MCQ Questions for Class 9 Maths Areas Of Parallelograms And Triangles Quiz 1

Two triangles on same base and are between same parallel lines have ____ .

0%
Equal perimeter
0%
Equal area
0%
Both
0%
None of these

Explanation

We know, two triangles lying on same base and between same parallel lines have equal area.

So,

$Op-B$ is correct.

In the figure given above, if the area of the triangle ABE =

${43 cm ^2}$ ; find the area of the parallelogram ABEF.

0%
${86\ cm ^2}$
0%
${88\ cm ^2}$
0%
${43\ cm ^2}$
0%
${54\ cm ^2}$

Explanation

Since, the triangle and the parallelogram are on the same base and between the same parallels,the area of a parallelogram is twice to the area of the triangle.
Hence, area of parallelogram ABEF =

$2 \times 43$

$cm^2$
area of parallelogram ABEF=

$86$

$cm^2$

Since, the triangle and the parallelogram are on the same base and between the same parallels,the area of a parallelogram is twice to the area of the triangle.
Hence, area of parallelogram ABCD =

$2 \times 43$

$cm^2$
area of parallelogram ABCD=

$86$

$cm^2$

Since, both the triangles and the parallelogram are on the same base and between the same parallels,the area will be the same for both of them.
Hence, area of triangle AEF =

$43$

$cm^2$

A rectangle and a rhombus are on the same base and between the same parallels. The ratio of their areas is :

0%
$1 : 2$
0%
$1 : 3$
0%
$1 : 4$
0%
$1 : 1$

Explanation

As we know parallelogram between same base and same parallels have the same area.

$\therefore$ area of rectangle to the area of rhombus is same.

$\Rightarrow \dfrac {Rectangle}{Rhombus}$

$=\dfrac{1}{1}$

If two triangles are on the same base and between the same parallels, then the ratio of
their areas is :

0%
$1 : 1$
0%
$1 : 2$
0%
$2 : 1$
0%
$1 : 4$

Explanation

$\triangle ADB$ and

$\triangle ABC$ are triangles on same base

$AB$ and between the same parallels

$AB$ and

$DC.$

We have to prove that

$ar\left(ADB\right)=ar\left(ABC\right)\Rightarrow 1:1$

When we construct a line through

$A$ parallel to

$BC$ meeting

$DC$ at

$E$ i.e,

$AB\parallel BC$ and construct a line through

$B$ parallele to

$AD$ meeting

$DC$ at

$F$ i.e,

$BF\parallel AD$

From

$ABFD,$ From

$ABCE,$

$DF\parallel AB$ (Given as

$AB\parallel CD$ )

$EC\parallel AB$ (Given as

$AB\parallel CD$ )

and

$BF\parallel AD$ ( By construction ) and

$AE\parallel BC$ ( By construction )

Thus, both pair of opposite sides are parallel. Thus, both pair of opposite sides are parallel.

Thus,

$ABFD$ and

$ABCE$ are two parallograme with the same base

$AB$ and between two parallel

$AB\;$ and

$\;EF$

$\therefore$

$ar\left (ABFD \right) = ar\left( ABCE\right)$

$\therefore ar(\Delta ADB)=ar(\Delta ABC)$

$\therefore$ The ratio of their area will be equal to

$1:1$

Hence, the answer is

$1:1.$

In the given figure, KL // AC // YZ. B and D are

equidistant from AC. If $5KL = YZ$ ,
find the ratio of areas of $\Delta$ BKL and
$\Delta$

DYZ.

0%
1: 25
0%
5 : 1
0%
1 : 5
0%
25 : 1

D and E are the mid points of the sides AB and AC of a triangle ABC. Then in what ratio are the areas of the triangles ADE and ABC ?

0%
1 : 4
0%
1 : 3
0%
1 : 2
0%
2 : 3

Two parallelograms are on the same base and between the same parallels. The ratio of their areas is :

0%
$1 : 1$
0%
$1 : 2$
0%
$2 : 1$
0%
$1 : 4$

Explanation

We know that,

Area of a parallelogram = base ⨯ height

Now, if both parallelograms are on the same base and between the same parallels, then their heights will be equal.

Hence, their areas will also be equal.

