Areas of Parallelograms and Triangles

Download CBSE Class 9th Standard CBSE Mathematics question papers, sample papers, important questions, and previous year solved papers in PDF format. Get free study materials, NCERT solutions, and exam preparation resources for Class 9th Standard CBSE Mathematics

Questions:

Generic 4

4 x 4 = 16

Areas of triangles on the same bases and between the same parallels are equal in. Prove it.

In the figure, diagonals AC and BD of a trapezium ABCD with AB || CD intersect each other at O. Show that ar (\(\Delta \) AOD)= ar(\(\Delta \) BOC).

In the given figure, AB 11DC. Show that ar(BDE) = ar(ACED).

ABCD is a quadrilateral and BD is one of its diagonals as shown in figure. Show that ABCD is a parallelogram and find its area.

Generic 2

5 x 2 = 10

In given figure, ABCD is a parallelogram and BE\(\bot \)AD. If BE=14 cm and AD=8 cm, find the area of \(\Delta \)DBC.

ABCD is a rectangle and BD is one of its diagonals. If ar(\(\Delta\)ABD) = 8cm², find ar(\(\Delta\) BCD).

In the figure, ABCD is a parallelogram. P is a point on AB produced and \(DN\bot AB\) . If AB = 8 cm and DN = 3 cm. Find the area of \(\Delta \) CPD.

In the given figure, ABC and DBC are triangles on the same base and between parallel lines I and m. If AB = 3 cm, BC = 5 cm, \(\angle A=90°\), find area of \(\Delta \) DBC.

In a parallelogram ABCD, AB = 8 cm. The altitudes corresponding to sides AB and AD are respectively 4 cm and 5 cm. Find measure of AD.

Generic 3

1 x 3 = 3

10.

In a triangle ABC, E is the midpoint of median AD. Show that ar(\(\Delta \) BED) = \(\frac { 1 }{ 4 } \) ar(\(\Delta \) ABC).

Multiple Choice Question

10 x 1 = 10

11.

In the figure, BC = 2BE and area (\(\Delta\)ABC) = 60 cm², then ar (\(\Delta\)AEC) is:

(a)

15 cm²

(b)

20 cm²

(c)

30 cm²

(d)

40 cm²

12.

In \(\Delta\)ABC, E is the mid-point of median AD. Then the ratio of areas of \(\Delta\)BED to area of \(\Delta\)ABC is:

(a)

1:2

(b)

2:1

(c)

4:1

(d)

1:4

13.

In the figure below, D is mid-point of side BC of \(\Delta\) ABC. O is mid-point of AD. If area of \(\Delta\)AOB 8 cm² , then area of \(\Delta\)ABC is:

(a)

16 cm²

(b)

24 cm²

(c)

32 cm²

(d)

4 cm²

14.

In the figure, ABCD is a parallelogram. If area (AOD) = 12 cm² then area (ABCD) is:

(a)

3 cm²

(b)

24 cm²

(c)

48 cm²

(d)

36 cm²

15.

If a parallelogram and a triangle are on the same base and between the same parallels, then

(a)

area of the triangle =\(\frac { 1 }{ 2 } \) area of the parallelogram

(b)

area of the triangle = area of the parallelogram

(c)

area of the triangle =\(\frac { 1 }{ 3 } \) area of the parallelogram

(d)

area of the triangle =\(\frac { 1 }{ 4 } \) area of the parallelogram

16.

Which of the following figures lie on the same base and between the same parallels?

(a)

(i) (iv)

(b)

(ii),(iii)

(c)

(ii),(iv)

(d)

(i),(iii)

17.

If P, Q, Rand S are the midpoints of a rectangle of area 36 sq. cm, then PQRS is a parallelogram of area

(a)

24 cm²

(b)

18cm²

(c)

12cm²

(d)

9cm²

18.

If length of the diagonal of a square is 8 cm, then its area will be

(a)

64cm²

(b)

32cm²

(c)

16cm²

(d)

48cm²

19.

In the following figure, find the area of quad. ABCD.

(a)

7 square units

(b)

12 square units

(c)

6 square units

(d)

24 square units

20.

Area of a triangle is equal to

(a)

\(\frac { 1 }{ 2 } \times Base\times Corresponding\quad altitude\)

(b)

\(\frac { 1 }{ 4 } \times Base\times Corresponding\quad altitude\)

(c)

\(\frac { 1 }{ 3 } \times Base\times Corresponding\quad altitude\)

(d)

\( Base\times Corresponding\quad altitude\)

1 Marks

2 x 1 = 2

21.

In \(\Delta\)ABC, E is the mid-point of median AD, then the ratio of area of \(\Delta\)BED to the area \(\Delta\)ABC is _______________

22.

Why we cannot construct a triangle of given sides as 5 cm, 5 cm and 10 cm?

Visit QB365-SLMS - https://www.qb365.in/slms/

Answers Key:

Generic 4

4 x 4 = 16

Let ABC and A'BC be two triangles on the same base BC and between the same parallels PQ and RS.

