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Areas of Parallelograms & Triangles Test - 7

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Areas of Parallelograms & Triangles Test - 7

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Weekly Quiz Competition

Question 1
1 / -0

In the given figure, ST is parallel to PQ, and PS is parallel to QR. Also, if area of ΔPQT = 70 cm², then find the area of PQRS.

A
120 cm²

B
140 cm²

C
150 cm²

D
160 cm²

Solution

If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.
Here, △PQT and parallelogram PQRS are on the same base PQ and between the same parallel ST and PQ.
Area of △ PQT = of area of PQRS
70 = ar(PQRS)
Area of PQRS = 140
Question 2
1 / -0

In the figure given below, ADCB and DGFA are two parallelograms. If FE = 5 cm and BC = 12 cm, the what will be the area of triangle GDF?

A
25 cm²

B
30 cm²

C
50 cm²

D
60 cm²

Solution

ADCB and DGFA are two parallelograms.
As they both are on same base and between same parallel lines,
Area of DGFA = Area of ADCB
In parallelogram ADCB,
BC = 12 cm, FE = 5 cm
Area of ADCB = b × h
Area of ADCB = 12 × 5 = 60 cm²
So,
Area of DGFA = Area of ADCB = 60 cm²Now,FD is the diagonal of parallelogram DGFA.
A diagonal divides the parallelogram in two equal triangles.
So,
Area of triangle GDF = × Area of parallelogram DGFA

Area of trianlge GDF = × 60 = 30 cm²
Question 3
1 / -0

In the given figure, GI is parallel to HL and GH is parallel to LI. If a median KN of the ΔKLH is drawn, then what will be the area of ΔKNH, given that the area of IJK = 72 cm²?

A
72 cm²

B
36 cm²

C
24 cm²

D
18 cm²

Solution

Given that ΔLGI and ΔLHI share the same base LI, and are between the same parallel lines GH and LI.
So,
Area of ΔLGI = Area of ΔLHI ... (1) (Triangles between the same parallel lines and on the same base are equal in area)
Area of ΔLGI - Area of ΔKLI = Area of ΔLHI - Area of ΔKLI (subtracting area of same triangle from the triangles)
Area of ΔKIG = Area of ΔKHL ... (2)
ΔKIG and ΔIJK are between HI and GJ and on same base IK.
So,
Area of ΔKIG = Area of ΔIJK … (3) (Triangles between the same parallel lines and on the same base are equal in area)
As,
Area of ΔIJK = Area of ΔKIG
And,
Area of ΔKIG = Area of ΔKHL
Hence,
Area of ΔIJK = Area of ΔKHL
According to the question, KN is the median.
So, ΔKHL is divided into two equal triangles, i.e. ΔKNH and ΔKNL (Median of a triangle divides it into two triangles of equal area)

Area of ΔKLH = Area of ΔKNH

As,

Area of ΔIJK = Area of ΔKHL

Area of ΔIJK = Area of ΔKNH

= 36 cm²
Question 4
1 / -0

The perimeter of regular hexagon MNOPQR is 60 cm, if TN and TS are 8 cm and 5 cm, respectively, find the area of quadrilateral MNUS.

A
18 cm²

B
30 cm²

C
46 cm²

D
64 cm²

Solution

Perimeter of hexagon = 60 cm
Each side of hexagon = 60 ÷ 6 = 10 cm
In ΔMTN, MT² = MN² – TN²
MT² = 100 – 64 = 36
MT = 6 cm
Area of ΔMTN = × base × height = × MT × TN
Area of ΔMTN = × 6 × 8 = 24 cm²
Area of quadrilateral MNUS = Area of ΔMTN + Area of rectangle TNUS
Area of rectangle TNUS = Length × Width = TS × TN
Area of rectangle TNUS = 5 × 8 = 40 cm²Area of quadrilateral MNUS = Area of ΔMTN + Area of rectangle TNUS = 24 + 40 = 64 cm²
Question 5
1 / -0

In parallelogram ABCD, FB is median and E is a point on DC. If ar(ΔDAE) = 35 cm², ar(ΔEAC) = 15 cm², then what is ar(ΔFBC)?

