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Triangles

Exercise 6.2

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Problems
10 total
Q1

(i) In △ABC\triangle ABC, DE∥BCDE \parallel BC with DD on ABAB and EE on ACAC. If AB=12 cmAB = 12\ \text{cm}, AC=9 cmAC = 9\ \text{cm} and AD=4 cmAD = 4\ \text{cm}, find ECEC.

(ii) In △ABC\triangle ABC, DE∥BCDE \parallel BC with DD on ABAB and EE on ACAC. If AB=8 cmAB = 8\ \text{cm}, BC=6 cmBC = 6\ \text{cm} and DE=3 cmDE = 3\ \text{cm}, find AD .AD\,.

Pending
Q2

E and F are points on the sides PQ and PR respectively of a △\triangle 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

Pending
Q3

In Fig. 6.18, if LM∥CBLM \parallel CB and LN∥CDLN \parallel CD, prove that AMAN=ABAD=LMLN .\displaystyle \frac{AM}{AN} = \frac{AB}{AD} = \frac{LM}{LN}\,.

Pending
Q4

In Fig. 6.19, DE∥ACDE \parallel AC and DF∥AEDF \parallel AE. Prove that BF×BE=FE×EC .BF \times BE = FE \times EC\,.

Pending
Q5

In Fig. 6.20, DE∥OQDE \parallel OQ and DF∥ORDF \parallel OR. Show that EF∥QR .EF \parallel QR\,.

Pending
Q6

In Fig. 6.21, AA, BB and CC are points on OPOP, OQOQ and OROR respectively such that AB∥PQAB \parallel PQ and AC∥PRAC \parallel PR. Show that BC∥QRBC \parallel QR.

Pending
Q7

Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Pending
Q8

Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Pending
Q9

ABCDABCD is a trapezium in which AB∥DCAB \parallel DC and its diagonals intersect each other at the point OO. Show that AOCO=BODO\dfrac{AO}{CO} = \dfrac{BO}{DO}.

Pending
Q10

The diagonals of a quadrilateral ABCDABCD intersect each other at the point OO such that AOCO=BODO\displaystyle \frac{AO}{CO}=\frac{BO}{DO}. Show that ABCDABCD is a trapezium.

Pending