus one side of the parallelogram = 17 cm and its adjacent side = 17 + 10 = 27 cm
and the other sides of parallelogram are 17 cm and 27 cm.
B
26. In the adjoining figure, P is a point in the interior of
M
∠ AOB. If PL ⊥ OA and
PM ⊥ OB and ∠ AOB = 40°, find the measure of ∠ LPM.
P
x=
∴
Ans. From the figure, we see that OLPM is a quadrilateral in which
∠ O = 40°, ∠ L = 90°, ∠ M = 90°
∴ ∠ O + ∠ OLP + ∠ LPM + ∠ PMO = 360°
⇒ 40° + 90° + ∠ LPM + 90° = 360°
O
L
(by property)
Z
B
220° + ∠ LPM = 360°
∠ LPM = 360° – 220°
⇒
= 140°
27. In the adjoining figure, ABCD is a parallelogram in which
∠ A = 115°. Find each of the remaining angles.
Ans. In the given parallelogram ABCD, ∠ A = 115° and AB || DC and
AD || BC
⇒
∴
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∠ A = ∠ C and ∠ B = ∠ D
A
C
D
115°
A
(Opposite angles of a
parallelogram)
B
∠ C = ∠ A = 115°
∴
But ∠ A + ∠ B = 180° (Co-interior angles of a parallelogram.)
115° + ∠ B = 180°
⇒
∠ B = 180° – 115 ° = 65°
⇒
But
∠ D = ∠ B = 65°
Hence ∠ B = 65°, ∠ C = 115° and ∠ D = 65°
28. In the adjoining figure, ABCD is a parallelogram in which
∠ BAO = 35°, ∠ DAO = 40° and ∠ COD = 95°. Calculate,
(i) ∠ ABO (ii)
∠ ODC
Ans. ABCD is a parallelogram in which AB || DC and AD || BC.
AC and BD are diagonals which intersect each other at O.
∠ AOB = ∠ COD. = 95°
(i)
D
C
95°
O
A
40°
35°
B
(Vertically opposite angles)
Now in ∆ AOB,
∠ OAB + ∠ OBA + ∠ AOB = 180°
ICSE Math Class VII
(Sum of angles of a triangle)
8
Question Bank
⇒ 35° + ∠ OBA + 95° = 180°
130° + ∠ OBA = 180°
⇒
∴ ∠ OBA = 180° – 130° = 50° or ∠ ABO = 50°
(ii) ∵ AB || DC
(Alternate Angles)
∠ ODC = ∠ ABO = 50°
∴
29. In each of the figures given below, ABCD is a rhombus. Find the values of x
and y in each case.
D
C
(i)
D
(ii)
y°
x°
C
y°
Z
B
O
118°
A
B
42°
x°
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A
B
Ans. (i) ABCD is a rhombus and AC is its diagonal
In ∆ ABC,
AB = BC
∴
∠ BAC = ∠ BCA
∴
∠ BAC = x
∴
But ∠ BAC + ∠ BCA + ∠ ABC = 180°
x + x + 118° = 180°
⇒
2x = 118° – 118° = 60°
⇒
⇒
∴
∴
⇒
(ii)
∴
∴
62
= 31°
2
Diagonal AC bisects ∠ A and ∠ C
x=y
y = 31°. Hence x = 31°, y = 31°
In rhombus ABCD
Diagonal bisect each other at right angles
[sides of a rhombus]
(Angles opposite to equal sides)
(Sum of angles of a triangle)
x=
∠ AOB = 90°
D
C
y°
O
Now in ∆ AOB,
∠ AOB + ∠ OAB + ∠ OBA = 180°
90° + 42° + x = 180°
⇒
132° + x = 180°
⇒
x = 180° – 132° = 48°
⇒
ICSE Math Class VII
9
42°
A
x°
B
Question Bank
In ∆ ABD,
AB = AD
∴
⇒
∠ ABD = ∠ ADB
x = y = 48°
Hence x = 48°, y = 48°
30. In the adjoining parallelogram ABCD, ∠ C = 80°, AB = 8
cm and BC = 5.5 cm. Find the remaining angles and the
sides of the parallelogram ABCD.
