(i) Given: ABCD is a rectangle ∠1 = ∠2 and ∠3 = ∠4. To prove ABCD is a square. Solution: ∠1 = ∠4 ...(1) [∵ Alternate angle] ∠3 = ∠4 ...(2) [∵ Given] Hence, ∠1 = ∠3 ...(3) [∵ From (1) and (2)] In ΔADC, ∠1 = ∠3 [∵ From (3)] DC = AD [∵ In a triangle, side opposite to equal angle are equal] A rectangle,Read more
(i) Given: ABCD is a rectangle ∠1 = ∠2 and ∠3 = ∠4.
To prove ABCD is a square.
Solution: ∠1 = ∠4 …(1) [∵ Alternate angle]
∠3 = ∠4 …(2) [∵ Given]
Hence, ∠1 = ∠3 …(3) [∵ From (1) and (2)]
In ΔADC,
∠1 = ∠3 [∵ From (3)]
DC = AD [∵ In a triangle, side opposite to equal angle are equal]
A rectangle, whose adjacent sides are equal, is a square.
Hence, ABCD is a square.
(ii) To prove: Diagonal BD bisects angle B as well as angle D.
Solution: ∠5 = ∠8 …(4) [ Alternate angle]
In ΔADB,
AB = AD [∵ ABCD is a square]
∠7 = ∠5 …(5) [∵ Angles opposite to equal sides are equal]
Hence, ∠7 = ∠8 …(6) [∵ From (4) and (5)]
and ∠7 = ∠6 …(7) [∵ Alternate angles]
Hence, ∠5 = ∠6 …(8) [∵ From (5) and (7)]
Hence, from (6) and (8), diagonal BD Bisects angle B as well as D.
In ΔABC, P is mid-point to AB [∵ Given] Q is mid-point of BC [∵ Given] Hence, PQ ∥ AC and PQ = (1/2)AC ... (1) [∵ Mid-point Theorem] Similarly, in ΔACD, S is mid-point of AD [∵ Given] R is mid-point of CD [∵ Given] Hence, SR ∥ AC and SR = (1/2) AC ...(2) [∵ Mid-Point Theorem] From (1) and (2), we haRead more
In ΔABC,
P is mid-point to AB [∵ Given]
Q is mid-point of BC [∵ Given]
Hence, PQ ∥ AC and PQ = (1/2)AC … (1) [∵ Mid-point Theorem]
Similarly, in ΔACD,
S is mid-point of AD [∵ Given]
R is mid-point of CD [∵ Given]
Hence, SR ∥ AC and SR = (1/2) AC …(2) [∵ Mid-Point Theorem]
From (1) and (2), we have
PQ ∥ SR …(3) [∵ PQ ∥ AC and SR ∥ AC]
and PQ = SR …(4) [∵ SR = (1/2)AC and PQ = (1/2)AC]
Hence, PQRS is a parallelogram.
Similarly, in ΔBCD,
Q is mid-point of BC [∵ Given]
R is mid-point of CD [∵ Given]
Hence, QR ∥ BD [∵ Mid-point Theorem]
⇒ QN ∥ LM …(5)
and LQ ∥ MN …(6) [∵ PQ ∥ AC]
From (5) and (6), we have
LMNQ is a parallelogram.
Hence, ∠LMN = ∠LQN [∵ Opposite angles of a parallelogram]
But, ∠LMN = 90° [∵ Digonals of a rhombus are perpendicular to each other]
Hence, ∠LQN = 90°
A parallelogram whose one angle is a rectangle. Hence, PQRS is a rectangle.
ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:
(i) Given: ABCD is a rectangle ∠1 = ∠2 and ∠3 = ∠4. To prove ABCD is a square. Solution: ∠1 = ∠4 ...(1) [∵ Alternate angle] ∠3 = ∠4 ...(2) [∵ Given] Hence, ∠1 = ∠3 ...(3) [∵ From (1) and (2)] In ΔADC, ∠1 = ∠3 [∵ From (3)] DC = AD [∵ In a triangle, side opposite to equal angle are equal] A rectangle,Read more
(i) Given: ABCD is a rectangle ∠1 = ∠2 and ∠3 = ∠4.
See lessTo prove ABCD is a square.
Solution: ∠1 = ∠4 …(1) [∵ Alternate angle]
∠3 = ∠4 …(2) [∵ Given]
Hence, ∠1 = ∠3 …(3) [∵ From (1) and (2)]
In ΔADC,
∠1 = ∠3 [∵ From (3)]
DC = AD [∵ In a triangle, side opposite to equal angle are equal]
A rectangle, whose adjacent sides are equal, is a square.
Hence, ABCD is a square.
(ii) To prove: Diagonal BD bisects angle B as well as angle D.
Solution: ∠5 = ∠8 …(4) [ Alternate angle]
In ΔADB,
AB = AD [∵ ABCD is a square]
∠7 = ∠5 …(5) [∵ Angles opposite to equal sides are equal]
Hence, ∠7 = ∠8 …(6) [∵ From (4) and (5)]
and ∠7 = ∠6 …(7) [∵ Alternate angles]
Hence, ∠5 = ∠6 …(8) [∵ From (5) and (7)]
Hence, from (6) and (8), diagonal BD Bisects angle B as well as D.
ABCD is a rhombus and P, Q, R and S are ©wthe mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.
In ΔABC, P is mid-point to AB [∵ Given] Q is mid-point of BC [∵ Given] Hence, PQ ∥ AC and PQ = (1/2)AC ... (1) [∵ Mid-point Theorem] Similarly, in ΔACD, S is mid-point of AD [∵ Given] R is mid-point of CD [∵ Given] Hence, SR ∥ AC and SR = (1/2) AC ...(2) [∵ Mid-Point Theorem] From (1) and (2), we haRead more
In ΔABC,
See lessP is mid-point to AB [∵ Given]
Q is mid-point of BC [∵ Given]
Hence, PQ ∥ AC and PQ = (1/2)AC … (1) [∵ Mid-point Theorem]
Similarly, in ΔACD,
S is mid-point of AD [∵ Given]
R is mid-point of CD [∵ Given]
Hence, SR ∥ AC and SR = (1/2) AC …(2) [∵ Mid-Point Theorem]
From (1) and (2), we have
PQ ∥ SR …(3) [∵ PQ ∥ AC and SR ∥ AC]
and PQ = SR …(4) [∵ SR = (1/2)AC and PQ = (1/2)AC]
Hence, PQRS is a parallelogram.
Similarly, in ΔBCD,
Q is mid-point of BC [∵ Given]
R is mid-point of CD [∵ Given]
Hence, QR ∥ BD [∵ Mid-point Theorem]
⇒ QN ∥ LM …(5)
and LQ ∥ MN …(6) [∵ PQ ∥ AC]
From (5) and (6), we have
LMNQ is a parallelogram.
Hence, ∠LMN = ∠LQN [∵ Opposite angles of a parallelogram]
But, ∠LMN = 90° [∵ Digonals of a rhombus are perpendicular to each other]
Hence, ∠LQN = 90°
A parallelogram whose one angle is a rectangle. Hence, PQRS is a rectangle.