Areas of Parallelograms & Triangles Test

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Areas of Parallelograms & Triangles Test - 6

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Question 1
1 / -0

There is a triangle ABC whose area is 264 cm². If we divide the triangle into two parts with a median AD and E is the mid pint of the median, then what will be the area of ΔBED?

A
33 cm²

B
64 cm²

C
66 cm²

D
74 cm²

Solution

AD is the median of ΔABC.
So, ar(△ABD) = ar(ΔACD)
ar(ΔABD) =× ar(ΔABC)
ar(ΔABD) = × 264 = 132 cm²---- (1)
In ΔABD, E is the mid point of AD. So, BE is the median.
ar(ΔBED) = ar(ΔABE)
ar(ΔBED) = ar(ΔABD)
ar(ΔBED) =× 132 ---- From eq (1)
ar(ΔBED) = 66 cm²
Question 2
1 / -0

In the given parallelogram ABCD, EFG is a triangle with area 10 cm², H and I are mid-points of EF and EG respectively and HI = 2 cm. Find the sum of areas of trapezium AEFD and EBCG, given that= FG.

A
70 cm²

B
60 cm²

C
50 cm²

D
40 cm²

Solution

In triangle EFG, we know that H and I are mid-points of EF and FG respectively.
Given, HI = 2 cm
We know that HI ∥ FG, therefore, HI =(Using Mid-Point theorem)
2 =
FG = 2 × 2
FG = 4 cm
Area of △EFG = 10 cm²- (i)

So, area of △EFG =
10 =
10 × 2 = 4 × height
Height = 20 ÷ 4
Height = 5 cm
Therefore, height of the parallelogram ABCD = 5 cm
Also,= FG
= 4
DC = 4 × 3
DC = 12 cm
Therefore, base of the parallelogram ABCD = 12 cm
Area of the parallelogram ABCD = base × height
= 12 × 5 = 60 cm²
Sum of areas of trapezium AEFD and EBCG = area of parallelogram ABCD – area of △EFG
= (60 – 10) cm²= 50 cm²
Question 3
1 / -0

If ABC ~ PQR, ar(ABC) = 16 cm² and ar(PQR) = 16 cm², which of the following statements is definitely true?

A
ABC and PQR are equilateral triangles.

B
ABC is isosceles.

C
ABC PQR

D
=

Solution

ABC ~ PQR

= =

1 =
AB = PQ, BC = QR, AC = PR

Hence, ABC PQR (By SSS similarity criterion)
Question 4
1 / -0

In parallelogram IJKL, IK is the diagonal. If the base of the △ILK is 6 cm and the height of △IJK is 4 cm, what will be the area of △ILK?

A
36 cm²

B
24 cm²

C
12 cm²

D
Cannot be determined

Solution

According to the given question, the parallelogram is as follows.

We know that the diagonal of a parallelogram divides it into two congruent triangles.
Therefore, height of ΔILK = height of ΔIJK = 4 cm
base of ΔILK = 6 cm
Area of ΔILK =
=
=
= 12 cm²
Question 5
1 / -0

In the triangle ABC, E and F are the midpoints of sides AB and AC, respectively.

Which of the following relationships is correct if the area of EOB and the area of FOC are equal?

A
EF ∦ BC

B
EF || BC

C
ar(BOC) = ar(EOF)

D
ar(EOB) = ar(BOC)

Solution

E and F are the respective midpoints of AB and AC of ABC.
Therefore, AE = EB and AF = FC.
Also, ar(EOB) = ar(FOC) (given)
ar(EOB + EOF) = ar(FOC + EOF)
ar(EBF) = ar(CEF)
Also, EBF and CEF lie on the same base EF.
Thus, EF || BC [Triangles on the same base and having equal areas lie between the same parallel lines.]
Question 6
1 / -0

In the given isosceles trapezoid MNOP, QR is the median and ST = 6 cm, and the area of trapezium QROP is 27 cm².Find the area of trapezium MNRQ, if PO = 7 cm.

A
33 cm²

B
36 cm²

C
39 cm²

D
49 cm²

Solution

In trapezium QROP, PO (a) = 7 cm
UT (Height) = 3 cm (Half of ST, because QR is the median.)
Let QR (b) be x cm.

