RS AGGARWAL CLASS 9 CHAPTER 11 AREAS OF PARALLELOGRAMS AND TRIANGLES MCQ

 MULTIPLE CHOICE QUESTIONS

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Question 1:

Out of the following given figures which are on the same base but not between the same parallels?

Answer 1:

In this figure, both the triangles are on the same base (QR) but not on the same parallels.

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Question 2:

In which of the following figures, you find polynomials on the same base and between the same parallels?

Answer 2:

In this figure, the following polygons lie on the same base and between the same parallel lines:
a) Parallelogram ABCD
b) ​​​Parallelogram ABPQ

Question 3:

The median of a triangle divides it into two
(a) triangles of equal areas
(b) congruent triangles
(c) isosceles triangles
(d) right triangles

Answer 3:

(a)  triangles of equal areas

Question 4:

The area of quadrilateral ABCD in the given figure is
(a) 57 cm²
(b) 108 cm²
(c) 114 cm²
(d) 195 cm²

Answer 4:

(c)114 cm²

ar (quad. ABCD) =  ar (∆ ABC)  +  ar (∆ ACD)
In right angle triangle ACD, we have:
AC  = $\sqrt{\left({17}^2-8^2\right)}=\sqrt{225}=15\mathrm{cm}$$\sqrt{\left({17}^2 - 8^2\right)} = \sqrt{225} = 15 \mathrm{cm}$
In right angle triangle ABC, we have:
BC = $\sqrt{\left({15}^2-9^2\right)}=\sqrt{144}=12\mathrm{cm}$$\sqrt{\left({15}^2 - 9^2\right)} = \sqrt{144} = 12 \mathrm{cm}$
Now, we have the following:
ar(∆ABC) = $\frac{1}{2}$$\frac{1}{2}$ × 12 × 9 = 54 cm²
ar(∆ADC) = $\frac{1}{2}$$\frac{1}{2}$ × 15 × 8 = 60 cm²
ar(quad. ABCD) =​ 54 + 60 = 114 cm²

Question 5:

The area of trapezium ABCD in the given figure is
(a) 62 cm²
(b) 93 cm²
(c) 124 cm²
(d) 155 cm²

Answer 5:

(c)124 cm²

In the right angle triangle BEC, we have:
EC  = $\sqrt{{17}^2-{15}^2}=\sqrt{289-225}=\sqrt{64}=8\mathrm{cm}$$\sqrt{{17}^2-{15}^2}=\sqrt{289-225}=\sqrt{64}=8 \mathrm{cm}$
ar(trapez. ABCD) = $\frac{1}{2}\times \left(\mathrm{sum}\mathrm{of}\mathrm{parallel}\mathrm{sides}\right)\times \mathrm{distance}\mathrm{between}\mathrm{them}=\frac{1}{2}\times 31\times 8=124$$\frac{1}{2}\times (\mathrm{sum} \mathrm{of} \mathrm{parallel} \mathrm{sides})\times \mathrm{distance} \mathrm{between} \mathrm{them}=\frac{1}{2}\times 31\times 8=124$ cm²

Question 6:

In the given figure, ABCD is a || gm in which AB = CD = 5 cm and BD ⊥ DC such that BD = 6.8 cm. Then, the area of || gm ABCD = ?
(a) 17 cm²
(b) 25 cm²
(c) 34 cm²
(d) 68 cm²

Answer 6:

(c) 34 cm²

ar(parallelogram ABCD) = base × height = 5 ​× 6.8 =  34 cm²

Question 7:

In the given figure, ABCD is a || gm in which diagonals AC and BD intersect at O. If ar(||gm ABCD) is 52 cm², then the ar(∆OAB) = ?
(a) 26 cm²
(b) 18.5 cm²
(c) 39 cm²
(d) 13 cm²

Answer 7:

(d) 13 cm²
The diagonals of a parallelogram divides it into four triangles of equal areas.
∴ Area of ∆OAB = $\frac{1}{4}$$\frac{1}{4}$ ⨯ ar(||gm ABCD)
⇒ ar(∆OAB) = ​​$\frac{1}{4}$$\frac{1}{4}$ ⨯​ 52 = 13 cm²