A
75 cm²

B
50 cm²

C
25 cm²

D
20 cm²

Solution

As the diagonal of a parallelogram divides the parallelogram into two triangles of equal areas,
ar(ΔADC) = ar(ΔABC)
ar(ΔDAE) + ar(ΔAEC) = ar(ΔADC)
Area of ΔDAE = 35 cm²
Area of ΔEAC = 15 cm²
ar(ΔADC) = 35 + 15 = 50 cm²
ar(ΔADC) = ar(ΔABC) = 50 cm²
As FB is the median,
ar(ΔABF) = ar(ΔFBC)
ar(ΔABC) = ar(ΔABF) + ar(ΔFBC)
50 = ar(ΔFBC) + ar(ΔFBC)
50 = 2 ar(ΔFBC)
ar(ΔFBC) = 50 ÷ 2
ar(ΔFBC) = 25 cm²
Question 6
1 / -0

MPQN is a parallelogram. Length of the perpendicular ON is 15 cm and divides QP in the ratio 1 : 4. If the area of ΔNOQ is 60 cm², then what will be the area of parallelogram MPQN?

A
600 cm²

B
540 cm²

C
420 cm²

D
320 cm²

Solution

ar(ΔNOQ) = 60 cm²ON = 15 cm
Area of ΔNOQ =× b × h
60 =× b × 15
15b = 60 × 2
b = 120 ÷ 15 = 8 cm
OQ = 8 cm

According to the question,
=
=

OP = 8 × 4 = 32 cm
PQ = OP + OQ
PQ = 32 + 8 = 40 cm
Area of parallelogram = b × h = 40 × 15 = 600 cm²
Question 7
1 / -0

In the given figure, AE is parallel to CD and the area of the parallelogram ABDC is 42 cm². Find the area of the ΔCED.

A
31.5 cm²

B
21 cm²

C
10.5 cm²

D
Cannot be determined

Solution

We know that if a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.
Therefore,
Area of the ΔCED = × Area of parallelogram ABDC
Area of ΔCED =
Area of ΔCED = 21 cm²
Question 8
1 / -0

In the given figure, MLON is the parallelogram. If LG = GO and OF is the median of ΔLOM, then which of the following equals the area of ΔLFG?

A
Area of parallelogram LMNO

B
Area of parallelogram LMNO

C
Area of parallelogram LMNO

D
Area of parallelogram LMNO

Solution

MO is the diagonal of parallelogram MLON.
Thus, parallelogram MLON is divided by the diagonal into two equal parts.
Therefore,
Area of ΔLMO = Area of ΔOMN = (Area of parallelogram MLON) ... (A)

Now as median of ΔLMO is OF, ΔLMO is divided into two triangles that have equal areas.

Area of ΔOFM = Area of ΔLFO
= (Area of ΔLMO)
= × Area of parallelogram ... (From A)
= (Area of parallelogram MLON) ... (B)

In ΔLFO, midpoint of LO is G.
Therefore, FG is the median.
Thus,
Area of ΔLFG = Area of ΔOFG
= Area of ΔLFO
= × Area of parallelogram MLON ... (From B)
= Area of parallelogram MLON
Hence,
Area of ΔLFG = Area of parallelogram MLON
Question 9
1 / -0

In the given figure, FGIH is a parallelogram and FH is parallel to IN. What is the area of ΔMGI?

A
Less than area of ΔNMH

B
Greater than area of ΔNMH

C
Equal to area of ΔNMH

D
Cannot be determined

Solution

FGIH is a parallelogram, GI is extended to the point N, and it is given that FH is parallel to IN.
ΔNFI and ΔNHI are on the same base NI and between the same parallel lines FH and GN.
Therefore,
ar(ΔNFI) = ar(ΔNHI) … (i) [Two triangles on the same base and same parallels are equal in areas.]
Now,
ar(ΔNFI) - ar(ΔNMI) = ar(ΔNHI) - ar(ΔNMI) ... [Subtracting the same area from the both the sides]
ar(ΔMFI) = ar(ΔNMH) ... (ii)
ΔMFI and ΔMGI are on the same base MI and between same parallels lines FG and IM.
Therefore,
ar(ΔMFI) = ar(ΔMGI) ... (iii) [Triangles between two parallel lines and on the same base have equal areas.]
From (ii) and (iii) we get,
ar(ΔMGI) = ar(ΔNMH)
Question 10
1 / -0

State 'T' for true and 'F' for false for the following statements.

