RD Sharma Solutions Chapter 4 Triangles Exercise 4.6 Class 10 Maths
Chapter Name  RD Sharma Chapter 4 Triangles 
Book Name  RD Sharma Mathematics for Class 10 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 4.5 Solutions
1. Triangles ABC and DEF are similar
(i) If area (Î”ABC) = 16 cm^{2} , area(Î”DEF) = 25cm^{2} and BC = 2.3 cm, find EF.
(ii) If area (Î”ABC) = 9cm^{2} , area(Î”DEF) = 64 cm^{2} and DE = 5.1 cm, find AB.
(iii) If AC = 19 cm and DF = 8cm, find the ratio of the area of two triangles.
(iv) If area(Î”ABC) = 36cm^{2} , area (Î”DEF) = 64 cm^{2} and DE = 6.2 cm, find AB.
(v) If AB = 1.2cm and DE = 1.4 cm, find the ratio of the areas of Î”ABC and Î”DEF.
Solution
(i) If area (Î”ABC) = 16 cm^{2} , area(Î”DEF) = 25cm^{2} and BC = 2.3 cm, find EF.
(ii) If area (Î”ABC) = 9cm^{2} , area(Î”DEF) = 64 cm^{2} and DE = 5.1 cm, find AB.
(iii) If AC = 19 cm and DF = 8cm, find the ratio of the area of two triangles.
(v) If AB = 1.2cm and DE = 1.4 cm, find the ratio of the areas of Î”ABC and Î”DEF.
2. In fig. below Î”ACB ～Î”APQ. If BC = 10 m, PQ = 5cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ. Also, find the area (Î”ACB): area(Î”APQ)
Solution
3. The areas of two similar triangles are 81 cm^{2} and 49cm^{2} respectively. Find the ratio of their corresponding heights. What is the ratio of their corresponding medians ?
Solution
Since, the ratio of the area of two similar triangles is equal to the ratio of the squares of the squares of their corresponding altitudes and is also equal to the squares of their corresponding medians.
Hence, ratio of altitudes = Ratio of medians = 9 : 7
4. The areas of two similar triangles are 169 cm^{2} and 121 cm^{2} respectively. If the longest side of the larger triangle is 26 cm, find the longest side of the smaller triangle.
Solution
(i) If DE = 4cm, BC = 6cm and Area (Î”ADE) = 16 cm^{2} , find the area of Î”ABC.
(ii) If DE = 4cm, BC = 8cm and Area (Î”ADE) = 25 cm^{2} , find the area of Î”ABC.
(iii) If DE : BC = 3 : 5. Calculate the ratio of the areas of Î”ADE and the trapezium BCED.
We have, DEBC, DE = 4cm, BC = 6 cm and area(Î”ADE) = 16cm^{2}
In Î”ADE and Î”ABC
∠A = ∠A [Common]
∠ADE = ∠ABC [Corresponding angles]
Then, Î”ADE ～Î”ABC [By AA similarity]
∴ By area of similar triangle theorem
10. In Î”ABC, D and E are the mid  Points of AB and AC respectively. Find the ratio of the areas of Î”ADE and Î”ABC
Since Î”ABC and Î”DBC are one same base.
Therefore ratio between their areas will be as ratio of their heights.
Let us draw two perpendiculars AP and DM on line BC>
(i) Î”AOB and Î”COD
(ii) If OA = 6 cm, OC = 8cm,
Find :
(a) area(Î”AOB)/area(Î”COD)
(b) area(Î”AOD)/area(Î”COD)
Î”ABC ～Î”DEF such that AB = 5cm,
Area(Î”ABC) = 20cm^{2} and area(Î”DEF) = 45cm^{2}
By area of similar triangle theorem