Download Polygons, Quadrilaterals and Area Propositions.pmd

10
POLYGONS, QUADRILATERALS
AND AREA PROPOSITIONS
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1. The angles of a quadrilateral are in the ratio 4 : 3 : 2 : 6. Find the
measure of each angle.
Ans.Let the angles of the quadrilateral are 4x, 3x, 2x and 6x.
[by property]
∴ 4x + 3x + 2x + 6x = 360°
⇒ 15x = 360°
360°
x =
⇒
⇒ x = 24°
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Hence, the angle of quadrilateral are (4 × 24)°, (3 × 24)°, (2 × 24)°
and (6 × 24)° i.e. 96°, 72°, 48° and 144°.
2. Three angles of a quadrilateral are in the ratio 2 : 5 : 6. If the fourth
angle measure 100°, find the other three angles of the quadrilateral.
Ans.Let the three angles of the quadrilateral are 2x, 5x and 6x.
∴ 2x + 5x + 6x + 100° = 360° (by property)
⇒ 13x + 100°= 360°
⇒ 13x = 360 – 100 ⇒ 13x = 260°
260°
x
=
⇒
⇒ x = 20°
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The three angles of the quadrilateral are (2 × 20)°, (5 × 20)° and
(6 × 20)° i.e., 40°, 100° and 120°.
3. Two angles of a quadrilateral are 116° and 84° and the remaining
two angles are equal. What is the measure of each of the equal
angles ?
Ans.Let the measure of each of the equal angles = x°
∴ x + x + 116° + 84° = 360°
⇒ 2x + 116° + 84° = 360° ⇒ 2x + 200° = 360°
2x = 360 – 200° ⇒ 2x = 160°
⇒
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160°
⇒ x = 80°
2
Hence, the measure of each of the equal angles is 80°.
4. Two angles of a quadrilateral are 68° and 76°. If the other two angles
are in the ratio 5 : 7; find the measure of each of them.
Ans. Two angles are 68° and 76°. Let other two angles be 5x and 7x
∴ 68° + 76° + 5x + 7x = 360°
⇒
12x + 144° = 360° ⇒ 12x = 360° – 144°
⇒
12x = 216° ⇒
x = 18°
Hence, required angles are 5x = 5 × 18° = 90° and 7x = 7× 18° = 126°.
5. Angles of a quadrilateral are (4x)°, 5(x + 2)°, (7x – 20)° and
6(x + 3)°. Find :
(i) the value of x, (ii) each angle of the quadrilateral.
Ans.Angles of quadrilateral are (4x)°, 5(x +2)°, (7x – 20)° and 6(x +3)°.
Thus, 4x + 5(x + 2)° + (7x – 20)° + 6(x + 3)° = 360°
⇒ 4x + 5x + 10° + 7x – 20° + 6x + 18° = 360°
⇒ 22x + 8° = 360° ⇒ 22x = 360° – 8° ⇒ 22x = 352°
Hence, the value of x = 16°.
(ii) each angle of quadrilateral are 4x = 4 × 16° = 64°
5(x + 2)° = 5x + 10° = 5 × 16° + 10° = 90°
(7x – 20)° = 7 × 16° – 20° = 112° – 20° = 92°
and
6(x + 3)° = 6x + 18° = 6 × 16° + 18 = 114°.
6. Use the informations given in the following figure to find :
(i) x
(ii) ∠B and ∠C
⇒ x=
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(3x – 5)°
D (8x–15)° (2x+4)° B
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Ans.
(Given)
∠A = 90°
Now, ∠A + ∠B + ∠C + ∠D = 360°
90° + (2x + 4)° + (3x – 5)° + (8x – 15)° = 360°
90° + 2x + 4° + 3x° – 5° + 8x – 15° = 360°
74° + 13x = 360° ⇒ 13x = 360° – 74°
⇒
13x = 286° ⇒ x = 22°
⇒
Therefore,
∠B = (2x + 4)° = 2 × 22° + 4° = 48°
And,
∠C = (3x – 5)° = 3 × 22° – 5° = 61°
7. In a quadrilateral ABCD, AB || DC. If ∠A : ∠D = 2 : 3 and
∠B : ∠C = 7 : 8, find the measure of each angle.
Ans.As AB || CD
∠A + ∠D = 180° and ∠B +∠C = 180°D
C
2x
+
3x
=
180°
and
7y
+
8y
=
180°
⇒
3x
8y
5x = 180° and 15y = 180°
x = 36° and y = 12°
7y
2x
∴ ∠A = 2 × 36° = 72°, ∠D = 3 × 36° = 108°,
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A
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∠B = 7y = 7 × 12° = 84°, ∠C = 8y = 8 × 12° = 96°
8. From the adjoining figure, find
(i) x
(ii) ∠DAB
(iii) ∠ADB
D
(5x+8)°
C
(3x+10)°
x°
(3x+4)° 50°
A
B
Ans. (i) ABCD is a quadrilateral.
∴ ∠A + ∠B + ∠C + ∠D = 360°
(3x + 4)° + (50 + x)° + (5x + 8)° + (3x + 10)° = 360°
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3x + 4° + 50° + x + 5x + 8° + 3x + 10° = 360°
12x + 72° = 360°
12x = 288°
⇒
⇒ x = 24
(ii) ∠DAB = (3x + 4) = 3 × 24 + 4 = 76°
(iii) ∠ADB = 180° – (76° + 50°) = 54°
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9. In quadrilateral ABCD; ∠C = 64°, ∠D = ∠C – 8°, ∠A = 5(a + 2)°
and ∠B = 2(2a + 7)°. Calculate ∠A.
Ans.∵
∴
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∠C = 64°
(Given)
∠D = ∠C – 8° = 64° – 8° = 56°
∠A = 5 (a + 2)°, ∠B = 2 (2a + 7)°
Now, ∠A + ∠B + ∠C + ∠D = 360°
5 (a + 2)° + 2 (2a + 7)° + 64° + 56° = 360°
5a + 10° + 4a + 14° + 64° + 56° = 360°
9a + 144° = 360°
⇒
9a = 360° – 144° = 216°
⇒
a = 24°
∠A = 5 (a + 2) = 5 (24 + 2)° = 130°.
