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NCERT exemplar Class 9 Maths solutions chapter 9 provides the fundamentals of the area of parallelogram and area of a triangle to develop the formulas of area for different structures. These fundamentals are essential and useful for higher mathematics. These NCERT exemplar Class 9 Maths chapter 9 solutions are prepared by experts with 6+ years of subject matter expertise. These NCERT exemplar Class 9 Maths chapter 9 solutions are designed to provide the students with detailed steps while studying NCERT Book for Class 9 Maths. They develop a better understanding of the concept of the area of parallelogram as they are pretty expressive. The NCERT exemplar Class 9 Maths solutions chapter 9 follows the recommended CBSE Syllabus for Class 9.
Question:1
The median of a triangle divides it into two
(A) triangles of equal area
(B) congruent triangles
(C) right triangles
(D) isosceles triangles
Question:2
Answer: [d]Question: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 24 cm ^{2}
(B) a square of area 25 cm ^{2}
(C) a trapezium of area 24 cm ^{2}
(D) a rhombus of area 24 cm ^{2}
Question:4
In Fig. 9.4, the area of parallelogram ABCD is :
(A) AB × BM
(B) BC × BN
(C) DC × DL
(D) AD × DL
Question:5
In Fig. 9.5, 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
(D) Perimeter of ABCD = (Perimeter of ABEM)
Answer:
Answer: [C]Question: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
(A)
(B)
(C)
(D)
Question: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
Question: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) or (C)
Question: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
Question:10
ABCD is a trapezium with parallel sides AB = a cm and DC = b cm (Fig.). E and F are the mid-points of the non-parallel sides. The ratio of ar (ABFE) and ar (EFCD) is
(A) a : b
(B) (3a + b) : (a + 3b)
(C) (a + 3b) : (3a + b)
(D) (2a + b) : (3a + b)
Answer:B
Given: ABCD is a trapezium in which ,Question:1
Write true or false and justify your answer.
ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm^{2}, then ar (ABC) = 24 cm^{2}.
= (AB) (GX) …(2)
Area of trapezium AXCD = (Sum of parallel sides) (Distance between them)
=
Now, and (Given)
Area of trapezium …(3)
Area of (base) (height)
…(4)
Now,
Area of trapezium
(given)
From equation (4),
Area of
Hence, the given statement is false.
Question:2
Write true or false and justify your answer.
PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ.
If PS = 5 cm,then ar (PAS) = 30 cm^{2}.
PS = 5 cm, PR = 13 cm
In D PSR, using Pythagoras theorem,
SR = 12 cm = PQ (opposite sides of a rectangle are equal)
Area of a triangle is given as
Now,
As,
or
But it is given that
Hence, the given statement is false.
Question:3
Write true or false and justify your answer.
PQRS is a parallelogram whose area is 180 cm^{2} and A is any point on the diagonal QS. The area of DASR = 90 cm^{2}.
Answer:
Answer: [False]Given that A is any point on SQ
So,
Hence,
But given that area of
Hence, the given statement is false.
Question:4
Answer:
Answer: [True]Hence,
Therefore the given statement is true.
Answer:[False]
Solution.Question:1
Answer:
Solution.Question:2
Answer:
Solution.Question:3
The area of the parallelogram ABCD is (see Fig.9.13). Find
(i) ar (ABEF)
(ii) ar (ABD)
(iii) ar (BEF)
Answer:
Answer:Question:4
In , D is the mid-point of AB and P is any point on BC. If meets AB in Q (Fig.), then prove that .
Answer:
Solution.Question:5
Answer:
Solution.Question:6
O is any point on the diagonal PR of a parallelogram PQRS (Fig.). Prove that ar (PSO) = ar (PQO).
Answer:
Solution.Question:7
Answer:
Answer:Question:8
Answer:
Solution.Question:9
Answer:
Solution.Question:1
Answer:
Solution.Hence proved.
Question:2
Answer:
Solution.Question:3
Answer:
Solution.Question:4
In Fig., CD || AE and CY || BA. Prove that ar (CBX) = ar (AXY)
Answer:
Solution.Question:6
In , if L and M are the points on AB and AC, respectively such that . Prove that ar (LOB) = ar (MOC)
Answer:
Solution.Question:5
Answer:
Solution.=
Question:7
Answer:
Solution.Question:10
In Fig., ABCD and AEFD are two parallelograms. Prove that ar (PEA) = ar (QFD)
Answer:
Solution.Question:9
Answer:
Solution.Question:8
If the medians of a intersect at G, show that ar (AGB) = ar (AGC) = ar (BGC) = ar (ABC)
Answer:
Solution.Highlights of the key topics covered in NCERT exemplar Class 9 Maths solutions chapter 9 are as follows:
These class 9 maths NCERT exemplar chapter 9 solutions provide fundamental knowledge of parallelograms and triangle’s area, which will be very useful for higher mathematics for competitive exams like JEE Mains and JEE Advanced. The Class 9 Maths NCERT exemplar solutions chapter 9 Area of parallelograms is a good starting point for practicing the problems based on this chapter and widening the knowledge base. Once the students are thorough with the problems and solutions of exemplar, it will be easier to solve other books such as NCERT Class 9 Maths, RD Sharma Class 9 Maths, RS Aggarwal Class 9 Maths, et cetera.
To view/download these solutions in an offline environment, students can use the NCERT exemplar Class 9 Maths solutions chapter 9 pdf download feature, enabling them to have a pdf version of the chapter handy while attempting the NCERT exemplar Class 9 Maths chapter 9.
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 |
If sides are in the ratio 1:2 that means base length will be in the ratio 1:2 as well as height will be in the ratio 1:2. Therefore, we can say that area will be in the ratio 1:4.
By various experiences with comparison of rectangular lands, the formula of area of rectangle is devised. It is the simplest formula we assume for the area of the rectangle.
Even if we had assumed that the area of the rectangle is twice the product of length and breadth, it does not harm our understanding of space and areas.
No, these two triangles must have the same height but their shapes may be different.
Generally, this chapter accounts for 8-10 % weightage of the final paper. These NCERT exemplar Class 9 Maths solutions chapter 9 are sufficient to attempt and excel in the chapter of Area of Parallelograms and Triangles.
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Application Date:31 May,2024 - 15 June,2024
Application Date:01 June,2024 - 15 June,2024
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