1. If a triangle and a parallelogram share the same base and are between the same parallels, then area of parallelogram is of the area of the triangle.
2. If a planar region formed by a figure P, is made up of two non-overlapping planar regions formed by figures U and B, then ar(P) = ar(U) × ar(B).
3. If a parallelogram and a rectangle have the same base and equal areas, then the ratio of their perimeters is 1 : 1.

A
T, F, F

B
F, F, T

C
F, F, F

D
F, T, F

Solution

(1)

Let ΔABP and a parallelogram ABCD be on the same base AB and between the same parallels AB and PC.
Therefore, ar(ABQP) = ar(ABCD)
But,
ΔPAB ≅ ΔBQP (Diagonal PB divides parallelogram ABQP into two congruent triangles.)
So,
ar(PAB) = ar(BQP) --- (2)
∴ ar(PAB) = ar(ABQP) --- (3) [ from (2)]
This gives,
ar(PAB) = ar(ABCD) [ from (1) and (3)]
So the statement "If a triangle and a parallelogram share the same base and are between the same parallels, then area of triangle is of area of the parallelogram" is false.

(2). If a planar region formed by a figure P is made up of two non-overlapping planar regions formed by figures U and B, then ar(P) = ar(U) + ar(B).
So, this is false.

(3)

PUBT is a rectangle and RUBA is a parallelogram.
Opposite sides of the parallelogram and rectangle are equal.
UB = RA (RUBA is a parallelogram.)
UB = PT (PUBT is a rectangle.)
RA = PT … (1)
UB + RA = UB + PT ... (2)
Since out of all the segments that can be drawn to a given line from a point not lying on it, the perpendicular segment is the shortest, thus

BT < BA and UP < UR
Perimeter of || gm RUBA > Perimeter of rectangle PUBT
So, the ratio of their perimeters cannot be 1 : 1.
Hence, it is false.
Question 11
1 / -0

A big parallelogram and a triangle are on the same base and between the same parallel lines. The area of that triangle is cut out from the area of the big parallelogram and a new parallelogram is made out of the remaining area. A new triangle is cut out from the new parallelogram such that the new triangle and the new parallelogram are on the same base and between the same parallel lines.

What would be the ratio of the area of the big parallelogram to that of the new triangle?

A
1 : 2

B
4 : 1

C
1 : 4

D
2 : 1

Solution

Given,
A big parallelogram and a triangle are on the same base and between the same parallel lines.
It means that the area of the triangle would be half the area of the big parallelogram.
Therefore, the area of the remaining part would be half the area of the big parallelogram too.
This can be shown as:

Now, a new parallelogram is made out of the remaining area.
Therefore, the area of that new parallelogram would be half the area of the big parallelogram too.
A new triangle is cut out from the new parallelogram such that the new triangle and the new parallelogram are on the same base and between the same parallel lines.
The new parallelogram and the new triangle in it can be shown by this figure:

Therefore, the area of the new triangle would be half the area of the new parallelogram, which means that the area of the new triangle would be one-fourth the area of the big parallelogram.
The ratio of the area of the big parallelogram to the area of the new triangle can be calculated as:

Hence, the ratio of the area of the big parallelogram to that of the new triangle is 4:1 which means that option 2 is the answer.
Question 12
1 / -0

From the given figure, find the ratio of the area of the parallelogram ABCD to that of the shaded region, if points E, F, G and H are exactly in the middle of each side of the parallelogram.

A
8 : 3

B
4 : 1

C
8 : 1

D
3 : 8

Solution

Given that ABCD is a parallelogram.
As we can see in the figure that EG is a line drawn exactly in the middle of the parallelogram, it divides the parallelogram into two small parallelograms with equal areas.
Also, points E, F, G and H are exactly in the middle of each side of the parallelogram.
Let the area of parallelogram ABCD = x
Therefore,
Area of parallelogram AEGD = Area of the parallelogram BCGE = × Area of the parallelogram ABCD
= × x
(Area of the triangle on the same base and between the same parallel lines to that of the parallelogram is half.)
Now, Area of ΔEHG = Area of ΔEFG = × Area of the parallelogram AEGD = × Area of the parallelogram BCGE =
= × x
(If a triangle and parallelogram are on the same base and have the same altitude, the area of the triangle will be half that of the parallelogram.)
Now,
Area of parallelogram BCGE = Area of ΔEFG + Area of ΔEBF + Area of ΔFGC
× x = × x + Area of ΔEBF + Area of ΔFGC
Area of ΔEBF + Area of ΔFGC = x
Area of ΔEBF + Area of ΔFGC = × x ... eq 1
Also as points E, F, G and H are exactly in the middle of each side of the parallelogram,
Area of ΔEBF = Area of ΔFGC
(Because the area of the small triangles on either side of the big triangle would be the same.)
Putting this in equation 1, we get:
2 × Area of ΔEBF = × x
Area of ΔEBF = × x
Now,
Area of the shaded region = Area of ΔEHG + Area of ΔEBF = x
= × x
Now,
Ratio of the area of the parallelogram ABCD to the area of the shaded region = x : x
= 8 : 3
Hence, option 1 is the answer.
Question 13
1 / -0

