It is given that △ABC ~ △DEF, ∠A = 30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF = 7.5 cm. Then, the following is true:
a. DE = 12 cm, ∠F = 50°
b. DE = 12 cm, ∠F = 100°
c. EF = 12 cm, ∠D = 100°
d. EF = 12 cm, ∠D = 30°
Solution:
Given, the triangles ABC and DEF are similar.
Also, ∠A =30° and ∠C = 50°
The length of the sides
AB = 5 cm
AC = 8 cm
DF = 7.5 cm
We have to determine the correct solution from the given options.
Since the two triangles are similar, the corresponding sides will be
AB/DF = AC/DE = BC/EF
5/7.5 = 8/DE = BC/EF
Taking 5/7.5 = 8/DE
On cross multiplication,
5(DE) = 8(7.5)
DE = 60/5
DE = 12 cm
In triangle ABC,
∠A + ∠B + ∠C = 180°
30° + ∠B + 50° = 180°
∠B = 180° - 80°
∠B = 100°
As the triangles are similar, the corresponding angles will be equal.
So, ∠B = ∠F = 100°
Therefore, DE = 12 cm and ∠F = 100°
✦ Try This: It is given that △ABC ~ △DEF, ∠A =40°, ∠C = 80°, AB = 8 cm, AC = 9 cm and DF= 8.5 cm. Find DE and ∠F
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 6
NCERT Exemplar Class 10 Maths Exercise 6.1 Problem 9
It is given that △ABC ~ △DEF, ∠A = 30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF = 7.5 cm. Then, the following is true, a. DE = 12 cm, ∠F = 50°, b. DE = 12 cm, ∠F = 100°, c. EF = 12 cm, ∠D = 100°, d. EF = 12 cm, ∠D = 30°
Summary:
It is given that △ABC ~ △DEF, ∠A =30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF= 7.5 cm then DE = 12 cm and ∠F = 100° is true
☛ Related Questions:
- If in triangles ABC and DEF, AB/DE = BC/FD , then they will be similar, when, a. ∠B = ∠E, b. ∠A = ∠D . . . .
- If △ABC ~ △QRP, ar(ABC)/ar(PQR) = 9/4, AB = 18 cm and BC = 15 cm, then PR is equal to, a.10 cm, b. 1 . . . .
- If S is a point on side PQ of a △PQR such that PS = QS = RS, then, a. PR . QR = RS², b. QS² + RS² = . . . .