F.3 Mathematics
M.C. Revision Exercise – Ch.4 Quadrilaterals
Solutions
Multiple-choice Questions
13083001
In the figure, ABCD is a parallelogram. Find x.
A. 27
B. 28
C. 29
D. 31
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Solution:
The answer is B.
∠DAB = ∠DCB (opp. ∠s of //gram)
3x = x + 56
2x = 56
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13083002
In the figure, PQRS is a rectangle. Find x.
A. 56°
B. 62°
C. 112°
D. 124°
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Solution:
The answer is D.
OS = OR (property of rectangle)
∠OSR = ∠ORS (base ∠s, isos. △)
= 28°
x =∠SOR (vert. opp. ∠s)
In △ORS,
∠SOR + ∠OSR + ∠ORS = 180° (∠ sum of △)
x + 28° + 28° = 180°
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13083003
In the figure, ABCD is a rhombus. Find x.
A. 20°
B. 30°
C. 35°
D. 45°
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Solution:
The answer is C.
As shown in the figure,
a = 90° (property of rhombus)
∠ABO = x (property of rhombus)
∠ABO + ∠BAO = a (ext. ∠ of △)
x + x + 20° = 90°
2x = 70°
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13083004
In the figure, ABCD is a square. Find x.
A. 97°
B. 128°
C. 143°
D. 166°
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Solution:
The answer is B.
∠EDO = 45° (property of square)
x = ∠EDO + 83° (ext. ∠ of △)
= 45° + 83°
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13083005
Which of the following conditions can be used to determine that ABCD is a parallelogram?
A. ∠A = ∠B, ∠C = ∠D
B. AB = AD, BC = CD
C. AD // BC, AD = BC
D. AB // DC, ∠B = ∠C
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Solution:
The answer is C.
AD // BC, AD = BC (given)
ABCD is a parallelogram. (2 sides equal and //)
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13083006
In the figure, ABCD is a parallelogram. ∠A :∠B :∠C :∠D can be
A. 1 : 2 : 3 : 4.
B. 1 : 2 : 2 : 1.
C. 2 : 1 : 2 : 1.
D. 2 : 2 : 1 : 1.
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Solution:
The answer is C.
In parallelogram ABCD,
∠A = ∠C and ∠B = ∠D
Only ∠A :∠B :∠C :∠D = 2 : 1 : 2 : 1 satisfies this condition.
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13083007
If a side of a rhombus is equal to one of its diagonal, then the large interior angle is
A. two times the small interior angle.
B. three times the small interior angle.
C. four times the small interior angle.
D. five times the small interior angle.
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Solution:
The answer is A.
As shown in the figure, if AB = BD (the length of a side = the length of a diagonal),
then △ABD is an equilateral triangle.
∠BAD = ∠ABD = 60° (property of equil. △)
and ∠DBC = 60° (property of rhombus)
∠ABC = ∠ABD + ∠DBC
= 60° + 60°
= 2∠BAD
The large interior angle is two times the small interior angle.
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13083008
In the figure, ABCD is a rectangle. AC = 4 cm. If , then AD =
A. 2 cm.
B. 4 cm.
C. .
D. .
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Solution:
The answer is A.
Let AD = x cm, then .
DC = AB, ∠ADC = 90° (property of rectangle)
In △ACD,
(Pyth. theorem)
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13083009
In parallelogram ABCD, AB = 10 cm. If the perimeter of the parallelogram is 30 cm, then AD =
A. 20 cm.
B. 10 cm.
C. 6 cm.
D. 5 cm.
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Solution:
The answer is D.
In parallelogram ABCD, AD = BC, AB = CD (opp. sides of //gram)
Perimeter of the parallelogram = AB + BC + CD + DA
= 2AB + 2AD
2AB + 2AD = 30 cm
10 cm + AD = 15 cm
AD
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13083010
In the figure, ABCD and DBFE are squares, then the ratio of the areas of square DBFE and square ABCD is
A. 2 : 1.
B. 1 : 2.
C. 1 : 1.
D. .
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Solution:
The answer is A.
Let the side of the square ABCD be x cm.
