RS Aggarwal Solutions: Exercise 6C - Triangles Class 10 Notes | EduRev

Mathematics (Maths) Class 10

Class 10 : RS Aggarwal Solutions: Exercise 6C - Triangles Class 10 Notes | EduRev

 Page 1


 
1. cm
2
and 121cm
2
. If EF = 15.4cm, find BC.
Sol:
Therefore, ratio of the areas of these triangles will be equal to the ration of squares of their 
corresponding sides.
Let BC be X cm.
(64×15.4 × 15.4)
121
=
= 11.2
Hence, BC = 11.2 cm
2. The areas of two similar triangles ABC and PQR are in the ratio 9:16. If BC = 4.5cm, find
the length of QR.
Sol:
Therefore, the ration of the areas of triangles will be equal to the ratio of squares of their
corresponding sides.
(4.5 ×4.5 ×16)
9
=
= 6 cm
Hence, QR = 6 cm
3.
Sol:
Given : 
Page 2


 
1. cm
2
and 121cm
2
. If EF = 15.4cm, find BC.
Sol:
Therefore, ratio of the areas of these triangles will be equal to the ration of squares of their 
corresponding sides.
Let BC be X cm.
(64×15.4 × 15.4)
121
=
= 11.2
Hence, BC = 11.2 cm
2. The areas of two similar triangles ABC and PQR are in the ratio 9:16. If BC = 4.5cm, find
the length of QR.
Sol:
Therefore, the ration of the areas of triangles will be equal to the ratio of squares of their
corresponding sides.
(4.5 ×4.5 ×16)
9
=
= 6 cm
Hence, QR = 6 cm
3.
Sol:
Given : 
 
PQR 
Hence, QR = 6 cm
4. The areas of two similar triangles are 169cm
2
and 121cm
2
respectively. If the longest side of
the larger triangle is 26cm, find the longest side of the smaller triangle.
Sol:
It is given that the triangles are similar.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of
their corresponding sides.
Let the longest side of smaller triangle be X cm.
= 22
Hence, the longest side of the smaller triangle is 22 cm.
5.
2
and 49cm
2
is 5cm, f
Sol:
Therefore, the ration of the areas of these triangles will be equal to the ratio of squares of 
their corresponding sides.
Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their 
corresponding altitudes.
Page 3


 
1. cm
2
and 121cm
2
. If EF = 15.4cm, find BC.
Sol:
Therefore, ratio of the areas of these triangles will be equal to the ration of squares of their 
corresponding sides.
Let BC be X cm.
(64×15.4 × 15.4)
121
=
= 11.2
Hence, BC = 11.2 cm
2. The areas of two similar triangles ABC and PQR are in the ratio 9:16. If BC = 4.5cm, find
the length of QR.
Sol:
Therefore, the ration of the areas of triangles will be equal to the ratio of squares of their
corresponding sides.
(4.5 ×4.5 ×16)
9
=
= 6 cm
Hence, QR = 6 cm
3.
Sol:
Given : 
 
PQR 
Hence, QR = 6 cm
4. The areas of two similar triangles are 169cm
2
and 121cm
2
respectively. If the longest side of
the larger triangle is 26cm, find the longest side of the smaller triangle.
Sol:
It is given that the triangles are similar.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of
their corresponding sides.
Let the longest side of smaller triangle be X cm.
= 22
Hence, the longest side of the smaller triangle is 22 cm.
5.
2
and 49cm
2
is 5cm, f
Sol:
Therefore, the ration of the areas of these triangles will be equal to the ratio of squares of 
their corresponding sides.
Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their 
corresponding altitudes.
 
DEF be DQ, drawn from D to meet EF at Q.
Then,
6. The corresponding altitudes of two similar triangles are 6cm and 9cm respectively. Find the
ratio of their areas.
Sol:
Let the two triangles be ABC and DEF with altitudes AP and DQ, respectively.
We know that the ration of areas of two similar triangles is equal to the ratio of squares of 
their corresponding altitudes.
=  
=
Hence, the ratio of their areas is 4 : 9
7. The areas of two similar triangles are 81cm
2
and 49cm
2
respectively. If the altitude of the
first triangle is 6.3cm, find the corresponding altitude of the other.
Sol:
It is given that the triangles are similar.
Therefore, the areas of these triangles will be equal to the ratio of squares of their
corresponding sides.
Page 4


 
1. cm
2
and 121cm
2
. If EF = 15.4cm, find BC.
Sol:
Therefore, ratio of the areas of these triangles will be equal to the ration of squares of their 
corresponding sides.
Let BC be X cm.
(64×15.4 × 15.4)
121
=
= 11.2
Hence, BC = 11.2 cm
2. The areas of two similar triangles ABC and PQR are in the ratio 9:16. If BC = 4.5cm, find
the length of QR.
Sol:
Therefore, the ration of the areas of triangles will be equal to the ratio of squares of their
corresponding sides.
(4.5 ×4.5 ×16)
9
=
= 6 cm
Hence, QR = 6 cm
3.
Sol:
Given : 
 
PQR 
Hence, QR = 6 cm
4. The areas of two similar triangles are 169cm
2
and 121cm
2
respectively. If the longest side of
the larger triangle is 26cm, find the longest side of the smaller triangle.
Sol:
It is given that the triangles are similar.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of
their corresponding sides.
Let the longest side of smaller triangle be X cm.
= 22
Hence, the longest side of the smaller triangle is 22 cm.
5.
2
and 49cm
2
is 5cm, f
Sol:
Therefore, the ration of the areas of these triangles will be equal to the ratio of squares of 
their corresponding sides.
Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their 
corresponding altitudes.
 
