Quadrilaterals — Practice

Question 1

Three angles of a quadrilateral are 80°, 100°, 110°. Find the fourth.

Accepted Answer

Step 1 — Apply the Quadrilateral Angle Sum Property.
  ∠A + ∠B + ∠C + ∠D = 360°.

Step 2 — Substitute known values.
  80° + 100° + 110° + ∠D = 360°.
  290° + ∠D = 360°.

Step 3 — Solve.
  ∠D = 70°.

✦ Answer: Fourth angle = 70°.

Question 2

Name the quadrilateral whose diagonals are equal AND bisect each other at right angles.

Accepted Answer

Step 1 — Diagonals bisect each other → parallelogram.

Step 2 — Diagonals are equal → among parallelograms, this is a rectangle.

Step 3 — Diagonals also perpendicular → it's also a rhombus.

Step 4 — A figure that is both a rectangle and a rhombus is a SQUARE.

✦ Answer: Square.

Question 3

In parallelogram ABCD, ∠A = 75°. Find ∠C.

Accepted Answer

Step 1 — Opposite angles of a parallelogram are equal.

Step 2 — ∠C is opposite to ∠A.
  So ∠C = ∠A = 75°.

✦ Answer: ∠C = 75°.

Question 4

In parallelogram ABCD, ∠A = 70°. Find ∠B.

Accepted Answer

Step 1 — Adjacent angles of a parallelogram are supplementary.

Step 2 — ∠A and ∠B are adjacent.
  ∠A + ∠B = 180° → 70° + ∠B = 180° → ∠B = 110°.

✦ Answer: ∠B = 110°.

Question 5

In △ABC, D and E are midpoints of AB and AC respectively. If BC = 10 cm, find DE.

Accepted Answer

Step 1 — Apply Mid-Point Theorem.
  DE is parallel to BC and DE = BC/2.

Step 2 — Compute.
  DE = 10/2 = 5 cm.

✦ Answer: DE = 5 cm.

Question 6

A rhombus has diagonals 16 cm and 30 cm. Find each side.

Accepted Answer

Step 1 — Rhombus diagonals bisect each other at right angles.
  So each diagonal is split into halves: 8 cm and 15 cm.

Step 2 — Half-diagonals form a right triangle with the side as hypotenuse.
  side² = 8² + 15² = 64 + 225 = 289.
  side = √289 = 17 cm.

✦ Answer: Each side of the rhombus = 17 cm.

Nice property: the half-diagonals (8, 15, 17) form a Pythagorean triple.

Question 7

In rectangle ABCD, AC and BD are diagonals. If AC = 26 cm and AB = 10 cm, find BC.

Accepted Answer

Step 1 — Diagonal of rectangle = hypotenuse of right triangle with sides AB and BC.
  AC² = AB² + BC².

Step 2 — Substitute.
  26² = 10² + BC²
  676 = 100 + BC²
  BC² = 576 → BC = 24.

✦ Answer: BC = 24 cm.

Note: 10-24-26 is a Pythagorean triple (twice 5-12-13).

Question 8

In quadrilateral ABCD, AB = CD and AB ∥ CD. Show that ABCD is a parallelogram.

Accepted Answer

Step 1 — This is Converse C4: one pair of opposite sides EQUAL AND PARALLEL → parallelogram.

Step 2 — Draw diagonal AC. In △ABC and △CDA:
  AB = CD (given).
  AB ∥ CD ⇒ ∠BAC = ∠DCA (alternate interior).
  AC = AC (common).

Step 3 — By SAS, △ABC ≅ △CDA.

Step 4 — By CPCT, BC = AD AND ∠BCA = ∠DAC.
  ∠BCA = ∠DAC are alternate interior angles for lines AD and BC with transversal AC. So AD ∥ BC.

Step 5 — Both pairs of opposite sides are parallel ⇒ ABCD is a parallelogram. ∎

✦ Answer: ABCD is a parallelogram by the converse criterion.

Question 9

In △ABC, P, Q, R are midpoints of BC, CA, AB respectively. Show that PQRS (where S is... actually a simpler version) — Show that △PQR has each side equal to half of one side of △ABC.

Accepted Answer

Step 1 — Apply Mid-Point Theorem to each pair.
  In △ABC with P, Q midpoints of BC, CA: PQ ∥ AB and PQ = AB/2.
  Similarly QR ∥ BC and QR = BC/2.
  Similarly RP ∥ CA and RP = CA/2.

Step 2 — Conclude.
  Each side of △PQR equals half the parallel side of △ABC.

✦ Answer: Each side of △PQR = half the corresponding side of △ABC.

Corollary: △PQR is similar to △ABC with ratio 1:2 (Class 10 concept). Its perimeter is half of △ABC's; its area is 1/4 of △ABC's.

Question 10

Prove that the diagonals of a rhombus bisect each other at right angles.

Accepted Answer

Step 1 — Setup. Rhombus ABCD with diagonals AC, BD meeting at O.

Step 2 — Since ABCD is a (special) parallelogram, the diagonals bisect each other (parallelogram property). So OA = OC and OB = OD.

Step 3 — Compare △AOB and △AOD.
  AB = AD (rhombus — all sides equal).
  OB = OD (Step 2).
  OA = OA (common).

Step 4 — By SSS, △AOB ≅ △AOD.

Step 5 — By CPCT, ∠AOB = ∠AOD.

Step 6 — ∠AOB + ∠AOD = 180° (linear pair on line BD).
  Combining with Step 5: 2·∠AOB = 180° → ∠AOB = 90°.

Step 7 — Therefore the diagonals meet at right angles AND bisect each other. ∎

✦ Answer: Diagonals of a rhombus bisect each other at right angles.

Question 11

In △PQR, S is the midpoint of PQ. A line through S parallel to QR meets PR at T. Prove that T is the midpoint of PR.

Accepted Answer

Step 1 — Setup. S midpoint of PQ; ST ∥ QR, T on PR.

Step 2 — Apply the Converse of the Mid-Point Theorem.
  Statement: a line drawn through the midpoint of one side, parallel to the second side, bisects the third side.

Step 3 — In our case:
  S is midpoint of side PQ ✓.
  Line ST is parallel to side QR ✓.
  Line ST meets side PR at T.
  By the Converse MPT, T is the midpoint of PR. ∎

✦ Answer: T is the midpoint of PR.

Length bonus: ST = QR / 2 (by Mid-Point Theorem applied to the now-known midpoints S, T).

Question 12

Prove: in any quadrilateral, the midpoints of the four sides form a parallelogram.

Accepted Answer

Step 1 — Setup. Quadrilateral ABCD. Let P, Q, R, S be the midpoints of AB, BC, CD, DA respectively.

Step 2 — Draw diagonal AC. Now △ABC and △ACD share AC.

Step 3 — In △ABC, P and Q are midpoints of AB and BC.
  By Mid-Point Theorem: PQ ∥ AC and PQ = AC/2.

Step 4 — In △ACD, S and R are midpoints of AD and CD.
  By Mid-Point Theorem: SR ∥ AC and SR = AC/2.

Step 5 — Combine.
  PQ and SR are both parallel to AC AND both equal to AC/2.
  So PQ ∥ SR AND PQ = SR.

Step 6 — One pair of opposite sides of PQRS is equal AND parallel → PQRS is a parallelogram (Converse C4). ∎

✦ Answer: PQRS is a parallelogram. (This is Varignon's Theorem.)

This works for ANY quadrilateral — convex, concave, or even self-intersecting (though the resulting 'parallelogram' may be degenerate in special cases).

Practice — Quadrilaterals

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