CBSE Class 9

These NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Exercise 12.2

Question 1.
A park in the shape of quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
Solution:
In ABCD,
CD = 5 cm, BC = 12 and ∠C = 90°

∴ By pythagoras theorem we know that
h² = P² + b²
⇒ BD² = CD² + BC²
⇒ BD² = 5² + 12²
⇒ BD² = 169
⇒ BD = √169 = 13 cm.
Area of triangle BCD = \(\frac {1}{2}\) × 12 × 5 = 30 cm.
In ∆ABD,
AD = 8 m, AB = 9 m, and BD = 13 cm
∴ s = \(\frac{8+9+13}{2}\) = 15
∴ By Heron’s formula we know that
Area of triangle

= 35.5 m² (approx)
Now,
Area perpendicular ABCD = Ar ΔBCD + Ar ΔABD
= 30 + 35.5
= 65.5 m² (approx)
Therefore, quadrilateral ABCD occupy 65.5 m² area.

Question 2.
Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
Solution:
In ΔABC,
AB = 3 cm, BC = 4 cm and AC = 5 cm
∴ s = \(\frac{3+4+5}{2}\) = 6
∴ By Heron’s formula we know that
Area of ΔABC

Again in ΔACD,
AC = 5 cm, AD = 5 cm, and CD = 4 cm
∴ s = \(\frac{5+5+4}{2}\) = 7
∴ By Heron’s formula we know that,

= 9.2 cm² (approx)
∴ Area of quadrilateral ABCD = Ar ΔABC + Ar ΔACD
= 6 + 9.2
= 15.2 cm² (approx)

Question 3.
Radha made a picture of an aeroplane with coloured paper .is shown in Fig. 12.15. Find the total area of the paper used.

Solution:
In part I, sides of triangle are 5 cm, 5 cm and 1 cm
∴ s = \(\frac{5+5+1}{2}\) = 5.5
∴ By Heron’s formula

The area of the IInd part which is in the form of the rectangle is
6.5 × 1 = 6.5 cm² (Area of rectangle = l × b)
Area of IIIrd part = 3 × \(\frac{\sqrt{3}}{4}\) cm = 1.3 cm² (approx)
Area of IVth part, which is in the form of triangle is = \(\frac {1}{2}\) × 15 × 6 = 4.5 cm²
Area of Vth part, which is also in Hie form of triangle is = \(\frac {1}{2}\) × 15 × 6 = 4.5 cm²
Therefore, total area of the paper used = 2.5 + 6.5 +1.3 + 4.5 + 4.5 = 19.3 cm² (approx)

Question 4.
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm, and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Solution:
Sides of triangle are 26 cm, 28 cm and 30 cm.
∴ s = \(\frac{26+28+30}{2}\) = 42
∴ By Heron’s formula we know that

According to question,
Area of parallelogram = Area of triangle
Base × altitude = 336
28 × altitude = 336
∴ Altitude = \(\frac{336}{28}\) = 12
∴ Height of the parallelogram stands on the base 28 cm is 12 cm.

Question 5.
A rhombus-shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much rea of grass field will each cow be getting.

Solution:
In ∆ABD
AB = 30 m, AD = 30 m and BD = 48 m.
∴ s = \(\frac{30+30+48}{2}\) = 54
∴ By Heron’s formula we know that

∴ Area of rhombus ABCD = 2 × 432 = 864 m²
Therefore, area of grass getting by each cow = \(\frac{864}{18}\) = 48 m²

Question 6.
An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig. 12.16), each piece measuring 20 cm, 50 cm, and 50 cm. How much cloth of each colour is required for the umbrella?

Solution:
For one triangular piece,
∴ s = \(\frac{20+50+50}{2}\) = 60
∴ By Heron’s formula
Area of one triangular piece

Therefore,
Area of five triangular piece = 5 × 200√6 = 1000√6 cm²
Hence, clothes of each colour are required for the umbrella is 1000√6 cm².

Question 7.
A kite in the shape of a square with a diagonal 32 cm and art isosceles triangle of base S cm and sides 6 cm each is lobe made of three diffident shades as shown in Fig. 12.17. How much paper of each shade has been used in it?

