Official MSBSHSE Book for Class 10 Maths: Part II Chapter 1 Similarity
Review the Maharashtra Board Class 10 Maths Part II Chapter 1 Similarity PDF Download designed for Class 10 Maths students. Published under the latest MSBSHSE guidelines for 2026-27, this chapter-wise resource supports daily study and targeted revision.
Chapter-wise Study Material: Part II Chapter 1 Similarity
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Similarity
Let's Study
Ratio of areas of two triangles
Basic proportionality theorem
Converse of basic proportionality theorem
Tests of similarity of triangles
Property of an angle bisector of a triangle
Property of areas of similar triangles
The ratio of the intercepts made on the transversals by three parallel lines
Let's Recall
We have studied Ratio and Proportion. The statement, 'the numbers a and b are in the ratio \(\frac{m}{n}\)' is also written as, 'the numbers a and b are in proportion m:n.'
For this concept we consider positive real numbers. We know that the lengths of line segments and area of any figure are positive real numbers.
We know the formula of area of a triangle.
Area of a triangle = \(\frac{1}{2} \times \text{Base} \times \text{Height}\)
Let's Learn
Ratio Of Areas Of Two Triangles
Let's find the ratio of areas of any two triangles.
In Triangle ABC, AD is the height and BC is the base.
In Triangle PQR, PS is the height and QR is the base.
\[\frac{A(\triangle ABC)}{A(\triangle PQR)} = \frac{\frac{1}{2} \times BC \times AD}{\frac{1}{2} \times QR \times PS}\]
Therefore the ratio of the areas of two triangles is equal to the ratio of the products of their bases and corresponding heights.
Base of a triangle is \(b_1\) and height is \(h_1\). Base of another triangle is \(b_2\) and height is \(h_2\). Then the ratio of their areas = \(\frac{b_1 \times h_1}{b_2 \times h_2}\)
Suppose some conditions are imposed on these two triangles.
Condition 1: If The Heights Of Both Triangles Are Equal
\[\frac{A(\triangle ABC)}{A(\triangle PQR)} = \frac{BC \times h}{QR \times h} = \frac{BC}{QR}\]
Property: The ratio of the areas of two triangles with equal heights is equal to the ratio of their corresponding bases.
Condition 2: If The Bases Of Both Triangles Are Equal
\[\frac{A(\triangle ABC)}{A(\triangle APB)} = \frac{AB \times h_1}{AB \times h_2} = \frac{h_1}{h_2}\]
Property: The ratio of the areas of two triangles with equal bases is equal to the ratio of their corresponding heights.
Teacher's Note
In real life, when you compare the size of two similar maps or photos, you can use this property to find missing measurements. For example, if two maps have the same height but different widths, the areas will be in the same ratio as the widths.
Exam Trick
Remember: Equal height means areas are in ratio of bases. Equal base means areas are in ratio of heights. This simple rule will solve most problems quickly in the exam.
Points To Remember
Area of triangle = half of base times height.
Two triangles with equal heights have areas in the ratio of their bases.
Two triangles with equal bases have areas in the ratio of their heights.
Ratio of areas equals ratio of (base × height) products.
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MSBSHSE Book Class 10 Maths Part II Chapter 1 Similarity
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