​

Question 8:

In the given figure, ABCD is a || gm in which DL ⊥ AB. If AB = 10 cm and DL = 4 cm, then the ar(||gm ABCD) = ?
(a) 40 cm²
(b) 80 cm²
(c) 20 cm²
(d) 196 cm²

Answer 8:

(a) 40 cm²
ar(||gm ABCD) = base × height =  10 ​× 4 =  40 cm²

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Question 9:

The area of ||gm ABCD is
(a) AB × BM
(b) BC × BN
(c) DC × DL
(d) AD × DL

Answer 9:

Area of a parallelogram is base into height.
Height = DL = NB
Base = AB = CD
So, area of parallelogram ABCD = DC $\times$$\times$ DL
Hence, the correct answer is option (c). 

Question 10:

Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is
(a) 1 : 2
(b) 1 : 1
(c) 2 : 1
(d) 3 : 1

Answer 10:

Parallelograms on the same base and between the same parallels are equal in area.
So, the ratio of their areas will be 1 : 1.  
Hence, the correct answer is option (b). 

Question 11:

In the given figure, ABCD and ABPQ are two parallelograms and M is a point on AQ and BMP is a triangle.
Then, ar(∆BMP) = $\frac{1}{2}$$\frac{1}{2}$ ar(||gm ABCD) is
(a) true
(b) false

Answer 11:

We know parallelogram on the same base and between the same parallels are equal in area. 
Here, AB is the common base and AB || PD
Hence, ar(ABCD) = ar(ABPQ)                             .....(1)
Also, when a triangle and a parallelogram are on the same base and between the same parallels then the
area of triangle is half the area of the parallelogram. 
Here, for the ∆BMP and parallelogrm ABPQ, BP is the common base and they are between the common parallels BP and AQ
So, ar(∆BMP) = $\frac{1}{2}$$\frac{1}{2}$ ar(||gm ABPQ)                      .....(2)
From (1) and (2) we have
ar(∆BMP) = $\frac{1}{2}$$\frac{1}{2}$ ar(||gm ABCD) 
Thus, the given statement is true.
Hence, the correct answer is option (a).    
 

Question 12:

The midpoints of the sides of a triangle along with any of the vertices as the fourth point makes a parallelogram of area equal to
(a) $\frac{1}{2}\left(\mathrm{ar}∆\mathrm{ABC}\right)$
(b) $\frac{{\displaystyle 1}}{{\displaystyle 3}}\left(\mathrm{ar}∆\mathrm{ABC}\right)$
(c) $\frac{{\displaystyle 1}}{{\displaystyle 4}}\left(\mathrm{ar}∆\mathrm{ABC}\right)$

Answer 12:

D, E and F are the midpoints of sides BC, AC and AB respectively. 
On joining FE, we divide $△$ABC into 4 triangles of equal area.
Also, median of a triangle divides it into two triangles with equal area
$$\begin{gathered} \mathrm{ar}\left(\mathrm{AFDE}\right)=\mathrm{ar}\left(△\mathrm{AFE}\right)+\mathrm{ar}\left(△\mathrm{FED}\right) \\ =2\mathrm{ar}\left(△\mathrm{AFE}\right) \\ =2\times \frac{1}{4}\mathrm{ar}\left(△\mathrm{ABC}\right)=\frac{1}{2}\mathrm{ar}\left(△\mathrm{ABC}\right) \end{gathered}$$
Hence, the correct answer is option (a). 

Question 13:

The lengths of the diagonals of a rhombus are 12 cm and 16 cm. The area of the rhombus is
(a) 192 cm²
(b) 96 cm²
(c) 64 cm²
(d) 80 cm²

Answer 13:

(b) 96 cm²
Area of the rhombus = $\frac{1}{2}$ × product of diagonals = $\frac{1}{2}$ ×​ 12 ​× 16 = 96 cm²

Question 14:

Two parallel sides of a trapezium are 12 cm and 8 cm long and the distance between them is 6.5 cm. The area of the trapezium is
(a) 74 cm²
(b) 32.5 cm²
(c) 65 cm²
(d) 130 cm²
Figure

Answer 14:

(c) 65 cm²
Area of the trapezium = $\frac{1}{2}$ × (sum of parallel sides) × distance between them
=  $\frac{1}{2}$ ×​ ( 12 + 8) ​× 6.5
= 65 cm²

Question 15:

In the given figure ABCD is a trapezium such that AL ⊥ DC and BM ⊥ DC. If AB = 7 cm, BC = AD = 5 cm and AL = BM = 4 cm, then ar(trap. ABCD) = ?
(a) 24 cm²
(b) 40 cm²
(c) 55 cm²
(d) 27.5 cm²

Answer 15:

(b) 40 cm²

In right angled triangle MBC, we have:
MC = $\sqrt{5^2 - 4^2} = \sqrt{9} = 3 \mathrm{cm}$
In right angled triangle ADL, we have:
DL = $\sqrt{5^2 - 4^2} = \sqrt{9} = 3 \mathrm{cm}$

Now, CD = ML + MC + LD = 7 + 3 + 3 = 13 cm
∴ Area of the trapezium = $\frac{1}{2}$ × (sum of parallel sides) × distance between them
=  $\frac{1}{2}$ ×​ ( 13 + 7) ​× 4
= 40 cm²

Question 16:

In a quadrilateral ABCD, it is given that BD = 16 cm. If AL ⊥ BD and CM ⊥ BD such that AL = 9 cm and CM = 7 cm, then ar(quad. ABCD) = ?
(a) 256 cm²
(b) 128 cm²
(c) 64 cm²
(d) 96 cm²

Answer 16:

(b)128 cm²

ar(quad ABCD) = ar (∆ ABD) + ​ar (∆ DBC)
​
We have the following:
ar(∆ABD) = $\frac{1}{2}$ × base ​× height ​= ​$\frac{1}{2}$× 16 ​× 9 = 72 cm²
ar(∆DBC) = $\frac{1}{2}$ × base ​× height ​= ​$\frac{1}{2}$ × 16 ​× 7 = 56 cm²

∴ ar(quad ABCD) =​ 72 + 56 = 128 cm²

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Question 17:

ABCD is a rhombus in which ∠C = 60°. Then, AC : BD = ?
(a) $\sqrt{3}:1$
(b) $\sqrt{3}:\sqrt{2}$
(c) 3 : 1
(d) 3 : 2

Answer 17:

(a)$\sqrt{3}:1$

ABCD is a rhombus. So all of its sides are equal.
Now, BC = DC 
⇒ ∠BDC = ∠DBC = x^o   (Angles opposite to equal sides are equal)
Also, ∠BCD = 60^o
∴ x^o + x^o + 60^o = 180^o 
⇒​ 2x^o = 120^o
⇒​ x^o = 60^o
i.e., ∠BCD = ∠BDC = ∠DBC =  60^o
So, ​∆BCD is an equilateral triangle.
∴ BD = BC = a
Also, OB = a /2
Now, in ∆OAB, we have:
 $A{B}^2= O{A}^2 + O{B}^2$ 
$$\begin{gathered} \Rightarrow O{A}^2=A{B}^2-O{B}^2=a^2-{\left({\displaystyle \frac{a}{2}}\right)}^2=a^2-\frac{a^2}{4}=\frac{3a^2}{4} \\ \Rightarrow O{A}^2=\frac{3a^2}{4} \\ \Rightarrow OA=\sqrt{\frac{3a^2}{4}} \\ \Rightarrow OA=\frac{\sqrt{3}a}{2} \\ \mathrm{Now}, AC=2\times OA=2\times \frac{\sqrt{3}a}{2}=\sqrt{3}a\therefore AC:BD=\sqrt{3}a:a =\sqrt{3} :1 \end{gathered}$$

Question 18:

In the given figure ABCD and ABFE are parallelograms such that ar(quad. EABC) = 17 cm² and ar(||gm ABCD) = 25 cm². Then, ar(∆BCF) = ?
(a) 4 cm²
(b) 4.8 cm²
(c) 6 cm²
(d) 8 cm²

Answer 18:

(d) 8 cm²

Since ||gm  ABCD and ||gm ABFE are on the same base and between the same parallel lines, we have:
ar(||gm ABFE​) = ar(||gm ABCD​) = 25 cm²
⇒ ar(∆BCF ) = ar(||gm ABFE​)​ $-$ ar(quad EABC​)​ = ( 25 $-$ 17) = 8 cm²

Question 19:

∆ABC and ∆BDE are two equilateral triangles such that D is the midpoint of BC. Then, ar(∆BDE) : ar(∆ABC) = ?
(a) 1 : 2
(b) 1 : 4
(c) $\sqrt{3}:2$
(d) 3 : 4

Answer 19:

(b) 1:4

∆ABC and ∆BDE are two equilateral triangles​ and D is the midpoint of BC.
Let AB = BC = AC =  a
Then BD = BE = ED =  $\frac{a}{2}$
∴ $\frac{\mathrm{ar}(∆BDE)}{\mathrm{ar}(∆ABC)}=\frac{{\displaystyle \frac{\sqrt{3}}{4}}A{B}^2}{{\displaystyle \frac{\sqrt{3}}{4}}B{E}^2} =\frac{{\left({\displaystyle \frac{a}{2}}\right)}^2}{a^2}= \frac{1}{4}$
So, required ratio = 1 : 4

Question 20:

In a || gm ABCD, if P and Q are midpoints of AB and CD respectively and ar(|| gm ABCD) = 16 cm², then ar(|| gm APQD) = ?
(a) 8 cm²
(b) 12 cm²
(c) 6 cm²
(d) 9 cm²

Answer 20:

(a) 8 cm²

Let the distance between AB and CD be h cm.
Then ar(||gm APQD​) = AP ×​ h
= $\frac{1}{2}$ ×​ AB ×​h               (AP = $\frac{1}{2}$AB )
= $\frac{1}{2}$ ×​ ar(||gm ABCD)                [ ar(|| gm ABCD) = AB ×​h )
∴ ar (||gm APQD​) = $\frac{1}{2}$ ×​ 16 = 8 cm²

Question 21:

The figure formed by joining the midpoints of the adjacent sides of a rectangle of sides 8 cm and 6 cm is a
(a) rectangle of area 24 cm²
(b) square of area 24 cm²
(c) trapezium of area 24 cm²
(d) rhombus of area 24 cm²

Answer 21:

(d) rhombus of 24 cm²

We know that the figure formed by joining the midpoints of the adjacent sides of a rectangle is a rhombus.
So, PQRS is a rhombus and SQ and PR are its diagonals.
i.e., SQ = 8 cm and PR = 6 cm
∴ ar(rhombus PQRS​) = $\frac{1}{2}$ ×​ product of diagonals = $\frac{1}{2}$ ×​ 8  ×​ 6 = 24 cm²               

Question 22:

In ∆ABC, if D is the midpoint of BC and E is the midpoint of AD, then ar(∆BED) = ?
(a) $\frac{1}{2}\mathrm{ar}\left(∆ABC\right)$
(b) $\frac{1}{3}\mathrm{ar}\left(∆ABC\right)$
(c) $\frac{1}{4}\mathrm{ar}\left(∆ABC\right)$
(d) $\frac{2}{3}\mathrm{ar}\left(∆ABC\right)$

Answer 22:

(c) $\frac{1}{4}$ar (∆ ABC )

Since D is the mid point of BC, AD is a median of ∆ABC and BE is the median of ∆ABD.
We know that a median of a triangle divides it into two triangles of equal areas.
 i.e., ar(∆ABD ) = $\frac{1}{2}$ ar(∆ABC)                      ...(i)

⇒ ar(∆BED) = $\frac{1}{2}$​ ar(∆ABD)                      ...(ii)

From (i) and (ii), we have:
ar(∆BED) = $\frac{1}{2}$⨯ $\frac{1}{2}$​ ⨯​ ar(∆ABC)
∴​ ar(∆BED)​ = $\frac{1}{4}$⨯ ar(∆ABC)  
 ar(∆ABC)

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Question 23:

The vertex A of ∆ABC is joined to a point D on BC. If E is the midpoint of AD, then ar(∆BEC) = ?
(a) $\frac{1}{2}\mathrm{ar}\left(∆ABC\right)$
(b) $\frac{1}{3}\mathrm{ar}\left(∆ABC\right)$
(c) $\frac{1}{4}\mathrm{ar}\left(∆ABC\right)$
(d) $\frac{1}{6}\mathrm{ar}\left(∆ABC\right)$

