Download Class 9 Quadrilaterals

ID : in-9-Quadrilaterals [1]
Class 9
Quadrilaterals
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Answer t he quest ions
(1)
T he diameter of circumcircle of a rectangle is 13 cm and rectangle's width is 5 cm. Find length of
the rectangle.
(2)
In a quadrilateral ABCD, the angles A, B, C and D are in ratio 3:4:5:6. Find the measure of each
angle of the quadrilateral.
(3)
In a parallelogram, prove that the ratio of any side to altitude of adjacent side is same f or all its
sides.
(4) In a rhombus, one of the diagonals is equal to a side of the rhombus. Find the angles of the
rhombus.
(5)
ABCD is a parallelogram with E and F as the mid-points of side AB and CD. Line segments CE
and AF intersect the diagonal BD at point P and Q respectively. If QD = 3 cm and BC = 8 cm, f ind
the length of diagonal BD.
(6) ABCD is a rectangle and point P is such that PB = 3√2 cm, PC = 4 cm and PD = 3 cm, f ind the
value of PA.
Choose correct answer(s) f rom given choice
(7) ABCD is a parallelogram. L and M are points on AB and DC respectively such that AL = CM. T he
diagonal BD intersects LM at point O. If AB = 15 cm, AD = 5 cm and LO = 3 cm, f ind the length of
OM.
(8)
a. 7 cm
b. 3 cm
c. 6 cm
d. can not be determined
T he perimeter of a parallelogram is 180 cm. If one side exceeds the other side by 12 cm, what
are the sides of the parallelogram.
a. 39 cm, 51 cm
b. 41 cm, 53 cm
c. 39 cm, 27 cm
d. 45 cm, 57 cm
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ID : in-9-Quadrilaterals [2]
(9) In a triangle ABC, AD is the median. F is the point on AC such that line BF bisect AD at E. T he
measure of AC = ___________
a. 3AF
b. 3AE
c. 2AF
d. 3EF
(10) T he length and breadth of a rectangle is 8 cm and 15 cm. Find the radius of circumcircle of this
rectangle.
a. 8.5 cm
b. 17 cm
c. 25.5 cm
d. 4.25 cm
(11) ABCD is a parallelogram and E is the midpoint of side BC. When DE and AB are extended, they
meet at point F. If AB = 10 cm and AD = 7 cm, f ind the measure of AF.
a. 15 cm
b. 17.5 cm
c. 30 cm
d. 20 cm
(12) In a parallelogram ABCD, f ind ∠CDB if ∠DAB = 51° and ∠DBC = 45°.
a. 84°
b. 45°
c. 51°
d. 39°
(13) In the quadrilateral ABCD, AO and DO are the bisectors of ∠A and ∠D respectively. If ∠B =
115° and ∠C = 97° than ∠AOD is:
a. 119°
b. 96°
c. 106°
d. 212°
(14) ABCD is a quadrilateral and ∠A = ∠B = ∠C = ∠D = 90°. T hen ABCD can be called as
a. Parallelogram
b. Square
c. Rectangle
d. Both rectangle and parallelogram
(15) In a parallelogram ABCD, the diagonals bisect at O. What kind of a triangle AOB is?
a. a right angled but not an isosceles triangle
b. an isosceles but not right angled triangle
c. neither isosceles nor a right angled triangle
d. an isosceles right angled triangle
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ID : in-9-Quadrilaterals [3]
Answers
(1)
12 cm
Step 1
Following f igure shows the rectangle ABCD with it's circumcircle,
AB, BC and AC are the length, breadth and diameter of circumcircle of the rectangle ABCD.
According to the question, AC = 13 cm,
BC = AD = 5 cm.
Step 2
Now, in right angled triangle ABC, AB2 = AC2 - BC2
⇒ AB = √[ AC2 - BC2 ]
= √[ 132 - 52 ]
= 12 cm
Step 3
Hence, the length of the rectangle is 12 cm.
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(2)
∠A = 60°, ∠B = 80°, ∠C = 100°, ∠D = 120°
Step 1
Let's assume x is the common f actor of the angles of the quadrilateral.
