Unit 24 Class 10 Math Sindh Board Solutions
The Unit 24 Class 10 Math Sindh Board Solutions provide complete, step-by-step answers for the chapter Proportionality and Similarity. This unit explains ratios in triangles, the basic proportionality theorem, the angle-bisector theorem, similar triangles, and the relationship between corresponding sides and areas.
The PDF covers Exercise 24.1, Exercise 24.2, and Review Exercise 24. Numerical questions, theorem proofs, diagrams, calculations, and final answers are explained in a student-friendly manner.
Students can use these solutions to:
- Understand important geometry theorems
- Complete textbook exercises
- Prepare for school and board examinations
- Revise numerical questions and proofs
- Check their working and final answers
Download Unit 24 Class 10 Math Sindh Board Solutions
The PDF available on this page contains complete solutions of Unit 24 according to the Class 10 Mathematics textbook prescribed by the Sindh Textbook Board.
It includes:
- Essential theorems and ratios
- Complete solutions of Exercise 24.1
- Complete solutions of Exercise 24.2
- Solved Review Exercise 24
- Diagrams for geometrical questions
- Detailed theorem proofs
- Correct answers to multiple-choice questions
Students should try to solve each question independently before consulting the complete solution.
About Unit 24: Proportionality and Similarity
Unit 24 introduces geometrical relationships involving triangles, parallel lines, ratios, angle bisectors, corresponding sides, and areas.
Two ratios form a proportion when they are equal:
\(\displaystyle \frac{a}{b}=\frac{c}{d}\)
In this proportion, \(b\) and \(c\) are called the means, while \(a\) and \(d\) are called the extremes.
By cross multiplication:
\(\displaystyle ad=bc\)
Two figures are similar when they have the same shape but may have different sizes.
In similar triangles:
- Corresponding angles are equal.
- Corresponding sides are proportional.
- The ratio of their areas is the square of the ratio of corresponding sides.
These ideas are applied throughout the Unit 24 Class 10 Math Sindh Board Solutions.
Exercise 24.1 Solutions
Exercise 24.1 mainly focuses on the basic proportionality theorem, its converse, and related results involving triangles and trapeziums.
Basic Proportionality Theorem
The basic proportionality theorem states:
If a line is drawn parallel to one side of a triangle and intersects the other two sides, it divides those two sides in the same ratio.
In triangle \(ABC\), suppose \(DE\parallel BC\), where \(D\) lies on \(AB\) and \(E\) lies on \(AC\).
Then:
\(\displaystyle \frac{AD}{DB}=\frac{AE}{EC}\)
This theorem is also known as Thales’ theorem.
Converse of the Basic Proportionality Theorem
The converse states that if:
\(\displaystyle \frac{AD}{DB}=\frac{AE}{EC}\)
then:
\(\displaystyle DE\parallel BC\)
This result is useful when students have to prove that a particular line is parallel to one side of a triangle.
Exercise 24.1 Question 1: Finding an Unknown Side
In one question, the given measurements are:
\(\displaystyle BD=6\text{ cm}\)
\(\displaystyle DC=12\text{ cm}\)
\(\displaystyle AE=5\text{ cm}\)
Since the line is parallel to one side of the triangle, the sides are divided proportionally:
\(\displaystyle \frac{CD}{DB}=\frac{CE}{EA}\)
Substitute the values:
\(\displaystyle \frac{12}{6}=\frac{CE}{5}\)
Simplify:
\(\displaystyle 2=\frac{CE}{5}\)
Multiply both sides by \(5\):
\(\displaystyle CE=2\times 5\)
Therefore:
\(\displaystyle CE=10\text{ cm}\)
The result can be checked by comparing the ratios:
\(\displaystyle \frac{CD}{DB}=\frac{12}{6}=2\)
and:
\(\displaystyle \frac{CE}{EA}=\frac{10}{5}=2\)
Both ratios are equal.