Solution:
We have given that, the shape of a kite is square and length of its diagonal is 32 cm.
Area of 1 triangular region
= \(\frac {1}{2}\) × base × height
= \(\frac {1}{2}\) × 32 × 16
= 256 cm²
(∵ We know that diagonals of square bisects each other at 90°)
Now, Area of IInd triangular region
= \(\frac {1}{2}\) × base × height
= \(\frac {1}{2}\) × 32 × 16
= 256 cm²
Again, we have given that IIIrd part of the given kite is in the shape of an isosceles triangle whose equal sides are 6 cm and the base is 8 cm.
s = \(\frac{a+b+c}{2}\)
= \(\frac{6+6+8}{2}\)
= 10
∴ By Heron’s formula,
Area of IIIrd triangular region

Question 8.
A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm, and 35 cm (see Fig. 12.18). Find the cost of polishing the tiles at the rate of 50 p per cm².

Solution:
The sides of one triangular tile are 9 cm, 28 cm, and 35 cm
s = \(\frac{9+28+35}{2}\) = 36 cm
By Heron’s formula, we know that
Area of one triangular tile

Therefore, area of 16 triangular tiles = 88.18 × 16 = 1410.88 cm²
Hence, the cost of polishing at the rate of 50 p per cm² is 1410.88 × 0.50 = Rs. 705.44 (approx)

Question 9.
A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

Solution:
Draw a line through A and parallel to BC cut CD at E. Again draw AM ⊥ CD.
Now, sides of ∆ADE are 13 m, 15 m and 14 m
s = \(\frac{13+14+15}{2}=\frac{42}{2}\) = 21 m
By Heron’s formula, we know that

But, Area of ∆ADE = \(\frac {1}{2}\) × DE × AM
or, 84= \(\frac {1}{2}\) × 15 × AM
or, AM = \(\frac{84 \times 2}{15}\) = 11.2 m
We know that
Area of trapezium
= \(\frac {1}{2}\) × (25 + 10) × 11.2
= \(\frac {1}{2}\) × 35 × 11.2
= 196 m²

These NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Exercise 12.1

Question 1.
A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using the Herons formula. If its perimeter is 180 cm, what will be the area of the signal board?
Solution:
In first case, length of each sides of an equilateral triangle is a
s = \(\frac{a+b+c}{2}=\frac{3 a}{2}\)
Therefore by Heron’s formula-

or, area of triangle = \(\frac{\sqrt{3} \cdot a^{2}}{4}\)
In case II,
Perimeter of an equilateral triange is 180 cm
∴ One side = \(\frac{180}{3}\) = 60 cm
s = \(\frac{60+60+60}{2}\) = 90 cm.
∴ Area of signal board = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{90(90-60)(90-60)(90-60)}\)
= \(\sqrt{90 \times 30 \times 30 \times 30}\)
= 900√3 cm²
or, Area of signal board = 900√3 cm².

Question 2.
The triangular side walls of a flyover have been used for advertisements. The sides of the wall are 122 m, 22 m, and 120 m (see fig. 12.9). The advertisements yield an earning of Rs. 5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?

Solution:
The sides of the triangular wall are 122 m, 22 m, and 120 m.
∴ s = \(\frac{120+22+120}{2}\) = 132 m.
Therefore, by Heron’s Formula.
Area of triangular wall
= \(\sqrt{132(132-122)(132-122)(132-122)}\)
= \(\sqrt{132 \times 10 \times 110 \times 12}\)
= 1320 m²
∴ Rent = 5000 × 1320 × \(\frac{3}{12}\) = 16,50,000
Therefore, rent of this wall for 3 months is Rs. 16,50,000

Question 3.
There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN” (see Fig. 12.10 ). If the sides of the wall are 15m, 11m, and 6m, find the area painted in colour.

Solution:
The sides of the wall are 15 m, 11 m, and 6 m.
∴ s = \(\frac{15+11+6}{2}\) = 16 m
∴ By Heron’s formula, we know that
Area of the wall

Therefore, the Area of the wall is 20√2 m².

Question 4.
Find the area of the triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Solution:
Let third side of triangle is x cm.
∴ Perimeter = 18 + 10 + x
⇒ 42 = 18 + 10 + x
⇒ x = 14 cm
s = \(\frac{18+10+14}{2}=\frac{42}{2}\) = 21
By Heron’s formula we know that
Area of triangle

Therefore, the area of the triangle = 2√11 cm².

