Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 10 Area of Triangles Questions and Answers Notes Pdf to clear their doubts.
SCERT Class 7 Maths Chapter 10 Solutions Area of Triangles
Class 7 Maths Chapter 10 Area of Triangles Questions and Answers Kerala State Syllabus
Area of Triangles Class 7 Questions and Answers Kerala Syllabus
Page 142
Now try these problems:
Question 1.
Find the areas of the triangles shown below:
Answer:
Area of the triangle = \(\frac{1}{2}\) × base × altitude
= \(\frac{1}{2}\) × 4 × 2
= 4 sq cm
Area of the triangle = \(\frac{1}{2}\) × base × altitude
= \(\frac{1}{2}\) × 6 × 3
= 9sq cm
Area of the triangle = \(\frac{1}{2}\) × base × altitude
= \(\frac{1}{2}\) × 5 × 4
= 10 sq cm
Area of the triangle = \(\frac{1}{2}\) × base × altitude
= \(\frac{1}{2}\) × 4 × 3
= 6 sq cm
Area of the triangle = \(\frac{1}{2}\) × base × altitude
= \(\frac{1}{2}\) × 3 × 4
= 6 sq cm
Question 2.
What is the area of the triangle shown below?
Answer:
Area of the triangle = \(\frac{1}{2}\) × base × altitude
= \(\frac{1}{2}\) × 7 × 4
= 14 sq cm
(i) Draw a right triangle of the same area
Answer:
(ii) Draw a triangle of the same area with one angle greater than a right angle
Answer:
Page 146
Now try these problems:
Question 1.
Draw a triangle of sides 3,4 and 6 centimetres. Draw three different right triangles of the same area.
Answer:
Question 2.
How many different triangles can be drawn with two sides 8 centimetres, 6 centimetres and area 12 square centimetres? What if the area is to be 24 square centimetres?
Answer:
Two triangles can be drawn with two sides 8 centimetres, 6 centimetres and area 12 square centimetres
For, area = 24 square centimetres
Question 3.
Draw the triangle below in the notebook:
Draw triangles ABP, BCQ and CAR of the same area with angles as given below:
i) ∠BAP = 90°
Answer:
This is the required figure of ∆ABP with ∠BAP = 90°
ii) ∠CBQ = 60°
Answer:
This is the required figure of ∆BCQ with ∠CBQ = 60°
iii) ∠ACR = 30°
Answer:
This is the required figure of ∆CAR with ∠ACR =30°
Intext Questions And Answers
Question 1.
Now, can’t you calculate the area of the triangles shown below?
Answer:
Area of the triangle = \(\frac{1}{2}\) × 3 × 5 = 7.5 sq cm
Answer:
Area of the triangle = \(\frac{1}{2}\) × 4 × 5 = 10 sq cm
Answer:
Here, the two small triangles are of the same size
Therefore, the area of a small triangle = \(\frac{1}{2}\) × 3 × 3 = 4.5 sq cm
The total area of the triangle = 2 × 4.5 = 9 sq cm
Question 2.
Now calculate the area of the triangle shown below?
Answer:
Let’s denote the length of the bottom side of the larger right triangle as x and that of the smaller triangle as y.
Then
Area of the larger right triangle = \(\frac{1}{2}\) × x × 4 = \(\frac{4}{2}\) = 2x square centimetres
Area of the smaller right triangle = \(\frac{1}{2}\) × y × 4 = \(\frac{4}{2}\)y = 2y square centimetres
Thus, the area of the original triangle = \(\frac{4}{2}\)x + \(\frac{4}{2}\)y = \(\frac{3}{2}\)(x + y) = 2x + 2y = 2 (x + y)
But x + y = 5
So, the area of the triangle = 2 × 5 = 10 square centimetres
Now, let’s explore how to calculate the area of the triangle as shown below:
Here, we must consider that a small right triangle cut away from a larger right triangle.
Let’s take the length of the bottom side of the larger right triangle as x centimetres and the length of the bottom side of the smaller right triangle as y centimetres
Then
Area of the larger right triangle = \(\frac{1}{2}\) × x × 3 = \(\frac{3}{2}\)x square centimetres
Area of the smaller right triangle = \(\frac{1}{2}\) × y × 3 = \(\frac{3}{2}\)y square centimetres
Thus, the area of the original triangle = \(\frac{2}{2}\)x – \(\frac{3}{2}\)y = \(\frac{3}{2}\)(x – y)
Since x – y = 8
So, the area of the triangle = \(\frac{3}{2}\) × 8 = 12 square centimetres
In general, we can make a statement about computing the area of a triangle as:
- The area of any triangle is half the product of one of the sides and the height from this side to the opposite vertex
- If we name the side used to compute the area, the base of the triangle and the height to the opposite vertex, the altitude, then this can be shortened a bit:
- The area of any triangle is half the product of the base and the altitude
Question 3.
Can you draw a right triangle of the same area with a base of 4 centimetres?
And another with base 5 centimetres?
Answer:
Class 7 Maths Chapter 10 Kerala Syllabus Area of Triangles Questions and Answers
Question 1.
