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The NCERT Solutions for Class 9 Maths exercise 9.3 deal with triangles rather than parallelograms. In Class 9 Maths chapter 9 exercise 9.3 we deal with triangles that have a common base and are in the same parallels.
The important concepts related to NCERT solutions for Class 9 Maths chapter 9 exercise 9.3 in order to solve this exercise are:
Triangles that have the same base or the equal base and are between the same parallel lines have equivalent areas. This is on the grounds that we realize that the space of the triangle is half of the product of base and tallness since the height (distance between the parallels) remains the same for the two triangles and they have the same bases thus their areas are equal.
The figure shown below is a set of triangles with the same base and between the same parallel
The triangles ABC and BCD have the same area.
Along with NCERT book exercise 9.3 Class 9 Maths, the following exercises are also present.
Q1 In Fig.
Answer:
We have
Therefore, ar(
Similarly, In triangle
ar(
On subtracting eq(ii) from eq(i), we get
ar(
Hence proved.
Q2 In a triangle ABC, E is the mid-point of median AD. Show that
Answer:
We have a triangle ABC and AD is a median. Join B and E.
Since the median divides the triangle into two triangles of equal area.
Now, in triangle
BE is the median [since E is the midpoint of AD]
From eq (i) and eq (ii), we get
ar (
ar (
Hence proved.
Q3 Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Answer:
Let ABCD is a parallelogram. So, AB || CD and AD || BC and we know that Diagonals bisects each other. Therefore, AO = OC and BO = OD
Since OD = BO
Therefore, ar (
Similarly, ar(
and, ar (
From eq (a), (b) and eq (c), we get
ar (
Thus, the diagonals of ||gm divide it into four equal triangles of equal area.
Answer:
We have,
Now, in
Similarly, in
Adding equation (i) and eq (ii), we get
ar (
Hence proved.
Answer:
We have a triangle
Now, in
F and E are the midpoints of the side AB and AC.
Therefore according to mid-point theorem, the line segment connecting the midpoints of two sides of a triangle is parallel to the third side and half of the third side.
also, EF = 1/2 (BC)
Similarly, ED || BF and ED = FB
Hence BDEF is a parallelogram.
Q5 (ii) D, E and F are respectively the mid-points of the sides BC, CA and AB of a
Answer:
We already proved that BDEF is a ||gm.
Similarly, DCEF and DEAF are also parallelograms.
Now, ||gm BDEF and ||gm DCEF is on the same base EF and between same parallels BC and EF
It is known that diagonals of ||gm divides it into two triangles of equal area.
and, Ar(
From equation(i), (ii) and (iii), we get
Ar(
Thus, Ar (
Ar (
Hence proved.
Q5 (iii) D, E and F are respectively the mid-points of the sides BC, CA and AB of a
Answer:
Since we already proved that,
ar(
So, ar(||gm BDEF) = ar(
= 2 . ar(
Hence proved.
Q6 (i) In Fig.
[ Hint: From D and B, draw perpendiculars to AC.]
Answer:
We have ABCD is quadrilateral whose diagonals AC and BD intersect at O. And OB = OD, AB = CD
Draw DE
In
OB = OD [given]
Therefore, by AAS congruency
and ar(
Now, In
DE = FB
DC = BA [given]
So, by RHS congruency
and, ar(
By adding equation(i) and (ii), we get
Hence proved.
Q6 (ii) In Fig.
[ Hint: From D and B, draw perpendiculars to AC.]
Answer:
We already proved that,
Now, add
Hence proved.
Q6 (iii) In Fig.
[ Hint : From D and B, draw perpendiculars to AC.]
Answer:
Since
Therefore, They lie between the same parallels BC and AD
also
So, AB || CD
Hence ABCD is a || gm
Q7 D and E are points on sides AB and AC respectively of
Answer:
We have
Since
Hence DE || BC.
Answer:
We have a
Since XY || BC and BE || CY
Therefore, BCYE is a ||gm
Now, The ||gm BCEY and
Similarly, ar(
Also, ||gm BEYC and ||gmBCFX are on the same base BC and between the same parallels BC and EF.
From eq (i), (ii) and (iii), we get
ar(
Hence proved.
Answer:
Join the AC and PQ.
It is given that ABCD is a ||gm and AC is a diagonal of ||gm
Therefore, ar(
Also, ar(
Since
Now, subtracting
ar(
ar(
From eq(i), (ii) and (iii) we get
Hence proved.
Q10 Diagonals AC and BD of a trapezium ABCD with
Answer:
We have a trapezium ABCD such that AB || CD and it's diagonals AC and BD intersect each other at O
Now, subtracting
ar (
Hence proved.
Q11 In Fig.
(i)
(ii)
Answer:
We have a pentagon ABCDE in which BF || AC and CD is produced to F.
(i) Since
(ii) Adding the ar (AEDC) on both sides in equation (i), we get
ar(
Hence proved.
Answer:
We have a quadrilateral shaped plot ABCD. Draw DF || AC and AF || CF.
Now,
On subtracting
ar(
The portion of
We need to prove that ar(
Now, adding ar(quad. ABCE) on both sides in eq (i), we get
ar (
ar (ABCD) = ar(
Answer:
We have a trapezium ABCD, AB || CD
XY ||AC meets AB at X and BC at Y. Join XC
Since
Therefore, ar(
Similarly ar(
From eq (i) and eq (ii), we get
ar(
Hence proved.
Answer:
We have, AP || BQ || CR
Therefore, ar (
Similarly, ar (
Add the eq(i) and (ii), we get
ar (
Hence proved.
Answer:
We have,
ABCD is a quadrilateral and diagonals AC and BD intersect at O such that ar(
Now, add ar (
ar(
ar (
Since the
Therefore, AB || CD
Hence ABCD is a trapezium.
Q16 In Fig.
Answer:
Given,
ar(
and ar(
from equation (i),
Since
Hence quadrilateral DCPR is a trapezium
Now, by subtracting eq(ii) - eq(i) we get
ar(
ar(
Hence ABCD is a trapezium.
From NCERT solutions for Class 9 Maths exercise 9.3, we get to know another concept (theorem) which deals with the converse of the above concept. We get to know that if two triangles have the same base (or if they have an equal base) and if they have equal areas then they lie on the same parallel line. Thus, a quick thing to remember from NCERT solutions for Class 9 Maths exercises 9.3 is that if the triangle has a median (a line that divides the base into two parts), then the median divides the triangle into two parts both on each side.
Also Read| Areas Of Parallelograms And Triangles Class 9 Notes
Exercise 9.3 Class 9 Maths, is based on the AREAS OF PARALLELOGRAMS AND TRIANGLES by providing us with a general (much simpler) approach to calculate the area of triangles and parallelograms and concepts of figures in the same base and same parallel lines.
NCERT syllabus Class 9 Maths chapter 9 exercise 9.3 introduces us to a new concept of triangles on the same base and between the same parallels and its properties, like calculating the area and height.
Also, See
In this exercise, we learn about dealings with areas and height related to triangles especially if they are on the common base and between the same parallels.
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