DEF be DQ, drawn from D to meet EF at Q.
Then,
6. The corresponding altitudes of two similar triangles are 6cm and 9cm respectively. Find the
ratio of their areas.
Sol:
Let the two triangles be ABC and DEF with altitudes AP and DQ, respectively.
We know that the ration of areas of two similar triangles is equal to the ratio of squares of 
their corresponding altitudes.
=  
=
Hence, the ratio of their areas is 4 : 9
7. The areas of two similar triangles are 81cm
2
and 49cm
2
respectively. If the altitude of the
first triangle is 6.3cm, find the corresponding altitude of the other.
Sol:
It is given that the triangles are similar.
Therefore, the areas of these triangles will be equal to the ratio of squares of their
corresponding sides.
 
Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their 
corresponding altitudes.
Let the two triangles be ABC and DEF with altitudes AP and DQ, respectively.
Hence, the altitude of the other triangle is 4.9 cm.
8. The areas of two similar triangles are 64cm
2
and 100cm
2
respectively. If a median of the
smaller triangle is 5.6cm, find the corresponding median of the other.
Sol:
Let the two triangles be ABC and PQR with medians AM and PN, respectively.
Therefore, the ratio of areas of two similar triangles will be equal to the ratio of squares of 
their corresponding medians.
= 7 cm
Hence, the median of the larger triangle is 7 cm.
Page 5


 
1. cm
2
and 121cm
2
. If EF = 15.4cm, find BC.
Sol:
Therefore, ratio of the areas of these triangles will be equal to the ration of squares of their 
corresponding sides.
Let BC be X cm.
(64×15.4 × 15.4)
121
=
= 11.2
Hence, BC = 11.2 cm
2. The areas of two similar triangles ABC and PQR are in the ratio 9:16. If BC = 4.5cm, find
the length of QR.
Sol:
Therefore, the ration of the areas of triangles will be equal to the ratio of squares of their
corresponding sides.
(4.5 ×4.5 ×16)
9
=
= 6 cm
Hence, QR = 6 cm
3.
Sol:
Given : 
 
PQR 
Hence, QR = 6 cm
4. The areas of two similar triangles are 169cm
2
and 121cm
2
respectively. If the longest side of
the larger triangle is 26cm, find the longest side of the smaller triangle.
Sol:
It is given that the triangles are similar.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of
their corresponding sides.
Let the longest side of smaller triangle be X cm.
= 22
Hence, the longest side of the smaller triangle is 22 cm.
5.
2
and 49cm
2
is 5cm, f
Sol:
Therefore, the ration of the areas of these triangles will be equal to the ratio of squares of 
their corresponding sides.
Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their 
corresponding altitudes.
 
DEF be DQ, drawn from D to meet EF at Q.
Then,
6. The corresponding altitudes of two similar triangles are 6cm and 9cm respectively. Find the
ratio of their areas.
Sol:
Let the two triangles be ABC and DEF with altitudes AP and DQ, respectively.
We know that the ration of areas of two similar triangles is equal to the ratio of squares of 
their corresponding altitudes.
=  
=
Hence, the ratio of their areas is 4 : 9
7. The areas of two similar triangles are 81cm
2
and 49cm
2
respectively. If the altitude of the
first triangle is 6.3cm, find the corresponding altitude of the other.
Sol:
It is given that the triangles are similar.
Therefore, the areas of these triangles will be equal to the ratio of squares of their
corresponding sides.
 
Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their 
corresponding altitudes.
Let the two triangles be ABC and DEF with altitudes AP and DQ, respectively.
Hence, the altitude of the other triangle is 4.9 cm.
8. The areas of two similar triangles are 64cm
2
and 100cm
2
respectively. If a median of the
smaller triangle is 5.6cm, find the corresponding median of the other.
Sol:
Let the two triangles be ABC and PQR with medians AM and PN, respectively.
Therefore, the ratio of areas of two similar triangles will be equal to the ratio of squares of 
their corresponding medians.
= 7 cm
Hence, the median of the larger triangle is 7 cm.
 
9. In the given figure, ABC is a triangle and PQ is a straight line meeting AB in P and AC in
of the 
Sol:
We have :
Also,
-
Hence proved.
10.
2
, find the area
Sol:
It is given that DE || BC
By AA similarity, we can c
=
Hence, area of triangle ABC is 60 
11. ABC is right angled at A and AD BC. If BC = 13cm and AC = 5cm, find the ratio of the
Sol:
Hence, the ratio of the areas of these triangles is equal to the ratio of squares of their
corresponding sides.
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