Download Grade 9 Quadrilaterals

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Grade 9
Quadrilaterals
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Answer t he quest ions
(1)
In a quadrilateral ABCD, the angles A, B, C and D are in ratio 2:3:4:6. Find the measure of each
angle of the quadrilateral.
(2)
In the parallelogram ABCD, ∠D is 50°, then the measurement of ∠A is _____.
(3)
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.
(4) If angles P, Q, R and S of the quadrilateral PQRS taken in order are in ratio 8:12:11:9, then what
kind of shape does PQRS represent.
(5)
ABCD is a parallelogram and the diagonals of it intersect at point O. If ∠DAO = 37 °, ∠BAO =
33° and ∠COD = 97°, f ind ∠ODC.
(6) In the parallelogram PQRS, PQ = 10 cm. T he altitude corresponding to the side PQ and PS are 5
and 6 respectively. Find PS.
(7) In the parallellogram ABCD, f ind the measurement of ∠BAC and ∠DAC, when ∠BCA is 41° and
∠DCA is 45°.
(8)
In a triangle, ABC, D is a point on AB such that AB = 4AD and E is a point on AC such that AC =
4AE. Prove that BC = 4ED.
(9) In a parallelogram, prove that the ratio of any side to altitude of adjacent side is same f or all its
sides.
Choose correct answer(s) f rom given choice
(10) In a parallelogram ABCD, f ind ∠CDB if ∠DAB = 54° and ∠DBC = 59°.
a. 31°
b. 121°
c. 54°
d. 67°
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(11) ABCD and PQRS are parallelograms as shown in the f igure below:
a. x + y + z + w = l + m + n + o
b. x + y = l + o
c. x + w = l + m
d. all of above
(12) ABCD is a parallelogram and P and R are the midpoints of side DC and BC respectively. If line PR
intersect diagonal AC at Q, than AC = _______________ .
a. 2CQ
b. (AB + BC)/2
c. 4CQ
d. 3CQ
(13) T he three angles of a quadrilateral are 64°, 77° and 111° respectively. Find the f ourth angle.
a. 252°
b. 72°
c. 18°
d. 108°
(14) In a quadrilateral ABCD, O is a point inside the quadrilateral such that AO and BO are the
bisectors of ∠A and ∠B respectively. T he value of ∠AOB is
a.
1
(∠C + ∠D)
2
b.
1
(∠A + ∠B)
2
c. 180° - (∠A + ∠B)
d. ∠C + ∠D
(15) ABCD is a parallelogram. T he angle bisectors of ∠A and ∠D meet at O. What is the measure of
∠AOD?
a. sum of ∠A and ∠D
b. 90°
c. 45°
d. Can not be determined
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Answers
(1)
∠A = 48°, ∠B = 72°, ∠C = 96°, ∠D = 144°
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 2:3:4:6.
T heref ore,
∠A = 2x,
∠B = 3x,
∠C = 4x 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°
⇒ 2x + 3x + 4x + 6x = 360
⇒ 15x = 360
⇒x=
360
15
⇒ x = 24
Step 3
Hence, ∠A = 2x = 2 × 24 = 48°,
∠B = 3x = 3 × 24 = 72°,
∠C = 4x = 4 × 24 = 96° and
∠D = 6x = 6 × 24 = 144°.
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ID : ae-9-Quadrilaterals [4]
(2)
130°
Step 1
Following f igure shows the parallelogram ABCD,
Step 2
According to the question, ∠D = 50°.
If we look at the f igure caref ully, we notice that, the ∠A and the ∠D are the consecutive
angles, the consecutive angles of a parallelogram are supplementary.
T heref ore, ∠A + ∠D = 180°
⇒ ∠A + ∠50° = 180°
⇒ ∠A = 180° - 50°
⇒ ∠A = 130°.
Step 3
T hus, the measurement of ∠A is 130°.
(3)
9 cm
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(4) T rapezium
Step 1
Following f igure shows the quadrilateral PQRS,
Let's assume x is the common ratio of the angles of the quadrilateral PQRS.
T heref ore, the ∠P, ∠Q, ∠R and ∠S of the quadrilateral are 8x, 12x, 11x and 9x
respectively.
Step 2
We know that the sum of all the angles of a quadrilateral is equal to 360°.
T heref ore, 8x + 12x + 11x + 9x = 360
⇒ 40x = 360
⇒ x = 360/40
⇒x=9
Step 3
Now, ∠P = 8x = 8 × 9 = 72°,
∠Q = 12x = 12 × 9 = 108°,
∠R = 11x = 11 × 9 = 99° and
∠S = 9x = 9 × 9 = 81°.
Step 4
T hus, ∠P + ∠Q = 72° + 108° = 180°,
∠R + ∠S = 99° + 81° = 180°
Since, the adjacent angles of the quadrilateral PQRS are supplementary and hence, the
PQRS is a T rapezium.
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(5)
50°
Step 1
Following f igure shows the parallelogram ABCD,
According to the question, ∠DAO = 37 °, ∠BAO = 33° and ∠COD = 97°.
