R.d Sharma 2022 Mcqs Solutions for Class 9 Maths Chapter 14 Areas Of Parallelogram And Triangles are provided here with simple step-by-step explanations. These solutions for Areas Of Parallelogram And Triangles are extremely popular among class 9 students for Maths Areas Of Parallelogram And Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the R.d Sharma 2022 Mcqs Book of class 9 Maths Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s R.d Sharma 2022 Mcqs Solutions. All R.d Sharma 2022 Mcqs Solutions for class 9 Maths are prepared by experts and are 100% accurate.

Page No 150:

Question 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:

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).

 

Page No 150:

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:

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).

 

Page No 150:

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:

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).

Page No 150:

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:

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).

 

Page No 150:

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:

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).

Page No 150:

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:

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).



Page No 151:

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:

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).

Page No 151:

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:

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).

Page No 151:

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:

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).

Page No 151:

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:

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).

Page No 151:

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:

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).

Page No 151:

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:

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).

Page No 151:

Question 13:

In the given figure, ABCD is a parallelogram. If AB = 12 cm, AE = 7.5 cm, CF = 15 cm, then AD =
(a) 3 cm
(b) 6 cm
(c) 8 cm
(d)10.5 cm
Graphic
 

Answer:

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).

Page No 151:

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 (||gmXQRY) : ar (ΔQSR) =

(i) 1 : 4

(ii) 2 : 1

(iii) 1 : 2

(iv) 1 : 1

Answer:

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).

Page No 151:

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:

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).

Page No 151:

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:

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).

Page No 151:

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:

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)

 



Page No 152:

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:

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).

Page No 152:

Question 19:

In the given figure, a âˆ¥gm 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 = 12 perimeter of ABEF

Answer:

AB = EF  [sides of rectangle] in rectangle ABEF, and CD = AB in parallelogram ABCD.

When both are added together, we get
AB + CD = EF + AB    .....(1)

The perpendicular distance between two parallel sides of a parallelogram is always shorter than the length of the other parallel sides.

BE < BC and AF < AD  [∵ The hypotenuse of a right-angled triangle is longer than the other side]

On adding both above inequalities, we get
BE + AF < BC + AD 
⇒ BC + AD >  BE + AF

On adding AB + CD both sides, we get
AB + CD + BC + AD > AB + CD + BE + AF 
⇒ AB + BC + CD + AD > AB + BE + EF + AF  [∴ CD = AB = EF]
⇒ Perimeter of parallelogram ABCD > Perimeter of rectangle ABEF

Hence, the correct answer is option (c).


 

Page No 152:

Question 20:

In the given figure, the area of parallelogram ABCD is
(a) AB × BM
(b) BC × BN
(c) DC × DL
(d) AD × DL

Answer:

The area of a parallelogram is equal to the product of either side and the corresponding altitude (or height).

When AB is the base, then DL is the height.
∴ Area of parallelogram = AB × DL 
                                        = AB × BN                   (∵ DL = BN)

Similarly, when AD is the base, the height is equal to BM.
∴ Area of parallelogram = AD × BM

Similarly, when DC equals base, height equals BN.
∴ Area of parallelogram = DC × BN 
                                       = DC × DL                         (∵ â€‹DL = BN)

But, when BC is base, then height is not given.

Hence, the correct answer is option (c).

Page No 152:

Question 21:

In the given figure, ABCD and FECG are parallelograms equal in area. If ar (ΔAQE) = 12 cm2, then ar (||gmFGBQ) =
(a) 12 cm2

(b) 20 cm2

(c) 24 cm2

(d) 36 cm2

Answer:

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).

Page No 152:

Question 22:

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:

The diagonal of a parallelogram, rectangle, rhombus, or 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 rhombus as it can be a square as well.

Hence, the correct answer is option (d).

Page No 152:

Question 23:

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) 12ar (ΔABC)

(c) 13ar (ΔABC)

(d) 14ar (ΔABC)

Answer:

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).