$ar\left( \parallel ^{gm}\;1 \right) =ar\left( \parallel^ {gm}\;2 \right)$

Since both areas are equal they get cancelled on both sides,

i.e,

$1:1.$

Hence, the answer is

$1:1.$

From a point P which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is

0%
$60 cm^2$
0%
$65 cm^2$
0%
$30 cm^2$
0%
$32.5 cm^2$

Explanation

${\textbf{Step - 1: Find the length of PQ and PR}}$

${\text{In right angled triangle OPQ ,we have}}$

${\text{PQ = }}\sqrt {{{(OP)}^2} - {{(OQ)}^2}}$

$= \sqrt {169 - 25} = \sqrt {144} = 12\ cm$

$\Rightarrow PR = 12\ cm\ \ \ {\text{(Two tangents from the same external point to a circle are equal)}}$

${\textbf{Step - 2: Find Area of quadrilateral}}$

${\text{ = (2}} \times \dfrac{1}{2} \times 12 \times 5)\ c{m^2} = 60\ c{m^2}$ $\textbf{[area of PQOR = 2} \times {\textbf{Area of }}\Delta {\textbf{POQ}]}$

${\textbf{Hence , area of the required quadrilateral is (A) 60 c}}{{\textbf{m}}^2}$

If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of the parallelogram is:

0%
$1 : 3$
0%
$1 : 2$
0%
$3 : 1$
0%
$1 : 4$

Explanation

$\textbf{Step -1: Find the area of required triangle and parallelogram.}$

$\text{As, the }\triangle ABD\text{ and parallelogram }ABCD\text{ are on same base and between same parallels.}$

$\therefore \text{Height of the triangle and parallelogram is same.}$

$\text{Area of }\triangle ABD=\dfrac{1}{2}\times\text{Base}\times\text{Height}$

$=\dfrac{1}{2}\times AB\times DE$

$\text{Area of parallelogram}=\text{Base}\times\text{Height}$

$=AB\times DE$

$\textbf{Step -2: Find the required ratio.}$

$\dfrac{\text{Area of }\triangle ABD}{\text{Area of parallelogram }ABCD}=\dfrac{\dfrac{1}{2}\times AB\times DE}{AB\times DE}$

$=\dfrac{1}{2}$

$\textbf{Hence, option B is correct.}$

Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is :

0%
$1 : 2$
0%
$1 : 1$
0%
$2 : 1$
0%
$3 : 1$

Explanation

Given:

$ABEF$ and

$ABCD$ are two paralleologram with same base

$AB$ and within same parallel lines

$AB$ &

$FC$ .

To find : The ratio of their areas.

Solution:

Let the perpendicular distance between

$AB$ and

$FC$ be

$h$ .

$\therefore area \ ABEF=$ base

$\times$ height

$=AB\times h$

and

$\quad area \ ABCD=$ base

$\times$ height

$=AB\times h$

$\therefore \quad area \ ABEF:area \ ABCD$

$=AB\times h:AB\times h=1:1$

In which of the following figures, find two polygons on the same base and between the same parallels?

Explanation

In options A, B and C there are no two polygons having same base and being within same parallels.

In option D, quadrilaterals

$PAQR$ and

$BQRS$ have same base

$QR$ .

Again

$PA$ and

$BS$ are on the same line

$PS$ which is parallel to

$QR$ .

$\therefore$

$PAQR$ and

$SBQR$ have

$QR$ and they are within the same parallels

$QR$ and

$PS$

The following diagram shows two parallelograms

$ABCD$ and

$BEFG.$ Is the area of parallelogram

$ABCD$ is equal to area of parallelogram

$BEFG$ ?

0%
True
0%
False

Explanation

Let us join

$AC$ and

$GE$ .

$\triangle ACG$ and

$\triangle AEG$ are on the same base

$AG$ and between the same parallels

$AG$ and

$CE$ .

$\therefore ar(\triangle ACG)=ar(\triangle AEG)$

$\Rightarrow ar(\triangle ACG)-ar(\triangle ABG)=$

$ar(\triangle AEG)-ar(\triangle ABG)$

$\Rightarrow ar(\triangle ABC)=ar(\triangle BEG)\quad\quad\dots(1)$

SInce

$AC$ and

$GE$ are diagonals of parallelograms

$ABCD$ and

$BGFE$ respectively.