∠DAB = 90° (property of square)
(Pyth. theorem)
= (x2 + x2) cm2
= 2x2 cm2
Area of square DBFE =
Area of square ABCD =
Ratio of the areas of square DBFE and square ABCD = 2x2 : x2
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13083011
In the figure, AC is a diagonal of rectangle ABCD and DE ^ AC. If AC = 4 cm, DC = 2 cm, then DE =
A. .
B. .
C. 1 cm.
D. 2 cm.
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Solution:
The answer is A.
∠ADC = 90° (property of rectangle)
In △ACD,
(Pyth. theorem)
Consider the area of △ACD:
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13083012
In the figure, DB and AC are the diagonals of the rectangle ABCD. How many pairs of congruent triangles are there in the figure?
A. 2 pairs
B. 4 pairs
C. 6 pairs
D. 8 pairs
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Solution:
The answer is D.
In rectangle ABCD,
△ADE @ △BCE, △ABE @ △CDE, △ABD @ △BAC, △ABD @ △CDB,
△ABC @ △CDA, △BAC @ △CDB, △ABD @ △DCA, △CDB @ △DCA
There are 8 pairs of congruent triangles.
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13083013
In the figure, PQRS is a rectangle., PQ = 4 cm. Find the area of the rectangle.
A. 8 cm2
B.
C.
D. 12 cm2
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Solution:
The answer is A.
∠PQR = 90° (property of rectangle)
In △PQR,
PR2 = PQ2 + QR2 (Pyth. theorem)
Area of the rectangle = PQ ´ QR
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13083014
A quadrilateral with a pair of opposite sides parallel and equal is a
A. square.
B. rhombus.
C. rectangle.
D. parallelogram.
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Solution:
The answer is D.
According to the test for parallelogram (2 sides equal and //), this quadrilateral is a parallelogram.
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13083015
In the figure, PQRS is a parallelogram. Find ∠TSP.
A. 110°
B. 140°
C. 145°
D. 150°
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Solution:
The answer is C.
∠PSR + ∠SRQ = 180° (int. ∠s, PS // QR)
∠PSR + 70° = 180°
∠PSR = 110°
∠RST = ∠RTS (base ∠s, isos. △)
∠RST + ∠RTS = ∠SRQ (ext. ∠ of △)
2∠RST = 70°
∠RST = 35°
∠TSP = ∠PSR + ∠RST
= 110° + 35°
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13083016
In the figure, ABCD is a parallelogram. Find x.
A. 36°
B. 52°
C. 58°
D. 64°
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Solution:
The answer is B.
∠DAB + ∠ABC = 180° (int. ∠s, AD // BC)
∠DAB + 52° = 180°
∠DAB = 128°
In △ABF,
∠BAF + ∠AFB + ∠ABF = 180° (∠ sum of △)
∠BAF + 90° + 52° = 180°
∠BAF = 38°
In △ADE,
∠ADE = 52° (opp. ∠s of //gram)
∠DAE + ∠AED + ∠ADE = 180° (∠ sum of △)
∠DAE + 90° + 52° = 180°
∠DAE = 38°
∠DAB = ∠DAE + ∠EAF + ∠FAB
128° = 38° + x + 38°
x
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13083017
In the figure, ABCD is a rectangle. BC = 6 cm. Find AB.
A. 6 cm
B.
C. 12 cm
D.
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Solution:
The answer is B.
OB = OC (property of rectangle)
∠OBC = ∠OCB (base ∠s, isos. △)
∠AOB = ∠OBC + ∠OCB (ext. ∠ of △)
120° = 2 ´ ∠OBC
∠OBC = 60°
△OBC is an equilateral triangle.
OB = BC
= 6 cm
BD = 2 ´ OB
= 2 ´ 6 cm
= 12 cm
∠BCD = 90° (property of rectangle)
In △BCD,
(Pyth. theorem)
AB = DC (property of rectangle)
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13083018
In the figure, ABCD is a parallelogram. Find CF.
A.
B.
C.
D. 2 cm
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Solution:
The answer is B.
AD = BC = 6 cm (opp. sides of //gram)
AB = DC = 5 cm (opp. sides of //gram)
Consider the area of parallelogram ABCD:
AD ´ CE = AB ´ CF
6 ´ 4 = 5 ´ CF
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13083019
In the figure, ABCD is a parallelogram and the perimeter of △ABE is 18 cm. Find the perimeter of △ADE.
