Areas of Parallelograms and Triangles - Test Papers

 CBSE Test Paper 01

CH-9 Areas of Parallelograms & Triangles

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1. The area of the parallelogram ABCD in the figure is :
   

   1. 12cm2$12c{m}^2$

   2. 10cm2$10c{m}^2$

   3. 9cm2$9c{m}^2$

   4. 15cm2$15c{m}^2$

2. Find the area of parallelogram in the adjoining figure.
   

   1. 60 square feet

   2. 48 square feet

   3. 1759 feet

   4. 84 square feet

3. PQRS is a parallelogram. If X and Y are mid-points of PQ and SR and diagonal SQ is joined, then ar(∥XQRY):ar(△QSR)$ar(\parallel XQRY):ar(\mathrm{△}QSR)$ is equal to
   

   1. it is 2 : 1.

   2. it is 1 : 2.

   3. it is 1 : 1.

   4. it is 1 : 4.

4. Points A, B, C, and D are collinear. AB = BC = CD. XY∥AD.$XY\parallel AD.$ If P and M lie on XY and ar(△MCD)=7cm2,$ar(\mathrm{△}MCD)=7c{m}^2,$ then ar(△APB)$ar(\mathrm{△}APB)$and ar(△APD)$ar(\mathrm{△}APD)$ respectively are
   

   1. 7cm2,21cm2.$7c{m}^2,21c{m}^2.$

   2. 7cm2,14cm2.$7c{m}^2,14c{m}^2.$

   3. 14cm2,14cm2.$14c{m}^2,14c{m}^2.$

   4. 14cm2,21cm2.$14c{m}^2,21c{m}^2.$

5. If the sum of the parallel sides of a trapezium is 7 cm and distance between them is 4 cm, then area of the trapezium is :

   1. 7cm2$7c{m}^2$

   2. 21cm2$21c{m}^2$

   3. 14cm2$14c{m}^2$

   4. 28cm2$28c{m}^2$

6. Fill in the blanks:

   The area of a rhombus is equal to ________ of the product of its two diagonals.

7. Fill in the blanks:

   A median of a triangle divides it into ________ triangles of equal areas.

8. In a given figure, it is given that AD || BC. Prove that ar(△$\mathrm{△}$CGD) = ar(△$\mathrm{△}$ABG).
   

9. Is the given figure lie on the same base and between a same parallels. In such a case, write the common base and the two parallels :
   

10. In Fig., X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar (△$\mathrm{△}$ABP) = ar (△$\mathrm{△}$ACQ).
    

11. ABCD is a parallelogram. P is any point on CD. If area(△$\mathrm{△}$DPA) = 15 cm2 and area (△$\mathrm{△}$APC) = 20 cm2, find the area(△$\mathrm{△}$APB).

12. In figure, AP || BQ || CR. Prove that ar(AQC) = ar(PBR).
    

13. D and E are mid-points of BC and AD respectively. If area of △$\mathrm{△}$ABC = 10 cm2, find area of △$\mathrm{△}$EBD.

    

14. In the given Fig., X and y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z. Prove that ar (△$\mathrm{△}$LZY) = ar (MZYX).
    

15. If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area. If G is the mid-point of median AD, prove that ar(△$\mathrm{△}$BGC) = 2ar(△$\mathrm{△}$AGC).

CBSE Test Paper 01
CH-9 Areas of Parallelograms & Triangles

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Solution

1. (a) 12cm2$12c{m}^2$
   Explanation:
   Area of parallelogram = Base ×$\times$ Height
   = AB ×$\times$ BD
   = 3 ×$\times$ 4
   = 12 cm2
2. (a) 60 square feet
   Explanation:
   In the figure, Base AB = 12 feet and Correspongin height DE = 5 feet
   Then
   Area of parallelogram ABCD = Base ×$\times$ Corresponding height
   = AB ×$\times$ DE
   = 12 ×$\times$ 5
   = 60 square feet
3. (c) it is 1 : 1.
   Explanation:
   Since parallelogram PQRS and triangle AQS are on the same base QS and between the same parallels, then
   ar(△PQS)=12×ar(∥gmPQRS)$\mathrm{a}\mathrm{r}\left(\mathrm{△}\mathrm{P}\mathrm{Q}\mathrm{S}\right)=\frac{1}{2}\times \mathrm{a}\mathrm{r}\left(\parallel gm\mathrm{P}\mathrm{Q}\mathrm{R}\mathrm{S}\right)$              .................(i)
   Also, if X and Y are mid-points of PQ and SR and diagonal SQ is joined, then
   area(∥gmXQRY)=12×area(∥gmPQRS)$\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\left(\parallel gm\mathrm{X}\mathrm{Q}\mathrm{R}\mathrm{Y}\right)=\frac{1}{2}\times \mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\left(\parallel gm\mathrm{P}\mathrm{Q}\mathrm{R}\mathrm{S}\right)$          ....................(ii)
   From eq.(i) and (ii), we get
   area(△PQS)=area(∥gmXQRY)$\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\left(\mathrm{△}\mathrm{P}\mathrm{Q}\mathrm{S}\right)=\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\left(\parallel gm\mathrm{X}\mathrm{Q}\mathrm{R}\mathrm{Y}\right)$ 
   ⇒$\Rightarrow$ area(∥gmXQRY):area(△PQS)=1:1$\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\left(\parallel gm\mathrm{X}\mathrm{Q}\mathrm{R}\mathrm{Y}\right):\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\left(\mathrm{△}\mathrm{P}\mathrm{Q}\mathrm{S}\right)=1:1$
4. (a) 7cm2,21cm2.$7c{m}^2,21c{m}^2.$