According to the question, the angles A, B, C and D are in ratio 3:4:5:6.
T heref ore,
∠A = 3x,
∠B = 4x,
∠C = 5x and
∠D = 6x.
Step 2
We know that the sum of all interior angles of a quadrilateral is equal to 360°.
T heref ore, ∠A + ∠B + ∠C + ∠D = 360°
⇒ 3x + 4x + 5x + 6x = 360
⇒ 18x = 360
⇒x=
360
18
⇒ x = 20
Step 3
Hence, ∠A = 3x = 3 × 20 = 60°,
∠B = 4x = 4 × 20 = 80°,
∠C = 5x = 5 × 20 = 100° and
∠D = 6x = 6 × 20 = 120°.
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(3)
Step 1
Let ABCD is a parallelogram as shown below. We have extended its sides to show the
altitudes as CP and CQ
Step 2
Lets f irst consider side AB. Its adjacent sides are AD and BC. From the picture we can see
that altitude f or AD and BC is CQ.
T heref ore required ratio is,
AB
CQ
Step 3
Similarly these ratios will be as f ollowing f or other sides
CD
AD
,
CQ
CP
,
BC
CP
Step 4
T heref ore we need to prove that,
AB
=
CQ
CD
CQ
=
AD
CP
=
BC
CP
Step 5
If we compare triangles ΔBPC and ΔDQC, we observe f ollowing,
- ∠DQC = ∠BPC ..... (Both are right angle)
- ∠DCQ = ∠BCP ..... (∠DCQ = 90° -∠BCD and ∠BCP = 90° -∠BCD)
Step 6
Since two angles are equal, triangles ΔBPC and ΔDQC are similar triangles
Step 7
Since ratio of corresponding sides of similar triangles is same,
BC
CP
=
CD
CQ
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ID : in-9-Quadrilaterals [6]
Step 8
Also since AB = CD and BC = AD,
AB
=
CQ
CD
CQ
=
AD
CP
=
BC
CP
(4) 60°, 120°
Step 1
Following f igure shows the rhombus ABCD,
Step 2
We know that, all sides of a rhombus have equal length.
According to the question, the diagonal AC of the rhombus ABCD is equal to the side of
the rhombus.
T heref ore, AB = BC = CD = DA = AC
Step 3
In ΔABC, AB = BC = CA,
Hence, the ΔABC is a equilateral triangle.
Step 4
Similarly, the ΔACD is a equilateral triangle.
Step 5
All angles of a equilateral triangle is equal to 60°.
T heref ore, ∠B = ∠D = 60°,
∠A = ∠BAC + ∠CAD = 60° + 60° = 120°,
Similarly, the ∠C = 120°
Step 6
Hence, the angles of the rhombus are 60° and 120°.
(5)
9 cm
(6) 5 cm
(7) b. 3 cm
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(8)
a. 39 cm, 51 cm
Step 1
Let's assume a and b are the sides of the parallelogram.
T he perimeter of the parallelogram = 2(a + b)
Step 2
According to the question, one side of the parallelogram exceeds the other side by 12 cm.
T heref ore, a = b + 12
Step 3
Since, the perimeter of the parallelogram = 180 cm
T heref ore, 2(a + b) = 180
⇒ 2(b + 12 + b) = 180
⇒ (2b + 12) = 180/2
⇒ 2b = 90 - 12
⇒b=
78
2
⇒ b = 39 cm,
a = b + 12 = 39 + 12 = 51 cm
Step 4
Hence, the sides of the parallelogram are 39 cm and 51 cm.