A. 14 cm
B. 16 cm
C. 18 cm
D. 20 cm
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Solution:
The answer is B.
AD = BC (opp. sides of //gram)
DE = BE (diagonals of //gram)
Perimeter of △ADE = AD + DE + EA
= BC + BE + EA
= (AB + BE + EA) – AB + BC
= (18 – 8 + 6) cm
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13083020
In the figure, ABCD is a square. Find the area of the square.
A. 144 cm2
B. 117 cm2
C. 90 cm2
D. 81 cm2
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Solution:
The answer is A.
△ABE ~ △FCE (AAA)
Let AB = x cm, then BE = (x – 3) cm.
Area of the square = 122 cm2
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13083021
In the figure, AB = BC. Find x.
A. 50°
B. 55°
C. 60°
D. 65°
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Solution:
The answer is D.
AB = BC (given)
∠BAC = ∠BCA = x (base ∠s, isos. △)
AD = DB (given)
AE = EC (given)
DE // BC (mid-pt. theorem)
∠AED = ∠ACB = x (corr. ∠s, DE // BC)
In △ADE,
∠ADE + ∠AED + ∠DAE = 180° (∠ sum of △)
50° + x + x = 180°
2x = 130°
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13083022
Find the unknowns in the figure.
A. b = 5, c = 18
B. b = 6, c = 16
C. b = 6.5, c = 15
D. b = 7, c = 18
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Solution:
The answer is D.
AD = DB (given)
DE // BC (given)
AE = EC (intercept theorem)
(mid-pt. theorem)
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13083023
Find the unknowns in the figure.
A. x = 7, y = 10
B. x = 8, y = 12
C. x = 9, y = 10
D. x = 10, y = 7
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Solution:
The answer is C.
AE = EC (given)
DE // BC (given)
DB = AD (intercept theorem)
(mid-pt. theorem)
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13083024
Referring to the figure, find x.
A. 6
B. 6.5
C. 7
D. 7.5
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Solution:
The answer is C.
Draw AG // BF, as shown in the figure.
ABFG and ABDH are parallelograms. (2 sides equal and //)
GF = HD (opp. sides of //gram)
= AB
= 5.5
EG = EF – GF
= 8.5 – 5.5
= 3
AC = CE and HC // GE (given)
AH = HG (intercept theorem)
(mid-pt. theorem)
x = CH + HD
= 1.5 + 5.5
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13083025
In the figure, AD is the height of △ABC. E is the mid-point of AB. EF ^ BC,
DC = BF and BC = 1, then FC =
A. .
B. .
C. .
D. .
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Solution:
The answer is D.
BE = EA (given)
EF // AD (corr. ∠s equal)
BF = FD (intercept theorem)
and DC = BF (given)
BF = FD = DC
BC = BF + FD + DC
1 = 3BF
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13083026
In △ABC, if D, E, F are the mid-points of BC, CA, AB respectively, then the perimeter of quadrilateral AFDE is
A. AD + BC.
B. AB + AC.
C. .
D. BC + AC.
-- ans --
Solution:
The answer is B.
According to the question, draw △ABC.
BF = FA and BD = DC (given)
FD = (mid-pt. theorem)
CE = EA and CD = DB (given)
DE = (mid-pt. theorem)
Perimeter of quadrilateral AFDE = AF + FD + DE + EA
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13083027
In the figure, ABCD is a square and AEFG is a rectangle. If JB = DG and AB = 2 cm, find the area of rectangle AEFG.
A.
B.
C.
D.
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Solution:
The answer is B.
Let DG = x cm, GI = y cm.
△DGI @ △JBE (ASA)
GI = BE (corr. sides, △s)
BE = y cm
AE = AB + BE
= (2 + y) cm
GF = GI + IF
IF = GF – GI
= AE – GI (property of rectangle)
= (2 + y – y) cm
= 2 cm
EF = AG (property of rectangle)
= AD – DG
= (2 – x) cm
△DGI ~ △EFI (AAA)
Area of rectangle AEFG = AE ´ EF
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13083028
Referring to the figure, which of the following must be correct?
I. AB // PQ
II. AB // XY
III. PQ // XY
A. I only
B. II only
C. I and II only
D. I, II and III
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Solution:
The answer is B.