Exercise 24.1 Question 2: Finding the Value of xxx
The given expressions are:
\(\displaystyle AP=5x-3\)
\(\displaystyle PB=2\)
\(\displaystyle AQ=2x+1\)
\(\displaystyle QC=3\)
Since \(PQ\parallel BC\), the basic proportionality theorem gives:
\(\displaystyle \frac{AP}{PB}=\frac{AQ}{QC}\)
Substitute the expressions:
\(\displaystyle \frac{5x-3}{2}=\frac{2x+1}{3}\)
Cross-multiply:
\(\displaystyle 3(5x-3)=2(2x+1)\)
Expand both sides:
\(\displaystyle 15x-9=4x+2\)
Move the variable terms to one side:
\(\displaystyle 15x-4x=2+9\)
\(\displaystyle 11x=11\)
Divide by \(11\):
\(\displaystyle x=1\)
Checking the lengths:
\(\displaystyle AP=5(1)-3=2\)
\(\displaystyle AQ=2(1)+1=3\)
Therefore:
\(\displaystyle \frac{AP}{PB}=\frac{2}{2}=1\)
and:
\(\displaystyle \frac{AQ}{QC}=\frac{3}{3}=1\)
Hence, the answer is correct.
Parallel Line in a Trapezium
Exercise 24.1 proves that a line drawn parallel to the parallel sides of a trapezium divides its non-parallel sides proportionally.
Suppose:
\(\displaystyle AB\parallel EF\parallel CD\)
The required result is:
\(\displaystyle \frac{AE}{ED}=\frac{BF}{FC}\)
A diagonal is drawn to divide the trapezium into two triangles. The basic proportionality theorem is then applied separately to both triangles.
From one triangle:
\(\displaystyle \frac{AE}{ED}=\frac{AG}{GC}\)
From the other triangle:
\(\displaystyle \frac{AG}{GC}=\frac{BF}{FC}\)
Therefore:
\(\displaystyle \frac{AE}{ED}=\frac{BF}{FC}\)
Hence, the line parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally.
Midpoint Theorem
Another important result proved in Exercise 24.1 is:
A line drawn through the midpoint of one side of a triangle and parallel to another side bisects the third side.
Suppose \(D\) is the midpoint of \(AB\) and \(DE\parallel BC\).
Since \(D\) is the midpoint:
\(\displaystyle AD=DB\)
Therefore:
\(\displaystyle \frac{AD}{DB}=1\)
By the basic proportionality theorem:
\(\displaystyle \frac{AD}{DB}=\frac{AE}{EC}\)
Substitute:
\(\displaystyle 1=\frac{AE}{EC}\)
Therefore:
\(\displaystyle AE=EC\)
Hence, \(E\) is the midpoint of \(AC\).
Converse Result for a Trapezium
The exercise also proves that if a line divides the non-parallel sides of a trapezium proportionally, then it is parallel to the parallel sides.
Suppose:
\(\displaystyle AB\parallel CD\)
and:
\(\displaystyle \frac{AE}{ED}=\frac{BF}{FC}\)
Through \(E\), draw a line parallel to \(AB\) meeting \(BC\) at \(G\).
By the result already proved:
\(\displaystyle \frac{AE}{ED}=\frac{BG}{GC}\)
But it is given that:
\(\displaystyle \frac{AE}{ED}=\frac{BF}{FC}\)
Therefore:
\(\displaystyle \frac{BG}{GC}=\frac{BF}{FC}\)
Since a line segment can be divided internally in a given ratio at only one point:
\(\displaystyle G=F\)
Therefore:
\(\displaystyle EF\parallel AB\parallel CD\)
Exercise 24.2 Solutions
Exercise 24.2 covers the angle-bisector theorem, corresponding sides of similar triangles, area ratios, and the similarity of right triangles.
Angle-Bisector Theorem
The angle-bisector theorem states:
The internal bisector of an angle of a triangle divides the opposite side in the ratio of the two adjacent sides.
If \(AD\) bisects \(\angle A\) of triangle \(ABC\), then:
\(\displaystyle \frac{BD}{DC}=\frac{AB}{AC}\)
This theorem is used to calculate unknown sides or variables when an angle bisector divides the opposite side.