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(9) a. 3AF
Step 1
Following picture shows triangle ΔABC with median AD at E, and another line BF, which
bisects AD. We have also drawn another line DG which is parallel to BF
Step 2
From proportionality theorem we know that if a line is drawn parallel to one side of a
triangle, it divides other two sides in the same ration
Step 3
For triangle ΔBFC, DG is parallel to BF and using proportionality theorem we can f ind that,
FG
=
GC
BD
DC
Step 4
Since we know that BD = DC, we can f ind that
GC = FG ................ (1)
Step 5
Similarly we can use proportionality theorem f or triangle ΔADG,
AF
=
FG
AE
ED
Step 6
But since AE = ED,
FG = AF ................ (2)
Step 7
T heref ore,
AC = AF + FG + GC
Using relation (1) and (2),
AC = AF + AF + AF,
AC = 3 AF
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(10) a. 8.5 cm
Step 1
Following f igure shows the rectangle ABCD with it's circumcircle,
AB, BC and AC are the length, breadth and diameter of circumcircle of the rectangle ABCD.
According to the question, AB = 15 cm,
BC = 8 cm.
Step 2
Now, in right angled triangle ABC, AC2 = AB2 + BC2
⇒ AC = √[ AB2 + BC2 ]
= √[ 152 + 82 ]
= 17 cm
Step 3
T he diameter of circumcircle of the rectangle = 17 cm.
T he radius of circumcircle of the rectangle = 17/2 = 8.5 cm.
(11) d. 20 cm
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(12) a. 84°
Step 1
Following f igure shows the parallelogram ABCD,
According to the question ∠DAB = 51° and ∠DBC = 45°.
∠A = ∠C = 51° [Since the opposite angles of a parallelogram are congruent.]
Step 2
In ΔBCD,
∠DBC + ∠BCD + ∠CDB = 180° [Since the sum of all the angles of a triangle is
180°]
⇒ 45° + 51° + ∠CDB = 180°
⇒ 96° + ∠CDB = 180°
⇒ ∠CDB = 180° - 96°
⇒ ∠CDB = 84°
Step 3
Hence, the value of the ∠CDB is 84°.
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(13) c. 106°
Step 1
Following f igure shows the quadrilateral ABCD,
According to the question, AO and DO are the bisectors of ∠A and ∠D respectively.
T heref ore, ∠DAO = ∠A/2,
∠ADO = ∠D/2,
∠B = 115°,
∠C = 97°
Step 2
We know that the sum of all angles of a quadrilateral is equals to 360°.
T heref ore, ∠A + ∠B + ∠C + ∠D = 360°
⇒ ∠A + 115° + 97° + ∠D = 360°
⇒ ∠A + ∠D + 212° = 360°
⇒ ∠A + ∠D = 360° - 212°
⇒ ∠A + ∠D = 148° -----(1)
Step 3
In ΔAOD,
∠DAO + ∠ADO + ∠AOD = 180° ...[Since, we know that the sum of all three angles of
a triangle is equals to 180°]
⇒ ∠A/2 + ∠D/2 + ∠AOD = 180°
⇒ ∠AOD = 180° - ∠A/2 - ∠D/2
⇒ ∠AOD =
360° - ∠A - ∠D
2
⇒ ∠AOD =
360° - (∠A + ∠D)
2
⇒ ∠AOD =
360° - 148°
...[From equation (1)]
2
⇒ ∠AOD = 106°
Step 4
Hence, the value of ∠AOD is 106°.
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(14) d. Both rectangle and parallelogram
Step 1
Following f igure shows the quadrilateral ABCD where all f our angles are 90°
Step 2
A quadrilateral with all f our angles of 90° is a rectangle.
We also know that all rectangles are parallelogram since opposite sides of rectangles are
parallel and equal to each other.
Step 3
T heref ore, the correct answer is 'Both rectangle and parallelogram'.
(15) c. neither isosceles nor a right angled triangle
Step 1
Following f igure shows the parallelogram ABCD with diagonals.
Step 2
We know that diagonals of a parallelogram bisects each other but they are not
perpendicular, theref ore
∠AOC ≠ 90° , and
AO = AC/2 and OB = BD/2
Step 3
Also since diagonals of parallelogram are not equal,
AC ≠ BD
⇒ AC/2 ≠ BD/2
⇒ AO ≠ OB
Step 4
Since AO ≠ OB and ∠AOC ≠ 90° , triangle ΔAOB is neither isosceles nor a right angled
triangle.
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