Find the areas of the triangles shown below:
Answer:
Area = \(\frac{1}{2}\) × Base × Height
= \(\frac{1}{2}\) × 8 × 6
= 24 sq. cm
Answer:
Area = \(\frac{1}{2}\) × Base × Height
= \(\frac{1}{2}\) × 10 × 12
= 60 sq. cm
Answer:
Area = \(\frac{1}{2}\) × Base × Height
= \(\frac{1}{2}\) × 5 × 7
= 17.5 sq. cm
Answer:
Area = \(\frac{1}{2}\) × Base × Height
= \(\frac{1}{2}\) × 15 × 9
= 67.5 sq. cm
Question 2.
What is the area of the triangle shown below?
i) Draw a right triangle of the same area
ii) Draw a triangle of the same area with one angle greater than a right angle
Answer:
Area = \(\frac{1}{2}\) × Base × Height
= \(\frac{1}{2}\) × 9 × 4
= 18 sq cm
Question 3.
Draw a triangle of sides 5,6 and 7 centimetres. Draw three different right triangles of the same area.
Answer:
This is the required triangle with sides 5, 6 and 7 cm.
Here are three different triangles, each with the same area:
- The area of any triangle is half the product of its base and altitude.
- Triangles with the same base and third vertex on a line parallel to the base, have the same area.
Class 7 Maths Chapter 10 Notes Kerala Syllabus Area of Triangles
In this chapter, we explore the fascinating world of triangles by diving into three important concepts: right triangles, base and altitude, and parallel lines. Before we dive into formulas, let’s think of a fun fact – did you know that all triangles, no matter how they look, can be broken down into a simple formula to find their area? Imagine cutting a piece of paper into various triangles; as long as they share the same base and height, their areas will be the samel
This chapter will teach us that the area of any triangle is half the product of its base and altitude. We’ll also discover that triangles with the same base and third vertex on a line parallel to the base share the same area, unlocking new ways to understand geometry in both real-world and theoretical . contexts.
Right Triangles
Consider the rectangle,
We know the area of the rectangle is the product of its width and height.
Therefore, the area of the given rectangle = 8 × 4 = 32 cm
Now let’s cut the rectangle diagonally, where we will get two triangles.
So, the area of each triangle is 32 × \(\frac{1}{2}\) = 16 square centimetres
Here, the adjacent sides of the rectangle are perpendicular to each other; so, in each of these triangles, two of the sides are perpendicular to each other.
Such a triangle is called a right triangle.
We observe that the area of a right triangle is half that of a rectangle. Therefore, we can calculate the area of the right triangle as.
- The width and height of the rectangle are the perpendicular sides of the triangle
- The area of the rectangle is the product of its width and height
- The area of the right triangle is half the area of the rectangle
In summary, we can conclude.
The area of a right triangle is half the product of its perpendicular sides.
Base And Altitude
Now, let’s look at how to calculate the area of the triangle shown below.
Consider this triangle as two smaller right triangles joined together,
Let’s denote the length of the bottom side of the larger right triangle as x and that of the smaller triangle as y.
Then
Area of the larger right triangle = \(\frac{1}{2}\) × x × 3= \(\frac{3}{2}\)x square centimetres
Area of the smaller right triangle = \(\frac{1}{2}\) × y × 3 = \(\frac{3}{2}\)y square centimetres
Thus, the area of the original triangle = \(\frac{3}{2}\) x + \(\frac{3}{2}\) y = \(\frac{3}{2}\) (x + y)
But x + y = 6.
So, the area of the triangle = \(\frac{3}{2}\) × 6 = 9 square centimetres
Parallel Lines
Let’s explore how to draw a triangle with one side measuring 5 cm and an area of 10 square centimetres.
Here we can take the base as 5 cm.
Since the product of base and altitude must be twice the area.
Thus, the altitude should be 4 cm.
That is, base = 5 cm and altitude = 4 cm Let’s learn how to draw a triangle…
- First, draw a line 5 centimetres long
- Draw a perpendicular of height 4 centimetres
- Joining the top of the perpendicular to the ends of the bottom line.
- This is the required triangle with area 10 sq cm.
Even if the position of the perpendicular or the lengths of the left and right sides change, the area remains the same because the base and altitude stay constant.
When we join the top vertices of all these triangles, we will get a line parallel to the base.
If we extend this line and join any point on it to the endpoints of the bottom line, we get a triangle of base 5 centimetres and altitude 4 centimetres; that is, a triangle of base 5 centimetres and area 10 square centimetres.
In general,
All triangles with the same base and third vertex on a line parallel to the base, have the same area.
If the base of the triangle is slanted, the result will still be the same. That is, the area remains the same.
Now let’s look at how to draw a triangle with one side 9 cm and another side 6 cm and area 18 sq cm.
Here any point on the top line can be joined to the endpoints of the bottom line to get a triangle of area 18sqcm. .
To get the other side of the triangle as 6 cm, measure 6 cm using a scale and compass and mark 6 cm on the top line from a point on the bottom line.
Join the other side to get the required triangle.
Now let’s look, at how to draw a triangle of sides 4 centimetres, 5 centimetres and 6 centimetres:
Without calculating the area, we can draw a right triangle of the same area by drawing a line through the vertex, parallel to the base, to make the height equal.
