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Textbook Solutions: Similarity | Mathematics Class 10 (Maharashtra SSC Board) PDF Download

Q: 3. What are the properties of similar triangles?

Ans. The properties of similar triangles include:1. Corresponding angles are equal.2. The lengths of corresponding sides are in proportion.3. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

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 Page 1


Similarity 
 
Practice Set 1.1 
Q. 1. Base of a triangle is 9 and height is 5. Base of another triangle is 10 and 
height is 6. Find the ratio of areas of these triangles. 
Answer : We know that area of triangle =  × Base× Height 
? Area (triangle 1) =  ×9× 5 
=  
? Area (triangle 2) =  ×10× 6 
= 30 
? The ratio of areas of these triangles will be =  
=  
=  
=  
Q. 2. If figure 1.13 BC ? AB, AD ? AB, BC = 4, AD = 8, then find 
?? (??????? )
?? (??????? )
. 
 
 
Page 2


Similarity 
 
Practice Set 1.1 
Q. 1. Base of a triangle is 9 and height is 5. Base of another triangle is 10 and 
height is 6. Find the ratio of areas of these triangles. 
Answer : We know that area of triangle =  × Base× Height 
? Area (triangle 1) =  ×9× 5 
=  
? Area (triangle 2) =  ×10× 6 
= 30 
? The ratio of areas of these triangles will be =  
=  
=  
=  
Q. 2. If figure 1.13 BC ? AB, AD ? AB, BC = 4, AD = 8, then find 
?? (??????? )
?? (??????? )
. 
 
 
Answer : Here, ?ABC and ?ADB has common Base. 
?  
(PROPERTY: Areas of triangles with equal bases are proportional to their 
corresponding heights.) 
?  
=  
=  
Q. 3. In adjoining figure 1.14 seg PS ? seg RQ, seg QT ? seg PR. If RQ = 6, PS = 6 
and PR = 12, then find QT. 
 
Answer : 
 
Considering, Area of (?PQR) with base QR 
? PS will be the Height 
 
 
Page 3


Similarity 
 
Practice Set 1.1 
Q. 1. Base of a triangle is 9 and height is 5. Base of another triangle is 10 and 
height is 6. Find the ratio of areas of these triangles. 
Answer : We know that area of triangle =  × Base× Height 
? Area (triangle 1) =  ×9× 5 
=  
? Area (triangle 2) =  ×10× 6 
= 30 
? The ratio of areas of these triangles will be =  
=  
=  
=  
Q. 2. If figure 1.13 BC ? AB, AD ? AB, BC = 4, AD = 8, then find 
?? (??????? )
?? (??????? )
. 
 
 
Answer : Here, ?ABC and ?ADB has common Base. 
?  
(PROPERTY: Areas of triangles with equal bases are proportional to their 
corresponding heights.) 
?  
=  
=  
Q. 3. In adjoining figure 1.14 seg PS ? seg RQ, seg QT ? seg PR. If RQ = 6, PS = 6 
and PR = 12, then find QT. 
 
Answer : 
 
Considering, Area of (?PQR) with base QR 
? PS will be the Height 
 
 
Now, consider the Area of (?PQR) with base PR 
? QT will be the Height 
? The triangle is the same 
? The area will be the same irrespective of the base taken. 
And we know that area of triangle =  × Base× Height 
? ×QR×PS  
=  × PR × QT 
?  × 6 × 6   
=  × 12 × QT 
? QT = 3 
Q. 4. In adjoining figure, AP ? BC, AD || BC, then find A(?ABC) : A (?BCD) 
 
 
Answer : 
 
Page 4


Similarity 
 
Practice Set 1.1 
Q. 1. Base of a triangle is 9 and height is 5. Base of another triangle is 10 and 
height is 6. Find the ratio of areas of these triangles. 
Answer : We know that area of triangle =  × Base× Height 
? Area (triangle 1) =  ×9× 5 
=  
? Area (triangle 2) =  ×10× 6 
= 30 
? The ratio of areas of these triangles will be =  
=  
=  
=  
Q. 2. If figure 1.13 BC ? AB, AD ? AB, BC = 4, AD = 8, then find 
?? (??????? )
?? (??????? )
. 
 
 
Answer : Here, ?ABC and ?ADB has common Base. 
?  
(PROPERTY: Areas of triangles with equal bases are proportional to their 
corresponding heights.) 
?  
=  
=  
Q. 3. In adjoining figure 1.14 seg PS ? seg RQ, seg QT ? seg PR. If RQ = 6, PS = 6 
and PR = 12, then find QT. 
 
