MCQS
Page-14.59Question 1:
Two parallelograms are on the same base and between the same parallels. The ratio of their areas is
(a) 1 : 2
(b) 2 : 1
(c) 1 : 1
(d) 3 : 1
Answer 1:
Given: Two parallelogram with the same base and between the same parallels.
To find: Ratio of their area of two parallelogram with the same base and between the same parallels.
Calculation: We know that “Two parallelogram with the same base and between the same parallels, are equal in area”
Hence their ratio is 
,
So the correct answer is,i.e. option (c).
Question 2:
A triangle and a parallelogram are on the same base and between the same parallels. The ratio of the areas of triangle and parallelogram is
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 3
Answer 2:
Given: A triangle and a parallelogram with the same base and between the same parallels.
To find: The ratio of the area of a triangle and a parallelogram with the same base and between the same parallels.
Calculation: We know that,” the area of a triangle is half the area of a parallelogram with the same base and between the same parallels.”
Hence the ratio of the area of a triangle and a parallelogram with the same base and between the same parallels is 
Therefore the correct answer is option is (b).
Question 3:
Let ABC be a triangle of area 24 sq. units and PQR be the triangle formed by the mid-points of the sides of Δ ABC. Then the area of ΔPQR is
(a) 12 sq. units
(b) 6 sq. units
(c) 4 sq. units
(d) 3 sq. units
Answer 3:
Given: (1) The Area of ΔABC = 24 sq units.
(2) ΔPQR is formed by joining the midpoints of ΔABC
To find: The area of ΔPQR
Calculation: In ΔABC, we have
Since Q and R are the midpoints of BC and AC respectively.
PQ || BA 
PQ || BP
Similarly, RQ || BP. So BQRP is a parallelogram.
Similarly APRQ and PQCR are parallelograms.
We know that diagonal of a parallelogram bisect the parallelogram into two triangles of equal area.
Now, PR is a diagonal of 
APQR.
∴ Area of ΔAPR = Area of ΔPQR ……(1)
Similarly,
PQ is a diagonal of PBQR
∴ Area of ΔPQR = Area of ΔPBQ ……(2)
QR is the diagonal of PQCR
∴ Area of ΔPQR = Area of ΔRCQ ……(3)
From (1), (2), (3) we have
Area of ΔAPR = Area of ΔPQR = Area of ΔPBQ = Area of ΔRCQ
But
Area of ΔAPR + Area of ΔPQR + Area of ΔPBQ + Area of ΔRCQ = Area of ΔABC
4(Area of ΔPBQ) = Area of ΔABC
∴ Area of ΔPBQ 
Hence the correct answer is option (b).
Question 4:
The median of a triangle divides it into two
(a) congruent triangle
(b) isosceles triangles
(c) right triangles
(d) triangles of equal areas
Answer 4:
Given: A triangle with a median.
Calculation: We know that a ,”median of a triangle divides it into two triangles of equal area.”
Hence the correct answer is option (d).
Question 5:
In a ΔABC, D, E, F are the mid-points of sides BC, CA and AB respectively. If ar (ΔABC) = 16cm2, then ar (trapezium FBCE) =
(a) 4 cm2
(b) 8 cm2
(c) 12 cm2
(d) 10 cm2
Answer 5:
Given: In ΔABC
(1) D is the midpoint of BC
(2) E is the midpoint of CA
(3) F is the midpoint of AB
(4) Area of ΔABC = 16 cm2
To find: The area of Trapezium FBCE
Calculation: Here we can see that in the given figure,
Area of trapezium FBCE = Area of ||gm FBDE + Area of ΔCDE
Since D and E are the midpoints of BC and AC respectively.
∴ DE || BA DE || BF
Similarly, FE || BD. So BDEF is a parallelogram.
Now, DF is a diagonal of ||gm BDEF.
∴ Area of ΔBDF = Area of ΔDEF ……(1)
Similarly,
DE is a diagonal of ||gm DCEF
∴ Area of ΔDCE = Area of ΔDEF ……(2)
FE is the diagonal of ||gm AFDE
∴ Area of ΔAFE = Area of ΔDEF ……(3)
From (1), (2), (3) we have
Area of ΔBDF = Area of ΔDCF = Area of ΔAFE = Area of ΔDEF
But
Area of ΔBDF + Area of ΔDCE + Area of ΔAFE + Area of ΔDEF = Area of ΔABC
∴ 4 Area of ΔBDF = Area of ΔABC
Area of ΔBDF = Area of ΔDCE = Area of ΔAFE = Area of ΔDEF = 4 cm2 …….(4)
Now
Hence we get
Area of trapezium FBCE 
There fore the correct answer is option (c).