Page No 152:

Question 24:

​Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) ​Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): Diagonals of a parallelogram divide it into four triangles of equal area.
Statement-2 (Reason): A median of a triangle divides it into two triangles of equal area.

Answer:

Statement-2 (Reason): A median of a triangle divides it into two triangles of equal area.

Let ABC be a triangle and let AD be one of its medians on side BC.
⇒ BD = DC

Since the formula for area involves altitude, let us draw AN ⊥ BC.

arABD=12×BD×AN=12×CD×AN         BD=CD=arACD

Thus, Statement-2 is true.

Statement-1 (Assertion): Diagonals of a parallelogram divide it into four triangles of equal area.


Let ABCD be a parallelogram with diagonals AC and BD intersecting at O.

Since the diagonals of a parallelogram bisect each other at the point of intersection.

Therefore, AO = OC and BO = OD

According to Statement-2, a median of a triangle divides it into two triangles of equal area.
Now,
In âˆ†ABC
ar(â–³AOB) = ar(â–³BOC)        .....(1)                  (∵ BO is the median)
In âˆ†BCD
ar(â–³BOC) = ar(â–³COD)        .....(2)                  (∵ CO is the median)
In âˆ†ACD
ar(â–³AOD) = ar(â–³COD)        .....(3)                  (∵ DO is the median)

From (1), (2), and (3)
ar(â–³AOB) = ar(â–³BOC) = ar(â–³AOD) = ar(â–³COD)

Thus, statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
 
Hence, the correct answer is option (a).

Page No 152:

Question 25:

​Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) ​Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): ABCD is a parallelogram and O is a point in its interior. If ar (∆AOB) = 4 cm2 and ar (∆BOC) = 6 cm2, then ar (∥gm ABCD) is 20 cm2.
Statement-2 (Reason): A diagonal of a parallelogram divides it into two triangles of equal area.

Answer:

Statement-2 (Reason): A diagonal of a parallelogram divides it into two triangles of the same area.

Here, ABCD is a parallelogram and BD is diagonal.
∴ AD ∥ BC and AB ∥ DC       (∵ Opposite sides of a parallelogram are parallel)
In â–³ADB and â–³CBD
∠ADB = ∠CBD               (Alternate angles)

∠ABD = ∠CDB                (Alternate angles)
 DB = BD                           (Common side)
∴  â–³ADB ≅ â–³CBD           ( By SAS congruence rule)
Since congruent figures have the same area.
∴  ar(ADB) = ar(CBD)                   
 

Statement-1 (Assertion): ABCD is a parallelogram and O is a point in its interior. If ar (∆AOB) = 4 cm2 and ar (∆BOC) = 6 cm2, then ar (∥gm ABCD) is 20 cm2.

Given that,  ar (∆AOB) = 4 cm2 and ar (∆DOC) = 6 cm2
Now,
ar (∆ABC) = ar (∆AOB) + ar (∆DOC)  
                  = 4 + 6
                  = 10 cm2
According to Statement-2, diagonal of a parallelogram divides it into two triangles of equal area.
∴ ar (∆ABC) = ar (∆ADC) 
Now, ar (∥gm ABCD) = ar (∆ABC) + ar (∆ADC) 
                             = 10 + 10
                            = 20 cm2

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a). 



Page No 153:

Question 26:

​Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) ​Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): In the given figure, if ABCD and EFGD are two parallelograms and G is the mid-point of CD, then ar (∆DPC) = ar (∥gm EFGD).

Statement-2 (Reason): Median of a triangle divides it into two triangles of equal area.

Answer:

Statement-2 (Reason): Median of a triangle divides it into two triangles of equal area.

Let ABC be a triangle and let AD be one of its medians on side BC.
⇒ BD = DC

Since the formula for area involves altitude, let us draw AN ⊥ BC.

arABD=12×BD×AN=12×CD×AN         BD=CD=arACD

Thus, Statement-2 is true.

Statement-1 (Assertion): In the given figure, if ABCD and EFGD are two parallelograms and G is the mid-point of CD, then ar (∆DPC) = ar (∥gm EFGD).


As we know, if a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is equal to one-half area of the parallelogram.