Two parallelograms and three triangles share the same base between two parallel sides. What will be the ratio of total area of the two parallelograms and one triangle to the total area of remaining two triangles?

A
4 : 3

B
5 : 2

C
5 : 4

D
3 : 2

Solution

The parallelograms on the same base and between same parallel lines have the same area.
So, let the area of the parallelograms be 2 cm²each.If a triangle and a parallelogram share the same base between two parallel sides, then the area of the triangle is half the area of the parallelogram.
Therefore,
Area of triangles = 1 cm²each
Required ratio = Area of (two parallelograms + a triangle) : Area of (two triangles)
= (2 + 2 + 1) : (1 + 1)
= 5 : 2
Question 14
1 / -0

BP is the median of the ∆ABC. It is given that ar(ΔCEP) = ar(ΔCBP). The ar(ΔPCE) : ar(‖gm ABCD) is

A
6 : 1

B
1 : 6

C
1 : 16

D
16 : 1

Solution

If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.
ΔACB and parallelogram ABCD are between the same parallel lines and on the same base.
So, ar(ΔACB) = ar(||gm ABCD)
Also,
As BP is the median, it divides the triangle into two equal parts having the same area.
ar ΔABP = ΔCBP = ar ∆ACB
As, ΔACB = ar ||gm ABCD
Accordingly,
ar ΔCBP = ar ||gm ABCD
We are given that:
ar ΔPCE = ar ∆CBP
Hence, ar ΔPCE = ar ||gm ABCD [ar ΔCBP = ar ||gm ABCD]
ar ΔPCE: ar ||gm ABCD = 1 : 16
Question 15
1 / -0

CBAD is a parallelogram. Area of ΔAPB is 20 cm². Find the area of ΔAED. Given that E is the midpoint of diagonal AC, such that DE divides ΔACD into two triangles of equal areas.

A
6 cm²

B
8 cm²

C
10 cm²

D
12 cm²

Solution

We know that area of triangles between the same base and between same parallel lines is equal.
So, ar(ΔAPB) = ar(ΔACB)
Also we know that a diagonal divides a parallelogram into two equal halves having same area.
So,
ar(ΔACB) = ar(ΔADC)
Or,
ar(ΔADC) = 20 cm²
Now, we are given that DE divides this triangle into two equal halves. So,
ar(ΔAED) = ar(∆ADC)
= 20 cm² ÷ 2

= 10 cm²
Question 16
1 / -0

The shape given below has an area of 700 cm². What is the area of the parallelogram DCBM?

A
280 cm²

B
320 cm²

C
360 cm²

D
380 cm²

Solution

From the given information it can be noted that the shape is made up of two trapeziums, i.e. ABCD and CDFE.
Now, let x be the height of the trapezium CDFE.
So, the height of the trapezium ABCD = 30 - x cm
According to question,
Total area = 700 cm²
So,
Area of trapezium ABCD + Area of trapezium CDFE = 700
(0.5) × (AB + CD) × (30 - x) + (0.5) × (CD + EF) × (x)
= 700
Putting the values:
(0.5) × (31 + 19) × (30 - x) + 0.5 × (19 + 21) × (x) = 700
50(30 - x) + 40x = 1,400
1,500 – 50x + 40x = 1,400
100 = 10x
x = 10 cm
Height of parallelogram DCBM = 30 - x
= 30 - 10
= 20 cm
Area of parallelogram DCBM = height × base
= 20 × 19
= 380 cm²
Question 17
1 / -0

The shortest distance between two parallel sides and the shorter parallel side of a trapezium is 16 units and 7 units, respectively. Each non-parallel side measures 10 units and the area of the trapezium is 160 sq. units. Find the length of the perpendicular drawn from the vertex of the larger side of the trapezium to the non-parallel side which divides that non-parallel side in two equal parts.