    

   Explanation:

   Since triangles MCD and PAB are on the same base and between the same parallels, then
   area(△MCD)=area(△PAB)=7$\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\left(\mathrm{△}\mathrm{M}\mathrm{C}\mathrm{D}\right)=\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\left(\mathrm{△}\mathrm{P}\mathrm{A}\mathrm{B}\right)=7$ sq. cm
   Now, since AB = BC = CD, then
   AB = 13$\frac{1}{3}$AD and BC = 23$\frac{2}{3}$AD
   Then, area(△APB)=13×area(△APD)$\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\left(\mathrm{△}\mathrm{A}\mathrm{P}\mathrm{B}\right)=\frac{1}{3}\times \mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\left(\mathrm{△}\mathrm{A}\mathrm{P}\mathrm{D}\right)$ 
   ⇒$\Rightarrow$ 13×area(△APD)=7$\frac{1}{3}\times \mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\left(\mathrm{△}\mathrm{A}\mathrm{P}\mathrm{D}\right)=7$ sq. cm
   ⇒$\Rightarrow$  area(△APD)=21$\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\left(\mathrm{△}\mathrm{A}\mathrm{P}\mathrm{D}\right)=21$ sq. cm

5. (c) 14cm2$14c{m}^2$

    

   Explanation:

   Given: Sum of the parallel sides of a trapezium is 7 cm and distance between them is 4 cm.

   Area of the trapezium = 12$\frac{1}{2}$(Sum of parallel sides) x Height

   = 12$\frac{1}{2}$x 7 x 4

   = 14 cm2

    

6. 12$\frac{1}{2}$

7. two

8. Since △$\mathrm{△}$ADC and △$\mathrm{△}$ADB are on the same base AD and between the same parallels AD and BC.
   ar(△$\mathrm{△}$ADC) = ar(△$\mathrm{△}$ADB)
   ⇒$\Rightarrow$ ar(△$\mathrm{△}$ADC) - ar(△$\mathrm{△}$ADG) = ar(△$\mathrm{△}$ADB) - ar(△$\mathrm{△}$ADG)
   ⇒$\Rightarrow$ ar(△$\mathrm{△}$CGD) = ar(△$\mathrm{△}$ABG)

9. △$△$TRQ and parallelogram SRQP lie on the same base RQ and between the same parallels RQ and SP.

10. From the given figure it is clear that X and Y are the mid-points of AC and AB respectively.
     Therefore,  XY ∥$\parallel$ BC
    Since △$\mathrm{△}$BYC and △$\mathrm{△}$BXC are on the same base BC and between the same parallels XY and BC, we can write
       ar (△$\mathrm{△}$BYC) = ar (△$\mathrm{△}$BXC)
    ⇒$\Rightarrow$ar (△$\mathrm{△}$BYC) - ar (△$\mathrm{△}$BOC) = ar (△$\mathrm{△}$BXC) - ar (△$\mathrm{△}$BOC)
    ⇒$\Rightarrow$ ar (△$\mathrm{△}$BOY) = ar (△$\mathrm{△}$COX)
    ⇒$\Rightarrow$ar (△$\mathrm{△}$BOY) + ar (△$\mathrm{△}$XOY) = ar (△$\mathrm{△}$COX) + ar (△$\mathrm{△}$XOY)
    ⇒$\Rightarrow$ar (△$\mathrm{△}$BXY) = ar (△$\mathrm{△}$CXY) .......................................(i)
    Since quadrilaterals XYAP and XYQA are on the same base XY and between the same parallels XY and PQ, we have
      ar(XYAP) = ar (XYQA) .........................................(ii)
    On adding (i) and (ii), we obtain
    ar (△$\mathrm{△}$BXY) + ar (XYAP) = ar (△$\mathrm{△}$CXY) + ar (XYQA)
    ⇒$\Rightarrow$ar (△$\mathrm{△}$ABP) = ar (△$\mathrm{△}$ ACQ)

    Hence proved.