Answer 23:

(a) $\frac{1}{2}\mathrm{ar}\left(∆ABC\right)$

Since E is the midpoint of AD, BE is a median of ∆ABD.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(∆BED) = $\frac{1}{2}$⨯ ar(∆ABD)                ...(i)
Since E is the midpoint of AD, CE is a median of ∆ADC.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(∆CED ) = $\frac{1}{2}$⨯ ar(∆ADC)               ...(ii)

Adding (i) and (ii), we have:
ar(∆BED ) + ar(∆CED ) = $\frac{1}{2}$ ⨯ ar(∆ABD) + $\frac{1}{2}$⨯ ar(∆ADC)
⇒ ar (∆ BEC ) = $\frac{1}{2}\left(∆ABD+ADC\right)=\frac{1}{2}∆ABC$    

Question 24:

In ∆ABC, it is given that D is the midpoint of BC; E is the midpoint of BD and O is the midpoint of AE. Then, ar(∆BOE) = ?
(a) $\frac{1}{3}\mathrm{ar}\left(∆ABC\right)$
(b) $\frac{1}{4}\mathrm{ar}\left(∆ABC\right)$
(c) $\frac{1}{6}\mathrm{ar}\left(∆ABC\right)$
(d) $\frac{1}{8}\mathrm{ar}\left(∆ABC\right)$

Answer 24:

(d) $\frac{1}{8}$ ar (∆ ABC)

Given: D is the midpoint of BC, E is the midpoint of BD and O is the mid point of AE.
Since D is the midpoint of BC, AD is the median of ∆ABC.
E is the midpoint of BC, so AE is the median of ∆ABD. O is the midpoint of AE, so BO is median of ∆ABE.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(∆ABD ) = $\frac{1}{2}$ ⨯ ​ar(∆ABC)                ...(i)
ar(∆ABE ) =$\frac{1}{2}$​ ⨯​ ar(∆ABD)                        ...(ii)
ar(∆BOE) = ​$\frac{1}{2}$⨯​ ar(∆ABE)                       ...(iii)

From (i), (ii) and (iii), we have:
ar(∆BOE ) =​ $\frac{1}{2}$ar(∆ABE)
ar(∆BOE ) = $\frac{1}{2}$ ⨯​ $\frac{1}{2}$​ ⨯​ $\frac{1}{2}$​ ⨯​ ar(∆ABC)​
∴​​  ar(∆BOE )​ = $\frac{1}{8}$ ar(∆ABC)18ar (∆ ABC)                    

Question 25:

If a triangle and a parallelogram are on the same base and between the same parallels, then the ratio of the area of the triangle to the area of the parallelogram is
(a) 1 : 2
(b) 1 : 3
(c) 1 : 4
(d) 3 : 4
Figure

Answer 25:

(a) 1:2

If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangles is half the area of the parallelogram.
i.e., area of triangle = $\frac{1}{2}$× area of parallelogram
∴ Required ratio = area of triangle : area of parallelogram =$\frac{1}{2}$ : 1 = 1 : 2

Question 26:

In the given figure ABCD is a trapezium in which AB || DC such that AB = a cm and DC = b cm. If E and F are the midpoints of AD and BC respectively. Then, ar(ABFE) : ar(EFCD) = ?
(a) a : b
(b) (a + 3b) : (3a + b)
(c) (3a + b) : (a + 3b)
(d) (2a + b) : (3a + b)

Answer 26:

(c) (3a +b) : (a +3b)

Clearly, EF = $\frac{1}{2}$ (a + b)                   [Mid point theorem]
Let d be the distance between AB and EF.
Then d is the distance between DC and EF.
 $$\begin{gathered} \mathrm{Now}, \mathrm{we} \mathrm{have}: \\ \mathrm{ar}(\mathrm{trap} ABEF)=\frac{1}{2}\left(a+\frac{a+b}{2}\right)d =\frac{\left(3a+b\right)d}{4} \\ \mathrm{ar}(\mathrm{trap} EFCD)=\frac{1}{2}\left(b+\frac{a+b}{2}\right)d=\frac{\left(a+3b\right)d}{4} \\ \therefore \mathrm{Required} \mathrm{ratio}= \frac{\left(3a+b\right)d}{4} : \frac{\left(a+3b\right)d}{4} = (3a+b) : ( a+3b) \end{gathered}$$ 

Question 27:

ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD is
(a) a rectangle
(b) a || gm
(c) a rhombus
(d) all of these

Answer 27:

(d) all of these
In all the mentioned quadrilaterals, a diagonal divides them into two triangles of equal areas. 