A
10 units

B
15 units

C
12 units

D
11 units

Solution

Given,
The shortest distance between two parallel sides and the shorter parallel side of the trapezium is 16 units and 7 units, respectively.
Area of the trapezium is 160 sq. units.
Let the larger parallel side be 'x' units.
We know that,
Area of the trapezium = × shortest distance between the parallel sides × (sum of the parallel sides)
160 = × 16 × (7 + x)
160 = 8(7 + x)
7 + x =
7 + x = 20
x = 20 – 7 = 13
Hence, the larger parallel side of the trapezium measures 13 units.
Now, as it is given that the perpendicular from the vertex of the larger side of the trapezium divides the non-parallel side into two equal lengths;
Length of DF = × length of DA
= × 10 = 5 units
Now, in the right-angled ΔCFD:
(CD)² = (DF)² + (CF)²(13)² = (5)² + (CF)²169 = 25 + (CF)² (CF)² = 169 – 25
(CF)² = 144
CF = 12 units
Hence, the answer is 12 units.
Question 18
1 / -0

In the given figure, area of parallelogram ABCD = 240 cm², point J divides the line IA in 2 : 1, IA = AB and CD = 12 cm. What would be the area of the right-angled triangle IDJ?

A
60 cm²

B
70 cm²

C
80 cm²

D
100 cm²

Solution

Given,

Area of the parallelogram ABCD = 240 cm²Point J divides the line IA in 2 : 1
IA = AB and CD = 12 cm
Now,
Area of the triangle DHC = (area of the parallelogram ABCD) [Parallelogram ABCD and triangle DHC lie on the same base, i.e. DC, and between the same parallel lines, i.e. AB and DC, have equal areas.]
Area of triangle DHC = 120 cm²
Now,
Area of triangle DHC = base × perpendicular height
120 cm² = × 12 × perpendicular height
Perpendicular height = 20 cm
Area of the right-angled triangle IDJ = × base × 20 ... Eq 1
Now,
Point J divides the line IA in the ratio 2 : 1, IA = AB
AB is parallel as well as equal to CD, as opposite sides of the parallelogram are equal.
Therefore, IA = AB = 12 cm
Now as point J divides line IA in 2x and 1x (let 'x' be the multiple):
x + 2x = 12
3x = 12
x = 4
Therefore, AJ = (1 × 4) cm = 4 cm
and IJ = (2 × 4) cm = 8 cm
As IJ is the base of the right-angled triangle, Eq 1 becomes:
Area of the right angled triangle = × 8 × 20 cm² = 80 cm²
Hence, option 3 is the answer.
Question 19
1 / -0

Given is a parallelogram ABCD and points P, Q, R and S are midpoints on side AB, BC, CD and DA respectively.

Which of the following is true ?

A
AD ⊥ CD

B
PQRS is a square.

C
ABQS is a parallelogram.

D
PQRS is a rhombus.

Solution

∠ADC cannot be 90° as the given figure is a parallelogram.
Therefore, statement A is false.

AB = AD is not given and interior angles of quadrilateral PQRS are not certainly 90°.
So, whether PQRS is a square or a rectangle, is not certain.
Therefore, statement B and E are false.

Also, AB = AD is not given which means that all the sides of PQRS are not equal in length.
So, PQRS is not surely a rhombus.
Therefore, statement D is false.

We know that S and Q are the midpoints of AD and BC, respectively and AB ∥ DC and AD ∥ BC
SQ is also parallel to AB and DC and AS is parallel to BQ.
Therefore, ABQS is a parallelogram.
Question 20
1 / -0

The centroid of a triangle divides the medians of the triangle in ratio of

A
2 : 1

B
3 : 2

C
5 : 4

D
7 : 2

Solution

Centroid is the point where all 3 medians intersect and is often described as the triangle's center of gravity or as the barycent. It is formed by the intersection of the medians. It is one of the points of concurrency of a triangle. The centroid divides each median in the ratio of 2 : 1.
Question 21
1 / -0

Two parallel sides of a trapezium are of lengths 26 cm and 13 cm, and the area of the trapezium is 0.02925 m². What is the difference between the two lines?

