Q7 experts plz answer

Q7 experts plz answer c 5. The figure, given below, shows a trapezium ABCD. M and N are the mid-points of the non- parallel sides AD and BC respectively. Find : l. In triangle ABC. M is mid-point of AB and a straight line through M and parallel to BC cuts AC at N. Find the lengths of AN and MN, if BC = 7 cm and AC = 5 cm. 2. Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus. 3. D, E and F are the mid-points of the sides AB, BC and CA of an isosceles A ABC in which AB = BC. Plove that A DEF is also isosceles. 4. The following figure shows a trapezium ABCD in which AB // DC. P is the mid-point of AD and PR // AB. Prove that : (ii) PR = — DB 6. 7. 8. 9. (ii) AB, if DC - 20 cm and MN = 27 cm. (iii) DC, if MN = 15 cm and AB = 23 cm. The diagonals of a quadrilateral intersect at right angles. Prove that the figure obtained b' joining the mid-points of the adjacent sides the quadrilateral is a rectangle. L and M are the mid-points of sides AB aryl DC respectively of parallelogram ABCb Prove that segments DL and BM trisect/ diagonal AC. ABCD is a quadrilateral in which AD = Be E, F, G and H are the mid-points of AB, BDJ CD and AC respectively. Prove that EFGH 2 a rhombus. A parallelogram ABCD has P the mid-point of DC and Q a point of AC Such that AC. PQ produced meets BC at R. D (i) MN, if AB 4 Prove that • (i) R is the mid-point of BC, 2 p B = II cm and DC = 8 cm. 150

ABCD is a parallelogram.

∴AB || CD

And hence, AE || FC

Again, AB = CD (Opposite sides of parallelogram ABCD)

AB =CD

AE = FC (E and F are mid-points of side AB and CD)

In quadrilateral AECF, one pair of opposite sides (AE and CF) is parallel and equal to each other. Therefore, AECF is a parallelogram.

⇒ AF || EC (Opposite sides of a parallelogram)

In ΔDQC, F is the mid-point of side DC and FP || CQ (as AF || EC). Therefore, by using the converse of mid-point theorem, it can be said that P is the mid-point of DQ.

⇒ DP = PQ ... (1)

Similarly, in ΔAPB, E is the mid-point of side AB and EQ || AP (as AF || EC). Therefore, by using the converse of mid-point theorem, it can be said that

Q is the mid-point of PB.

⇒ PQ = QB ... (2)

From equations (1) and (2),

DP = PQ = BQ

Hence, the line segments AF and EC trisect the diagonal BD.

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