∴
10. In a quadrilateral ABCD, it is being given that AB || DC, ∠A : ∠D
= 2 : 3 and ∠B : ∠C = 4 : 5. Find the measure of each angle of
quadrilateral ABCD.
Ans. Let ∠A = 2 x and ∠D = 3x and ∠B = 4 y and ∠C = 5 y
D
AB || DC, AD and BC are transversal
⇒
and
⇒
⇒
⇒
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∠A + ∠D = 180°
A
∠B + ∠C = 180°
2x + 3x = 180° and 4y + 5y = 180°
5x = 180° and
9y = 180°
x=
180°
and
5
4
y=
C
B
180°
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Question Bank
x = 36° and
y = 20°
∠A = 2 × 36° = 72°
∠B = 4 × 20° = 80°
∠ C = 5 × 20° = 100°
∠D = 3 × 36° = 108°.
and
11. In the adjoining figure, the sides BA and DC of a quadrilateral ABCD
have been produced to E and F respectively. Prove that
a + b = x + y.
⇒
Therefore,
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C
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E
Ans.
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B
F
y°
a°
b°
x°
A
D
A
C
B
B
∠EAD + ∠DAB = 180°
(Linear pair of angles)
b + ∠DAB = 180° ⇒ ∠DAB = 180° – b
⇒
and ∠FCB + ∠DCB = 180°
(Linear pair of angles)
a + ∠DCB = 180° ⇒ ∠DCB = 180° – a
⇒
In quadrilateral ABCD
(by property)
∠DAB + ∠ABC + ∠BCD + ∠CDA = 360°
180° – b + x + 180° – a + y = 360°
⇒
360° – b + x – a + y = 360°
⇒
x + y = 360° – 360° + a + b
⇒
Hence,
x + y = a + b.
12. Three angles of a quadrilateral are equal. If the Q
R
fourth angle is 69°; find the measure of equal 69°
x
angles.
Ans.Let each equal angle be x°
x + x + x + 69° = 360°
∴
3x = 360° – 69°
⇒
x
x
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3x = 291° ⇒ x = 97°
⇒
Hence, each equal angle is 97°.
13. Calculate each angle of the quadrilateral if ∠P : ∠Q : ∠R : ∠S = 3 :
4 : 6 : 7 and then prove that PQ and SR are parallel to each other.
(i) Is PS also parallel to QR?
(ii) Assign a special name to quadrilateral PQRS.
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Q
Ans. ∵ ∠P : ∠Q : ∠R : ∠S = 3 : 4 : 6 : 7
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72°
Let ∠P = 3x, ∠Q = 4x, ∠R = 6x
R
108°
and
∠S = 7x
Thus, ∠P + ∠Q + ∠R + ∠S = 360°
⇒ 3x + 4x + 6x + 7x = 360°
⇒
20x = 360°
126°
54°
⇒
x = 18°
S
P
∠P = 3x = 3 × 18° = 54°
Therefore,
∠Q = 4x = 4 × 18° = 72°
∠R = 6x = 6 × 18° = 108°
∠S = 7x = 7 × 18° = 126°
Also,
∠Q + ∠R = 72° + 108° = 180°
And,
∠P + ∠S = 54° + 126° = 180°
Hence PQ || SR
As ∠P + ∠Q = 72° + 54° = 126°. Which is ≠ 180°.
∴ PS and QR are not parallel.
(ii) PQRS is a Trapezium as its one pair of opposite side is parallel.
14. Use the informations given in the following figure to find the value of x.
C
D
56°
80°
E
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–6)
x
3
(
B
70°
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Question Bank
Ans. Take A, B, C, D as the vertices of quadrilateral and BA is produced to
E (say).
Since,
∠EAD = 70°
∠DAB = 180° – 70° = 110°
∴
[∵ EAB is a straight line and AD stands on it
∴ ∠EAD + ∠DAB = 180°]
∴ 110° + 80° + 56° + 3x – 6° = 360°
[∵ sum of interior angles of a quadrilateral = 360°]
⇒
3x = 360° – 110° – 80° – 56° + 6°
⇒
3x = 360° – 240° = 120°
x = 40°
∴
15. The following figure shows a trapezium in which sides AB and DC
are parallel.If ∠A : ∠D = 4 : 5, ∠B = (3x – 15)° and find each angle
of the trapezium ABCD.
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A
B
Ans. Let
∠A = 4y, ∠D = 5y
Since
∠A + ∠D = 180°
4y + 5y = 180°
∴
9y = 180°
⇒
⇒ y = 20°
∴ ∠A = 4 × 20° = 80°, ∠D = 5 × 20° = 100°
Again,
∠B + ∠C = 180°
∴ 3x – 15° + 4x + 20° = 180°
7x = 180° – 5° ⇒ 7x = 175° ⇒ x = 25°
∴
Math Class VIII
[∵ AB || DC]
[∵ AB || DC]
∠B = 75° – 15° = 60°
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and
∠C = 4 × 25° + 20° = 100° + 20° = 120°
16. PQRS is a parallelogram whose diagonals intersect at M.
If ∠PMS = 54°, ∠QSR = 25° and ∠SQR = 30°; find:
(i) ∠RPS
(ii) ∠PRS
(iii) ∠PSR .
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S
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Ans.
25°
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54°
M
30°
Q
P
Given: Parallelogram PQRS in which diagonals PR and QS
intersect at M.
Such that ∠PMS = 54°; ∠QSR = 25° and ∠SQR = 30°
To find: (i) ∠ RPS
Proof: QR || PS
⇒
But
Therefore,
(ii)
∠ PRS
(iii) ∠PSR
∠PSQ = ∠SQR
(Alternate angles)
∠SQR = 30°
(Given)
∠PSQ = 30°
In ∆SMP, ∠PMS + ∠PSM + ∠MPS = 180°
or 54° + 30° + ∠RPS = 180°
∠RPS = 180° – 84° = 96°
Now, ∠PRS + ∠RSQ = ∠PMS
∴
∠PRS + 25° = 54°
∠PRS = 54° – 25° = 29°
∠PSR = ∠PSQ + ∠RSQ
= 30° + 25° = 55°
Hence (i) ∠RPS = 96° (ii) ∠PRS = 29° (iii) ∠PSR = 55° .