Ans. Given : ABCD is a parallelogram
C
D
80°
5.5 cm
A
8 cm
B
[opposite angles are equal]
∴ ∠ C = ∠ A = 80°
[co-interior angles are supplementary]
∠ A + ∠ B = 180°
⇒ 80° + ∠ B = 180°
⇒ ∠ B = 180° – 80°
∠ B = 100°
[opposite angles of parallelogram are equal]
∠ B = ∠ D = 100°
DC = AB = 8 cm
[opposite angles of parallelogram are equal]
BC = AD = 5.5 cm
[opposite angles of parallelogram are equal]
31. In the adjoining isosceles trapezium, ∠ B = 75°. Find all the remaining angles of the
trapezium.
Ans. ∠ B = ∠ A = 75°
[Angles on the same base are equal]
C
D
∠ A + ∠ D = 180° [interior angles are supplementary]
75° + ∠ D = 180°
⇒
∠ D = 180° – 75°
⇒
∠ D = 105°
⇒
A
B
∠ C = ∠ D = 105°
∴
32. In the adjoining rhombus ABCD, ∠ CBD = 59°, find
(i) ∠ ACB (ii)
∠ DAB
D
C
Ans.
ABCD is a rhombus.
[Diagonals intersect at right angles]
∠ BOC = 90°
∴
O
BOC
+
OCB
+
CBO
=
180°
[Sum
of
three
angles
of triangle]
∠
∠
∠
59°
90° + ∠ OCB + 59° = 180°
⇒
B
A
∠ OCB = 180° – 149° = 31°
⇒
∠ OCB = ∠ ACB = 31°
∴
(ii) Class
[alternate angles
of BC ||Bank
AD]
∠ BCA
ICSE Math
VII = ∠ CAD = 31°
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As diagonal AC bisects ∠ DAB
∠ DAB = 2 ∠ CAD = 2 × 31° = 62°
∴
33. In the adjoining figure, ABCD is a kite. If ∠ ACB = 26° and ∠ ADB = 42° find:
(i) ∠ BCD (ii)
∠ BAD
Ans. Given :ABCD is a kite.
(i) As diagonal AC bisects ∠ BCD
∴ ∠ BCD = 2 ∠ ACD = 2 × 26° = 52°
[Diagonals intersect at right angle]
(ii) ∠ DOA = 90°
[Sum of angles in
∠ DOA + ∠ ODA + ∠ DAO = 180°
∆ AOD]
90° + 42° + ∠ DAO = 180°
⇒
⇒
⇒
Z
B
D
42°
A
O
26°
C
B
132° + ∠ DAO = 180°
∠ DAO = 180° – 132° = 48°
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As diagonal CA bisects ∠ BAD
∠ BAD = 2 ∠ DAO = 2 × 48° = 96°
∴
34. In the adjoining figure, ABCD is a trapezium in which AB || DC. If ∠ A = 50° and
∠ B = 65°, find the measures of ∠ C and ∠ D.
Ans. In trapezium ABCD, AB is parallel to DC
Now AB parallel to DC and AD is a transversal then ∠ A + ∠ D = 180°
D
C
50° + ∠ D = 180°
⇒
∠ D = 180° – 50° = 130°
∴
Similarly AB is parallel to DC an BC is a transver65°
50°
A
B
sal
then ∠ B + ∠ C = 180° [sum of co-interior angles]
65° + ∠ C = 180°
⇒
∠ C = 180° – 65° = 115°
∴
A
35. In the adjoining figure, ABCD is a kite in which AB = AD, CB
= CD and the diagonals AC and BD intersect at O at right
angles. If ∠ BAO = 25° and ∠ CDO = 50°. Find the measure
25° d°
of each of the angles marked a, b, c, d and e.