Draw \(BM\bot PQ\),then,
\(ar(\Delta ABC)\)
\( =\frac { 1 }{ 2 } \times Base\times Corresponding\ altitude\)
\(\\ =\frac { 1 }{ 2 } \times BC\times BM\quad \quad \quad \quad .....(i)\)
\(\\ ar(\Delta A'BC)\)
\(\\ =\frac { 1 }{ 2 } \times Base\times Corresponding\ altitude\)
\(\\ =\frac { 1 }{ 2 } \times BC\times BM\quad\quad \quad \quad \quad .....(ii)\)
Form (i) and (ii),we get
\(ar(\Delta ABC)=ar(\Delta ABC)\)
Hence, \(\Delta ABC\) and \(\Delta A'BC\) are equal in area.

Given: Diagonals AC and BD of a trapezium ABCD with AB || CD intersect each other at O.
To Prove: ar(\(\Delta \)AOD) = ar(\(\Delta \) BOC)
Proof: \(\Delta \)ADB and \(\Delta \)ACB are on the same base AB and between the same parallels AB and DC
ar(\(\Delta \)ADB) = ar(\(\Delta \) ACB)
|Two triangles on the same base and between the same parallels are equal in area
\(\Rightarrow \) ar(\(\Delta \) ADB) - ar(\(\Delta \) AOB)
\(\Rightarrow \) ar(\(\Delta \) ACB) - ar(\(\Delta \)AOB)
I Subtracting ar( \(\Delta \)AOB) from both sides
\(\Rightarrow \) ar(\(\Delta \) AOD) = ar(\(\Delta \)BOC)

Given: AB || DC in the given figure.
To Prove: ar(BDE) = ar(ACED)
Proof:\(\Delta \) ADC and \(\Delta \) BDC are on the same base DC and between the same parallels AB and DC
ar(\(\Delta \) ADC) = ar(\(\Delta \) BDC)
\(\Rightarrow ar(\Delta \ ADC)+ar(\Delta \ DCE)\)
\( =ar(\Delta \ BDC)+ar(\Delta \ DCE)\)
|Adding ar(\(\Delta \quad DCE\)) to both sides
\(\Rightarrow \ ar(ACED)=ar(BDE)\)
\(\\ \Rightarrow \ ar(BDE)\ =ar(ACED)\)

Given: ABCD is a quadrilateral and BD is one of its diagonals.
To Prove: ABCD is a parallelogram and to determine its area.
Proof:\(\angle ABD=\angle BDC(=90°)\) |Given
But these angles form a pair of equal alternate.interior angles for lines AB, DC and a transversal BD
AB || DC
Also, AD = DC (= 3 cm) I Given
Hence, quadrilateral ABCD is a parallelogram.
I A quadrilateral is a parallelogram if its one pair of opposite sides are parallel and equal
Now,
\(ar(||gm\ ABCD)=base\times Corresponding\ altitude\)
\(\\ =3\times 4\)
\(\\ =12{ cm }^{ 2 }\)

Generic 2

5 x 2 = 10

Given: BE = 14 cm, AD = 8 cm
\(\therefore \ Area(\triangle ADB)=\frac { 1 }{ 2 } \times 8\times 14\)
= 56 cm²
\(\because\) ABCD is a parallelogram.
\(\therefore\) ar(\(\Delta \)DBC) = ar(\(\Delta \)ADB) = 56 cm².

A diagonal of a parallelogram divides it into two triangles of equal area
\(\therefore\) ar(\(\Delta\)ABD)= ar(\(\Delta\)BCD)
Since, given, ar(\(\Delta\)ABD) = 8 cm²
\(\therefore\) ar(\(\Delta\)BCD) = 8 cm².

12 cm²

7.5 cm²

6.4cm

Generic 3

1 x 3 = 3

10.

AD is the median of ΔABC. Therefore, it will divide ΔABC into two triangles of equal areas.
∴ Area (ΔABD) = Area (ΔACD)
⇒ Area (ΔABD) = 1/2Area (ΔABC)... (1)
In ΔABD, E is the mid-point of AD. Therefore, BE is the median.
∴ Area (ΔBED) = Area (ΔABE)
⇒ Area (ΔBED) = 1/2Area (ΔABD)
⇒ Area (ΔBED) = 1/2 x 1/2Area (ΔABC) [From equation (1)]
⇒ Area (ΔBED) = 1/4Area (ΔABC)

Multiple Choice Question

10 x 1 = 10

11.

12.

A median of a triangle divides it into two triangles of equal area.

13.

A median of a triangle divides it into two triangles of equal areas

14.

A diagonal of a parallelogram divides it into two congruent triangles. Two congruent triangles have equal areas. The diagonals of a parallelogram bisect each other.A median of a triangle divides it into two triangles of equal areas.

15.

Theorem

16.

\(\Delta\)TQR and parallelogram SRQP lie on the same base RQ and between the same parallels RQ and SP.
\(\Delta\)PDC and trapezium ABCD lie on the same base DC and between the same parallels AB and De.

17.