Math Class VIII
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17. ABCD is a parallelogram. What kind of quadrilateral is it if :
(i) AC = BD and AC is perpendicular to BD ?
(ii) AC is perpendicular to BD but is not equal to it ?
(iii) AC = BD but AC is not perpendicular to BD ?
D
Ans. (i)
AC = BD (Given)
and
AC ⊥ BD (Given)
i.e. Diagonals of quadrilateral are
equal and they are perpendicular
to each other.
A
D
Thus, ABCD is square.
(ii)
AC⊥ BD (Given)
But AC and BD are not equal.
∴ ABCD is a rhombus.
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(iii) AC = BD but AC and BD are not
perpendicular to each other.
Hence, ABCD is a rectangle.
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18. Prove that the diagonals of a parallelogram bisect each other.
Ans. Given:Parallelogram ABCD so A | | CD & AB = CD.
To prove: OA = OC and OB = OD
In ∆ODC and ∆OAB
Proof:
AB | | CD (ABCD || gm)
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∠1 = ∠2 (alternate angles)
D
3
∠3 = ∠4 (alternate angles)
O
AB = CD (opposite sides of || gm)
∴ ∆ODC ≅ ∆OAB (ASA axiom)
4
2
∴
OD = OB, OC = OA
A
∴
Diagonals of || gm bisect each other.
19. If O is a point within a quadrilateral ABCD, prove that:
OA + OB + OC + OD > AC + BD.
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B
C
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O
B
A
Ans.Given: In quadrilateral ABCD,
O is any point within it OA, OB, OC and OD are joined.
To prove: OA + OB + OC + OD > AC + BD
Proof: In ∆OAC, OA + OC > AC
…(i)
(Sum of two sides of a triangle in greater than its third side)
Similarly in ∆OBD,
OB + OD > BD
…(ii)
Adding (i) and (ii), we get
OA + OC + OB + OD > AC + BD
⇒ OA + OB + OC + OD > AC + BD.
20. If the diagonals of a parallelogram are of equal lengths, the
parallelogram is a rectangle. Prove it.
Ans.Given: Parallelogram ABCD in which AC = BD
To Prove: ABCD is rectangle.
Proof: In ∆ABC and ∆ABD
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AB = AB (Common)
D
C
AC = BD (Given)
BC = AD (opposite sides of parallelogram)
∴ ∆ABC ≅ ∆ABD(S.S.S. Rule)
∴ ∠A = ∠B
A
B
But, AD || BC (∴ opposite sides of parallelogram are parallel)
∠A + ∠B = 180°
∴
Thus,
∠A = ∠B = 90°
and
∠D = ∠C = 90°
Hence, ABCD is a rectangle.
21. In a parallelogram ABCD, E is mid-point of AD and F is mid-point
of BC. Prove that BFDE is a parallelogram.
Ans. Parallelogram ABCD in which E and F are mid-points of AD and
BC respectively.
D
C
To Prove: BFDE is a Parallelogram
Proof: ∵ E is mid-point of AD. (Given) E
F
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AD
∴
2
Also, F is mid-point of BC
DE =
∴
But
A
B
(Given)
1
BC
2
AD = BC (opposite sides of parallelogram)
BF = DE
BF =
∴
Again, AD || BC
DE || BF
⇒
Now
DE || BF and DE = BF
Hence, BFDE is a parallelogram.
22. In parallelogram ABCD, E is mid-point of side AB and CE bisects
angle BCD. Prove that :
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(i) AE = AD,
(ii) DE bisects ∠ADC and
(iii) Angle DEC is a right angle.
Ans. Given: Parallelogram ABCD in which D
4
E is mid-point of AB and CE bisects
∠BCD.
A
To Prove: (i) AE = AD
(ii) DE bisects ∠ADC
(iii) ∠DEC = 90°
Const. Join DE
Proof: (i) AB || CD and CE bisects it.
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(Given)
(alternate angles)
…(i)
∠1 = ∠3
∴
But
(Given)
…(ii)
∠1 = ∠2
From (i) and (ii) we have
∠2 = ∠3
BC = BE
(sides opposite to equal angles)
∴
But
BC = AD
(opposite sides of parallelogram)
and
BE = AE
AD = AE
(ii) Now, AD = AE ⇒ ∠4 = ∠5 (angles opposite to equal sides)
But
∠5 = ∠6
(alternate angles)
∠4 = ∠6
⇒
∴ DE bisects ∠ADC.
(iii) Now AD || BC ⇒ ∠D + ∠C = 180°
2∠6 + 2∠1 = 180°
∴
∵ DE and CE are bisectors
180°
∠6 + ∠1 =
2
∠6 + ∠1 = 90°
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But In ∆DEC ∠DEC + ∠6 + ∠1 = 180°
∴ ∠DEC + 90° = 180°
∠DEC = 180° – 90° = 90°
Hence, the result.
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C
23. In the adjoining figure, ABCD is a
Q
parallelogram and AP || CQ. Prove that
(i) ∆OAP ≅ ∆OCQ
O
(ii) AP = CQ
(iii) APCQ is a parallelogram.
P
Ans.(i) In ∆OAP and ∆OCQ
B
A
∠OAP = ∠OCQ (alternate angles)
OA = OC
(diagonals of parallelogram
bisect each other)
∠AOP = ∠COQ (vertically opposite angles)
∆OAP ≅ ∆OCQ (A.S.A.)
∴
(ii) As,
∆OAP ≅ ∆OCQ
AP = CQ
(c.p.c.t.)
∴
(iii) As AP = CQ and AP || CQ
∴ APCQ is a parallelogram.