A
12 cm

B
16 cm

C
18 cm

D
15 cm

Solution

Given,
Length of two parallel sides = 26 cm and 13 cm
Area of trapezium = 0.02925 m²
0.02925 × 10,000 = 292.5 cm²
Let the distance between the two sides be a.
Area of the trapezium = 1/2 × (sum of parallel sides) × (difference between them)
292.5 = {1/2 × (26 + 13) × a}
292.5 = 19.5 × a

= a

a = 15 cm
Question 22
1 / -0

In the given figure ABDC, ratio of parallel sides is 5 : 4. If the distance between two parallel sides is 36 cm and area is 486 cm², find the length of the shortest parallel side.

A
18 cm

B
12 cm

C
8 cm

D
35 cm

Solution

Let parallel sides be a and b.
a = 5x
b = 4x
height = 36 cm
Area of trapezium = 486 cm²
× h = 486
× 36 = 486
(9x) ÃÂ× 18 = 486
x = 3
a = 15 cm
b = 12 cm
Hence, the length of the shortest parallel side is 12 cm.
Question 23
1 / -0

Michael is a farmer. He has a piece of land which is shaped like a parallelogram. He divided his land as given in the figure, where DB∥PQ. Can you help him to find area of ABCD figure, if area of PQRS = 4,098 square yards?

A
2,049 sq yards

B
2,460 sq yards

C
2,437 sq yards

D
2,567 sq yards

Solution

In parallelogram PQRS,
PQ = RS and PQ || RS
QR = SP and QR || SP (Opposite sides of parallelogram are equal and parallel.)
PQBD is a parallelogram. [Given, DB || PQ]
Since ΔABD and parallelogram PQBD are on the same base BD and between the same parallel lines PQ and BD,
ar(ABD) = ar(PQBD) --- (1)
Similarly, it can be proved that
ar(BCD) = ar(DBRS) --- (2)
On adding (1) and (2),

ar(ABD) + ar(BCD) = ar(PQBD) + ar(DBRS)
ar(ABCD) =
Area of PQRS = 4,098 sq yards
So,
Area of ABCD = 2,049 sq yards
Question 24
1 / -0

In given right-angled triangle BAE, CF is perpendicular to AD, AD is perpendicular to BE and DC is perpendicular to AB. If AE = 14 cm, AD and AC are 25 cm and 7 cm, respectively, and CF = AC, find the ratio of area of quadrilateral ACDE to area of triangle ADC.

A
77 : 56

B
75 : 23

C
82 : 54

D
96 : 25

Solution

AE = 14 cm

CF = AC = 3.5 cm

AD = 25 cm
And,
AC = 7 cm

In triangle ACD, which is a right-angled triangle,
AD² = AC²+ CD²(25)²= (7)²+ CD²
625 = 49 + CD²
CD² = 625 - 49
CD² = 576
CD = 24 cm
Area of trapezium ACDE = [ × (AE + CD) × AC]
= × (24 + 24) × 7
= 168 cm²
Area of triangle ADC = × AD × CF = × 25 × 3.5 = 43.75 cm²
Required ratio = 168 : 43.75 = 96:25, which is the same as 96 : 25
Question 25
1 / -0

The area of the given parallelogram ABCD is 40 cm². Also, line DF measures the difference between the lines AD and DE if line AD = 8 cm. CDFG is also a parallelogram. What will be the area of the shaded part if the un-shaded parts are right-angled triangles?

A
27.5 cm²

B
30.5 cm²

C
36.5 cm²

D
38.5 cm²

Solution

Given,
Area of the parallelogram ABCD = 40 cm²
As AD = 8 cm,
DE = = 5 cm (AD × DE = area of the parallelogram ABCD)
Area of the triangle AED = × 40 cm² (We know that the triangle would have half of the area of the parallelogram if the triangle and the parallelogram are on the same base and between the same parallel lines.)
= 20 cm²
Area of (triangle BAE + triangle EDC) = Area of the parallelogram ABCD - Area of the triangle AED = (40 - 20) cm² = 20 cm²
Now,
Line DF measures the difference between the lines AD and DE = (8 - 5) cm = 3 cm
Area of the right-angled triangle CDF = × 3 × Perpendicular ... Eq 1

As we know that DE = CF, So perpendicular distance is 5cm.
Area of the right-angled triangle CDF = × 3 × 5 = 7.5 cm²
Area of the parallelogram CDFG = Area of right-angled triangle CDF = Area of triangle CFG = 7.5 cm²
Therefore, total area of the shaded figure is : (20 + 7.5 ) cm² = 27.5 cm²
Hence, option 1 is the answer.

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