Exercise 24.2 Question 1: Finding xxx and the Type of Triangle
The given expressions are:
\(\displaystyle AB=x+9\)
\(\displaystyle AC=11-x\)
\(\displaystyle CD=2x+3\)
\(\displaystyle BD=3x+2\)
By the angle-bisector theorem:
\(\displaystyle \frac{BD}{DC}=\frac{AB}{AC}\)
Substitute the expressions:
\(\displaystyle \frac{3x+2}{2x+3}=\frac{x+9}{11-x}\)
Cross-multiply:
\(\displaystyle (3x+2)(11-x)=(2x+3)(x+9)\)
Expand the left side:
\(\displaystyle 33x-3x^2+22-2x\)
\(\displaystyle =31x-3x^2+22\)
Expand the right side:
\(\displaystyle 2x^2+18x+3x+27\)
\(\displaystyle =2x^2+21x+27\)
Therefore:
\(\displaystyle 31x-3x^2+22=2x^2+21x+27\)
Bring all terms to one side:
\(\displaystyle 5x^2-10x+5=0\)
Divide by \(5\):
\(\displaystyle x^2-2x+1=0\)
Factorize:
\(\displaystyle (x-1)^2=0\)
Therefore:
\(\displaystyle x=1\)
Now calculate the side lengths:
\(\displaystyle AB=1+9=10\)
\(\displaystyle AC=11-1=10\)
\(\displaystyle BD=3(1)+2=5\)
\(\displaystyle CD=2(1)+3=5\)
Therefore:
\(\displaystyle BC=BD+CD\)
\(\displaystyle BC=5+5=10\)
Thus:
\(\displaystyle AB=AC=BC=10\)
Since all three sides are equal, triangle \(ABC\) is an equilateral triangle.
Exercise 24.2 Question 2: Similar Triangles
Suppose triangles \(PQR\) and \(ABC\) are similar.
The corresponding sides are:
\(\displaystyle PQ=9\)
\(\displaystyle PR=6\)
\(\displaystyle QR=12\)
\(\displaystyle AB=3\)
\(\displaystyle AC=x\)
\(\displaystyle BC=y\)
The correspondence is:
\(\displaystyle P\leftrightarrow A\)
\(\displaystyle Q\leftrightarrow B\)
\(\displaystyle R\leftrightarrow C\)
Therefore:
\(\displaystyle \frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}\)
Use the known pair:
\(\displaystyle \frac{AB}{PQ}=\frac{3}{9}=\frac{1}{3}\)
For \(x\):
\(\displaystyle \frac{x}{6}=\frac{1}{3}\)
Cross-multiply:
\(\displaystyle 3x=6\)
Therefore:
\(\displaystyle x=2\)
For \(y\):
\(\displaystyle \frac{y}{12}=\frac{1}{3}\)
Cross-multiply:
\(\displaystyle 3y=12\)
Therefore:
\(\displaystyle y=4\)
Hence:
\(\displaystyle x=2,\qquad y=4\)
Areas of Similar Triangles
For two similar triangles, the ratio of their areas is equal to the square of the ratio of corresponding sides.
If:
\(\displaystyle \triangle ABC\sim\triangle PQR\)
then:
\(\displaystyle \frac{[\triangle ABC]}{[\triangle PQR]}=\left(\frac{AB}{PQ}\right)^2\)
This formula is used in several questions of Exercise 24.2.