(∵ when one pair of sides of a quadrilateral is equal and
parallel then, it is a parallelogram.)
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C
24. In the adjoining isosceles trapezium ABCD,
102°
∠C = 102°. Find all the remaining angles of
the trapezium.
Ans. AB || CD ∴ ∠B + ∠C = 180°
A
B
(∵ adjacent angles on the same side of transversal are
supplementary)
⇒ ∠B + 102°= 180° ⇒ ∠B = 180° – 102° = 78°
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As
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AD = BC
(given)
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∠A = ∠B = 78°
∴
∠A + ∠B + ∠C + ∠D = 360°
⇒ 78° + 78° + 102° + ∠D = 360°
∠D + 258° = 360° ⇒ ∠D = 102°
25. In the adjoining figure, ABCD is a rhombus and DCFE is a square, if
∠ABC = 56°, find
(i) ∠DAG,
(ii) ∠FEG,
E
F
(iii) ∠GAC,
(iv) ∠AGC
Ans. (i) ABCD is a rhombus,
∴ AB = BC = CD = DA …(i)
G
C
D
Also, DCFE is a square
∴ DC = CF = FE = ED…(ii)
From (i) and (ii), we have,
56°
ED = AD
B
A
In ∆EDA,
∠EDA + ∠DAE + ∠DEA = 180°
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(90° + 56°) + 2∠DAE = 180° (∵ ∠DAE = ∠DEA as ED = AD)
2∠DAE = 180° – 146° = 34°
34°
= 17°
2
∠DAE = ∠DAG = 17°
∴
(ii)
∠FEG = ∠FED – ∠DEG
= 90° – 17° = 73° (∵ DCFE is a square)
(iii) In ∆DAC,
AD = DC
∠DAC = ∠DCA = x
∴
Also, ∠D = ∠B = 56°
(opposite angles of rhombus)
56° + 2x = 180°
∴
2x = 180° – 56°
2x = 124
⇒
⇒
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∠DAE =
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Question Bank
x = 62°
⇒
⇒ ∠DAC = 62°
Now,
∠GAC = ∠DAC – ∠DAG = 62° – 17° = 45°
(iv) In ∆DEG, ∠D + ∠DEG + ∠DGE = 180°
90° + 17° + ∠DGE = 180°
⇒ ∠DGE= 180° – 107° = 73°
(vertically opposite angle to ∠DGE)
∴ ∠AGC = 73°
26. In the diagonals of a parallelogram are equal in length, show that the
C
D
parallelogram is a rectangle.
Ans. Given: ABCD is a parallelogram whose
diagonals are AC and BD and AC = BD.
To prove: ABCD is a rectangle.
Proof: In ∆ABC and ∆ABD
AB = AB (common) A
B
BC = AD (opposite sides of a parallelogram)
AC = BD (given)
∆ABC ≅ ∆ABD (SSS axiom)
∴
∠ABC = ∠BAD (c.p.c.t)
∴
But ∠ABC + ∠BAD = 180°
(co-interior angles)
∠ABC = ∠BAD = 90°
∴
Hence, a parallelogram whose each angles is 90° is a rectangle.
27. In the adjoining figure, ABCD is
D
C
a square and CDE is an
equilateral triangle. Find:
(i) ∠AED (ii) ∠EAB(iii) reflex ∠AEC
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Ans. Given: ABCD is a square and CDE is an
equilateral triangle. AE is joined.
To find: (i) ∠AED
(ii) ∠EAB
(iii) reflex ∠AEC
∆EDC is an equilateral triangle
ED = DC = EC
...(i)
∴
and
AD = DC = AB = BC (ABCD is a square) ...(ii)
From (i) and (ii) ∴ ED = AD
(i) In ∆ADE,
AD = ED
(equal sides)
∠AED = ∠EAD
∴
But
∠ADE = ∠ADC – ∠EDC = 90° – 60° = 30°
∴ ∠AED + ∠EAD = 180° – 30° = 150°
1
and
∠AED = ∠EAD = × 150° = 75°
2
(ii)
∠DAB = 90°
∠EAB = ∠DAB – ∠AED = 90° – 75° = 15°
∴
(iii)
Reflex ∠AEC = 360° – (∠AED + ∠DEC)
= 360° – (75° + 60°)
= 360° – 135° = 225°
28. In the adjoining figure, ABCD is a parallelogram and AX || CY. Prove
that
(i) AX = CY
(ii) AXCY is a parallelogram
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Ans. Given: In parallelogram ABCD, AX || CY
AC and BD are joined AY and CX are also joined.
To prove: (i) AX = CY
(ii) AXCY is a parallelogram
Proof: In ∆AOX and ∆COY
AO = CO
(Diagonals of a parallelogram bisect each other)
∠AOX = ∠COY (vertically opposite angles)
∠XAO = ∠YCO (alternate angles)
∆AOX ≅ ∆COY (ASA axiom)
∴
AX = CY
(c.p.c.t)
∴
But AX || CY (given)
D
C
Hence AXCY is a parallelogram.
P
29. In the adjoining figure, ABCD is a
parallelogram and ∠A = 120°. If the
bisector of ∠A and ∠B meet at a point
P. Show that ∠APB is a right angle.
A
B
Ans.In parallelogram ABCD
∠A = 120°
(given)
But
∠A + ∠B = 180°
(co-interior angles)
120° + ∠B = 180°
⇒
∠B = 180° – 120° ⇒ ∠Β = 60°
⇒
1
1
Now
∠PAB = ∠A = × 120° = 60°
2
2
1
1
∠PBA = ∠B = × 60° = 30°
2
2
In ∆APB, ∠APB + ∠PAB + ∠PBA = 180° (by property)
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⇒ ∠APB + 60° + 30° = 180°
∠APB + 90° = 180°
⇒
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∠APB = 180° – 90° = 90°
⇒
F
Hence, ∠APB is a right angle.
C
D 26°
30. In the adjoining figure, ABCD is a
E 32°
parallelogram and EF is a line segment
such that EF || AC. If ∠ADE = 32° and
∠CDF = 26°. Find the measure of
∠ABC.
