>
ii E and F are points on the sides PQ and PR respectively of a Δ PQR. For the following case, state whether EF || QR. ii PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
2
Leave a Comment / / By mathemerize
Leave a Comment / / By mathemerize
Leave a Comment / / By mathemerize
Leave a Comment / / By mathemerize
Leave a Comment / / By mathemerize
Leave a Comment / / By mathemerize
Leave a Comment / / By mathemerize
Solution : Let ladder be AB, B be the window and CB be the wall. Then, ABC is a right triangle, angled at C. \(\therefore\) \({AB}^2\) = \({AC}^2\) + \({BC}^2\) So, \({10}^2\) = \({AC}^2\) + \(8^2\) or \({AC}^2\) = 100 – 64 \(\implies\) \({AC}^2\) = 36 \(\implies\) AC = 6 m
Leave a Comment / / By mathemerize
Solution:
We know according to the basic proportionality theorem, if a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, the other two sides are divided in the same ratio.
Consider the trapezium ABCD as shown below.
In trapezium ABCD,
AB || CD
Also, AC and BD intersect at ‘O’
Construct XY parallel to AB and CD (XY || AB, XY || CD) through ‘O’
In ΔABC
OY || AB (construction)
According to theorem 6.1 (Basic Proportionality Theorem)
BY/CY = AO/OC................. (1)
In ΔBCD
OY || CD (construction)
According to theorem 6.1 (Basic Proportionality Theorem)
BY/CY = OB/OD................. (2)
From equations (1) and (2)
OA/OC = OB/OD
⇒ OA/OB = OC/OD
Hence proved.
☛ Check: NCERT Solutions for Class 10 Maths Chapter 6
Video Solution:
Class 10 Maths NCERT Solutions Chapter 6 Exercise 6.2 Question 9
Summary:
Hence it is proved that AO/BO = CO/DO if ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O.
☛ Related Questions:
- In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii)
- E and F are points on the sides PQ and PR respectively of a ∆ PQR. For each of the following cases, state whether EF || QR : (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
- In Fig. 6.18, if LM || CB and LN || CD, prove that AM/AB = AN/AD.
- In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC.
Math worksheets and
visual curriculum
E and F are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether EF || QR:(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm.(ii) PE = 4 cm, QE = 4.5 cm. PF = 8 cm and RF = 9 cm.(iii) PQ = 1.28 cm. PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.
(i) We have,
From (i) and (ii), we have
Therefore, EF is not parallel to QR [By using converse of Basic proportionality theorem]
(ii) We have,
From (i) and (ii), we have
Therefore,
[Using converse of Basic proportionality theorem]
(iii) We have,
From (i) and (ii), we have
Therefore,
[Using converse of Basic proportionality theorem]