Area of trapezium QROP =
27 =

27 =

27 × 2 = (7 + x) × 3
54 = 21 × 3x
3x = 54 – 21
3x = 33
x = 33 ÷ 3
x = 11
QR = 11

MNOP is an isosceles trapezoid and QR is the median.
We know that the median of a trapezoid is one-half of the sum of measures of the bases.

Therefore, QR =
11× 2 = 7 + MN
22 = 7 + MN
MN = 22 – 7
MN = 15 cm.

In trapezium MNRQ, QR (a) = 11 cm
MN (b) = 15 cm
SU (Height) = 3 cm

Area of trapezium MNRQ =
=
= 39 cm²
Question 7
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The area of the given parallelogram MNOP is 96 cm². MT meets QR at point S such that QS = SR. The area of △SRT is 6 cm²and of △QMS is 12 cm². Find the length of MT, if PO = 12 cm.

A
(5 + ) cm

B
(5 + ) cm

C
(5 + ) cm

D
(5 + ) cm

Solution

Area of parallelogram MNOP = base × height
96 = PO × QR
96 = 12 × QR
QR = 96 ÷ 12
QR = 8 cm
We know that, QS = SR
Therefore, QS and SR = 4 cm
Now, area of △SRT = 6 cm²
Area of △SRT =
6 =

SR × RT = 6 × 2
4 × RT = 12
RT = 12 ÷ 4
RT = 3 cm
Now, in △SRT,
(Hypotenuse)² = (perpendicular)² + (base)²(Pythagoras theorem)
(ST)² = (SR)² + (RT)²
(ST)² = (4)² + (3)²
(ST)² = 16 + 9
(ST)² = 25
ST =
ST = 5 cm
Now, area of △QMS = 12 cm²
Area of △QMS =
12 =
QS × QM = 12 × 2
4 × QM = 24
QM = 24 ÷ 4
QM = 6 cm
Similarly, in △QMS,
(Hypotenuse)² = (perpendicular)² + (base)²(MS)²= (QS)² + (QM)²(MS)² = (4)² + (6)²
(MS)² = 16 + 36
(MS)² = 52
MS =
MS =
Required length = ST + MS
= (5 + ) cm
Question 8
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If the area of a trapezium-shaped field is 720 m², the distance between the two parallel sides is 20 m and the length of one of the parallel sides is 35 m, then the length of the other parallel side is

A
35 m

B
36 m

C
37 m

D
40 m

Solution

Area of trapezium = (Sum of parallel sides) x Distance between the parallel sides

720 = 0.5 x (35 + y) x 20

72 = 35 + y

y = 72 - 35

y = 37 m
Question 9
1 / -0

In the given figure, two lines DE and FG are parallel to ABand the three regions CDE, DFGE and FABG have equal areas. The ratio of CD to FA is equal to _____.

A

B

C

D

Solution

Since DE || FG || AB,

⇒ Δ CDE ~ ΔCFG ~ ΔCAB

Also, =
Area of similar Δs is proportional to the square of sides, =
Similarly, =

Or =

Now, = = = ×

=
Question 10
1 / -0

What is the ratio of the areas of trapezium ABCE and △BCD, if the area of △ACE is one-third of the area of △ADE and the area of parallelogram ABCD is 96 sq. units?

A
2 : 1

B
3 : 1

C
5 : 4

D
4 : 5

Solution

Given,
Area of the parallelogram ABCD = 96 sq. units
Therefore, the area of △ADC = × 96 = 48 sq. units
(Area of a triangle on the same base and between the same parallel lines to that of a parallelogram is half.)
Let the area of △ADE = 'x' sq. units
Therefore, the area of △ACE =× x
Also, area (△ADE + △ACE) = area (△ADC)
x += 48
= 48
4x = 48 × 3
4x = 144
x = 36 sq. units.
Therefor,e the area of △ADE = 36 sq. units
And area of △ACE = × 36 = 12 sq. units
Area of △ABC = area of △ BCD = × 96 = 48 sq. units
(Area of a triangle on the same base and between the same parallel lines to that of a parallelogram is half.)
Now, area of the trapezium ABCE = Area of △ACE + Area of △ ABC = (12 + 48) sq. units = 60 sq. units
Now, the ratio of the areas of the trapezium ABCE and △BCD = 60 : 48
or 5 : 4
Hence, option 3 is the answer.
Question 11
1 / -0

In the given figure, line is parallel to DC. ADC and BCD are two triangles on the same base DC. If ADC = BCD = 90°, which of the following relationships is true?