Question 28:

In the given figure, a || 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 $ABCD=\frac{1}{2}(\mathrm{perimeter} \mathrm{of} ABEF)$

Answer 28:

(c) perimeter of ABCD > perimeter of ABEF

Parallelogram ABCD and rectangle ABEF lie on the same base AB, i.e., one side is common in both the figures.
In ||gm ABCD, we have:
AD is the hypotenuse of right angled triangle ADF.
So, AD > AF
∴ Perimeter of ABCD > perimeter of ABEF

Question 29:

In the given figure, ABCD is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If $AD=2\sqrt{5}$ cm, then area of the rectangle is
(a) 32 cm²
(b) 40 cm²
(c) 44 cm²
(d) 48 cm²

Answer 29:

(b) 40 cm²

Radius of the circle, AC = 10 cm
Diagonal of the rectangle, AC = 10 cm

$$\begin{gathered} \mathrm{Now}, AB=\sqrt{A{C}^2-B{C}^2}=\sqrt{{10}^2-{\left(2\sqrt{5}\right)}^2}=\sqrt{80}=4\sqrt{5} \mathrm{cm} \\ \therefore \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{rectangle}=AB\times AD=2\sqrt{5}\times 4\sqrt{5}=40 {\mathrm{cm}}^2 \end{gathered}$$

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Question 30:

Which of the following is a false statement?
(a) A median of a triangle divides it into two triangles of equal areas.
(b) The diagonals of a || gm divide it into four triangles of equal areas.
(c) In a ∆ABC, if E is the midpoint of median AD, then $\mathrm{ar}(∆BED)=\frac{1}{4}\mathrm{ar}(∆ABC)$.

(d) In a trap. ABCD, it is given that AB || DC and the diagonals AC and BD intersect at O. Then, ar(∆AOB) = ar(∆COD).

Answer 30:

(d) In a trap. ABCD, it is given that AB || DC and the diagonals AC and BD intersect at O. Then ar(∆AOB) = ar(∆COD).


Consider ∆ADB and ∆ADC, which do not lie on the same base but lie between same parallel lines.
i.e., ar(∆ADB)$\ne$ ar(∆ADC)
Subtracting ar(∆AOD) from both sides, we get:
ar(∆ADB) $-$ ar(∆AOD) $\ne$ ar(∆ADC) $-$ ar(∆AOD)
Or ar(∆ AOB)$\ne$ ar(∆ COD)

Question 31:

Which of the following is a false statement?
(a) If the diagonals of a rhombus are 18 cm and 14 cm, then its area is 126 cm².
(b) Area of $\mathrm{a}\left|\right|\mathrm{gm}=\frac{1}{2}\times \mathrm{base}\times \mathrm{corresp}o\mathrm{nding} \mathrm{height}.$
(c) A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
(d) If the area of a || gm with one side 24 cm and corresponding height h cm is 192 cm², then h = 8 cm.

Answer 31:

(b) $\mathrm{Area} \mathrm{of} \mathrm{a} \left|\right|\mathrm{gm}=\frac{1}{2}\times \mathrm{base}\times \mathrm{corresponding} \mathrm{height}$

Area of a parallelogram  = ​base ​× corresponding height

Question 32:

Look at the statements given below:
I. A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
II. In a || gm ABCD, it is given that AB = 10 cm. The altitudes DE on AB and BF on AD being 6 cm and 8 cm respectively, then AD = 7.5 cm.
III. Area of a || gm $=\frac{1}{2}\times \mathrm{base}\times \mathrm{altitude}.$
Which is true?
(a) I only
(b) II only
(c) I and II
(d) II and III

Answer 32:

(c) I and II

Statement I is true, because if a parallelogram and a rectangle lie on the same base and between the same parallel lines, then they have the same altitude and therefore equal areas.
Statement II is also true as area of a parallelogram = base × height
$\Rightarrow$AB × DE = AD × BF
$\Rightarrow$10 × 6 = 8 × AD
$\Rightarrow$ AD = 60 ÷ 8 = 7.5 cm
Hence, statements I and II are true.

