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Quadrilaterals | Quadrilaterals Exercise 10B

Question 13
  1. In each of the figures given below, ABCD is a rectangle. Find the values of x and y in each case.

(i)

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(i) From the figure we know that the diagonals are equal and bisect at point O.

Consider △ AOB

We get AO = OB

We know that the base angles are equal

∠ OAB = ∠ OBA = 35

By using the sum property of triangle

∠ AOB + ∠ OAB + ∠ OBA = 180

By substituting the values we get

∠ AOB + 35+ 35 = 180

On further calculation

∠ AOB = 180 - 35 - 35

By subtraction

∠ AOB = 180 - 70

∠ AOB = 110

From the figure we know that the vertically opposite angles are equal

∠ DOC = ∠ AOB = y = 110

In △ ABC

We know that ∠ ABC = 90

Consider △ OBC

We know that

∠ OBC = x = ∠ ABC - ∠ OBA

By substituting the values

∠ OBC = 90 - 35

By subtraction

∠ OBC = 55

Therefore, x = 55 and y = 110.

(ii) From the figure we know that the diagonals of a rectangle are equal and bisect each other.

Consider △ AOB

We get

OA = OB

We know that the base angles are equal

∠ OAB = ∠ OBA

By using the sum property of triangle

∠ AOB + ∠ OAB + ∠ OBA = 180

By substituting the values

110 + ∠ OAB + ∠ OBA = 180

We know that ∠ OAB = ∠ OBA

So we get

2 ∠ OAB = 180 - 110

By subtraction

2 ∠ OAB = 70

By division

∠ OAB = 35

We know that AB || CD and AC is a transversal

From the figure we know that ∠ DCA and ∠ CAB are alternate angles

∠ DCA = ∠ CAB = y = 35

Consider △ ABC

We know that

∠ ACB + ∠ CAB = 90

So we get

∠ ACB = 90 - ∠ CAB

By substituting the values in above equation

∠ ACB = 90 - 35

By subtraction

∠ ACB = x = 55

Therefore, x = 55 and y = 35.

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Question 13

  1. In each of the figures given below, ABCD is a rectangle. Find the values of x and y in each case.

(i)

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

(i) From the figure we know that the diagonals are equal and bisect at point O.

Consider △ AOB

We get AO = OB

We know that the base angles are equal

∠ OAB = ∠ OBA = 35

By using the sum property of triangle

∠ AOB + ∠ OAB + ∠ OBA = 180

By substituting the values we get

∠ AOB + 35+ 35 = 180

On further calculation

∠ AOB = 180 - 35 - 35

By subtraction

∠ AOB = 180 - 70

∠ AOB = 110

From the figure we know that the vertically opposite angles are equal

∠ DOC = ∠ AOB = y = 110

In △ ABC

We know that ∠ ABC = 90

Consider △ OBC

We know that

∠ OBC = x = ∠ ABC - ∠ OBA

By substituting the values

∠ OBC = 90 - 35

By subtraction

∠ OBC = 55

Therefore, x = 55 and y = 110.

(ii) From the figure we know that the diagonals of a rectangle are equal and bisect each other.

Consider △ AOB

We get

OA = OB

We know that the base angles are equal

∠ OAB = ∠ OBA

By using the sum property of triangle

∠ AOB + ∠ OAB + ∠ OBA = 180

By substituting the values

110 + ∠ OAB + ∠ OBA = 180

We know that ∠ OAB = ∠ OBA

So we get

2 ∠ OAB = 180 - 110

By subtraction

2 ∠ OAB = 70

By division

∠ OAB = 35

We know that AB || CD and AC is a transversal

From the figure we know that ∠ DCA and ∠ CAB are alternate angles

∠ DCA = ∠ CAB = y = 35

Consider △ ABC

We know that

∠ ACB + ∠ CAB = 90

So we get

∠ ACB = 90 - ∠ CAB

By substituting the values in above equation

∠ ACB = 90 - 35

By subtraction

∠ ACB = x = 55

Therefore, x = 55 and y = 35.

Our top 5% students will be awarded a special scholarship to Lido.

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