A
B
Ans.
∠DCA = ∠CDF
(alternate anlges)
∠DCA = 26° and ∠CAD = ∠ADE
(alternate angles)
⇒
∠CAD = 32°
⇒
(by property)
In ∆ADC ∠ADC + ∠ACD + ∠CAD = 180°
⇒ ∠ADC + 26° + 32° = 180°
∠ADC + 58° = 180°
⇒
∠ADC = 180° – 58° = 122°
⇒
∠ADC = ∠ABC
(opposite angle)
Hence,
∠ABC = 122°
31. The diagonals AC and BD of a rhombus intersect each other at O.
Prove that:
AB2 + BC2 + CD2 + DA2 = 4 (OA2 + OB2)
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B
A
Ans. We know that the diagonals of a rhombus bisect each other at right
angles.
In right ∆AOB
By Pythagoras theorem
AB2 = AO 2 + BO 2
…(i)
Similarly
BC2 = BO 2 + CO 2
…(ii)
CD2 = CO 2 + DO 2
…(iii)
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DA2 = DO 2 + AO 2
…(iv)
Adding (i), (ii), (iii) and (iv), we have,
AB2 + BC2 + CD2 + DA2
= AO2 + BO2 + BO2 + CO2 + CO2 + DO2 + DO2 + AO2
= 2 [AO2 + BO2 + CO2 + DO2]
= 2 [AO2 + BO2 + AO2 + BO2]
= 2 [2 AO2 + 2 BO2] = 4 [AO2 + BO2]
Hence proved.
32. In an isosceles trapezium, prove that the opposite angles are
supplementary.
Ans.Given: ABCD is an isosceles trapezium in which AD = BC and
AB || DC
To prove: ∠A + ∠C = 180° and ∠B + ∠D = 180°
Construction: From C, draw CE || DA D
C
meeting AB at E
Proof: ∵ DC || AB and AD || CE
∴ AECD is a parallelogram
AD = CE
∴
But
AD = CB (given)
⇒
CE = CB
A
E
B
Then, ∠CEB = ∠CBE
But
∠CEB = ∠D
(corresponding angles)
⇒
∠D = ∠CBE or ∠B
(co-interior angles)
∴ DC || AB Then, ∠A + ∠D = 180°
∠B + ∠D = 180°
(∵ ∠D = ∠B)
⇒
Similarly, we can prove that
∠A + ∠C = 180°.
Hence proved
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33. In the adjoining figure, ABCD is a
parallelogram E is the midpoint of AB
and CE bisects ∠BCD. Prove that.
(i) AE = AD
(ii) DE bisects ∠ADC
(iii) ∠DEC = 90°
C
D
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B
A
B
E
Ans. Given: In parallelogram ABCD, E is mid point of AB and EC bisects
∠BCD.
To prove: (i) AE = AD
(ii) DE bisects ∠ADC
(iii) ∠DEC = 90°
Proof: (i) ∵ ECB is the bisector of ∠BCD
∠BCE = ∠ECD
…(i)
∴
But AB || CD
∠ECD = ∠CEB (alternate angles)
…(ii)
∴
From (i) and (ii), we have
∠BCE = ∠CEB
EB = BC
(sides opposite to equal angles)
∴
But
EB = AE
(∵ E is mid point of AB)
And
BC = AD (opposite sides of a parallelogram)
Thus,
AE = AD
(ii) ∵
∠AED = ∠ADE (angles opposite to equal sides)
But
∠AED = ∠EDC (alternate angles)
∠ADE = ∠EDC
∴
Hence, DE bisects ∠ADC.
(iii) In parallelogram ABCD,
∠ADC + ∠BCD = 180°
(co-interior angles)
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⇒
1
1
∠ADC + ∠BCD = 90°
2
2
⇒
∠EDC + ∠EDC = 90°
Now in ∆DEC, ∠EDC + ∠ECD + ∠DEC = 180°
Z
B
(Sum of angles of a triangle is 180°)
⇒
⇒
90° + ∠DEC = 180°
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∠DEC = 180° – 90° = 90°
Hence, ∠DEC = 90°.
34. P and Q are points of trisection of the diagonal BD of a parallelogram
ABCD. Prove that CQ || AP.
Ans. Given: In parallelogram ABCD, diagonals AC and BD intersect each
other at O. P and Q are the points which intersect BD. Join CQ and
AP.
To prove: CQ || AP
Construction: Join CP and AQ
Proof: P and Q bisect BD
∴
But,
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BD
3
1
OD = OB = BD
2
PB = DQ =
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Question Bank
∴
OD – DQ = OB – PB
1
1
1
1
BD – BD = BD – BD
2
3
2
3
1
1
BD
=
BD
⇒
6
6
OQ = OP and OA = OC
⇒
∴ diagonals AC and PQ bisect each other at O.
∴ APCQ is a parallelogram.
or
CQ || AP.
∴ AP || CQ
Hence proved.
35. Calculate the sum of angles of a polygon with :
(i) 10 sides
(ii) 12 sides
(iii) 20 sides
(iv) 25 sides.
Ans. (i) Number of sides n = 10
Sum of angles of polygon = (n – 2) × 180°
= (10 – 2) × 180° = 1440°
(ii) Number of sides n = 12
Sum of angles of polygon = (n – 2) × 180°
= (12 – 2) × 180°
= 10 × 180° = 1800°
(iii) Number of sides n = 20
∴ Sum of angles of Polygon = (n – 2) × 180°
= (20 – 2) × 180° = 3240°
(iv) Number of sides n = 25
∴ Sum of angles of polygon = (n – 2) × 180°
= (25 – 2) × 180° = 4140°.
36. Find the number of sides in a polygon if the sum of its interior angles
is :
(i) 900°
(ii) 1620°
(iii) 16 right-angles
(iv) 32 right-angles.