Ans. In ∆ ABC, AB = AD and AO ⊥ BD
a° O
c°
∴ From right angled ∆ AOB
D
B
50°
⇒ a + 90° + 25° = 180° ⇒ a + 115° = 180°
e°
(by property)
C
a = 180° – 115° = 65°
∴
ICSE Math Class VII
11
Question Bank
36. In the adjoining figure, ABCD is a parallelogram. AM and CN are drawn perpendicular from A and C respectively on the diagonal BD. Prove that AM = CN.
Ans. In ∆ ABM and ∆ CDN
AB = CD ∠ M = ∠ N and ∠ B = ∠ D
D
C
(∵Diagonal BD bi∠ ABM = ∠ CDN
sects the angles of al parallelogram)
M
∆ ABM ≅ ∆ CDN (By A.A.S. congruence
∴
rule)
N
AM = CN
(Corresponding parts A
∴
B
of congruent triangles)
Z
B
37. In the figure along side, P is a point in the interior of ∠ AOB, PM ⊥ OA and PN
⊥ OB. If ∠ AOB = 35°, what is the measure of ∠ MPN ?
Ans. P is a point interior of ∠ AOB, PM ⊥ OA and PN ⊥ OB
∠ AOB = 35° or ∠ NOM = 35°
B
N
∵ OM PN is a quadrilateral
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∴ ∠ OMP + ∠ MPN + ∠ PNO + ∠ NOM = 360°
(∴ PM O 35°
⇒ 90° + ∠ MPN + 90° + 35° = 360°
⊥ OA and PN ⊥ OB)
215° + ∠ MPN = 360°
⇒
⇒ ∠ MPN = 360° – 215° = 145°
38. In the figure alongside. ABCD is a trapezium, AB||DC.
If ∠ A = ∠ B = 40°, what are the measure of the other two angles?
Ans. ABCD is a trapezium in which AB||DC. ∠ A = ∠ B = 40°.
But ∠ A + ∠ B + ∠ C + ∠ D = 360°
⇒ 40° + 40° + ∠ C + ∠ D = 360°
80° + ∠ C + ∠ D = 360°
⇒
A
⇒ ∠ C + ∠ D = 360° – 80° = 280°
But ∠ C + ∠ D = 280°
Hence, ∠ C = ∠ D =
D
P
M
A
C
B
280°
= 140°
2
39. The adjacent angles of a parallelogram are equal. What is the measure of each ?
Ans. Let ABCD be a parallelogram then ∠ A = ∠ C, ∠ B = ∠ D (Opposite angles)
Then, ∠ A + ∠ B = 180° [Adjacent angles of parallelogram are supplementary]
If adjacent angles of parallelogram are equal, then
ICSE Math Class VII
12
Question Bank
∠A = ∠B
D
C
∠ A + ∠ A = 180°
∴
⇒
2 ∠ A = 180°
∠A =
⇒
180°
= 90°
2
A
B
Hence, ∠ B = 90°, ∠ C = 90° and ∠ D = 90°
40. Two adjacent angles of a parallelogram are in the ratio 2 : 3. Find measures of all the
angles.
Ans. Let ABCD be a parallelogram then ∠ A = ∠ C, ∠ B = ∠ D and ∠ A : ∠ B = 2 : 3
Let ∠ A = 2x and ∠ B = 3x then 2x + 3x = 180°
Z
B
180°
= 36°
5
⇒
5x = 180° ⇒ x =
∴
∠ A = 2x = 2 × 36° = 72° and ∠ B = 3x = 3 × 36° = 108°
Hence ∠ A = 72°, ∠ B = 108°, ∠ C = 72° and ∠ D = D
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108°
41. Show that a diagonal of a parallelogram divides it into two
congruent triangles.