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Question 33:

Assertion: In a trapezium ABCD we have AB || DC and the diagonals AC and BD intersect at O. Then, ar(∆AOD) = ar(∆BOC)

Reason: Triangles on the same base and between the same parallels are equal in areas.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer 33:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

In trapezium ABCD, ∆ABC and ∆ABD  are on the same base and between the same parallel lines.
∴ ar(∆ABC) = ar(∆ABD)
⇒ ar(∆ABC) $-$ ar(∆AOB) = ar(∆ABD) $-$ ar(∆AOB)
⇒ ​ar(∆BOC) = ar(∆AOD) 
∴ Assertion (A) is true and, clearly, reason (R) gives (A).

Question 34:

Assertion: If ABCD is a rhombus whose one angle is 60°, then the ratio of the lengths of its diagonals is $\sqrt{3}:1$.
Reason: Median of triangle divides it into two triangles of equal area.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer 34:

(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.

Reason (R) is clearly true.
The explanation of assertion (A)​ is as follows:
ABCD is a rhombus. So, all of its sides are equal.
Now, BC = DC
⇒∠BDC = ∠DBC = x$°$
Also, ∠BCD = 60$°$
∴ x$°$ + x$°$ + 60$°$ = 180$°$
⇒​2x$°$ = 120$°$
⇒​ x$°$ = 60$°$
∴ ∠BCD = ∠BDC = ∠DBC =  60$°$
So, ​∆BCD is an equilateral triangle.
i.e., BD = BC = a
∴ OB = $\frac{a}{2}$
Now, in ∆ OAB, we have:
$$\begin{gathered} O{A}^2=A{B}^2-O{B}^2=a^2-{\left(\frac{a}{2}\right)}^2=\frac{3a^2}{4} \\ \Rightarrow OA=\frac{\sqrt{3}a}{2} \\ \Rightarrow AC=2\times \frac{\sqrt{3}a}{2}=\sqrt{3}a \end{gathered}$$
$\therefore AC:BD=\sqrt{3}a:a=\sqrt{3}:1$
Thus, assertion (A)​ is also true, but reason (R) does not give (A).
​Hence, the correct answer is (b).

Question 35:

Assertion: The diagonals of a || gm divide it into four triangles of equal area.
Reason: A diagonal of a || gm divides it into two triangles of equal area.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer 35:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Question 36:

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.
Reason: The area of an equilateral triangle of side 8 cm is $16\sqrt{3} {\mathrm{cm}}^2$.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer 36:

(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not a correct explanation of Assertion (A).

Explanation:
Reason (R):
∴ ar(∆ABC ) = $\frac{\sqrt{3}}{4}\times (\mathrm{side}{)}^2=\left(\frac{\sqrt{3}}{4}\times 8\times 8\right) = 16\sqrt{3} {\mathrm{cm}}^2$
Thus, reason (R) is true.

Assertion (A):
Area of trapezium = $\frac{1}{2}\times \left(25+15\right)\times 6 = 120 {\mathrm{cm}}^2$
Thus, assertion (A) is true, but reason (R) does not give assertion (A).

Question 37:

Assertion: In the given figure, ABCD is a || gm in which DE ⊥ AB and BF ⊥ AD. If AB = 16 cm, DE = 8 cm and BF = 10 cm, then AD is 12 cm.

Reason: Area of a || gm = base × height.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer 37:

(d) Assertion is false and Reason is true.

Clearly, reason (R) is true.
Assertion: Area of a parallelogram = base × height
$\Rightarrow$AB ×​ DE = AD × BF
$\Rightarrow$AD = (16 × 8) ÷ 10 = 12.8 cm

So, the assertion is ​false.