Circle

Exercise 10.5 Part 2

Question 5: In the given figure; A. B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.

circle exercise 10.5 question 5 figure

Answer:In ΔEDC;
∠EDC + ∠ECD = 130°
Because external angle of a triangle is equal to sum of two opposite angles
Or, ∠EDC = 130° - 20° = 110°
Angles made on one side of a chord are equal
So, ∠BAC = ∠BDC = 110°


Question 6: ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further if AB = BC, find ∠ECD.

Answer: Given ABCD is a cyclic quadrilateral in which AB = BC, and ∠DBC = 70° and ∠BAC = 30°

circle exercise 10.5 answer 6 figure

∠DAC = ∠DBC = 70°
Because angles made one side of a chord are equal.
Now, ∠DAB = ∠DAC + ∠BAC
Or, ∠DAB = 70° + 30° = 100°
Opposite angles of cyclic quadrilateral are supplementary
Hence, ∠BCD = 180° - ∠DAB
Or, ∠BCD - 180° - 100° = 80°

In ΔABC we have AB = BC
So, ∠BAC = ∠BCA = 30°
So, ∠ECD = ∠BCD - ∠BCA
Or, ∠ECD = 80° - 30° = 50°


Question 7: If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Answer: Given circle with centre O in which diagonals of cyclic quadrilateral ABCD intersect at O. This means AC = BD = diameter
To Prove: ABCD is a rectangle

circle exercise 10.5 answer 7 figure

In ΔAOB and ΔOBC;
AO = OC
BO = BO
All are radii
Hence, ΔAOB ≈ ΔOBC
So, ∠AOB = ∠BOC

As these angles make linear pair
So, ∠AOB + ∠BOC = 180°
Or, 2∠AOB = 180°
Or, ∠AOB = 90°

Now, in ΔAOB; as AO = OB
So, ∠OAB = ∠OBA
Or, ∠OAB + ∠OBA = 90°
Or, 2∠OAB = 90°
Or, ∠OAB = 45°

Similarly, it can be proved that ∠OBC = ∠OCB = ∠OCD = ∠ODC = 45°
This means that ∠OBA + ∠OBC = 90°
Similarly, it can be proven that all angles of quadrilateral ABCD are right angles
Hence, ABCD is a rectangle is proved

Question 8: If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Answer: Given trapezium ABCD in which AB||DC and AD = BC

circle exercise 10.5 answer 8 figure

Let us name the angles as 1, 2, 3, 4, 5, 6, 7 and 8; as shown in the figure.
In ΔADC and ΔBCD;
AD = BC (given)
DC = DC (common side)
Hence, ΔADC ≈ ΔBCD
So, ∠1 = ∠4 .............(1)
And ∠7 = ∠6 ……………(2)

In ΔDAB and ΔCBA
AD = BC (Given)
AB = AB (common side
Hence, ΔDAB ≈ ΔCBA
So, ∠8 = ∠5 …………..(3)
And ∠2 = ∠3 ……………(4)

Now, ∠1 + ∠4 + ∠7 + ∠6 + ∠8 + ∠5 + ∠2 + ∠3 = 360° (angle sum of quadrilateral)
From equations (1), (2), (3) and (4)
Or, ∠1 + ∠1 + ∠6 + ∠6 + ∠5 + ∠5 + ∠2 + ∠2 = 360°
Or, 2(∠1 + ∠2 + ∠6 + ∠5) = 360°
Or, ∠1 + ∠2 + ∠6 + ∠5 = 180°
Or, ∠DAB + ∠BCD = 180°

Similarly, following can be proven
∠ABC + ∠CDA = 180°
Here, angles of opposite vertices are supplementary
Hence, ABCD is a cyclic quadrilateral proved.



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