11. 
    From the given figure it is clear that area (△$\mathrm{△}$ADC) = area (△$\mathrm{△}$ADP)+ area(△$\mathrm{△}$APC)  = 15 + 20 = 35 cm2
    Now in parallelogram opposite sides are equal i.e., CD = AB,

    Also AB ∥$\parallel$ CD.
    Now, we know that triangles having same base and lie with the same parallels have the same area.

    ∴$\therefore$  area of △$\mathrm{△}$ABP = area of △$\mathrm{△}$ADC =  35 cm2

12. Given : AP || BQ || CR
    To Prove : ar(△$\mathrm{△}$AQC) = ar(△$\mathrm{△}$PBR)
    Proof : ar(△$\mathrm{△}$BAQ) = ar(△$\mathrm{△}$BPQ) . . . .[△$\mathrm{△}$s on the base BQ and between the same parallels BQ and AP] . . . (1)
    ar(△$\mathrm{△}$BCQ) = ar(△$\mathrm{△}$BQR) . . . .[△$\mathrm{△}$s on the base BQ and between the same parallels BQ and CR] . . . (2)
    ar(△$\mathrm{△}$BAQ) + ar(△$\mathrm{△}$BCQ) = ar(△$\mathrm{△}$BPQ) + ar(△$\mathrm{△}$BQR) . . . [Adding corresponding sides of (1) and (2)]
    ⇒$\Rightarrow$ ar(△$\mathrm{△}$AQC) = ar(△$\mathrm{△}$PBR)

13. It is given that D is the midpoint of BC, therefore we can say that AD is the median.
    
    Now, Area of △$\mathrm{△}$ABD = 12$\frac{1}{2}$ ×$\times$ area of △$\mathrm{△}$ABC   ............[Median divides a triangle into two triangles of the equal area]
    Thus, Area of △$\mathrm{△}$ABD = 12×10$\frac{1}{2}\times 10$ cm2 = 5 cm2
    Again, BE is the median of △$\mathrm{△}$ABD.
    ∴$\therefore$ Area of △$\mathrm{△}$EBD = 12$\frac{1}{2}$ ×$\times$ area of △$\mathrm{△}$ABD
                                = 12×5$\frac{1}{2}\times 5$ = 2.5 cm2

14. From the given figure,
    
    it is clear that △$\mathrm{△}$LXZ and △$\mathrm{△}$MXZ  (adding x and M) lie on the same base XZ and between the same parallels XZ and LM.
    ∴$\therefore$ ar(△$\mathrm{△}$LXZ) = ar(△$\mathrm{△}$MXZ)
    Adding ar(△$\mathrm{△}$XYZ) on both sides, we obtain
    ar(△$\mathrm{△}$LXZ) + ar(△$\mathrm{△}$XYZ) = ar(△$\mathrm{△}$MXZ) + ar(△$\mathrm{△}$XYZ)
    ⇒$\Rightarrow$ar (△$\mathrm{△}$LYZ) = ar(MZYX)
    Hence proved

15. 
    Draw AM ⊥$\perp$ BC.
    Since, AD is the median of △$\mathrm{△}$ABC,
    ∴$\therefore$ BD = DC
    ⇒$\Rightarrow$ BD ×$\times$ AM = DC ×$\times$ AM
    ⇒$\Rightarrow$ 12$\frac{1}{2}$(BD ×$\times$ AM) = 12$\frac{1}{2}$(DC ×$\times$ AM)
    ⇒$\Rightarrow$ ar(△$\mathrm{△}$ABD) = ar(△$\mathrm{△}$ACD) .....(i)
    In △$\mathrm{△}$BGC, GD is the median,
    ∴$\therefore$ ar(△$\mathrm{△}$BGD) = ar(△$\mathrm{△}$CGD) ....(ii)
    In △$\mathrm{△}$ACD, CG is a median,
    ∴$\therefore$ ar(△$\mathrm{△}$AGC) = ar(△$\mathrm{△}$CGD) .....(iii)
    From (ii) and (iii), we have,
    ar(△$\mathrm{△}$BGD) = ar(△$\mathrm{△}$AGC)
    But, ar(△$\mathrm{△}$BGC) = 2ar(△$\mathrm{△}$BGD)
    ∴$\therefore$ ar(△$\mathrm{△}$BGC) = 2 ar(△$\mathrm{△}$AGC)