Answer : 
 
Considering, Area of (?PQR) with base QR 
? PS will be the Height 
 
 
Now, consider the Area of (?PQR) with base PR 
? QT will be the Height 
? The triangle is the same 
? The area will be the same irrespective of the base taken. 
And we know that area of triangle =  × Base× Height 
? ×QR×PS  
=  × PR × QT 
?  × 6 × 6   
=  × 12 × QT 
? QT = 3 
Q. 4. In adjoining figure, AP ? BC, AD || BC, then find A(?ABC) : A (?BCD) 
 
 
Answer : 
 
We can re-draw the fig. 1.15 (as shown above) where we add DO 
Which will be height of ?BCD. 
Now,  
(PROPERTY: Areas of triangles with equal bases are proportional to their 
corresponding heights.) 
?  
?  
(? the distance between the two parallel lines is always equal ? AP = DO) 
?  = 1:1 
Q. 5. In adjoining figure PQ ? BC, AD ? BC then find following ratios. 
 
 
Page 5


Similarity 
 
Practice Set 1.1 
Q. 1. Base of a triangle is 9 and height is 5. Base of another triangle is 10 and 
height is 6. Find the ratio of areas of these triangles. 
Answer : We know that area of triangle =  × Base× Height 
? Area (triangle 1) =  ×9× 5 
=  
? Area (triangle 2) =  ×10× 6 
= 30 
? The ratio of areas of these triangles will be =  
=  
=  
=  
Q. 2. If figure 1.13 BC ? AB, AD ? AB, BC = 4, AD = 8, then find 
?? (??????? )
?? (??????? )
. 
 
 
Answer : Here, ?ABC and ?ADB has common Base. 
?  
(PROPERTY: Areas of triangles with equal bases are proportional to their 
corresponding heights.) 
?  
=  
=  
Q. 3. In adjoining figure 1.14 seg PS ? seg RQ, seg QT ? seg PR. If RQ = 6, PS = 6 
and PR = 12, then find QT. 
 
Answer : 
 
Considering, Area of (?PQR) with base QR 
? PS will be the Height 
 
 
Now, consider the Area of (?PQR) with base PR 
? QT will be the Height 
? The triangle is the same 
? The area will be the same irrespective of the base taken. 
And we know that area of triangle =  × Base× Height 
? ×QR×PS  
=  × PR × QT 
?  × 6 × 6   
=  × 12 × QT 
? QT = 3 
Q. 4. In adjoining figure, AP ? BC, AD || BC, then find A(?ABC) : A (?BCD) 
 
 
Answer : 
 
We can re-draw the fig. 1.15 (as shown above) where we add DO 
Which will be height of ?BCD. 
Now,  
(PROPERTY: Areas of triangles with equal bases are proportional to their 
corresponding heights.) 
?  
?  
(? the distance between the two parallel lines is always equal ? AP = DO) 
?  = 1:1 
Q. 5. In adjoining figure PQ ? BC, AD ? BC then find following ratios. 
 
 
Answer : We know that area of triangle =  × Base × Height 
(i)  
(PROPERTY: Areas of triangles with equal heights are proportional to their 
corresponding bases.) 
(ii) 
(PROPERTY: Areas of triangles with equal bases are proportional to their 
corresponding heights.) 
(iii)  
(PROPERTY: Areas of triangles with equal heights are proportional to their 
corresponding bases.) 
(iv)  
=  
 
Practice Set 1.2 
Q. 1. Given below are some triangles and lengths of line segments. Identify in 
which figures, ray PM is the bisector of ?OPR. 
 
 
Answer :

	Mathematics Class 10 (Maharashtra SSC Board) 26 videos\|208 docs\|38 tests

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FAQs on Textbook Solutions: Similarity - Mathematics Class 10 (Maharashtra SSC Board)

1. What is similarity in geometry?

Ans. Similarity in geometry refers to a relationship between two shapes where they have the same shape but may differ in size. This means that corresponding angles are equal, and the lengths of corresponding sides are in proportion.

2. How can we determine if two triangles are similar?

Ans. Two triangles are similar if they satisfy any of the following criteria: 1. Angle-Angle (AA) criterion: If two angles of one triangle are equal to two angles of another triangle. 2. Side-Angle-Side (SAS) criterion: If one angle of a triangle is equal to one angle of another triangle, and the sides including these angles are in proportion. 3. Side-Side-Side (SSS) criterion: If the corresponding sides of the two triangles are in proportion.

3. What are the properties of similar triangles?

Ans. The properties of similar triangles include: 1. Corresponding angles are equal. 2. The lengths of corresponding sides are in proportion. 3. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

4. How do we find the scale factor of similar figures?

Ans. The scale factor of similar figures can be found by measuring the lengths of corresponding sides from the two figures. The scale factor is the ratio of the lengths of one side of a figure to the length of the corresponding side of the other figure. For example, if one side of the first figure is 4 units and the corresponding side of the second figure is 8 units, the scale factor is 4:8 or simplified to 1:2.

5. Can similarity be applied to real-life situations?

Ans. Yes, similarity can be applied in various real-life situations such as in architecture, engineering, and design. For instance, architects use similar shapes when creating scale models of buildings, ensuring that the proportions are maintained while the size is reduced. Similarly, maps represent larger areas in a smaller, similar format.

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