In this page 10th grade geometry solution1 we are going to see solutions of some practice questions.
(1) In a triangle ABC,D and E are points on the sides AB and AC respectively such that DE is parallel to BC.
(i) If AD = 6 cm, DB = 9 cm and AE = 8 cm, then find AC.
Solution:
In the given triangle ABC the sides DE is parallel to the side BC. By using “Thales theorem”, we get
(AD/DB) = (AE/EC) (6/9) = (8/EC) EC = (9 x 8)/6 EC = (3 x 4) EC = 12 cm |
(ii) If AD = 8 cm, AB = 12 cm and AE =12 cm, then find CE.
Solution:
In the given triangle ABC the sides DE is parallel to the side BC. By using “Thales theorem”, we get
(AD/DB) = (AE/EC) AB = AD + DB 12 = 8 + DB
DB = 4 cm (AD/DB) = (AE/EC) (8/4) = (12/EC) EC = (12 x 4)/8 EC = 48/8 EC = 6 cm |
(iii) If AD = 4 x – 3, BD = 3 x – 1, AE = 8 x – 7 and EC = 5 x – 3, then find the value of x.
Solution:
In the given triangle ABC the sides DE is parallel to the side BC. By using “Thales theorem”, we get
(AD/DB) = (AE/EC)
(4 x – 3)/(3 x – 1) = (8 x – 7)/(5 x – 3)
(4 x – 3) (5 x – 3) = (8 x – 7) (3 x – 1)
20 x ² – 12 x – 15 x + 9 = 24 x ² – 8 x – 21 x + 7
20 x ² – 24 x + 9 = 8 x ² – 29 x + 7
20 x ² - 24 x ² – 27 x + 29 x + 9 – 7 = 0
- 4 x ² + 2 x + 2 = 0
Multiply by - 2
2 x ² – x - 1 = 0
(x – 1) (2 x + 1) = 0
x – 1 = 0 2 x + 1 = 0
x = 1 2 x = -1
x = -1/2
Since -1/2 is negative it is not acceptable. So the value of x is 1.
10th grade geometry solution1 10th grade geometry solution1