Ans. (i) Let number of sides be n
Sum of angles of polygon be 900°
⇒
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∴ (n – 2) × 180° = 900°
900°
⇒ n–2=
⇒ n–2=5 ⇒ n =5+2=7
180°
(ii) Let number of sides = n
Sum of angles of polygon = 1620°
∴ (n – 2) × 180° = 1620°
1620°
⇒
n–2=
⇒ n – 2 = 9 ⇒ n = 9 + 2 = 11
180°
(iii) Let number of sides = n
Sum of angles of polygon = 16 right angles
= 16 × 90° = 1440°
Hence, (n – 2) × 180° = 1440°
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1440°
⇒ (n – 2) = 8 ⇒ n = 8 + 2 = 10
180°
(iv) Let number of sides = n
Sum of angles of polygon = 32 right angles
= 32 × 90 = 2880°
Hence, (n – 2) × 180° = 2880°
⇒
(n – 2) =
2880°
⇒ n – 2 = 16 ⇒ n = 16 + 2 = 18
180°
37. Is it possible to have a polygon; the sum of whose interior angles is :
(i) 870°
(ii) 2340°
(iii) 7 right-angles
(iv) 4500°
Ans. (i) Let number of sides = n
Sum of angles = 870°
(n – 2) × 180° = 870°
⇒
n–2=
29
29
41
870°
+2=
⇒n–2=
⇒n=
6
6
6
180°
Which is not a whole number.
Hence, it is not possible to have a polygon.
⇒
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n–2=
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Question Bank
(ii) Let number of sides = n
Sum of angles = 2340°
Hence, (n – 2) × 180° = 2340°
2340°
⇒ n – 2 = 13 ⇒ n = 13 + 2 = 15
⇒
n – 2=
180°
Which is a whole number.
Hence, it is possible to have a polygon.
(iii) Let number of sides be n
Sum of angles = 7 right angles = 7 × 90° = 630°
Hence, (n – 2) × 180° = 630°
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630°
7
7
11
⇒ n–2=
⇒n = +2=
180°
2
2
2
Which is not a whole number. Hence it is not a polygon.
(iv) Let number of sides be n
∴ (n – 2) × 180° = 4500°
⇒
n–2=
4500°
⇒ n – 2 = 25 ⇒ n = 25 + 2 = 27
180°
Which is a whole number.
Hence, it is possible to have a polygon.
38. The angles of a pentagon are x°, (x – 10°), (x + 20°), (2x – 44°) and
(2x – 70°). Calculate x.
Ans.Sum of interior angles of pentagon is 540°.
(∵ n = 5)
∴ x + (x – 10)° + (x + 20)° + (2x – 44)° + (2x – 70)° = 540°
x + x – 10° + x + 20° + 2x – 44° + 2x – 70° = 540°
7x – 104° = 540° ⇒ 7x = 644°
⇒
644°
x=
= 92°.
⇒
7
39. The exterior angles of a pentagon are in the ratio 1 : 2 : 3 : 4 : 5. Find
all the interior angles of the pentagon.
Ans. Let the exterior angles of pentagon are x, 2x, 3x, 4x and 5x.
⇒
Math Class VIII
n – 2=
24
Question Bank
We know that sum of exterior angles of polygon is 360°.
∴ x + 2x + 3x + 4x + 5x = 360° ⇒ 15x = 360°
360°
x
=
⇒
⇒ x = 24°
15
Thus, exterior angles are 24°, 48°, 72°, 96°, 120°.
Interior angles are 180° – 24°, 180° – 48°, 180° – 72°, 180° – 96°,
180° – 120° i.e., 156°, 132°, 108°, 84°, 60°.
41. The ratio between an exterior angle and the interior angle of a regular
polygon is 1 : 5. Find
(i) the measure of each exterior angle
(ii) the measure of each interior angle
(iii) the number of sides in the polygon
Ans. (i) Let exterior angle be x and interior angle be 5x.
x + 5x = 180° ⇒ 6x = 180° or x = 30°
∴
Thus, the measure of each exterior angle is 30°.
(ii) The measure of each interior angle is 5x = 5 × 30° = 150°
(2n – 4)
right angles
(iii) Each interior angle =
n
150n = (2n – 4) × 90°
∴
150n = 180n – 360° ⇒ 30n = 360°
⇒
Thus,
n = 12.
42. Find the sum of exterior angles obtained on producing, in order, the
sides of a polygon with :
(i) 7 sides
(ii) 10 sides
(iii) 250 sides.
Ans. (i) Number of sides n = 7
Sum of interior and exterior angles at one vertex = 180°
Sum of all interior and exterior angles = 7 × 180° = 1260°
Sum of interior angles = (n – 2) × 180°
= (7 – 2) × 180° = 900°
∴ Sum of exterior angles = 1260° – 900° = 360°
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(ii) Number of sides n = 10
Sum of interior and exterior angles = 10 × 180° = 1800°
But sum of interior angles = (n – 2) × 180°
= (10 – 2) × 180° = 1440°
Sum of exterior angles = 1800° – 1440° = 360°
(iii) Number of side n = 250
Sum of all interior and exterior angles = 250 × 180° = 45000°
But sum of interior angles = (n – 2) × 180° = (250 – 2) × 180°
= 248 × 180° = 44640°
Hence, sum of exterior angles = 45000° – 44640° = 360°
43. The sides of a hexagon are produced in order. If the measures of
exterior angles so obtained are (6x – 1)°, (10x + 2)°, (8x + 2)°
(9x – 3)°, (5x + 4)° and (12x + 6)°; find each exterior angle.
Ans.Sum of exterior angles of hexagon formed by producing sides of
order = 360°
∴ (6x – 1)° + (10x + 2)° + (8x + 2)° + (9x – 3)° + (5x + 4)° +
(12x + 6)° = 360°
⇒
50x + 10° = 360° ⇒ 50x= 360° – 10° = 350°
350°
= 7°
x=
50
Thus, angles are (6x – 1)°; (10x + 2)°; (8x + 2)°; (9x – 3)°;
(5x + 4)° and (12x + 6)°
i.e., (6 × 7 – 1)°; (10× 7 + 2)°; (8 × 7 + 2)°; (9 × 7 – 3)°;
(5 × 7 + 4)°; (12 × 7 + 6)°
i.e., 41°; 72°, 58°; 60°; 39° and 90°
44. The figure, given below, shows a pentagon ABCDE with sides AB
and ED parallel to each other, and ∠B : ∠C : ∠D = 5 : 6 : 7.
