NCERT Class 9 Solutions: Quadrilaterals (Chapter 8) Exercise 8.2 – Part 2

D and E are midpoints of sides AB and AC respectively. DE is parallel to BC and half its length.

Mid Point Theorem

D and E are midpoints of sides AB and AC respectively. DE is parallel to BC and half its length.

Mid-point Theorem: In a triangle, the line segment that joins the midpoints of the two sides of a triangle is parallel to the third side and is half of it.

Q-3 ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Solution:

Quadrilateral poly1 Quadrilateral poly1: Polygon A, B, C, D Segment a Segment a: Segment [A, B] of Quadrilateral poly1 Segment b Segment b: Segment [B, C] of Quadrilateral poly1 Segment c Segment c: Segment [C, D] of Quadrilateral poly1 Segment d Segment d: Segment [D, A] of Quadrilateral poly1 Segment f Segment f: Segment [P, Q] Segment g Segment g: Segment [Q, R] Segment h Segment h: Segment [R, S] Segment i Segment i: Segment [S, P] Segment j Segment j: Segment [B, D] Segment k Segment k: Segment [A, C] Point A A = (-0.64, 2.94) Point A A = (-0.64, 2.94) Point A A = (-0.64, 2.94) Point B B = (3.34, 2.94) Point B B = (3.34, 2.94) Point B B = (3.34, 2.94) Point C C = (3.34, 0.98) Point C C = (3.34, 0.98) Point C C = (3.34, 0.98) Point D D = (-0.64, 0.98) Point D D = (-0.64, 0.98) Point D D = (-0.64, 0.98) Point P Point P: Point on a Point P Point P: Point on a Point P Point P: Point on a Point Q Point Q: Point on b Point Q Point Q: Point on b Point Q Point Q: Point on b Point R Point R: Point on c Point R Point R: Point on c Point R Point R: Point on c Point S Point S: Point on d Point S Point S: Point on d Point S Point S: Point on d

ABCD Is a Rectangle and PQRS Are Midpoints of the Sides

ABCD is a rectangle PQRS is a rhombus where P, Q, R and S are mid-points of the sides AB, BC, CD, DA respectively

Given,

ABCD is ABCD is a rectangle and PQRS is a quadrilateral formed by joining midpoints of its sides.

In ΔABC

  • P and Q are the mid-points of AB and BC respectively

  • Thus, PQAC and PQ=12AC (Mid-point theorem) …….equation (1)

In ΔADC ,

  • S, R are the mid-points of AD and DC respectively

  • Therefore, SRAC and SR=12AC (Mid-point theorem)……….equation (2)

From equation (1) and (2)

  • PQSR and PQ=SR

Therefore, in quadrilateral PQRS pair of opposite sides is equal and parallel, so, it is a parallelogram.

  • Therefore, PSQR and PS=QR (Opposite sides of parallelogram) ……..equation (3)

Now, In ΔBCD ,

  • Q and R are mid points of side BC and CD respectively.

  • Thus, QRBD and QR=12BD (Mid-point theorem) ……..equation (4)

  • AC=BD (Diagonals of a rectangle are equal)………..equation (5)

From equations (1), (2), (3), (4) and (5),

  • PQ=QR=SR=PS

Since PQRS is a parallelogram, with equal sides. Therefore, PQRS is a rhombus.

Q-4 ABCD is a trapezium in which, ABDC , BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersects BC at F (see Fig.). Show that F is the mid-point of BC.

Segment f Segment f: Segment [D, C] Segment g Segment g: Segment [D, E] Segment g Segment g: Segment [D, E] Segment h Segment h: Segment [E, F] Segment i Segment i: Segment [C, F] Segment j Segment j: Segment [E, A] Segment j Segment j: Segment [E, A] Segment k Segment k: Segment [F, B] Segment l Segment l: Segment [B, A] Segment m Segment m: Segment [D, B] Point D D = (-1.7, 1.7) Point D D = (-1.7, 1.7) Point D D = (-1.7, 1.7) Point C C = (2.26, 1.7) Point C C = (2.26, 1.7) Point C C = (2.26, 1.7) Point E E = (-1.7, 0.64) Point E E = (-1.7, 0.64) Point E E = (-1.7, 0.64) Point F F = (1.72, 0.64) Point F F = (1.72, 0.64) Point F F = (1.72, 0.64) Point A A = (-1.7, -0.34) Point A A = (-1.7, -0.34) Point A A = (-1.7, -0.34) Point B B = (1.22, -0.34) Point B B = (1.22, -0.34) Point B B = (1.22, -0.34) Point G Point G: Intersection point of h, m Point G Point G: Intersection point of h, m Point G Point G: Intersection point of h, m

ABCD Is a Trapezium

ABCD is a trapezium. AB DC, BD is a diagonal and E is the mid-point of AD. E parallel to AB

Solution:

Given,

  • ABCD is a trapezium in which ABDC ,

  • BD is a diagonal and E is the mid-point of AD.

  • A line is drawn through E parallel to AB intersecting BC at F

To prove: F is the mid-point of BC.

Proof,

Now, BD intersected EF at G.

In ΔBAD ,

  • E is the mid-point of AD and also EGAB.

  • Thus, G is the mid-point of BD (Converse of midpoint theorem)

Now, in ΔBDC,

  • G is the mid-point of BD and also GF AB DC.

  • Thus, F is the mid-point of BC (Converse of midpoint theorem)

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