Download Grade 9 Quadrilaterals

ID : cn-9-Quadrilaterals [1]
Grade 9
Quadrilaterals
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Answer t he quest ions
(1)
ABCD is a rectangle and point P is such that PB = 3 cm, PC = 5 cm and PD = 3√2 cm, f ind the
value of PA.
(2)
ABCD is a parallelogram. T he angle bisectors of ∠A and ∠D meet at O. What is the measure of
∠AOD?
(3)
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 = 13 cm and AD = 8 cm, f ind the measure of AF.
(4) In a triangle, ABC, P, Q and R are the midpoints of sides BC, CA, and AB respectively. If AC = 25
cm, BC = 32 cm and AB = 28 cm, f ind the perimeter of the quadrilateral ARPQ.
(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 BP = 6 cm and BC = 14 cm,
f ind the length of diagonal BD.
(6) T he three angles of a quadrilateral are 63°, 85° and 91° respectively. Find the f ourth angle.
Choose correct answer(s) f rom given choice
(7) T he a parallelogram ABCD, the bisector of ‹A also bisects the side BC. If AB = 9 cm, f ind the
length of side AD.
(8)
a. 9 cm
b. 21 cm
c. 18 cm
d. can not be determined
In a rectangle ABCD, E, F, G, and H are the mid points of the f our side, what kind of shape is
represented by EFGH.
a. T rapezium
b. Rhombus
c. Square
d. Can not be determined
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(9) In the parallelogram PQRS, f ind the measurement of ∠PSQ and ∠PQS when ∠QPS is 46° and
∠QSR is 79°.
a. 79°, 55°
b. 55°, 79°
c. 55°, 44°
d. 11°, 79°
(10) In a parallelogram the ratio of any side to altitude of adjacent side is
a. not same f or all its sides and corresponding b. twice the ratio of bigger side and its altitude
altitudes
c. equal to the ratio of other sides and their
corresponding adjacent side's altitudes
d. equal to the ratio of other sides and their
corresponding altitudes
(11) In a quadrilateral PQRS, the diagonals PR and QS are equal and are perpendicular to each other.
T he quadrilateral PQRS is a
a. Rectangle
b. Square
c. Rhombus
d. All of above
(12) 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. 180° - (∠A + ∠B)
b.
1
(∠C + ∠D)
2
c. ∠C + ∠D
d.
1
(∠A + ∠B)
2
(13) 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.
a. ∠A = 60°, ∠B = 80°, ∠C = 100°, ∠D = 120° b. ∠A = 20°, ∠B = 23°, ∠C = 24°, ∠D = 25°
c. ∠A = 80°, ∠B = 60°, ∠C = 120°, ∠D = 100° d. ∠A = 120°, ∠B = 100°, ∠C = 80°, ∠D = 60°
(14) 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 + m
c. x + w = l + m
d. all of above
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(15) If angles P, Q, R and S of the quadrilateral PQRS taken in order are in ratio 7:13:12:8, then what
kind of shape does PQRS represent.
a. Rhombus
b. Square
c. T rapezium
d. None of above
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Answers
(1)
4 cm
(2)
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°.
(3)
26 cm
(4) 53 cm
(5)
18 cm
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(6) 121°
Step 1
According to the question, the three angles of the quadrilateral are 63°, 85° and 91°
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, 63° + 85° + 91° + x = 360°
⇒ 239° + x = 360°
⇒ x = 360° - 239°
⇒ x = 121°.
Step 3
Hence, the f ourth angle of the quadrilateral is 121°.
(7) c. 18 cm
(8)
b. Rhombus
Step 1
Following f igure shows the rectangle ABCD,
Let's assume the length and breadth of the rectangle ABCD be a and b respectively.
Step 2
In ΔGDH, DG = CD/2 = a/2,
DH = DA/2 = b/2 [Since, G and H are the midpoints of the sides CD and DA
respectively.]
∠D = 90° [Since, ABCD is a rectangle]
GH2 = DG2 + DH2 [By the pythagorean theorem]
⇒ GH2 = (a/2)2 + (b/2)2
Similarly, HE2 = EF2 = FG2 = (a/2)2 + (b/2)2
and hence, HE2 = EF2 = FG2 = GH2,
or HE = EF = FG = GH
Step 3
We know that quadrilateral with all f our sides equal is rhombus. T heref ore EFGH is a
Rhombus.
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(9) b. 55°, 79°
Step 1
According to the question, ∠QPS = 46°,
∠QSR = 79°
Step 2
T he ∠QPS and the ∠PSR are consecutive angles of the parallelogram ABCD, we know
that, the consecutive angles of a parallelogram are supplementary.
T heref ore, ∠QPS + ∠PSR = 180°
⇒ ∠QPS + ∠PSQ + ∠QSR = 180° [Since ∠PSR = ∠PSQ + ∠QSR]
⇒ 46° + ∠PSQ + 79° = 180°
⇒ ∠PSQ + 125° = 180°
⇒ ∠PSQ = 180° - 125°
⇒ ∠PSQ = 55°
Step 3
∠PQS = ∠QSR [Alternate interior angles]
⇒ ∠PQS = 79°
Step 4
Hence, the measurement of the ∠PSQ and ∠PQS are 55° and 79° respectively.
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(10) c. equal to the ratio of other sides and their corresponding adjacent side's altitudes
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
(11) b. Square
Step 1
Following f igure shows the quadrilateral PQRS,
Step 2
Since diagonals are equal and bisects each other,
OP = OQ = OR = OS
Step 3
Since diagonals intersects perpendicularly,
∠PQQ = ∠QOR = ∠ROS = ∠SOP = 90°
Step 4
T heref ore triangles ΔPQQ , ΔQOR , ΔROS and ΔSOP are identical triangles. Hence,
PQ = QR = RS = SP .................. (1)
Step 5
∠OPQ = ( 180° - ∠POQ )/2 = ( 180° - 90° )/2 = 45°
Similarly, ∠OPS = 45°
T herf ore ∠SPQ = 90°
Step 6
Simiarly we can f ind all angles of quadrilateral,
∠SPQ = ∠PQR = ∠QRS = ∠RSP = 90° .................. (2)
Step 7
Form equations (1) and (2), we can inf er that quadrilateral PQRS is a square
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(12)
b.
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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(13) a. ∠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°.
(14) 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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(15) c. 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 7x, 13x, 12x and 8x
respectively.
Step 2
We know that the sum of all the angles of a quadrilateral is equal to 360°.
T heref ore, 7x + 13x + 12x + 8x = 360
⇒ 40x = 360
⇒ x = 360/40
⇒x=9
Step 3
Now, ∠P = 7x = 7 × 9 = 63°,
∠Q = 13x = 13 × 9 = 117°,
∠R = 12x = 12 × 9 = 108° and
∠S = 8x = 8 × 9 = 72°.
Step 4
T hus, ∠P + ∠Q = 63° + 117° = 180°,
∠R + ∠S = 108° + 72° = 180°
Since, the adjacent angles of the quadrilateral PQRS are supplementary and hence, the
PQRS is a T rapezium.
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