Parallelograms


   
 
Question (1): Prove that the area of a trapezium is half the product of its height and the sum of the parallel sides.
Answer:  Given:
ABCD is a trapezium in which AB || DC.

Let AB = a, DC = b and CE = AF = h.
To prove:

Construction:
Join AC.
Proof:
AC is the diagonal of quad. ABCD.

area of addition axiom.


By adding (2) and (3), we get


Question (2): Show that a median of a triangle divides it into two triangles of equal area.

Answer:  Given:
In D ABC, AD is one of the medians of D ABC.
To prove:

Construction:
Draw AE ^ BC.
Proof:


Since BD = DC and h is the same for both the triangles.

Question (3):

Answer:  To prove:

Construction:

Proof:
( given) ( common side)
AE = DF ( the constructed altitudes)
Consider D AEO and D DFO,

( vertically opposite angles)
( proved above)

( using AAS congruence
theorem)

( corresponding sides
congruent triangles)

Question (4):

Answer:  Given:

To prove:

Proof:
( given)
In D BCE and D BCD
( common side)


(Triangles having equal area and having one side equal, have their corresponding altitudes equal)
i.e., alt DM = alt EN
i.e., the perpendicular distance between DE and BC at different points is the same.

Question (5): The diagonals AC, BD of quadrilateral ABCD intersect at O, and separate the quadrilateral ABCD into four triangles of equal area. Show that ABCD is a parallelogram.
Answer:  Given:

To prove:
ABCD is a parallelogram.
Proof:
From D ABC,



()
(given)


(given)
Now D ABC and D ABD are equal in area and have the same base .
Hence D ABC and D ABD are between the same parallels.

Similarly,
From figure D ADB,


(given)


Now D ABD and D ADC are equal in area and have the
Hence D ADB and D ABD are between the same parallels.


Hence quadrilateral ABCD is a parallelogram.
Question (6):

Answer:  Given:
In parallelogram ABCD, E and F are any two points on AB and BC respectively.
To prove:

Construction:

Proof:
Consider D AFH,


Consider D DFH,

parallelogram DCFH)
By adding (i) and (ii),
ar (D AFH) + ar (D DFH)


(Area addition axiom)
Consider D DGE,


Consider D GCE,


By adding (iv) and (v), we get


(Area addition axiom)
From (iii) and (vi), we get



Question (7):

Answer:  Given:


To prove:

Proof:





Question (8): In parallelogram ABCD, R is the midpoint of DC and S is the midpoint of AD. If BR and BS are drawn, prove that D BSA and D BCR are equal in area.
Answer:  Given:
ABCD is a parallelogram. R is the mid-point of DC. S is the mid-point of AD.
To prove:

Proof:
Draw the diagonal BD.

(diagonal BD divides the parallelogram
into two triangles of equal area)
BR is the median of D DBC.
(R is the mid point of DC - data)

(median divides a triangle into two
equal triangles)
Similarly median of D ABD divides D BDA into two equal triangles.

From (i), (ii) and (iii),

Question (9):

Answer:  Given:

To prove:
ar (D ABP) = ar (quad. ABCD)
Proof:


Adding ar (D ABC) to both sides,
ar (D ACP) + ar (D ABC) = ar (D ACD) + ar (D ABC)

Question (10): at E and F resp. Show that arc (D ABE) = arc (D ACF)

Answer:  Given:

To prove:
arc (D ABE) = arc (D ACF)
Proof:

..... (i)
BCQE are on the same base and between same parallels)

Similarly,

(D ACF and parallelogram BCFP is on the same base
But
(parallelogram BCQE) = arc (parallelogram BCFP)......(iii)
( Parallelograms on equal
Hence from (i), (ii) and (iii),
arc (D AEB) = arc (D ACE)
Question (11):
PBQR completed figure. Show that ar (||gm ABCD) = ar (||gm BPRQ).
Answer:  Given:
ABCD is a parallelogram.
To prove:
ar (||gm ABCD) = ar (||gm BPRQ)
Construction:
Join A to C and P to Q.
Proof:
Consider D ABC and ||gm ABCD.


\ 2 ar (D ABC) = ar (||gm ABCD) ..... (i)
Similarly,
2 ar (D PBQ) = ar (||gm BPRQ) .....(2)




\ ar (D ABC) = ar (D PBQ) .....(iii)
From (i), (ii) and (iii),
ar (parallelogram ABCD) = ar (parallelogram BPRQ)
Question (12):

Answer:  Given:

To prove:

Proof:
( given)

.... (i)
( a median divides a triangle into two triangles of equal area.
Similarly in D ACE,

..... (ii)
Adding (i) and (ii), we get


Question (13):
[Hint: Join E to D]

Answer:  Given:
D is the mid-point of .

To prove:

Construction:
Join C to D.
Proof:
Consider D ABC,
.... (i)

(a median divides a triangle into two triangles of equal area)


From (i),



(using (ii))

Question (14): Prove that the area of a parallelogram is the product of its base and corresponding altitude.
Answer:  Given:
ABCD is a parallelogram. AF is the corresponding altitude to AB.
To prove:
ar (parallelogram ABCD) = AB x AF
Construction:
Complete the rectangle ABEF.
Proof:
Parallelogram ABCD and rectangle ABEF are on the same base AB and between the same parallels AB and CF.
ar (parallelogram ABCD) = ar (rectangle ABEF)
Now ar (rectangle ABEF) = AB x AF
( rectangular area axiom)
\ar (parallelogram ABCD) = AB x AF.
Question (15): Prove that the area of a triangle is half the product of any of its sides and the corresponding altitude.
Answer:  Given:
ABC is a triangle . AE is the altitude to the side BC.
To prove:

Construction:
Through C and A draw parallels to BA and BC respectively to intersect at D.
Proof:
Quadrilateral ABCD is a parallelogram.
AC is a diagonal of parallelogram ABCD.