A
ar(ADC) = ar(BCD)

B
ar(ADC) = 2ar(BCD)

C
ar(ADC) ar(BCD)

D
ar(ACD) = ar(ABCD)

Solution

Given: || DC and ADC and BCD are triangles on the same base DC.

Therefore, ar(ADC) = ar(BCD)
[Triangles on the same base and between the same parallels are equal in area]
Question 12
1 / -0

In an equilateral triangle ABC, a line DE is drawn parallel to AB that meets sides BC and AC at points D and E, respectively. The line DE divides BC in 3 : 2 i.e. BD : DC = 3 : 2. What is the ratio of area of CDE to that of area of quadrilateral ABDE?

A
4 : 25

B
21 : 25

C
4 : 21

D
25 : 8

Solution

ABC ~ EDC
⇒ = =
=
⇒ 1 + = 1 +
So, ar(CDE) : ar(ABDE) = 4 : 21
Question 13
1 / -0

In the given figure, F is the mid-point of side AB of the parallelogram ABCD. DF || CE and CF || DG. Find the ratio of the areas of the shaded regions to that of the trapezium GDCE.

A
2 : 3

B
1 : 2

C
2 : 1

D
1 : 3

Solution

Given,
F is the mid-point of side AB of the parallelogram ABCD.
Therefore,
Area of △ADF = Area of △BCF
Let,
Area of parallelogram ABCD = area of parallelogram FDCE = area of parallelogram FCDG = x sq. units
(Parallelograms on the same base and between the same parallel lines are equal in areas.)
Now,
Area of △ DFC =× area of parallelogram ABCD
=× x sq. units
(Area of a triangle on the same base and between the same parallel lines to that of a parallelogram is half.)
Also, area of △ADF = area of △FCB
We know that,
Area of parallelogram ABCD = area of △ADF + area of △FCB + area of △DFC
x = 2 × area of △ADF +× x
2 × area of △ADF = x –× x
2 × area of △ADF = × x
Area of △ADF = area of △FCB = × x
Area of △FCE = × area of parallelogram FDCE
Area of △FCE = × x
Area of △BCE = area of △FCE – area of △FCB
= × x – × x
= × x
Similarly, area of △ ADG = × x
Area of shaded region = area of △BCE + area of △ADG
=× x
= × x
Area of trapezium GDCE = area of parallelogram ABCD + area of △ ADG + area of △ BCE
= x + × x
=
Therefore, the ratio of the shaded region to that of the area of the trapezium GDCE = (× x) : () = 1 : 3
Hence, option 4 is the answer.
Question 14
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Given in the figure, UVWX is a parallelogram, VY ⊥ UX and VZ ⊥ XW, XU = 3.5 cm and VZ = 2.5 cm. If area of the parallelogram UVWX is 56 cm², find the length of VY and XW.

A
16 cm, 22.4 cm

B
17 cm, 25.4 cm

C
25.6 cm, 34 cm

D
35.4 cm, 47 cm

Solution

Given that, XU = 3.5 cm and VZ = 2.5 cm and area of the parallelogram UVWX = 56 cm².
So, Area of the parallelogram UVWX = VZ × XW = VY × XU
56 cm² = VY × 3.5 cm

VY = 16 cm.

Now, to find XW:
56 cm² = 2.5 cm × XW

XW = 22.4 cm
Question 15
1 / -0

In the adjoining figure, PQRS is a parallelogram and X is mid point of side QR. If SX and PQ when produced meet at point Y, then

A
PQ = 2RS

B
PY = 2PQ

C
XY = SR

D
PR = SX

Solution

Given,
X is the mid point QR.
Then, RX = QX
∠RSX = ∠QYX [Alternate angles]
i.e. By AAS congruency
△RXS ≌ △QXY
i.e. RS = QY
And PQ = RS = QY
i.e. PQ = QY
PY = PQ + QY
PY = 2PQ
Question 16
1 / -0

In PQR given below,

What is the ratio of

A
9 : 4

B
14 : 9

C
1 : 2

D
1 : 12

Solution

= QS = K, QV = 3K (say)

SV = QV – QS = 2K

Similarly, PR = 4r

PT = 3r

= = =
Question 17
1 / -0

In the given figure, ar(ABC) = ar(ABD). Which of the following relationships is correct for the lines p and q?

