Tag: perimeter and area of rectilinear figures

Questions Related to perimeter and area of rectilinear figures

One side of a parallelogram is 8 cm If the corresponding altitude is 6 cm then its area is given by

maths perimeter and area of rectilinear figures parallelogram and rectangle area of parallelogram area of a parallelogram

  1. 24 $ \displaystyle cm ^{2} $

  2. 36 $ \displaystyle cm ^{2} $

  3. 40 $ \displaystyle cm ^{2} $

  4. 48 $ \displaystyle cm ^{2} $

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Correct Option: D
Explanation:

Given One side of parallelogram is 8 cm and altitude is 6 cm 

Here base=8 cm and height(altitude)=6 cm
We know that area of parallelogram=$base \times hieght$

Here base=8 cm and height(altitude)=6 cm
Then area of  parallelogram=$8\times 6=48 cm^{2}$

The height of a parallelogram of area $ \displaystyle   350 cm^{2}   $ and base 25 cm is

maths perimeter and area of rectilinear figures parallelogram and rectangle area of parallelogram area of a parallelogram

  1. 12 cm

  2. 13 cm

  3. 14 cm

  4. 15 cm

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Correct Option: C
Explanation:

Given the area of parallelogram is 350 sq cm and base is 25 cm

Then area of  parallelogram if height h and base 25
$\therefore 350=base \times height\Rightarrow 25\times h=350\Rightarrow h=14$ cm

Two adjacent sides of a parallelogram are x and y and the included angle is $\displaystyle \alpha  $ then the area of the parallelogram is

maths perimeter and area of rectilinear figures parallelogram and rectangle area of parallelogram area of a parallelogram

  1. xy $\displaystyle \cot \alpha $

  2. xy $\displaystyle \cos \alpha $

  3. xy $\displaystyle \sin \alpha $

  4. none of these

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Correct Option: C
Explanation:

Given the two adjacent sides of a parallelogram is x and y and angle is $\alpha $

let the height of  parallelogram is h
Then $sin \alpha=\frac{h}{y}\Rightarrow h=y sin \alpha  $

So area of   parallelogram=$base\times heigth=x\times ysin\alpha =xy sin\alpha $

What will be area of a parallelogram with base 6 cm and altitude 3.5 cm---

maths perimeter and area of rectilinear figures parallelogram and rectangle area of parallelogram area of a parallelogram

  1. 28 square cm

  2. 20 square cm

  3. 21 square cm

  4. None of these

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Correct Option: C
Explanation:

Area = Base $\times$ Altitude
         = 6 cm $\times$ 3.5 cm
         = 21 square cm

In a trapezium whose parallel sides measure 12 cm and 10 cm and the distance between them is 8 cm. Find the area of trapezium---

maths perimeter and area of rectilinear figures parallelogram and rectangle area of parallelogram area of a parallelogram

  1. 84 $cm^2$

  2. 48 $cm^2$

  3. 88 $cm^2$

  4. 188 $cm^2$

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Correct Option: C
Explanation:

Area of trapezium = $\displaystyle \frac{h}{2} (a + b)$
                                = $\displaystyle \frac{8}{2} (12 + 10)$
                                = 4 $\times$ 22 = 88 $cm^2$

Find the area of the parallelogram whose base is 16 cm and height 0.4 m ?

maths perimeter and area of rectilinear figures parallelogram and rectangle area of parallelogram area of a parallelogram

  1. 440 $cm^2$

  2. 740 $cm^2$

  3. 640 $cm^2$

  4. None of these

Show answer
Correct Option: C
Explanation:

Base = 16 cm
Height = 0.4 m
            = 0.4 $\times$ 100 = 40 cm
Area of parallelogram = b $\times$ h
                                      = 16 $\times$ 40
                                      = 640 $cm^2$

In a parallelogram the base and height are is in the ratio of $5 : 2$. If the area of the parallelogram is $360\ m^{2}$, find its base and height.

maths perimeter and area of rectilinear figures parallelogram and rectangle area of parallelogram area of a parallelogram

  1. $30\ m, 12\ m$.

  2. $20\ m, 12\ m$.

  3. $30\ m, 10\ m$.

  4. None of these

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Correct Option: A

One side of a parallelogram is $8$ cm. If the corresponding altitude is $6$ cm, then its area is given by

maths perimeter and area of rectilinear figures parallelogram and rectangle area of parallelogram area of a parallelogram

  1. $24: cm^2$

  2. $36: cm^2$

  3. $40: cm^2$

  4. $48: cm^2$

Show answer
Correct Option: D
Explanation:

Area of parallelogram $=$ length $\times$ height
Here altitude is nothing but the height
$\therefore $ Area of parallelogram $= 8 \times 6$
$= 48: cm^2$

If the base and altitude of a parallelogram are doubled, what happens to the area compared to the original one?

maths perimeter and area of rectilinear figures parallelogram and rectangle area of parallelogram area of a parallelogram

  1. $8$ times original

  2. $2$ times original

  3. $4$ times original

  4. Remains same

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Correct Option: C
Explanation:

Area of parallelogram $S _1=bh$


New Area, $S _2=2b\times 2h=4bh=4S _1$

$A, B, C, D$ are mid points of sides of parallelogram $PQRS$. If $ar(PQRS)=36\ cm^{2}$ then $ar(ABCD)$

maths perimeter and area of rectilinear figures parallelogram and rectangle area of parallelogram area of a parallelogram

  1. $24\ cm^{2}$

  2. $18\ cm^{2}$

  3. $30\ cm^{2}$

  4. $36\ cm^{2}$

Show answer
Correct Option: B
Explanation:
$A,B,C,D$ are point of sides of parallelogram $PQRS$ area $(PQRS)=36\ cm^2 ar (ABCD)=?$
$AP=\dfrac {1}{2}SP,\ BP=\dfrac {1}{}QR$
$PS=QR, \dfrac {1}{2}PS=\dfrac {1}{2} QR$
area $(\triangle ABC)=\dfrac {1}{2} (\triangle PQCQ)$
area $(\triangle ADC)=\dfrac {1}{2} ar (ACRS)$
area $\triangle ABC+area (\triangle ADC)=\dfrac {1}{2} area (APCQ)=\dfrac {1}{2}area (ACRS)$
$area (ADCD)=\dfrac {1}{2}area (APCQ)+(ACRS)area $
$area (ABCD)=\dfrac {1}{2} area (PQRS)$
$area (ABCD)=\dfrac {1}{2} (36)$
$=18\ cm^2$