∴ ar (∆DPC) = 12ar (∥gm ABCD).          .....(1)

Let h be the height (distance between the parallel sides AB and DC) of the parallelogram ABCD.
∴ ar (∥gm ABCD) = CD × h                .....(2)   
Also,
ar (∥gm EFGD) = DG × h 
⇒ ar (∥gm EFGD) = 12 × CD × h     (∵ G is the mid-point of DC)
⇒ ar (∥gm EFGD) =12 × ar (∥gm ABCD)               .....(3)         [From (2)]    

From (1) and (3), we have
ar (∆DPC) = ar (∥gm EFGD)
Thus, Statement-1 is true.

So, Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
Hence, the correct answer is option (b).
 
 

Page No 153:

Question 27:

​Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) ​Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): PQRS is a parallelogram whose area is 180 cm2. If A is any point on the diagonal QS, then ar (∆ASR) = 90 cm2.
Statement-2 (Reason): A diagonal of a parallelogram divides it into two triangles of same area.

Answer:

Statement-2 (Reason): A diagonal of a parallelogram divides it into two triangles of the same area.

Here, ABCD is a parallelogram and BD is diagonal.
∴ AD ∥ BC and AB ∥ DC       (∵ Opposite sides of a parallelogram are parallel)
In â–³ADB and â–³CBD
∠ADB = ∠CBD               (Alternate angles)

∠ABD = ∠CDB                (Alternate angles)
 DB = BD                           (Common side)
∴  â–³ADB ≅ â–³CBD           ( By SAS congruence rule)
Since congruent figures have the same area.
∴  ar(ADB) = ar(CBD)                   
Thus, Statement-2 is true.
 
Statement-1 (Assertion): PQRS is a parallelogram whose area is 180 cm2. If A is any point on the diagonal QS, then ar (∆ASR) = 90 cm2.

PQRS is a parallelogram whose area is 180 cm2. If A is any point on the diagonal QS, then ar (∆ASR) = 90 cm2.

Given that, PQRS is a parallelogram whose area is 180 cm2 and  A is any point on the diagonal QS.
Since QS is the diagonal of the parallelogram PQRS. So, according to Statement-2  QS divides the parallelogram PQRS into two triangles of equal area, so
∴  ar (∆QRS) = 12 ×  ar (PQRS)
                      = 12×180
                     = 90 cm2

As, A is a point on SQ.
∴ ar (∆ASR) < ar (∆QRS)
⇒ ar (∆ASR) is less than 90 cm2

Thus, statement-1 is false.
So, Statement-1 is false, Statement-2 is true.

Hence, the correct answer is option (d).    

Page No 153:

Question 28:

​Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) ​Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): Let ABCD be a trapezium such that AB âˆ¥ DC. If diagonals AC and BD intersect at O, then ar (∆AOD) = ar (∆BOC).
Statement-2 (Reason): Triangles on the same base and between the same parallels are equal in area.

Answer:

Statement-2 (Reason): Triangles on the same base and between the same parallels are equal in area.
Thus, Statement-2 is true.

Statement-1 (Assertion): Let ABCD be a trapezium such that AB âˆ¥ DC. If diagonals AC and BD intersect at O, then ar (∆AOD) = ar (∆BOC).



From the figure, we know that â–³CDA and â–³CDB lie on the same base CD and between two parallel lines DC and AB.

According to Statement-2 ,the triangles lying on the same base and between the same parallels are equal in area.

∴ ar(â–³CDA) = ar(â–³CDB)                              .…(i)
 
If equals are subtracted from equals then the remainders are also equal. 
Subtracting ar(â–³DOC) on both the sides from (1),
ar(â–³CDA) − ar(â–³DOC) = ar(â–³CDB) − ar(â–³DOC)
⇒ ar(â–³AOD) = ar(â–³BOC)
Thus, statement-1 is true.

Statement-2 True, Triangles on the same base and between the same parallels are equal in area.

Hence, the correct answer is an option (a).