Exercise 24.2 Question 3: Finding the Area of a Similar Triangle
The given values are:
\(\displaystyle A_1=28\text{ cm}^2\)
\(\displaystyle AB=4x\)
\(\displaystyle PQ=2x\)
Using the area formula:
\(\displaystyle \frac{A_1}{A_2}=\left(\frac{AB}{PQ}\right)^2\)
Substitute:
\(\displaystyle \frac{28}{A_2}=\left(\frac{4x}{2x}\right)^2\)
Cancel \(x\):
\(\displaystyle \frac{28}{A_2}=2^2\)
\(\displaystyle \frac{28}{A_2}=4\)
Cross-multiply:
\(\displaystyle 4A_2=28\)
Divide by \(4\):
\(\displaystyle A_2=7\text{ cm}^2\)
The side ratio is \(2:1\), so the area ratio is:
\(\displaystyle 2^2:1^2=4:1\)
This agrees with:
\(\displaystyle 28:7=4:1\)
Exercise 24.2 Question 4: Finding a Side Ratio from an Area Ratio
The ratio of corresponding sides is:
\(\displaystyle 2:(x-5)\)
The ratio of the areas is:
\(\displaystyle 1:9\)
The side ratio is the positive square root of the area ratio:
\(\displaystyle \sqrt{1}:\sqrt{9}=1:3\)
Therefore:
\(\displaystyle 2:(x-5)=1:3\)
Write the ratios as fractions:
\(\displaystyle \frac{2}{x-5}=\frac{1}{3}\)
Cross-multiply:
\(\displaystyle 2(3)=x-5\)
\(\displaystyle 6=x-5\)
Add \(5\) to both sides:
\(\displaystyle x=11\)
Check:
\(\displaystyle 2:(11-5)=2:6=1:3\)
Squaring the side ratio gives:
\(\displaystyle 1^2:3^2=1:9\)
Therefore, the answer is correct.
Similarity of Two Right Triangles
Exercise 24.2 proves that two right triangles are similar if an acute angle of one triangle is equal to an acute angle of the other.
Suppose triangles \(ABC\) and \(PQR\) are right-angled at \(B\) and \(Q\) respectively.
Then:
\(\displaystyle \angle B=\angle Q=90^\circ\)
It is also given that:
\(\displaystyle \angle A=\angle P\)
Since two corresponding angles are equal:
\(\displaystyle \triangle ABC\sim\triangle PQR\)
by the AA similarity criterion.
Therefore:
\(\displaystyle \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}\)
Perpendicular from the Right-Angle Vertex to the Hypotenuse
Suppose triangle \(ABC\) is right-angled at \(B\) and \(BD\perp AC\).
The perpendicular divides the original triangle into two smaller triangles.
The required results are:
\(\displaystyle \triangle ABD\sim\triangle ABC\)
and:
\(\displaystyle \triangle CBD\sim\triangle ABC\)
Similarity of △ABD\triangle ABD△ABD and △ABC\triangle ABC△ABC
Since:
\(\displaystyle BD\perp AC\)
we have:
\(\displaystyle \angle ADB=90^\circ\)
The original triangle is right-angled at \(B\):
\(\displaystyle \angle ABC=90^\circ\)
Therefore:
\(\displaystyle \angle ADB=\angle ABC\)
Also, since \(D\) lies on \(AC\):
\(\displaystyle \angle BAD=\angle BAC\)
Thus, by AA similarity:
\(\displaystyle \triangle ABD\sim\triangle ABC\)
Similarity of △CBD\triangle CBD△CBD and △ABC\triangle ABC△ABC
Since:
\(\displaystyle BD\perp AC\)
we have:
\(\displaystyle \angle CDB=90^\circ\)
Also:
\(\displaystyle \angle ABC=90^\circ\)
Therefore:
\(\displaystyle \angle CDB=\angle ABC\)
Since \(D\) lies on \(AC\):
\(\displaystyle \angle BCD=\angle BCA\)
Thus, by AA similarity:
\(\displaystyle \triangle CBD\sim\triangle ABC\)
Therefore, both smaller triangles are similar to the original triangle.
They are also similar to each other.
Review Exercise 24 Solutions
The Review Exercise contains multiple-choice questions based on all the concepts studied in Unit 24.
The questions test students’ understanding of:
- Proportions
- Means and extremes
- Similarity symbols
- Equilateral triangles
- Basic proportionality theorem
- Converse of the proportionality theorem
- Angle-bisector theorem
- Corresponding sides
- Area ratios
- Side ratios
The PDF includes explanations for the correct answers rather than providing only an answer key.
Product of Means and Extremes
For the proportion:
\(\displaystyle \frac{a}{b}=\frac{c}{d}\)
cross multiplication gives:
\(\displaystyle ad=bc\)
The product of the extremes is equal to the product of the means.