A
p and q are coincident lines.

B
p is parallel to q.

C
p is not parallel to q.

D
p and q are intersecting lines.

Solution

Given: ar(ABC) = ar(ABD) and the triangles are on the same base AB.
Therefore, p || q [Triangles on the same base and having equal area lie between the same parallels]
Question 18
1 / -0

In the given figure, A, B, D, C, E and F are points on three lines l, m and n. If ar(ABC) = ar(ADC) = ar(AFC) = ar(AEC), then

A
BD || EF

B
BD ∦ EF

C
AD || CF

D
AD || CE

Solution

Given: ar(ABC) = ar(ADC) = ar(AFC) = ar(AEC)

Since ar(ABC) = ar(ADC) and ABC and ADC are on the same base, BD || AC. …(1)
Since ar(AEC) = ar(AFC) and AEC and AFC are on the same base, AC || EF. … (2)
[As the triangles on the same base and having the same area lie between the same parallels]
From equations (1) and (2), we get BD || EF.
Question 19
1 / -0

PQRS is a quadrilateral and PQ = RS. Which of the following options is correct if ar(APQ) = ar(BPQ)?

A
PQRS is a parallelogram.

B
PQRS is a rhombus.

C
RS is not parallel to PQ.

D
PS is not parallel to RQ.

Solution

PQRS is a quadrilateral in which PQ = RS and ar(APQ) = ar(BPQ).

Since ar(APQ) = ar(BPQ) and the triangles have common base PQ, PQ || RS. [Triangles on the same base and having equal areas lie between the same parallels]
Also, PQ = RS (given)
Thus, PQRS is a parallelogram.
Question 20
1 / -0

The area of the trapezium ABCD is 126 cm². The ratio of AB and DC is 4 : 3 and of DC and ED is 4 : 3. If and AE = , then find the area of the trapezium EBCF.

A
70.1 cm²

B
71.1 cm²

C
61.1 cm²

D
60.1 cm²

Solution

Ratio of AB to DC = 4 : 3
= 16 : 12 - (i)
Ratio of DC to ED = 4 : 3
= 12 : 9 - (ii)
From (i) and (ii), ratio of AB to ED = 16 : 9.
According to this, let AB, DC and ED be 16x, 12x and 9x, respectively.
In the given trapezium ABCD, AB (a) = 16x
DC (b) = 12x
ED (height) = 9x
Area of the trapezium ABCD = 126 cm²
Area of the trapezium ABCD =
126 =
126 × 2 = 28x × 9x
252 = 252x²252 ÷ 252 = x²
x² = 1
x =
x = 1

Therefore, AB, DC and ED equals 16cm, 12cm and 9cm, respectively.

Now,

CF = 3 cm
So, DF = DC - CF
DF = 12 - 3
DF = 9 cm
Therefore, area of △ EDF =
=

=

=
= 40.5 cm²
Also, AE =
AE =
AE = 3.2
Area of △ EDA =
=

=

=
= 14.4 cm²Areaof trapezium EBCF = (area of trapezium ABCD) - (area of △EDA + area of △EDF)=(126) - (40.5 + 14.4) cm²
= (126 - 54.9) cm²= 71.1 cm²
Question 21
1 / -0

ABC and ABP are on the same base AB. Points C and P are on a line such that AB || . Find ar(ABC) if ar(ABP) = 12 cm².

A
6 cm²

B
12 cm²

C
24 cm²

D
36 cm²

Solution

Given: || AB, ABC and APB are on the same base AB and lie between the same parallels and AB.

So, ar(ABC) = ar(APB) [Triangles on the same base and lying between the same parallels are equal in area.]
ar(ABC) = 12 cm²[∵ar(APB) = 12 cm²]
Question 22
1 / -0

PQRS is a parallelogram formed by joining the midpoints of the sides of quadrilateral ABCD. Which of the following relationships is correct?

A
ar(PQRS) = ar(ABCD)

B
ar(PRQ) = ar(PSR)

C
ar(PQR) = ar(DSP)

D
None of the above

Solution

PQRS is a parallelogram formed by joining the midpoints of the sides of quadrilateral ABCD.
Now, parallelogram PQRS and PQR are on the same base QR and lie between the same parallel lines RQ and PS.
So, ar(PRQ) = ar(PQRS) --- (1)
Similarly, for PSR and PQRS:
ar(PSR) = ar(PQRS) --- (2)
[If a parallelogram and a triangle are on the same base and lie between the same parallel lines, then the area of triangle is half the area of parallelogram.]

