Case Study Based Question (CSBQ) on Similar Triangles
Two observation towers A1B1 and A2B2, 8 m and 10 m tall respectively are on the either side of 50 m high observation tower PQ. Points C and D are chosen on the same level of ground for the base tent such that they are 12 m and 18 m away from the towers A1B1 and A2B2, respectively, in opposite directions, as shown in the figure.
Refer to the above figure and answer the following questions :
i) B1P in metres is equal to
a) 63
b) 36
c) 96
d) 60
ii) B2P in metres is equal to
a) 90
b) 80
c) 100
d) 72
iii) Distance between the towers A1B1 and A2B2 is
a) 136 m
b) 160 m
c) 126 m
d) 135 m
iv) Slant height of point C from the top of the tower A1B1 is
a) 4√13 m
b) 6√13 m
c) 8√13 m
d) 13√4 m
Doubt by Rachit
Solution :
i) B1P in metres is equal to
a) 63
b) 36
c) 96
d) 60
Ans :
In ΔA1B1C and ΔQPC
∠A1B1C=∠QPC (Each 90°)
∠C=∠C (Common)
ΔA1B1C ~ ΔQPC (by AA)
A1B1/QP = CB1/CP (Corresponding sides of similar triangles are proportional)
8/50 = 12 / (12+B1P)
2/50 = 3 / (12+B1P)
1/25 = 3 / (12+B1P)
12+B1P = 25×3
12+B1P = 75
B1P = 75-12
B1P = 63
Hence, a) 64 would be the correct option.
ii) B2P in metres is equal to
a) 90
b) 80
c) 100
d) 72
Ans : Similarly
ΔA2B2D ~ ΔQPD
A2B2/QP = DB2/DP
10/50 = 18/(PB2+18)
1/5 = 18/(PB2+18)
PB2 + 18 = 18×5
PB2 + 18 = 90
PB2 = 90-18
PB2 = 72
Hence, d) 72 would be the correct option.
iii) Distance between the towers A1B1 and A2B2 is
a) 136 m
b) 160 m
c) 126 m
d) 135 m
Ans : Distance between the tower A1B1 and A2B2
= B1P+B2P
= 63+72
= 135 m
Hence, d) would be the correct option.
iv) Slant height of point C from the top of the tower A1B1 is
a) 4√13 m
b) 6√13 m
c) 8√13 m
d) 13√4 m
Ans :
A1B1⊥CB1 (Towers always stand vertically on the ground)
In Rt. ΔA1B1C
A1C²=A1B1²+CB1² (By Pythagoras Theorem)
A1C² = (8)²+(12)²
A1C² = 64 +144
A1C² = 208
A1C = 4√13
Hence, a) 4√13 would be the correct option.