Question 6:
ABCD is a parallelogram. P is any point on CD. If ar (ΔDPA) = 15 cm2 and ar (ΔAPC) = 20 cm2, then ar (ΔAPB) =
(a) 15 cm2
(b) 20 cm2
(c) 35 cm2
(d) 30 cm2
Answer 6:
Given: (1) ABCD is a parallelogram
(2) P is any point on CD
(3) Area of ΔDPA = 15 cm2
(4) Area of ΔAPC = 20 cm2
To find: Area of ΔAPB
Calculation: We know that , “If a parallelogram and a a triangle are on the base between the same parallels, the area of triangle is equal to half the area of the parallelogram.”
Here , ΔAPB and ΔACB are on the same base and between the same parallels.
(since AC is the diagonal of parallelogram ABCD, diagonal of a parallelogram divides the parallelogram in two triangles of equal area)
Hence the correct answer is option (c).
Question 7:
The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 16 cm and 12 cm is
(a) 28 cm2
(b) 48 cm2
(c) 96 cm2
(d) 24 cm2
Answer 7:
Given: Rhombus with diagonals measuring 16cm and 12 cm.
To find: Area of the figure formed by lines joining the midpoints of the adjacent sides.
Calculation: We know that, ‘Area of a rhombus is half the product of their diagonals’.
H and F are the midpoints of AD and BC respectively.
Now ABCD is a parallelogram which means
……..(1)
……(2)
From 1 and 2 we get that ABFH is a parallelogram.
Since Parallelogram FHAB and ΔFHE are on the base FH and between the same parallels HF and AB.
……(3)
Similarly ,
……(4)
Adding 3 and 4 we get,
Hence the correct answer is option (b).
Question 8:
A, B, C, D are mid-points of sides of parallelogram PQRS. If ar (PQRS) = 36 cm2, then ar (ABCD) =
(a) 24 cm2
(b) 18cm2
(c) 30 cm2
(d) 36 cm2
Answer 8:
Given:
(1) PQRS is a parallelogram.
(2) A, B, C, D are the midpoints of the adjacent sides of Parallelogram PQRS.
(3) 
To find: 
Calculation:
A and C are the midpoints of PS and QR respectively.
Now PQRS is a parallelogram which means
AP = CQ ……..(1)
Also, PS || QR
AP || CQ ……(2)
From 1 and 2 we get that APCQ is a parallelogram.
Since Parallelogram APCQ and ΔABC are on the base AC and between the same parallels AC and PQ.
……(3)
Similarly ,
……(4)
Adding 3 and 4 we get,
Hence the correct answer is option (b).
Question 9:
The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is
(a) a rhombus of area 24 cm2
(b) a rectangle of area 24 cm2
(c) a square of area 26 cm2
(d) a trapezium of area 14 cm2
Answer 9:
Given: Rectangle with sides 8 cm and 6cm
To find: Area of the figure which is formed by joining the midpoints of the adjacent sides of rectangle.
Calculation: Since we know that 
For rhombus EFGH, EG is the one diagonal which is equal to DA
FH is the other diagonal which is equal to AB
Hence the result is option (a).
Question 10:
If AD is median of ΔABC and P is a point on AC such that
ar (ΔADP) : ar (ΔABD) = 2 : 3, then ar (Δ PDC) : ar (Δ ABC)
(a) 1 : 5
(b) 1 : 5
(c) 1 : 6
(d) 3 : 5
Answer 10:
Given: (1) AD is the Median of ΔABC
(2) P is a point on AC such that ar (ΔADP) : ar (ΔABD) = 
To find: ar (ΔPDC) : ar (ΔABC)
We know that” the medians of the triangle divides the triangle in two two triangles of equal area.”
Since AD is the median of ΔABC,
ar (ΔABD) = ar (ΔADC) ……(1)
Also it is given that
ar (ΔADP) : ar (ΔABD) = ……(2)
Now,
Therefore,
Hence the correct answer is option (c).
Question 11:
Medians of ΔABC intersect at G. If ar (ΔABC) = 27 cm2, then ar (ΔBGC) =
(a) 6 cm2
(b) 9 cm2
(c) 12 cm2
(d) 18 cm2
Answer 11:
Given: (1) Median of ΔABC meet at G.
(2) Area of ΔABC = 27 cm2
To find: Area of ΔBCG.
We know that the medians of the triangle divides each other in the ratio of 2:1
Hence,
Hence the correct answer is option (b).
