## 9 AREA of PARALLELOGRAMS and TRIANGLES

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Exercise 9.1 Multiple Choice Questions (MCQs):

Q-1: The media of a triangle divides it into two

(a) Triangles of equal areas                                                            (b) Congruent triangles

(c) Right angled triangles                                                                (d) Isosceles triangles

Solution: (a)

We know that, a median of a triangle is a line segment joining a vertex to the mid-point of the opposite side. Thus, a median of a triangle of equal area.

Q-2: In which of the following figures, you find two polygons on the same base and between the same parallels?

(a)

Solution: (d)

In figures, (a), (b) and (c) three are two polygons on the same base but they are not between the same parallels. In figure (d), there are two polygons (PQRA and BQRS) on the same base and between the same parallels

Solution: (d)

In figures, (a), (b) and (c) three are two polygons on the same base but they are not between the same parallels. In figure (d), there are two polygons (PQRA and BQRS) on the same base and between the same parallels.

Q-3: The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm, is

(a) A rectangle of area 42 cm2                                             (b) A square of area 25 cm2

(c) A trapezium of area 24 cm2                                           (d) A rhombus of area 24 cm2

Solution: (d)

We know that, on joining the mid-points of the adjacent sides of a rectangle, we get a rhombus.

Here,                          length of rectangle ABCD = 8 cm

And                        breadth of a rectangle ABCD = 6 cm

Let E, F, G and H are the mid-points of the sides of rectangle ABCD, then EFGH is a rhombus.

Then, diagonal of rhombus EFGH are EG and HF.

Here,                                        EG = BC = 8 cm

And                                            HF = AB = 6 cm

Area of rhombus = Product of diagonals/2

Hence, joining the mid-points of the adjacent sides of a rectangle forms a rhombus of area 24 cm2

Q-4: In the figure, the area of parallelogram ABCD is

Use the formula, area of parallelogram = Base x Altitude to get the required result.

Solution: (c)

We know that, we know that, area of parallelogram is the product of its any side and the corresponding altitude (or height).

Here, When AB is base, then height is DL.

Area of parallelogram = AB x DL

And when AD is base, then height is BM.

Area of parallelogram = AD x BM

When DC is base, then height is DL.

Area of parallelogram = DC x DL

And when BC is the base, then height is not given.

Hence, option (c) is correct.

Q-5: In the figure, if parallelogram ABCD and rectangle ABEM are of equal area, then

(a) Perimeter of ABCD = Perimeter of ABEM

(b) Perimeter of ABCD < Perimeter of ABEM

(c) Perimeter of ABCD > Perimeter of ABEM

(a) Perimeter of ABCD = Perimeter of ABEM

(b) Perimeter of ABCD < Perimeter of ABEM

(c) Perimeter of ABCD > Perimeter of ABEM

(d) Perimeter of ABCD = 1/2 (Perimeter of ABEM)

Solution: (c)

In rectangle ABEM,                                     AB = EM                                       [Sided of rectangle]

And in parallelogram ABCD,                       CD = AB

On adding, both equations, we get

AB + CD = EM + AB                                                     …(i)

We know that, the perpendicular distance between two parallel sides of a parallelogram is always less than the length of the other parallel sides.

BE < BC and AM < AD

[Since, in a right angle triangle, the hypotenuse is greater than the other side]

On adding both above inequalities, we get

BE + AM < BC + AD or BC + AD > BE + AM

On adding AB + CD both sides, we get

AB + CD + BC + AD > AB + CD + BE + AM

AB + BC + CD + AD > AB + BE + EM + AM          [ CD = AB = EM]

Perimeter of parallelogram ABCD > Perimeter of rectangle ABEM

Q-6: The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to

Solution: (a)

We know that, if D, E and F are respectively the mid-points of the sides

BC, CA and AB of a triangle ABC then all four triangles has equal area i.e.,

If we take D as fourth vertex, then area of the parallelogram AFDE

Q-7: 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: (b)

We know that, parallelogram on the equal bases and between the same parallels are equal in area. So, ratio of their areas is 1 : 1.

Q-8: ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD

(a) is a rectangle                                                              (b) is always a rhombus

(c) is a parallelogram                                                       (d) need not be any of (a), (b) and (c)

Solution: (d)

Here, ABCD need not be any rectangle, rhombus and parallelogram because if ABCD is a square, then its diagonal AC also divides it into two parts which are equal in area.

Q-9: 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 parallelogram is

(a) 1 : 3        (b) 1 : 2            (c) 3 : 1          (d) 1 : 4

Solution: (b) We know that, if a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.

Area of triangle : Area of parallelogram = 1 : 2