Q. ΔABC and ΔBDE are two equilateral triangles such that D is the
midpoint of BC. Ratio of the areas of triangles ABC and BDE is
(A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4.
Sol.
ΔABC
~
ΔBDE
Equilateral triangles are similar
ar (ABC)
ar (BDE)
∴
=
BC2
BD2
BC
=
2BD
…(ii) [D is mid point of BC]
ar (ABC)
ar (BDE)
∴
=
ar (ABC)
ar (BDE)
∴
=
BC
BD
ar (ABC)
ar (BDE)
∴
=
4
1
[From (i) and (ii)]
[The ratio of the areas of two similar
triangles is equal to the ratio of the
squares of the corresponding sides]
C
B
A
E
D
2
…(i)
To find : -
ar (ABC) : (BDE)
∴ ar (ABC) : (BDE) = 4 : 1
2BD
BD
2
EX.6.4 (Q.8)
Sol.
Let A1 and A2 be the areas of two similar triangles and
S1 and S2 be their corresponding sides.
∴
S1
S2
=
4
9
[Given]
A1
A2
=
S1
S2
2
∴
A1
A2
=
4
9
2
∴
A1
A2
∴
The two triangles are similar
[Given]
=
16
81
Q. If sides of two similar triangles are in the ratio 4 : 9 then
areas of these triangles are in the ratio ………….
(A) 2 : 3 (B) 4 : 9 (C) 81 : 16 (D) 16 : 81
[The ratio of the areas of two similar
triangles is equal to the ratio of the
squares of the corresponding sides]
…(i)
[From (i)]
A1 : A2 = 16 : 81
∴
EX.6.4 (Q.9)