Ans. ABCD is a parallelogram in which AB||DC and AD||BC and
AB = DC, AD = BC
A
C
B
∠ A = ∠ C and ∠ B = ∠ D
Diagonal BD is joined
Now in ∆ ABD and ∆ BCD
AB = DC (opposite sides of a parallelogram)
AD = BC
BD = BD
(common)
∴ ∆ ABD ≅ CDB
[by sss congruent]
Hence diagonal BD bisects the parallelogram.
Similarly we can prove that diagonal AC also bisects
the parallelogrm. Hence proved.
D
A
C
B
42. In the given figure, diagonals of parallelogram ABCD intersect at O.XY passes
through O as shown.
Give reasons for each of the following statements :
(a) OB = OD
(b)
(c) ∠ BOY = ∠ DOX
ICSE Math Class VII
∠ OBY = ∠ ODX
(d)
∆ BOY ≅ ∆ DOX
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C
D
Now, state if XY is bisected at O.
Ans. (a) Diagonal Ac and BD of the parallelogram ABCD bisects
X
each other at O.
∴ OB = OD.
A
(Alternate angles)
(b) ∠ OBY = ∠ ODX
(Vertically opposite angles)
(c) ∠ BOY = ∠ DOX
(d) In ∆ BOY and ∆ DOX, OB = OD (Diagonal BD is bisected O)
(Vertically opposite angle)
∠ BOY = ∠ DOX
(alternate angles proved)
∠ OBY = ∠ ODX
∴ ∆ BOY ≅ ∆ DOX (ASA Axiom)
(CPCT)
∴ OY = OX
Hence, XY is bisected at O
43. Construct a square ABCD of a side = 3.8 cm
Ans. Since all the sides of a square are equal
∴ AB = BC CD = DA = 3.8 cm.
Steps of construction:
(i) Draw AB = 3.8 cm
Z
B
Y
O
B
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P
C
D
(ii) At A, contruct ∠ BAP = 90°
A
B
3.8 cm
(iii) With A as centre and radius 3.8 cm, draw an arc to meet AP at D.
(iv) With B as centre and radius 3.8 cm, draw an arc.
(v) With D as centre and radius 3.8cm, draw an arc to meet the previous arc at C.
(vi) Join BC and DC. Then ABCD is the required square of side 3.8 cm.
44. Construct a square ABCD with diagonal BD = 7.2cm.
Ans. Since the diagonals of a square are equal, AC = BD = 7.2
P
cm
A
Steps of construction:
(i) Draw a line segment BD = 7.2 cm.
D
B
O 7.2cm
(ii) Draw perpendicular bisector PQ of BD to meet it
at O.
1
AC i.e. 3.6 cm
2
draw two arcs to meet OP at A and OQ at C.
(iv) Join AB, AD, BC and CD.
(v) Now, ABCD is the required square
(iii) With O as centre and radius OA =
ICSE Math Class VII
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45. Construct a parallelogram ABCD such that BC = 4.5 cm, BD = 4 cm and AC = 5.6
cm.
Ans. (i) Construct ∆ OBC with BC = 4.5 cm. BO =
1
4
BD =
= 2 cm
2
2
A
cm
D
2
1
5.6
CA =
= 2.8 cm
2
2
(i) Produce BO to D such that BO = OD.
O
2.8
cm
2
cm
CO =
2.8
cm
C
B
(ii) Produce CO to A such that CO = OA.
4.5 cm
(iii) Join AB, AD and DC, then ABCD is the required parallelogram.
46. Construct a parallelogram, one of whose sides is 4.4 cm and whose diagonals are
5.6 cm and 7 cm measure the other side.
Ans. Steps of construction:
(i) Draw a line segment AB = 4.4 cm
C
D
1
3
(ii) With A as a centre and radius 2.8 cm ( of 5.6
.5
cm
2
cm) and with centre B as centre and radius
O
1
3 .5
cm
3.5cm ( of 7.0 cm) draw two arcs intersect2
ing each other at O.
4.4 cm
A
B
(iii) Join AO and produce to C such that OC = 2.8 cm
(iv) Join BO and produce it to D such that OD = 3.5 cm
(v) Join AD, CD and BC.