$ar(\triangle ABC)=\dfrac{1}{2}ar(ABCD)\quad\quad...(2)$

$ar(\triangle BEG)=\dfrac{1}{2}ar(BGFE)\quad\quad...(3)$

From equations

$(1),(2)$ and

$(3)$ , we have

$\dfrac12 ar(ABCD)= \dfrac{1}{2}ar(BGFE)$

$\Rightarrow ar(ABCD)=ar(BGFE)$

Hence, the given statement is true.

1934859_195547_ans_c9c42ed6334b44788cd5721e8b8d384f.png

If two triangles are on the same base and between the same parallels then the ratio of their areas is

0%
1 : 2
0%
1 : 3
0%
2 : 3
0%
1 : 1

ABCD is a rectangle, ABEF is a parallelogram with area of

$60 \ cm^2$ and AB and CF are parallel. Find area of rectangle ABCD

0%
$60$ $cm^2$
0%
$50$ $cm^2$
0%
$40$ $cm^2$
0%
$75$ $cm^2$

Explanation

As we can see, ABCD and ABEF lie on the same base and between the same parallels.

$\therefore$

$Area(ABCD) = Area(ABEF) = 60cm^2$

ABCD is a rectangle, ABEF is a parallelogram with area of

$30 \ cm^2$ and AB and CF are parallel. Find area of rectangle ABCD

0%
$30\ sq\ units$
0%
$40\ sq\ units$
0%
$50\ sq\ units$
0%
$60\ sq\ units$

Explanation

Given ABCD is a rectangle ABEF is a parallelogram

area of

$ABEF = 30cm^2$

base

$\times$ height

$= 30cm^2$

$AB \times CB = 30cm^2$

In rectangle

$ABCD(AD = BC)$

area

$= AB \times CB = 30cm^2$

Length

$\times$ breadth

$=30cm^2$

area of rectangle

$=30cm^2$

Option A is correct

2104252_194871_ans_85dbc44d2ff849e9ac8a06639d4c0b9a.png

In the given figure, area of the parallelogram

$ABCD$ is

$48$

$cm^2$ and

$FC \parallel AB$ . Find the area of the parallelogram

$ABEF$ .

0%
$12\ cm^2$
0%
$24\ cm^2$
0%
$48\ cm^2$
0%
$8\ cm^2$

Explanation

$\Rightarrow$ In given figure, we can see that parallelogram

$ABCD$ and parallelogram

$ABEF$ both are lies between same parallel line that is

$FC$ and

$AB$ .

$\Rightarrow$ And parallelogram

$ABCD$ and

$ABEF$ contains same base that is

$AB$ .

$\Rightarrow$ So, we know that parallelogram on same base and between same parallels are equal in area.

$\Rightarrow$ So,

$Area(ABCD)=Area(ABEF)$ ------ ( 1 )

$\Rightarrow$ We have given

$Area(ABCD)=48\ cm^2$

From

$( 1 )$ we get,

Area of parallelogram

$ABEF=48\ cm^2$

For

$3$ parallelograms,

$A$ ,

$B$ and

$C$ , parallelograms

$A$ and

$B$ shares same base

$l$ and lie along same parallels

$l$ and

$m$ . Similarly the parallelograms

$B$ and

$C$ shares the same base

$m$ and between the same parallels. The area of all three parallelograms will be equal.

0%
True
0%
False

Explanation

Given three

$\parallel\;gm$

$A,B,C$

$A$ and

$B$

$\parallel\;gm$ share the same base as

$\Rightarrow ar\left( A \right) =ar\left( B \right)$

Similarily

$ar\left( B \right) =ar\left( C \right)$

$\Rightarrow ar\left( A \right) =ar\left( B \right) =ar\left( C \right)$

Hence, the answer is true.

The areas of parallelogram between the parallels are equal.

0%
True
0%
False

For figures on same base and same parallels, the figures have to lie between same _______

0%
parallels
0%
bases
0%
heights
0%
none of these

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

CBSE Questions for Class 9 Maths Areas Of Parallelograms And Triangles Quiz 1 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Areas Of Parallelograms And Triangles Quiz 1 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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