Page No 153:

Question 29:

​Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) ​Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): The diagonals of a parallelogram divide it into four triangles of equal area.
Statement-2 (Reason): A diagonal of a parallelogram divides it into two triangles of equal area.

Answer:

Statement-2 (Reason): A diagonal of a parallelogram divides it into two triangles of equal area. 

Here, ABCD is a parallelogram and BD is diagonal.
∴ AD ∥ BC and AB ∥ DC       (∵ Opposite sides of a parallelogram are parallel)
In â–³ADB and â–³CBD
∠ADB = ∠CBD               (Alternate angles)

∠ABD = ∠CDB                (Alternate angles)
 DB = BD                           (Common side)
∴  â–³ADB ≅ â–³CBD           ( By SAS congruence rule)
Since congruent figures have the same area.
∴  ar(ADB) = ar(CBD)                   
Thus, Statement-2 is true.

Statement-1 (Assertion): The diagonals of a parallelogram divide it into four triangles of equal area.


Let ABCD be a parallelogram with diagonals AC and BD intersecting at O.

Since the diagonals of a parallelogram bisect each other at the point of intersection.

Therefore, AO = OC and BO = OD

According to Statement-2, a median of a triangle divides it into two triangles of equal area.
Now,
In âˆ†ABC
ar(â–³AOB) = ar(â–³BOC)        .....(1)                  (∵ BO is the median)
In âˆ†BCD
ar(â–³BOC) = ar(â–³COD)        .....(2)                  (∵ CO is the median)
In âˆ†ACD
ar(â–³AOD) = ar(â–³COD)        .....(3)                  (∵ DO is the median)

From (1), (2), and (3)
ar(â–³AOB) = ar(â–³BOC) = ar(â–³AOD) = ar(â–³COD)

Thus, statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
 
Hence, the correct answer is option (b).

Page No 153:

Question 30:

​Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) ​Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): In the given figure, ABCD is a parallelogram in which DE ⊥ AB and BF ⊥ AD. If AB = 16 cm, DE = 8 cm and BF = 10 cm, then AD = 12 cm.

Statement-2 (Reason): Area of a parallelogram = Base × Height.

Answer:

Statement-2 (Reason): Area of a parallelogram = Base × Height.
As we know, the area of a parallelogram is the product of the base and its corresponding height.

Thus, Statement-2 is true.

Statement-1 (Assertion): In the given figure, ABCD is a parallelogram in which DE ⊥ AB and BF ⊥ AD. If AB = 16 cm, DE = 8 cm and BF = 10 cm, then AD = 12 cm.

According to Statement-2: Area of a parallelogram = Base × Height
∴ Area of parallelogram ABCD = AB × DE                       .....(1)                                            
                                                   
Also, area of parallelogram ABCD = AD × BF                 .....(2)
From (1) and (2)
AB × DE = AD × BF 
⇒ 16 × 8 = AD × 10
⇒ AD =12.8 cm
                                                   
Thus, Statement-1 is false.
So, Statement-1 is false, Statement-2 is true.

Hence, the correct answer is option (d).

 

Page No 153:

Question 31:

​Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) ​Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): The area of a trapezium whose parallel sides measure 25 cm and 15 cm respectively and the distance between them is 6 cm, is 120 cm2.
Statement-2 (Reason): The area of an equilateral triangle of side 8 cm is 83 cm2.

Answer:

Statement-2 (Reason): The area of an equilateral triangle of side 8 cm is 83 cm2.
Given that, side = 8 cm
Arae of an equilateral triangle = 34×side2
                                            = 34×8×8
                                           =163
Thus, Statement-2 is false.

Statement-1 (Assertion): The area of a trapezium whose parallel sides measure 25 cm and 15 cm respectively and the distance between them is 6 cm, is 120 cm2.
Given that, parallel sides of a trapezium measure 25 cm and 15 cm respectively and the distance between them is 6 cm.
Arae of trapezium = 12sum of parallel sides×height
                              = 1225+15×6
                              = 120 cm2
Thus, statement-1 is true.
So, Statement-1 is true, Statement-2 is false.

Hence, the correct answer is option (c).



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