Question 12:
In a ΔABC if D and E are mid-points of BC and AD respectively such that ar (ΔAEC) = 4cm2, then ar (ΔBEC) =
(a) 4 cm2
(b) 6 cm2
(c) 8 cm2
(d) 12 cm2
Answer 12:
Given: In ΔABC
(1) D is the midpoint of BC
(2) E is the midpoint of AD
(3) ar (ΔAEC) = 4 cm2
To find: ar (ΔBEC)
Calculation: We know that”the median of the triangle divides the triangle into two triangle of equal area”
Since AD is the median of ΔABC,
ar (ΔABD) = ar (ΔADC) …… (1)
EC is the median of ΔADC,
ar (ΔAEC) = ar (ΔDEC) …… (2)
⇒ ar (ΔDEC) = 4 cm2
EC is the median of ΔBED
ar (ΔBED) = ar (ΔDEC) …… (3)
From 2 and 3 we get,
ar (ΔBED) = ar (ΔAEC) …… (4)
⇒ ar (ΔBED) = 4 cm2
Now,
Hence the correct answer is option (c).
Question 13:
In the given figure, ABCD is a parallelogram. If AB = 12 cm, AE = 7.5 cm, CF = 15 cm, then AD =
Answer 13:
Given: (1) ABCD is a parallelogram.
(2) AB = 12 cm
(3) AE = 7.5 cm
(4) CF = 15cm
To find: AD
Calculation: We know that,
Area of a parallelogram = base × height
Area of a parallelogram ABCD = DC ×AE (with DC as base and AE as height) ……(1)
Area of a parallelogram ABCD = AD ×CF (with DC as base and AE as height) ……(2)
Since equation 1 and 2 both are Area of a parallelogram ABCD
Hence the correct answer is option (b).
Question 14:
In the given figure, PQRS is a parallelogram. If X and Y are mid-points of PQ and SR respectively and diagonal Q is joined. The ratio ar (||gm XQRY) : ar (ΔQSR) =
(i) 1 : 4
(ii) 2 : 1
(iii) 1 : 2
(iv) 1 : 1
Answer 14:
Given: (1) PQRS is a parallelogram.
(2) X is the midpoint of PQ.
(3) Y is the midpoint of SR.
(4) SQ is the diagonal.
To find: Ratio of area of ||gm XQRY : area of ΔQRS.
Calculation: We know that the triangle and parallelogram on the same base and between the same parallels are equal in area.
∴ Ar (||gm PQRS) = Ar (ΔQRS)
(since X is the mid point of PQ and Y is the midpoint of SR)
Hence the correct answer is option (d).
Question 15:
Diagonal AC and BD of trapezium ABCD, in which AB || DC, intersect each other at O. The triangle which is equal in area of ΔAOD is
(a) ΔAOB
(b) ΔBOC
(c) ΔDOC
(d) ΔADC
Answer 15:
Given: (1) ABCD is a trapezium, with parallel sides AB and DC
(2) Diagonals AC and BD intersect at O
To find: Area of ΔAOD equals to ?
Calculation: We know that ,” two triangles with the same base and between the same parallels are equal in area.”
Therefore,
Hence the correct answer is option (b).
Question 16:
ABCD is a trapezium in which AB || DC. If ar (ΔABD) = 24 cm2 and AB = 8 cm, then height of ΔABC is
(a) 3 cm
(b) 4 cm
(c) 6 cm
(d) 8 cm
Answer 16:
Given: (1) ABCD is a trapezium, with parallel sides AB and DC
(2) Area of ΔADB = 24 cm2
(3) AB = 8 cm
To find: Height of ΔABC.
Calculation: We know that ,” two triangles with the same base and between the same parallels are equal in area.”
Here we can see that, ΔADB and ΔACB are on the same base AB.
Hence,
Area of ΔACB = Area of ΔADB
Area of ΔACB = 24
Hence the correct answer is option (c).
Question 17:
ABCD is a trapezium with parallel sides AB =a and DC = b. If E and F are mid-points of non-parallel sides AD and BC respectively, then the ratio of areas of quadrilaterals ABFE and EFCD is
(a) a : b
(b) (a + 3b): (3a + b)
(c) (3a + b) : (a + 3b)
(d) (2a + b) : (3a + b)
Answer 17:
Given: (1) ABCD is a trapezium, with parallel sides AB and DC
(2) AB = a cm
(3) DC = b cm
(4) E is the midpoint of non parallel sides AD.
(5) G is the midpoint of non parallel sides BC.
To find: Ratio of the area of the Quadrilaterals ABFE and EFCD.