The extremes are \(a\) and \(d\).
The means are \(b\) and \(c\).
Equilateral Triangles and Similarity
Every equilateral triangle has three angles of:
\(\displaystyle 60^\circ\)
Therefore, any two equilateral triangles have equal corresponding angles and are similar by the AAA criterion.
However, they are not necessarily congruent because their side lengths may be different.
Angle-Bisector Ratio
If:
\(\displaystyle AB=6\text{ cm}\)
and:
\(\displaystyle AC=8\text{ cm}\)
then the angle-bisector theorem gives:
\(\displaystyle BD:DC=AB:AC\)
\(\displaystyle BD:DC=6:8\)
Simplify:
\(\displaystyle BD:DC=3:4\)
Side Ratio and Area Ratio
If the ratio of corresponding sides is:
\(\displaystyle 5:7\)
then the ratio of areas is:
\(\displaystyle 5^2:7^2\)
\(\displaystyle 25:49\)
Similarly, if the ratio of areas is:
\(\displaystyle 36:121\)
then the corresponding-side ratio is:
\(\displaystyle \sqrt{36}:\sqrt{121}\)
\(\displaystyle 6:11\)
Finding xxx from a Side and Area Ratio
Suppose the side ratio is:
\(\displaystyle 2:x\)
and the area ratio is:
\(\displaystyle 4:9\)
Then:
\(\displaystyle \left(\frac{2}{x}\right)^2=\frac{4}{9}\)
\(\displaystyle \frac{4}{x^2}=\frac{4}{9}\)
Cross-multiply:
\(\displaystyle 4x^2=36\)
Divide by \(4\):
\(\displaystyle x^2=9\)
Algebraically:
\(\displaystyle x=\pm 3\)
However, a side ratio must be positive.
Therefore:
\(\displaystyle x=3\)
Important Formulas of Unit 24
Basic Proportionality Theorem
If \(DE\parallel BC\), then:
\(\displaystyle \frac{AD}{DB}=\frac{AE}{EC}\)
Converse of the Basic Proportionality Theorem
If:
\(\displaystyle \frac{AD}{DB}=\frac{AE}{EC}\)
then:
\(\displaystyle DE\parallel BC\)
Angle-Bisector Theorem
If \(AD\) bisects \(\angle A\), then:
\(\displaystyle \frac{BD}{DC}=\frac{AB}{AC}\)
Corresponding Sides of Similar Triangles
If:
\(\displaystyle \triangle ABC\sim\triangle PQR\)
then:
\(\displaystyle \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}\)
Areas of Similar Triangles
\(\displaystyle \frac{[\triangle ABC]}{[\triangle PQR]}=\left(\frac{AB}{PQ}\right)^2\)
Side Ratio from Area Ratio
If the area ratio is:
\(\displaystyle m:n\)
then the corresponding-side ratio is:
\(\displaystyle \sqrt{m}:\sqrt{n}\)
Similarity Criteria for Triangles
Triangles can be proved similar using the following criteria.
AA Similarity
If two corresponding angles of two triangles are equal, the triangles are similar.
SAS Similarity
If two pairs of corresponding sides are proportional and the included angles are equal, the triangles are similar.
SSS Similarity
If all three pairs of corresponding sides are proportional, the triangles are similar.
Important Concepts to Remember
Students should always write corresponding vertices in the correct order.
For example, if:
\(\displaystyle \triangle ABC\sim\triangle PQR\)
then the correspondence is:
\(\displaystyle A\leftrightarrow P\)
\(\displaystyle B\leftrightarrow Q\)
\(\displaystyle C\leftrightarrow R\)
Therefore:
\(\displaystyle AB\leftrightarrow PQ\)
\(\displaystyle BC\leftrightarrow QR\)
\(\displaystyle AC\leftrightarrow PR\)
The correct proportional relationship is:
\(\displaystyle \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}\)
Writing corresponding sides in the wrong order results in an incorrect proportion.
Students should also remember:
- Congruent triangles always have equal corresponding sides.
- Similar triangles may have different corresponding side lengths.
- All equilateral triangles are similar.