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D
E
C
A
(i)
(ii)
(iii)
Ans. (i)
B
Z
B
Using formula, find the sum of interior angles of the pentagon.
Write the value of ∠A + ∠E
Find angles B, C and D.
Sum of interior angles of the pentagon = (5 – 2) × 180°
= 3 × 180° = 540°
[∵ sum for a polygon of n sides = (n – 2) × 180°]
(ii) Since AB || ED ∴ ∠A + ∠E = 180°
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(iii) Let ∠B = 5 x, ∠C = 6 x, ∠D = 7 x
∴ 5x + 6x + 7x + 180° = 540°( ∠A + ∠E = 180° )
⇒ 18x = 540° – 180° ⇒ 18x = 360° ⇒ x = 20°
Hence, ∠B = 5 × 20° = 100°, ∠C = 6 × 20° = 120°,
∠D = 7 × 20° = 140° .
45. In a regular pentagon ABCDE, draw a diagonal BE and then find the
measure of :
(i) ∠BAE
(ii) ∠ABE
(iii) ∠BED
Ans. (i) Since number of sides in the pentagon = 5
360°
= 72°
5
∴ ∠BAE = 180° – 72° = 108°
(ii) In ∆ABE, AB = AE
∠ABE = ∠AEB
∴
But ∠BAE + ∠ABE + ∠AEB = 180°
∴ Each exterior angle =
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Question Bank
∴ 108° + 2∠ABE = 180°
2∠ABE = 180° – 108° = 72° ⇒ ∠ABE = 36°
⇒
(iii) Since
∠AED = 108°
[∵ each interior angle = 108°]
∠AEB = 36°
⇒
∠BED = 108° – 36° = 72°
∴
46. Find the number of sides in a regular polygon if its interior and
exterior angles are equal.
Ans. Let interior and exterior angle of polygon be x.
∴ x + x = 180° ⇒ 2x = 180° ⇒ x = 90°
(2n – 4)
× 90°
Hence, each interior angle =
n
(2n – 4)
× 90° ⇒ n = 2n – 4 ⇒ n = 4
90°
=
⇒
n
47. The sum of interior angles of a regular polygon is twice the sum of
its exterior angles. Find the number of sides of the polygon.
Ans.Sum of interior angles = (2n – 4) × 90°
Sum of exterior angles = 360°
According to the given condition
(2n – 4) × 90°= 2 × 360 ⇒ 2n – 4 = 8
n=6
∴
48. An exterior angle of a regular polygon is one-fourth of its interior
angle. Find the number of sides in the polygon.
Ans.Let measure of interior angle = x°
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Then,
∴
exterior angle =
x+
1
x°
4
1
x = 180°
4
4
⇒
5
Therefore, each interior angle is 144°
x = 180° ×
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⇒
5
x = 180°
4
⇒
x = 144°
Question Bank
(2n – 4)
× 90
⇒
n
144n = 180n – 360 ⇒ 180n – 144n = 360
⇒
36n = 360 ⇒ n = 10
⇒
49. In a hexagon ABCDEF, side AB is parallel to side FE and ∠B : ∠C :
∠D : ∠E = 6 : 4 : 2 : 3. Find ∠B and ∠D .
144° =
Ans.
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B
C
D
Given: Hexagon ABCDEF in which AB || EF and
∠B : ∠C : ∠D : ∠E = 6 : 4 : 2 : 3.
To find: ∠B and ∠D
Proof: Number of sides n = 6
∴ Sum of interior angles = (n – 2) × 180° = (6 – 2) × 180° = 720°
∵ AB || EF (Given) ∠A + ∠F = 180°
(Proved)
But ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 720°
∠B + ∠C + ∠D + ∠E + ∠180° = 720°
∴ ∠B + ∠C + ∠D + ∠E = 720° – 180° = 540°
Ratio = 6 : 4 : 2 : 3, Sum of parts = 15
6
∠B
=
× 540° = 216°
∴
15
2
× 540° = 72°
∠D =
15
Hence, ∠B = 216°; ∠D = 72°.
50. In the adjoining figure, ABCD is a parallelogram and P is a point on
BC. If area of ∆APD = 17.6 cm2, find
(i) the area of parallelogram ABCD
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Question Bank
(ii) the sum of the areas of triangles ABP and DPC.
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Ans. (i) Area of parallelogram ABCD = 2 × (area of ∆APD)
= 2 × 17.6 cm2 = 35.2 cm2
(on the same base AD and between the same parallels AD and BC)
(ii) area of parallelogram ABCD = ar (∆DPC)+ ar (∆DPA) + ar (∆PAB)
35.2 = ar (∆DPC) + ar (∆ABP) + 17.6
⇒ ar (∆DPC) + ar (∆ABP) = 35.2 – 17.6 = 17.6 cm2.
G
C
51. In the adjoining figure, ABCD is a rectangle D
with sides AB = 4 cm and AD = 6 cm. Find
(i) the area of parallelogram DEFC.
(ii) the area of ∆EFG.
Ans.Area of rectangle ABCD = 4 × 6 = 24 cm2 A
B E
F
(i) Area of parallelogram DEFC = area of rectangle ABCD = 24 cm2
(∵ on the same base CD and between the same parallels AF
and DG)
1
area of parallelogram DEFC
2
1
= × 24 = 12 cm2
2
(on the same base EF and between the same parallels AF and DG)
52. In the adjoining figure, ABCD is a parallelogram and its diagonals
AC and BD intersect at O. Prove that area of ∆OAB = area of
∆OBC = area of ∆OCD = area of ∆OAD.
(ii) area of triangle EFG =
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Ans. ∵ ABCD is a parallelogram. Therefore, diagonal bisect each other.
Hence O is the mid-point of AC as well as BD.
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A
B
Now, in ∆ABC, OB is the median.
