ANGLE SUM PROPERTY OF A QUADRILATERAL
THEOREM-I : The sum of the four angles of a quadrilateral is 360°.
Given : A quadrilateral ABCD.
To Prove : ∠A + ∠B + ∠C + ∠D = 360°
Construction : Join AC.
Proof :
STATEMENT | REASON |
1. In ΔABC | Sum of the all angles of triangle is equal to 180° |
2. In ΔADC ∠2 + ∠3 + ∠5 = 180° | Sum of the all angles of triangle is equal to 180° |
3. (∠1 +∠2) + (∠3 +∠4) ∠5 +∠6 = 180° + 180° | Adding (1) & (2), we get |
4. ∠A + ∠C + ∠D + ∠B = 360° | |
5. ∠A + ∠B + ∠C +∠D = 360° |
Hence, proved.
Ex.1 Three angles of a quadrilateral measure 56°, 100° and 88°. Find the measure of the fourth angle.
Sol. Let the measure of the fourth angle be x°.
∴ 56° + 100° + 88° + x° = 360° ∴ [Sum of all the angles of quadrileteral is 360°]
⇒ 244+x=360
⇒ x = 360 – 244 = 116
Hence, the measure of the fourth angle is 116°.
Ex.2 The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Sol. Let the four angles of the quadrilateral be 3x, 5x, 9x and 13x . [NCERT]
∴ 3x + 5x + 9x + 13x = 360° ∴ [Sum of all the angles of quadrileteral is 360°]
⇒ 30x = 360°
⇒ x =12°
Hence, the angles of the quadrilateral are 3 × 12° = 36°, 5 × 12° = 60°, 9 × 12° = 108
PARALLELOGRAM:- A quadrilateral in which both pairs of opposite Csides are parallel is called a parallelogram.
In the figure, ABCD is a quadrilateral in which AB ⊥ DC, BC ⊥ AD.
∴ quadrilateral ABCD is a parallelogram.
PROPERTIES OF A PARALLELOGRAM
THEOREM-1 A diagonal of a parallelogram divides it into two congruent triangles.
Given : ABCD is a parallelogram and AC is a diagonal which forms two triangles CAB and ACD.
To Prove : ΔACD ≌ ΔCAB
Proof :
STATEMENT | REASON |
1. ∵ AB ⊥ DC and AD ⊥ BC | ABCD is a parallelogram |
2. (i) ∵ AB ⊥DC and AC is a transversal | Alternate angles Alternate angles |
3. In ΔACD and ΔCAB, |
From (2) Common From (3) By ASA congruence rule |
Hence, proved.
THEOREM-2 In a parallelogram, opposite sides are equal.
Given : ABCD is a parallelogram.
To Prove : AB = CD and BC = DA
Construction : Join AC.
Proof :
STATEMENT | REASON |
1. AB ⊥ DC and AD ⊥ BC | Since ABCD is a parallelogram |
2. In ΔABC and ΔCDA ∠BAC = ∠DCA | Alternate angles Common Alternate angles By ASA congruence rule C.P.C.T. |
Hence, proved.
THEOREM-3 In a parallelogram, opposite angles are equal.
Given : ABCD is a parallelogram.
To Prove : ∠A = ∠C and ∠B = ∠D
Proof :
STATEMENT | REASON |
1. AB ⊥ DC and AD ⊥ BC | Since ABCD is a parallelogram |
2. AB ⊥ DC and AD is a transversal | Sum of consecutive interior angles is 180° |
3. AD⊥BC and DC is a transversal | Sum of consecutive interior angles is 180° |
4. ∠A + ∠D = ∠D + ∠C |
From (2) & (3)
|
5. Similarly, ∠B = ∠D |
Hence proved.
THEOREM-4 : The diagonals of a parallelogram bisect each other.
Given : ABCD is a parallelogram, diagonals AC and BD intersect at O. O
To Prove : OA = OC and OB = OD
Proof :
STATEMENT | REASON |
1. AB ⊥ DC and AD ⊥ BC | ABCD is a parallelogram |
2. In ΔAOB and ΔCOD , | Alternate angles |
3. ΔAOB ΔCOD | By ASA congruence rule |
4. ∴ OA = OC and OB = OD | [C.P.C.T.] |
Hence, proved.
1 videos|228 docs|21 tests
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1. What are the properties of a parallelogram? |
2. What is the difference between a parallelogram and a rectangle? |
3. What is the formula for the area of a parallelogram? |
4. What is the Midpoint Theorem for a parallelogram? |
5. How do you prove that a quadrilateral is a parallelogram? |
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