Then ABCD is the required parallelogram. On measuring AD = BC = 4.6 cm.
47. Construct a rhombus ABCD ; if :
(a) AC = 8 cm and BD = 6 cm. Measure AB.
(b) Two diagonals are 4.6 cm and 3.4 cm respectively. Measure each side of the
rhombus.
D
(c) One side = 5 cm one diagonal = 5 cm.
(d) AD = 4.5 cm and BD = 5 cm. Measure AC.
8 cm
A
C
Ans. (a) Steps of Construction:
O
(i) Draw a line segment AC = 8 cm.
(ii) Draw the perpendicular bisector of AC meeting AC at O.
2.
8
cm
2.
8
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cm
Z
B
(iii) From O, cut off OB = OD =
6
= 3 cm
2
(iv) Join AB, BC, CD and DA.
Then ABCD is the required rhombus. On measuring its side is 5 cm.
ICSE Math Class VII
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Question Bank
cm
9
2.
(b) Steps of Construction:
D
(i) Draw a line segment AC = 4.6 cm.
(ii) Draw the perpendicular bisector of AC intersecting it
at O.
O
A
C
4.6 cm
3.4
= 1.7 cm
(iii) From D, cut off OB = OD =
2
(iv) Join AB, BC, CD and DA. Then ABCD is the required
B
rhombus. On measuring each side = 2.9 cm.
(c) Steps of Construction :
5 cm
D
C
(i) Draw a line segment AB = 5 cm.
8c
(ii) With centre A and radius 5 cm and with centre B
m
5 cm
and radius 8 cm, draw arcs intersecting each other 5 cm
at D.
(iii) Join AD and BD.
B
A
5 cm
(iv) With D and B as centres and radius 5 cm draw
arcs intersecting each other at C.
(v) Join DC and BC.
Then ABCD is the required rhombus.
4.5cm
D
(d) Steps of Construction :
C
(i) Draw a line segment AB = 4.5 cm.
(ii) With centre A and radius 4.5 cm and with 4.5cm
4.5cm
centre B and radius 5 cm, arcs intersectcm O
5
.
7
ing each other at D.
(iii) Join AD and BD
A
45.cm
B
(iv) With centre D and B and radius 4.5 cm,
draw arcs intersecting each other at C.
(v) Join BC and CD.
Then ABCD is the required rhombus.
(vi) Join AC and on measuring, it is 7.5 cm.
48. Construct a quadrilateral ABCD such that AB = 4.5 cm, BC = 4 cm, CD = 3.9 cm,
AD = 3.2 cm and ∠ B = 60°.
P
3.9 cm
D
Ans. Steps of construction :
C
(i) Draw AB = 4.5 cm.
(ii) At B, construct ∠ ABP = 60°
(iii) Cut off BC = 4 cm.
60°
(iv) Taking C as centre and radius
A
B
4.5 cm
CD = 3.9 cm, draw an arc
(v) With A as centre and radius 3.2 (=AD), draw an arc to cut the previous arc at D.
(vi) Join CD and AD, then ABCD is the required quadrilateral.
cm
2.
9
cm
cm
2.
9
9
2.
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B
ICSE Math Class VII
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4 cm
3.2 cm
m
5c
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49. Construct a quadrilateral ABCD in which AB = 3.5 cm, BC = 5 cm, CD = 5.6 cm, DA
= 4 cm and BD = 5.4 cm.
Ans. Steps of construction :
(i) Draw AB = 3.5 cm.
(ii) With A as centre and radius = 4 draw an arc. With B as centre and radius 5.4cm,
draw an arc to meet the previous arc at D. Join AD and BD
(iii) With B as centre and radius 5 cm draw an arc. With D as centre and radius 5.6
cm, draw an are to meet the previous arc at C.
(iv) Join BC and DC, then ABCD is the required quadrilateral.
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B
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ICSE Math Class VII
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