1
area of ∆ABC
…(i)
2
1
and
area of ∆OBC = area of ∆ABC
…(ii)
2
(∵ median divides the triangle into two parts of equal area)
Similarly, In ∆ADC, OD is the median
1
…(iii)
∴ area of ∆OCD = area of ∆OAD = area of ∆ADC
2
But area of ∆ABC = area of ADC
(∵ diagonal divides the parallelogram into two triangles of equal
area)
From (i), (ii) and (iii), we have
⇒ area of ∆OAB = area of ∆OBC = area of ∆OCD = area of ∆OAD.
53. In trapezium ABCD, it is being given that AB || DC and diagonals AC
and BD intersect at O. Prove that:
C
D
(i) Area (∆DAB) = Area (∆CAB)
(ii) Area (∆AOD) = Area (∆BOC).
O
Ans. (i) Triangles on the same base AB and
between the same parallel (AB || DC)
are equal in area
A
B
i.e., area of (∆DAB) = Area of (∆CAB)
(ii) Also, Area of (∆ADC) = Area of (∆BCD)
∴
Math Class VIII
area of ∆OAB =
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Question Bank
⇒ Area of (∆AOD) + Area (∆ODC)
= Area of (∆BOC) + Area of (∆ODC)
Hence, Area of (∆AOD) = Area of (∆BOC)
54. In the adjoining figure, ABCD is a parallelogram, P is point on DC
and Q is a point on BC. Prove that ∆APB and ∆AQD are equal in
area.
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Q
B
A
Ans.Area of (∆APB) =
Area of (∆AQD) =
1
area of parallelogram ABCD …(i)
2
(by theorem)
1
area of parallelogram ABCD
2
…(ii)
(by theorem)
From (i) and (ii), we have
Area of (∆APB) = Area of (∆AQD)
Proved
55. In the adjoining figure, ABCD is a quadrilateral. A line through D,
parallel to AC, meets BC produced in P.
Prove that Area (∆ABP) = Area (quad. ABCD)
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A
B
Ans. Given: ABCD is a quadrilateral. A line through D is drawn parallel
to AC meeting BC produced in P, PA is joined.
To prove: area (∆ABP) = area (quad. ABCD)
Proof: ∆APC and ∆ADC are on the same base AC and between
the same parallel lines
Area (∆APC) = area (∆ADC)
∵
Adding area (∆ABC) to both sides,
area (∆APC) + area (∆ABC) = area (∆ADC) + area (∆ABC)
area (∆APB) = area (quad. ABCD)
⇒
Hence proved.
56. Given: Area of parallelogram ABCD = 60 cm2. Find, giving reasons,
(i) area of parallelogram ABEF;
(ii) area of triangle ABC.
F
D
E
A
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Ans. Parallelogram ABCD and Parallelogram ABEF are on same base
and between same parallel lines AB and FC.
ar (|| gm ABCD) = ar (|| gm ABEF)
∴
But
ar(|| gm ABCD) = 60 cm2
(Given)
ar(|| gm ABEF) = 60 cm2
∴
Again ∆ABC and || gm ABCD are on same base AB and between same
parallel lines.
1
1
ar(∆ABC)
=
ar
(||
gm
ABCD)
=
× 60 = 30 cm2
∴
2
2
2
2
Hence (i) 60 cm (ii) 30 cm .
57. In the adjoining figure, AB || DC and area of ∆ABD is 24 sq. units. If
C
D
AB = 8 units, find the height of ∆ABC.
Ans. Area of ∆ABD = 24 sq. units. AB = 8 units
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area of triangle =
1
× base × height
2
B
A
24
1
= 6 units
24 = × 8 × h ⇒ h =
4
2
∴ Height of triangle ABC = 6 units.
58. In the adjoining figure, AB || DC || EF, DA || EB and DE || AF. Prove
that area of parallelogarm gm DEFH = area of || gm ABCD.
B
A
D
H
E
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Ans. area of parallelogram ABCD = area of || gm ADEG
…(i)
(∵ on the same base AD and between the
same parallels AD and BE)
Also, area of || gm ADEG = area of || gm DEFH
…(ii)
(∵ on the same base base DE and between the
same parallels DE and AF)
A
From (i) and (ii), we have
area of || gm DEFH = area of || ABCD
59. In the adjoining figure, DE is parallel to the
E
D
side BC of ∆ABC, BE and CD intersect at
O. Prove that
Q
(i) area of ∆BED = area of ∆CED
(ii) area of ∆BOD = area of ∆COE
(iii) area of ∆ABE = area of ∆ADC
B
C
Ans.(i) area of ∆BED = area of ∆CED
(∵ Triangles on the same base DE and between the same
parallels DE and BC are equal in area)
(ii) As, area of ∆BED = area of ∆CED
(from (i))
Subtract area of ∆DOE from both sides,
ar (∆BED) – ar (∆DOE) = ar (∆CED) – ar (∆DOE)
we get, ar (∆BOD), ar (∆COE)
(iii) We know that,
ar (∆BDE) = are (∆CED) (from (i))
Add or (∆ADE) to both sides,
ar (∆BDE) + ar (∆ADE) = ar (∆CDE) = ar (∆ADE)
⇒
Hence, ar (∆ABE) = ar (∆ADC).
60. In ∆ABC, P and Q are mid-points of the sides AB and AC respectively.
If BC = 6 cm, AB = 5.4 cm and AC = 5 cm, calculate the perimeter of
the quadrilateral PBCQ.
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B
L
A
A
N
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Ans. P, Q are mid-points of AB and AC
∴ By mid-point theorem,
1
1
PQ = BC = × 6 = 3 cm.
2
2
1
1
Now, BP = AB = × 5.4 = 2.7 cm. (∵ P is mid-point)
2
2
1
1
QC = AC = × 5 = 2.5 cm. (∵ Q is mid-point)
2
2
Hence, perimeter of quadrilateral PBCQ
= PB + BC + CQ + PQ
= 2.7 + 6 + 2.5 + 3 = 14.2 cm.
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