Areas of Parallelograms and Triangles Test

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Areas of Parallelograms and Triangles Test - 15

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Weekly Quiz Competition

Question 1
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In the figure given above, if the area of the triangle ABE = $${43 cm ^2}$$; find the area of the parallelogram ABEF.

A
$${86\ cm ^2}$$

B
$${88\ cm ^2}$$

C
$${43\ cm ^2}$$

D
$${54\ cm ^2}$$

Solution

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 = $$43$$ $$cm^2$$
Question 2
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Two triangles on same base and are between same parallel lines have ____ .

A
Equal perimeter

B
Equal area

C
Both

D
None of these

Solution

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

So, $$Op-B$$ is correct.
Question 3
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A rectangle and a rhombus are on the same base and between the same parallels. The ratio of their areas is :

A
$$1 : 2$$

B
$$1 : 3$$

C
$$1 : 4$$

D
$$1 : 1$$

Solution

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.
$$\Rightarrow \dfrac {Rectangle}{Rhombus}$$ $$=\dfrac{1}{1}$$
Question 4
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Two parallelograms are on the same base and between the same parallels. The ratio of their areas is :

A
$$1 : 1$$

B
$$1 : 2$$

C
$$2 : 1$$

D
$$1 : 4$$

Solution

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 $$1:1.$$
Question 5
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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

A
$$60 cm^2$$

B
$$65 cm^2$$

C
$$30 cm^2$$

D
$$32.5 cm^2$$

Solution

$$ (i)\quad Tangents\quad from\quad a\quad point\quad to\quad a\quad circle\quad are\quad equal.\\ (ii)\quad Angles\quad made\quad by\quad the\quad tangent\quad and\quad the\quad radius\quad at\quad the\\ point\quad of\quad contact\quad is={ 90 }^{ o }\\ $$

$$ Given-\\ PQ\quad \& \quad PR\quad are\quad tangents\quad drawn\quad from\quad P\quad to\quad a\quad circle,\quad with\quad O\quad as\quad center,\\ at\quad Q\quad \& \quad R.\quad OP=13CM,\quad OQ=5cm=OR\quad (radius)\\ To\quad find\quad out-\\ arPQOR\\ Solution-PQ\quad \& \quad PR\quad are\quad tangents\quad drawn\quad from\quad P\quad to\quad the\quad circle\\ at\quad Q\quad \& \quad R.\\ \therefore \quad PQ\quad =\quad PR\quad and\quad \angle PQO=\angle PRO={ 90 }^{ o }.......(i)\\ So\quad between\quad \Delta PQO\quad \& \quad \Delta PRO\quad we\quad have\\ PQ\quad =\quad PR,\quad OQ\quad =\quad OR\quad (radius\quad of\quad the\quad same\quad circle),\quad and\quad side\\ OP\quad is\quad common.\\ \therefore \quad By\quad SSS\quad test\quad \Delta PQO\quad is\quad congruent\quad to\quad \Delta PRO\\ \therefore \quad ar\Delta PQO\quad =\quad ar\Delta PRO\\ \Longrightarrow arPQOR=ar\Delta PQO\quad +\quad ar\Delta PRO\\ \Longrightarrow arPQOR=2ar\Delta POQ\quad or\quad 2ar\Delta PRO.........(ii)\\ Now\quad \Delta POQ\quad is\quad a\quad right\quad angled\quad one\quad with\quad hypotenuse\quad OP.(from\quad i)\\ \therefore \quad { PQ }^{ 2 }+{ OQ }^{ 2 }={ PO }^{ 2 }\\ \Longrightarrow { PQ }^{ 2 }+5^{ 2 }=13^{ 2 }\quad \Longrightarrow PQ=12cm\\ \therefore \quad ar\Delta POQ=\frac { 1 }{ 2 } \times OQ\times PQ\quad =\frac { 1 }{ 2 } \times 5\times 12=30cm.\\ arPQOR=2ar\Delta POQ=2\times 30{ cm }^{ 2 }=60{ cm }^{ 2 }(from\quad ii\quad )\\ Ans\quad Option\quad A $$
Question 6
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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:

A
$$1 : 3$$

B
$$1 : 2$$

C
$$3 : 1$$

D
$$1 : 4$$

Solution

Given:
$$ \Delta ABC$$ & parallelogram $$ADEC$$ are on the same base $$AC$$ and within the same parallels $$AC$$ & $$BE$$.

To find the ratio of area of the triangle to the area of the parallelogram.

Let $$h$$ be the perpendicular distance between $$AC$$ & $$BE$$.
Then, $$ar( ABC)=\cfrac { 1 }{ 2 } \times$$ base $$\times$$ height $$=\cfrac { 1 }{ 2 } AC\times h$$.
And, $$ar( ADEC)=$$ base $$\times$$ height $$=AC \times h$$.
$$\therefore ar( ABC) : ar (ADEC) = \cfrac { 1 }{ 2 } AC \times h : AC\times h$$
$$=\dfrac{1}{2}:1$$ $$= 1:2$$.

Therefore, option $$B$$ is correct.
Question 7
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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
$$3 : 1$$

Solution

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 $$=AB\times h$$

$$\therefore \quad area \ ABEF:area \ ABCD$$
$$=AB\times h:AB\times h=1:1$$
Question 8
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In which of the following figures, find two polygons on the same base and between the same parallels?

A

B

C

D

Solution

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 $$PS$$
Question 9
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If two triangles are on the same base and between the same parallels, then the ratio of
their areas is :

A
$$1 : 1$$

B
$$1 : 2$$

C
$$2 : 1$$

D
$$1 : 4$$

Solution

$$\triangle ADB$$ 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 $$1:1.$$
Question 10
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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.

A
1: 25

B
5 : 1

C
1 : 5

D
25 : 1