(theorem 1)
BC is a side of parallelogram BCDA and AE is the corresponding altitude.
ar (parallelogram ABCD) = BC x AE ....(2)
From (1) and (2), we get

Question (16): If a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is equal to half that of the parallelogram.
Answer:  Given:
D AEB and parallelogram ABCD are on the same base AB and between the same parallels AB and DC.
To prove:

Construction:
Draw CH ^ AB produced at H and EF ^ AB.
Proof:
Since E, D, C are collinear and DC || AB.
EF || CH .... (1)

EF = CH (distance between the parallel lines)


From (1) and (2), we get



Question (17): ABCD is a quadrilateral and is one of its diagonals as shown in figure. Show that ABCD is a parallelogram and find its area.

Answer:  Given:

To prove:
ABCD is a parallelogramand to find its area.
Proof:
(Each equals 90o and are alternate angles)


Since one pair of opposite sides are equal and parallel ABCD is a parallelogram.
Area of parallelogram = base x altitude
= 3 cm x 4 cm
= 12 sq. cm.
Question (18): In the figure, if AB = 10cm, find AD.
Answer:  Area of parallelogram ABCD = base x altitude
=
= 10 cm x 7 cm
= 70 cm2
Area of parallelogram ABCD is also =

But area = 70 cm2

= 8.75 cm.
Question (19): Show that the line segment joining the mid-points of a pair of opposite sides of a parallelogram, divide it into two equal parallelograms.
Answer:  Given:
ABCD is a parallelogram E and F are mid-points of
To prove:
AEFD and EBCF are parallelograms of equal area.
Proof:

parallelogram are equal and parallel.




Quadrilateral AEFD is a parallelogram.
Similarly,


E is the midpoint of AB and F is the midpoint of CD.
Quadrilateral BEFC is a parallelogram.
AEFD and EBCF are of equal area.
Both are on the equal base and between the same parallels.
Question (20):

Answer:  Given:

To prove:

Proof:


(A median divides a triangle into two triangles of equal area.)
Consider D XBC,


(i) - (ii)


Question (21): Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Answer:  Given:
diagonals.
To prove:

Proof:
(Diagonals of a parallelogram
bisect each other)
Consider D ABC,


(median divides a triangle units two triangles of equal area)
Consider D BCD,


Consider D ADC,


From (i), (ii) and (iii) we arrive at

Question (22):

Answer:  Given:

To prove:

Proof:
( given)



( median divides a triangle into two triangles of equal
area)
Consider D CBD,
.... (ii)

Adding (i) and (ii), we get


Question (23): A point O inside a rectangle PQRS is joined to the vertices. Prove that the sum of the area of a pair of opposite triangles so formed is equal to the sum of the other pair of triangles.
Answer:  Given:
O is a point in PQRS.
To prove:

Proof:


Adding (i) and (ii),









Similarly,


Adding (iv) and (v)







From (iii) and (vi),

Question (24): Show that the area of a rhombus is half the product of the length of its diagonals.
Answer:  To prove:

Proof:

( the diagonals of a rhombus rhombus intersect at right angles)




Question (25): Parallelogram ABCD and rectangle ABEF have the same base AB and also have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
Answer:  To prove:

Proof:

( opposite sides of a parallelogram are equal)

( opposite sides of a rectangle are equal)

,





Adding, we get, i.e., (iii) + (iv)


Or BC + AD > BE + AF .... (v)
By adding (ii) and (v), we get
AB + CD + BC + AD > AB + EF + BE + AF
(Adding equals to both sides of an inequality does not change the inequation)
Note: AB + CD = AB + EF (from (ii))
Question (26): The diagonals of a parallelogram ABCD intersect at a point O. Through O, a line is drawn to intersect AD at P and BC at Q. Show that PQ divides the parallelogram into two parts of equal area.
Answer:  Data:
ABCD is a parallelogram in which diagonals AC and BD intersect at O. A line drawn through O intersects AD and BC at P and Q respectively.
To prove:
PQ divides the parallelogram into two parts of equal area.

Proof:
Compare D AOP and D COQ,
AO=OC (diagonal BD bisects diagonal AC at O)
(alternate angles)
(vertically opposite angles)


From the figure,



proved above]


(diagonal AC divides)
PQ divides parallelogram into two parts of equal area.
Question (27): The medians BE and CF of D ABC intersect at G. Prove that area of
DGBC= area of quadrilateral AFGE.
Answer:  Data:
BE and CF are medians of D ABC.
To prove:
ar(D GBC) = ar (quad. AFGE)
Proof:
The median of a triangles divides the triangle into two
triangles of equal area.




Question (28): A parallelogram and a rectangle have common base and equal areas. Show that the perimeter of the rectangle is smaller than the perimeter of the parallelogram.
Answer:  Data:
Parallelogram ABCD and rectangle EBCF are equal in area
and stand on the same base and between the same parallels.
To prove:
Perimeter of rectangle ABCD is smaller than the perimeter
of Parallelogram EBCF.
Proof:
Perimeter of rect ABCD = 2AB+2BC
Perimeter of parallelogram BCFE = 2BC + 2BE
or 2AB+2BC < 2BC+2BE
2AB < 2 BE
(by removing common 2BC
from both sides)
AB < BE

BE is the hypotenuse.
AB < BE
Perimeter of rectangle ABCD < perimeter of parallelogram BCFE.
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