- All right triangles are not necessarily similar.
- Side lengths and side ratios must be positive.
- The area ratio is the square of the side ratio.
- The side ratio is the positive square root of the area ratio.
Common Mistakes in Unit 24
Reversing One Ratio
A common mistake is writing:
\(\displaystyle \frac{AD}{DB}=\frac{EC}{AE}\)
This is incorrect because the order of the corresponding segments is reversed.
The correct relationship is:
\(\displaystyle \frac{AD}{DB}=\frac{AE}{EC}\)
Using the Area Ratio as the Side Ratio
If the area ratio is:
\(\displaystyle 25:49\)
the side ratio is not \(25:49\).
The side ratio is:
\(\displaystyle \sqrt{25}:\sqrt{49}\)
\(\displaystyle 5:7\)
Squaring the Wrong Ratio
If the side ratio is:
\(\displaystyle 3:4\)
the area ratio is:
\(\displaystyle 3^2:4^2\)
\(\displaystyle 9:16\)
Students should square both terms of the ratio.
Assuming Triangles Are Similar from Their Appearance
Triangles should not be declared similar only because they appear to have the same shape in a diagram.
Similarity must be proved using AA, SAS, or SSS similarity.
Writing Corresponding Sides in the Wrong Order
If:
\(\displaystyle \triangle ABC\sim\triangle PQR\)
then \(AB\) corresponds to \(PQ\), not \(QR\).
Correct correspondence is essential when forming proportions.
How These Solutions Help Students
The Unit 24 Class 10 Math Sindh Board Solutions explain both the working and the final answers.
The theorem questions are arranged with:
- Given information
- Statement to be proved
- Required construction
- Step-by-step proof
- Final conclusion
The numerical questions include:
- Relevant formula or theorem
- Substitution of the given values
- Cross multiplication
- Algebraic simplification
- Final answer
- Verification where required
This approach helps students learn how to write complete solutions in board examinations.
Frequently Asked Questions
What is the name of Unit 24 in Class 10 Sindh Board Mathematics?
Unit 24 is titled Proportionality and Similarity.
Which exercises are included in the solutions?
The PDF includes:
- Exercise 24.1
- Exercise 24.2
- Review Exercise 24
What is the main theorem used in Exercise 24.1?
The main theorem is the basic proportionality theorem and its converse.
What is the main theorem used in Exercise 24.2?
Exercise 24.2 mainly uses the angle-bisector theorem and the properties of similar triangles.
What is the basic proportionality theorem?
If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.
In symbols:
\(\displaystyle \frac{AD}{DB}=\frac{AE}{EC}\)
What is the angle-bisector theorem?
The internal angle bisector of a triangle divides the opposite side in the ratio of the two adjacent sides:
\(\displaystyle \frac{BD}{DC}=\frac{AB}{AC}\)
What is the relationship between the sides and areas of similar triangles?
If the ratio of corresponding sides is:
\(\displaystyle a:b\)
then the ratio of their areas is:
\(\displaystyle a^2:b^2\)
Are all equilateral triangles similar?
Yes. Every equilateral triangle has three angles of \(60^\circ\), so any two equilateral triangles are similar.
Are all right triangles similar?
No. Two right triangles are similar only if another pair of corresponding acute angles is equal or another similarity condition is satisfied.
What does the symbol ∼\sim∼ mean?
The symbol:
\(\displaystyle \sim\)
means “is similar to.”
For example:
\(\displaystyle \triangle ABC\sim\triangle PQR\)
What does the symbol ≅\cong≅ mean?
The symbol:
\(\displaystyle \cong\)
means “is congruent to.”
Congruent figures have the same shape and the same size.
Conclusion
The Unit 24 Class 10 Math Sindh Board Solutions provide complete preparation for the chapter Proportionality and Similarity. The PDF covers numerical questions, theorem proofs, angle bisectors, proportional sides, similar triangles, area ratios, and review questions.
Students should revise the important formulas, practise writing ratios in the correct order, and carefully identify corresponding sides before solving similarity questions. Regular practice with these solutions will make the concepts easier to understand and apply in board examinations.