Comparing (1) and (2), we get
ar(PRQ) = ar(PSR)

Question 23

1 / -0

In a parallelogram MNOP, MQ is the bisector of ∠PMN and OR is the bisector of ∠NOP. PN is the diagonal and meets MQ at S, and OR at T. Height QU is 5 cm, PQ + QO + NO = 18 cm and MU - UR = 7 cm. If the ratio of NO, QO and PQ is 2 : 3 : 1, then match the following.

Shape	Area
a. Parallelogram MROQ	1. 45 cm²
b. Parallelogram MNOP	2. 7.5 cm²
c. △MPQ	3. 60 cm²
d. △MUQ	4. 20 cm²

a - 3, b - 1, c - 4, d - 2

a - 1, b - 3, c - 2, d - 4

a - 3, b - 1, c - 2, d - 4

a - 1, b - 3, c - 4, d - 2

Solution

According to the given information, the picture is as follows.

Part - a
Let PQ, QO and NO be x, 3x and 2x respectively.
According to the question, PQ + QO + NO = 18
x + 3x + 2x = 18
6x = 18
x = 18 ÷ 6
x = 3
Therefore, PQ = 3 cm, QO = 9 cm and NO = 6 cm
Now, area of parallelogram MROQ = base × height
= QO × QU
= 9 × 5 = 45 cm²

Part - b
Area of parallelogram MNOP = base × height
= (PQ + QO) × QU
= (3 + 9) × 5 = 60 cm²

Part - c
Now, we know that triangles MPQ and ONR are congruent.
Therefore, area of △ MPQ = area of △ ONR
Area of △ MPQ + area of △ ONR = area of parallelogram MNOP – area of parallelogram MROQ
= (60 – 45) = 15 cm²
Therefore, area of △ MPQ = 15 ÷ 2 = 7.5 cm²
Area of △ ONR = 15 ÷ 2 = 7.5 cm²

Part - d
We know that MR + RN = 12 cm (Same as PQ + QO)
MR + 3 = 12
MR = 12 -3
MR = 9 cm
Now, MU = MR - RU and given that MU - UR = 7
So,
MU = MR - (MU -7)
2MU = MR + 7
2MU = 9 + 7
MU = 8 cm
Area of △MUQ =
=

=

= 20 cm²

Question 24
1 / -0

In the given figure, ED = 4 cm and area of △ECD = 12cm². If EA =and the ratio of EA to CB is 1: 3, then find the area of trapezium EABC.

A
24 cm²

B
20 cm²

C
18 cm²

D
12 cm²

Solution

In △ECD, ED = 4 cm.
Area of ECD = 12 cm²

Area of △ECD =
12 =

12 =
24 = EC × 4
EC = 24 ÷ 4
EC = 6 cm
Since, EA =
EA =
EA = 2 cm

Given that EA : CB = 1 : 3 and EA = 2 cm, this implies that CB = 6 cm

Area of trapezium EABC =
=

= 24 cm²
Question 25
1 / -0

Two squares namely ABCD and DCFE are placed on top of each other such that the line DC is common between the two squares which are equal in area. Two different triangles with same height as that of the first square are placed on sides AD and CF via their perpendiculars. The obtained two squares make two different trapeziums namely GABC and DCHE. If the length GD is 'x cm' long, the length HF is 'y cm' long, and the area of the second square is exactly 169 cm² and 'x - y > 4', the difference in the areas of the two trapeziums is

A
> 19 cm²

B
> 26 cm²

C
> 28 cm²

D
> 30 cm²

Solution

Area of the trapezium GABC =
Since GC = AB + x and AD = AB
So, area of the trapezium GABC =

Area of the trapezium DCHE =
Since EH = DC + y and CF = DC
So, area of the trapezium DCHE =
Area of the square = 169 cm²
Side of the square = = 13 cm
Since the squares have overlapping sides and same areas, the lengths of the sides would also be the same, i.e. 13 cm.
The difference in the areas of the two trapeziums:

=
=
=
Since the difference in x and y is always more than 4, the area when the difference is exactly 4 is 26 cm² which makes the answer according to the inequality to be > 26 cm².