Access Answers for SAT Maths Chapter 8 – Quadrilaterals
Exercise 8.1
1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Solution:
Let the common ratio between the angles be = x.
We know that the sum of the interior angles of the quadrilateral = 360°
Now,
3x+5x+9x+13x = 360°
⇒ 30x = 360°
⇒ x = 12°
, Angles of the quadrilateral are:
3x = 3×12° = 36°
5x = 5×12° = 60°
9x = 9×12° = 108°
13x = 13×12° = 156°
2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Solution:
Given that,
AC = BD
To show that, ABCD is a rectangle if the diagonals of a parallelogram are equal
To show ABCD is a rectangle we have to prove that one of its interior angles is right angled.
Proof,
In ΔABC and ΔBAD,
AB = BA (Common)
,BC = AD (Opposite sides of a parallelogram are equal)
AC = BD (Given)
Therefore, ΔABC ≅ ΔBAD [SSS congruency]
∠A = ∠B [Corresponding parts of Congruent Triangles]
also,
∠A+∠B = 180° (Sum of the angles on the same side of the transversal)
⇒ 2∠A = 180°
⇒ ∠A = 90° = ∠B
Therefore, ABCD is a rectangle.
Hence Proved.
3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a
rhombus.
Solution:
Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.
Given that,
OA = OC
OB = OD
and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90°
To show that,
if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
i.e., we have to prove that ABCD is parallelogram and AB = BC = CD = AD
Proof,
In ΔAOB and ΔCOB,
OA = OC (Given)
∠AOB = ∠COB (Opposite sides of a parallelogram are equal)
, OB = OB (Common)
Therefore, ΔAOB ≅ ΔCOB [SAS congruency]
Thus, AB = BC [CPCT]
Similarly we can prove,
BC = CD
CD = AD
AD = AB
, AB = BC = CD = AD
Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.
, ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle.
Hence Proved.
4. Show that the diagonals of a square are equal and bisect each other at right angles.
Solution:
Let ABCD be a square and its diagonals AC and BD intersect each other at O.
To show that,
AC = BD
AO = OC
and ∠AOB = 90°
Proof,
In ΔABC and ΔBAD,
AB = BA (Common)
∠ABC = ∠BAD = 90°
BC = AD (Given)
ΔABC ≅ ΔBAD [SAS congruency]
Thus,
Exercise 8.1
1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Solution:
Let the common ratio between the angles be = x.
We know that the sum of the interior angles of the quadrilateral = 360°
Now,
3x+5x+9x+13x = 360°
⇒ 30x = 360°
⇒ x = 12°
, Angles of the quadrilateral are:
3x = 3×12° = 36°
5x = 5×12° = 60°
9x = 9×12° = 108°
13x = 13×12° = 156°
2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Solution:
Given that,
AC = BD
To show that, ABCD is a rectangle if the diagonals of a parallelogram are equal
To show ABCD is a rectangle we have to prove that one of its interior angles is right angled.
Proof,
In ΔABC and ΔBAD,
AB = BA (Common)
,BC = AD (Opposite sides of a parallelogram are equal)
AC = BD (Given)
Therefore, ΔABC ≅ ΔBAD [SSS congruency]
∠A = ∠B [Corresponding parts of Congruent Triangles]
also,
∠A+∠B = 180° (Sum of the angles on the same side of the transversal)
⇒ 2∠A = 180°
⇒ ∠A = 90° = ∠B
Therefore, ABCD is a rectangle.
Hence Proved.
3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a
rhombus.
Solution:
Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.
Given that,
OA = OC
OB = OD
and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90°
To show that,
if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
i.e., we have to prove that ABCD is parallelogram and AB = BC = CD = AD
Proof,
In ΔAOB and ΔCOB,
OA = OC (Given)
∠AOB = ∠COB (Opposite sides of a parallelogram are equal)
, OB = OB (Common)
Therefore, ΔAOB ≅ ΔCOB [SAS congruency]
Thus, AB = BC [CPCT]
Similarly we can prove,
BC = CD
CD = AD
AD = AB
, AB = BC = CD = AD
Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.
, ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle.
Hence Proved.
4. Show that the diagonals of a square are equal and bisect each other at right angles.
Solution:
Let ABCD be a square and its diagonals AC and BD intersect each other at O.
To show that,
AC = BD
AO = OC
and ∠AOB = 90°
Proof,
In ΔABC and ΔBAD,
AB = BA (Common)
∠ABC = ∠BAD = 90°
BC = AD (Given)
ΔABC ≅ ΔBAD [SAS congruency]
Thus,
