Free Areas of Triangles and Parallelograms 02 Practice Test - 9th Grade
If two parallelograms have the same (or equal) bases and are between the same parallel lines, their areas will be equal.
Solution : A
The above statement is true. This is because the perpendicular distance between two parallel lines is constant thus making the altitude for both parallelograms same. As they already have same or equal bases, their areas will also be same.
Two triangles have the same base and equal areas have their vertex lying on the same side of their base. The line joining the two apexes will be _______ to their base.
Solution : A
Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.
When two triangles have the same base and equal areas, it means they have equal altitudes or heights. So the third vertex will lie on the line parallel to their base. Thus, on joining them, the line produced will be parallel to the base.
The area of parallelogram ABCD is:
Solution : C and D
The area of a parallelogram is the product of base and height. Hence if we consider AD as the base then its corresponding height is BM and the area will be AD×BM
ii) If we consider DC as the base then the corresponding height will be BN or DL . and the area will be given by DC x BN or DC×DL
AE || BC and D is the mid-point of BC. If area △ABC = 84 cm2, what is area of △BDE?
Solution : B
If we draw EC, we can see that △ABC and △BEC will have equal area. Because triangles on same base and between same parallels are equal in area.
In △ABC and △BDE the altitude is same but the length of the base is different.
Base length (△ABC) = 2 × Base length (△BDE) (∵ D is the mid point)
So, Area (△ABC) = 2 × Area (△BDE)
Area (△BDE) = 42 cm2
The part of the plane enclosed by a simple closed figure is called a
The part of the plane enclosed by a simple closed figure is called a planar region. The magnitude or measure of this planar region is called its area.
The mid-points of the sides BC, CA and AB of a ΔABC are D, E and F respectively. Which of the following is true?
Area of (DEF)=12Area of (ΔABC)
Solution : B
EF is parallel to BC (Midpoint theorem)
In ΔCED and ΔEDF,
∠FED =∠EDC (Alternate angles)
EF = EF (common side)
∠CED =∠EDF (Alternate angles)
Hence, ΔCED and ΔEDF are congruent by ASA condition.
Area of ΔCED = Area of ΔEDF.
Therefore, EFDC is a parallelogram
(Diagonals divide a parallelogram into two congruent triangles)
Similarly, FD is parallel to AC and ED is parallel to AB. So, it can be proved that AEFD and EFDB are also parallelograms.
Thus, Area of ΔCED = Area of ΔEDF = Area of ΔAEF = Area of ΔFDB
∴ Area of (DEF)=14Area of (ΔABC)
ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD cannot be
Solution : D
Diagonals of a trapezium do not divide the trapezium into two equal parts, but in parallelogram, it is divided into two congruent triangles of equal areas.
If parallelogram ABCD and rectangle ABEM are of equal area, then:
Perimeter of AMEB = Perimeter of ADCB
Perimeter of AMEB < Perimeter of ADCB
Perimeter of AMEB >Perimeter of ADCB
Perimeter of AMEB = 12 Perimeter of ADCB
Solution : B
In right angled ΔAMD
AD2 = AM2 + MD2 (by Pythagoras theorem)
Similarly, In right angled ΔBEC
BC2 = BE2 + EC2
AB = DC = ME (opposite sides of rectangle and parallelogram)
Perimeter of rectangle = AM + ME + EB + AB.
Perimeter of parallelogram = AB + BC + DC + AD.
Hence Perimeter of ADCB > Perimeter of AMEB.
ABCD is a parallelogram whose area is 54 cm2. The area of triangle AEB is
If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram.
Thus, area of triangle AEB is 12×54=27 cm2.
In Δ triangle ABC, D and E are the midpoints of side AB and AC. State which of the following are correct option(s).
Solution : A, C, and D
D and E are the midpoints of AB and AC.
DE will be parallel to BC (Midpoint theorem).
Area of ∆BEC and ∆BDC will be equal as they are on the same base and between the same parallel lines.
Also, Area of ∆BEC - Area of ∆BOC = Area of ∆BDC - Area of ∆BOC.
Thus, Area of ∆DOB = Area of ∆EOC.