Calculation: We know that, ‘Area of a trapezium is half the product of its height and the sum of the parallel sides.’
Since, E and F are mid points of AD and BC respectively, so h1 = h2
Area of trapezium ABFE
Now, Area (trap ABCD) = area (trap EFCD) + Area (ABFE)
Therefore,
Thus,
Hence, the correct option is (c)
Question 18:
ABCD is a rectangle with O as any point in its interior. If ar (ΔAOD) = 3 cm2 and ar (ΔABOC) = 6 cm2, then area of rectangle ABCD is
(a) 9 cm2
(b) 12 cm2
(c) 15 cm2
(d) 18 cm2
Answer 18:
Given: A rectangle ABCD , O is a point in the interior of the rectangles such that
(1) ar (ΔAOB) = 3 cm2
(2) ar (ΔBOC) = 6 cm2
To find: ar (rect.ABCD)
Construction: Draw a line LM passing through O and parallel to AD and BC.
Calculation: We know that ,” If a triangle and a parallelogram are on the same base and between the same parallels the area of the triangle is equal to half the area of the parallelogram”
Here we can see that ΔAOD and rectangle AMLD are on the same base AD and between the same parallels AD and LM.
Hence ,
Similarly, we can see that ΔBOC and rectangle BCLM are on the same base BC and between the same parallels BC and LM
Hence,
We known that
Hence the correct answer is option (d).
Question 19:
In the given figure, a parallelogram ABCD and a rectangle ABEF are of equal area. Then,
(a) perimeter of ABCD = perimeter of ABEF
(b) perimeter of ABCD < perimeter of ABEF
(c) perimeter of ABCD > perimeter of ABEF
(d) perimeter of ABCD = perimeter of ABEF
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Answer 19:
We know, opposite sides of a rectangle are equal.
Therefore, in rectangle ABEF,
AB = EF ...(1)
Also, opposite sides of a parallelogram are equal.
Therefore, in parallelogram ABCD,
AB = DC ...(3)
From (1) and (3),
DC = EF
⇒ AB + EF = AB + DC ...(5)
Now, we know that, of all the line segments, perpendicular segment is the shortest.
∴ AF < AD
BE < BC
⇒ AF + BE < AD + BC ...(6)
Adding (5) and (6), we get
AB + EF + AF + BE < AB + DC + AD + BC
⇒ Perimeter of rectangle < perimeter of parallelogram
Hence, the correct option is (c).
Question 20:
In the given figure, the area of parallelogram ABCD is
(a) AB × BM
(b) BC × BN
(c) DC × DL
(d) AD × DL
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Answer 20:
We know, area of the parallelogram = Base × Altitude
Therefore,
The area of parallelogram ABCD = AB × DL
= DC × DL
Hence, the correct option is (c).
Question 21:
ABCD is quadrilateral whose diagonal AC divided it into two parts, equal in area, then ABCD
(a) is a rectangle
(b) is always a rhombus
(c) is a parallelogram
(d) need not be any of (a), (b) or (c)
Answer 21:
The diagonal of a parallelogram, rectangle, rhombus or a square divides them into two parts, equal in area.
Therefore, if ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD need not be a rectangle, parallelogram or a rhombus. It can be a square as well.
Hence, the correct option is (d).
Question 22:
The mid-points of the sides of a triangle ABC along with any of the vertices as the fourth point make a parallelogram of area equal to
(a) ar (ΔABC)
(b) ar (ΔABC)
(c) ar (ΔABC)
(d) ar (ΔABC)
Answer 22:
Given: (1) ABCD is a triangle.
(2) mid points of the sides of ΔABC with any of the vertices forms a parallelogram.
To find: Area of the parallelogram
Calculation: We know that: Area of a parallelogram = base × height
Hence area of ||gm DECF = EC × EG
area of ||gm DECF = EC × EG
area of ||gm DECF = 
(E is the midpoint of BC)
area of ||gm DECF = 
area of ||gm DECF = 
Hence the result is option (b).
Question 23:
In the given figure, ABCD and FECG are parallelograms equal in area. If ar (ΔAQE) = 12 cm2, then ar (||gm FGBQ) =
(a) 12 cm2
(b) 20 cm2
(c) 24 cm2
(d) 36 cm2
Answer 23:
Given: (1) Area of parallelogram ABCD is equal to Area of parallelogram FECG.
(2) If Area of ΔAQE is 12cm.
To find: Area of parallelogram FGBQ
Calculation: We know that diagonal of a parallelogram divides the parallelogram into two triangles of equal area.
It is given that,
Hence the correct answer is option (c).
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