∠OCD = ∠BAO = 33° [Alternate interior angles]
Step 2
In ΔCOD,
∠COD + ∠OCD + ∠ODC = 180° [Since the sum of all the angles of a triangle is 180°]
⇒ 97° + 33° + ∠ODC = 180°
⇒ 130° + ∠ODC = 180°
⇒ ∠ODC = 180° - 130°
⇒ ∠ODC = 50°
Step 3
Hence, the value of the ∠ODC is 50°.
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(6) PS = 8.33 cm
Step 1
Following f igure shows the parallelogram PQRS,
We know that, the area of a parallelogram = Base × Height
According to the question, DS = 5 cm,
T he area of the parallelogram PQRS = PQ × DS
= 10 × 5
= 50 cm2
Step 2
Similarly, QE = 6 cm
T he area of the parallelogram PQRS = PS × QE
⇒ 50 = PS × 6
⇒ PS =
50
6
⇒ PS = 8.33 cm2
Step 3
Hence, PS = 8.33 cm2
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(7) ∠BAC = 45°
∠DAC = 41°
Step 1
In the parallelogram ABCD, AB||DC,
T heref ore, ∠BAC = ∠DCA [Alternate interior angles]
Similarly, since AD||BC,
∠DAC = ∠BCA [Alternate interior angles]
Step 2
According to the question,
∠BCA = 41°,
∠DCA = 45°
T heref ore, ∠BAC = ∠DCA = 45°,
∠DAC = ∠BCA = 41°
Step 3
Hence, the measurement of the ∠BAC and ∠DAC is 45° and 41° respectively.
(8)
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(9)
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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Step 8
Also since AB = CD and BC = AD,
AB
=
CQ
CD
CQ
=
AD
=
CP
BC
CP
(10) d. 67°
Step 1
Following f igure shows the parallelogram ABCD,
According to the question ∠DAB = 54° and ∠DBC = 59°.
∠A = ∠C = 54° [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°]
⇒ 59° + 54° + ∠CDB = 180°
⇒ 113° + ∠CDB = 180°
⇒ ∠CDB = 180° - 113°
⇒ ∠CDB = 67°
Step 3
Hence, the value of the ∠CDB is 67°.
(11) d. all of above
Step 1
We know that the sum of the all angles of a parallelogram is 360°.T heref ore,
x + y + z + w = 360° and
l + m + n + o = 360°
T heref ore x + y + z + w = l + m + n + o
Step 2
We also know that, the sum of the consecutive angles of a parallelogram is supplementary.
T heref ore,
x + y = x + w = l + m = l + o = 180°
Step 3
As we can see that all options are correct, the correct answer is, 'all of above'.
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(12) c. 4CQ
(13) d. 108°
Step 1
According to the question, the three angles of the quadrilateral are 64°, 77° and 111°
respectively.
Let's assume, the f ourth angle of the quadrilateral be x.
Step 2
We know that, the sum of all interior angles of a quadrilateral is 360°.
T heref ore, 64° + 77° + 111° + x = 360°
⇒ 252° + x = 360°
⇒ x = 360° - 252°
⇒ x = 108°.
Step 3
Hence, the f ourth angle of the quadrilateral is 108°.
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(14)
a.
1
(∠C + ∠D)
2
Step 1
Following f igure shows the quadrilateral ABCD,
According to the question, AO and BO are the bisectors of ∠A and ∠B respectively.
T heref ore, ∠BAO = ∠A/2,
∠ABO = ∠B/2
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°
⇒ ∠C + ∠D = 360° - (∠A + ∠B) -----(1)
Step 3
In ΔAOB,
∠BAO + ∠ABO + ∠AOB = 180° [Since, we know that the sum of all three angles of a
triangle is equals to 180°]
⇒ ∠A/2 + ∠B/2 + ∠AOB = 180°
⇒ ∠AOB = 180° - ∠A/2 - ∠B/2
⇒ ∠AOB =
360° - ∠A - ∠B
2
⇒ ∠AOB =
360° - (∠A + ∠B)
2
⇒ ∠AOB =
(∠C + ∠D)
[From equation (1)]
2
Step 4
Hence, ∠AOB =
1
(∠C + ∠D)
2
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(15) b. 90°
Step 1
Following f igure shows the parallelogram ABCD,
AO and DO are the bisectors of ∠DAB and ∠ADC respectively.
T heref ore, the ∠DAB = 2∠DAO ,
the ∠ADC = 2∠ADO
Step 2
T he ∠DAB and the ∠ADC are consecutive angles of the parallelogram ABCD, we know
that, the consecutive angles of a parallelogram are supplementary.
T heref ore, ∠DAB + ∠ADC = 180°
⇒ 2∠DAO + 2∠ADO = 180°
⇒ 2(∠DAO + ∠ADO) = 180°
⇒ ∠DAO + ∠ADO = 90° ------(1)
Step 3
We know that, the sum of all the angles of a triangle is equal to 180°.
In ΔAOD, ∠DAO + ∠ADO + ∠AOD = 180°
⇒ 90° + ∠AOD = 180° [From equation (1), ∠DAO + ∠ADO = 90°]
⇒ ∠AOD = 180° - 90°
⇒ ∠AOD = 90°
Step 4
Hence, the measure of ∠AOD is 90°.
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