Question 5.
Sides of triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.
Solution:
Let the first side of triangle = 12x
Second side of triangle = 17x
and third side of triangle = 25x
Therefore,
perimeter of triangle = 12x + 17x + 25x
⇒ 540 = 54x
⇒ x = 10
First side of triangle = 12x = 12 × 10 = 120 cm
Second side of triangle = 17x = 17 × 10 = 170 cm
and Third side of triangle = 25x = 25 × 10 = 250 cm
∴ s = \(\frac{120+170+250}{2}=\frac{540}{2}\) = 270
∴ By Heron’s formula we know that
Area of triangle

Therefore, area of triangle = 9000 cm²

Question 6.
An isosceles triangle has a perimeter of 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Solution:
Let the third side of an isosceles triangle is x.
∴ Perimeter of triangle = 12 + 12 + x
or, 30 = 24 + x
or, x = 6x m.
The third side of an isosceles triangle is 6 cm.
∴ s = \(\frac{12+12+6}{2}\) = 15
By Heron’s formula, we know that
Area of triangle

Therefore, the area of an isosceles triangle is 9√15 cm².

These NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 11 Constructions Exercise 11.2

Question 1.
Construct a triangle ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm.
Solution:

Steps of construction:
Step-I: Draw the base BC = 7 cm and at point B make an angle say ∠XBC = 75°.
Step-II: Cut the line segment BD = AB + AC = 13 cm from die ray BX.
Step-III: Join DC and draw perpendicular bisector of DC. Which intersect DB at A.
Step-IV: Join AC.
Step-V: Triangle ABC is the required triangle.

Question 2.
Construct a triangle ABC in which BC = 8 cm, ∠B = 45° and AB – AC = 3.5 cm.
Solution:

Steps of Construction:
Step-I: Draw the base BC = 8 cm and make ∠XBC = 45°.
Step-II: Cut the Sine segment BD = AB – AC = 3.5 cm from ray BX.
Step-III: Join DC and draw the perpendicular bisector of DC.
Step-IV: Let the perpendicular bisector of DC intersect XB at A.
Step-V: Join AC.
Step-VI: Triangle ABC is the required triangle.

Question 3.
Construct a triangle PQR in which QR = 6 m, ∠Q = 60° and PR – PQ = 2 cm.
Solution:

Steps of Construction:
Step-I: Draw the base QR = 6 cm and at point Q make an angle ∠XQR = 60°.
Step-II: Cutline segment QM = PR – PQ = 2 cm from the line XQ extended on the opposite side of line segment QR.
Step-III: Join MC and draw the perpendicular bisector of MR which intersect XQ at P.
Step-IV: Join PR.
Step-V: Triangle PQR is the required triangle.

Question 4.
Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.
Solution:

Steps of Construction:
Step-I: Draw the line segment PQ = 11 cm. (= XY + YZ + ZX)
Step-II: At P construct an angle of 30° and at Q, an angle of 90°.
Step-III: Bisect these angles. Let the bisector of these angles intersect at a point X.
Step-IV: Draw perpendicular bisectors of XP and XQ which intersect PQ at Y and Z respectively.
Step-V: Join XY and XZ.
Step-VI: Triangle XYZ is the required triangle.

Question 5.
Construct a right triangle whose base is 12 cm and the sum of its hypotenuse and other side is 18 cm.
Solution:

Steps of Construction:
Step-I: Draw the base BC = 12 cm and at a point B makes an angle say ∠XBC = 90°.
Step-II: Cut the line segment BD = 18 cm (= AB + AC).
Step-III: Join DC and make a ∠DCY = ∠BDC.
Step-IV: CY intersect BX at A.
Step-V: Join AC.
Step-VI: Triangle ABC is the required triangle.

These NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 11 Constructions Exercise 11.1

Question 1.
Construct an angle of 90° at the initial point of a given ray and justify the construction.
Solution:
Steps of construction:
1. Draw a straight line AB.
2. Taking A as centre, draw an arc, which intersects AB at a point D.

3. Now, taking D as a centre, the same radius AD, intersect at E and F on the succession arc. (here AD = \(\widehat{\text { DE }}\) = \(\widehat{\text { EF }}\))
4. From point E take a radius (which should be greater than \(\frac {1}{2}\) \(\widehat{\text { EF }}\)) then with the same radius from F intersect the arc at point C.
5. Now, join C to A.
6. The required angle ∠ABC = 90°.

Question 2.
Construct an angle of 45° at the initial point of a given ray and justify the construction.
Solution:

Steps of Constructions:
1. Follow the instruction of the previous question upto ∠ABC = 90°.
2. Take an arc from the points H and G each which Intersect at I.
3. Here ∠FBC is half of ∠ABC, FB is the angle bisector.
4. So, the required ∠FBC = 45°

Question 3.
Construct the angles of the following measurements:
(i) 30°
(ii) 22½°
(iii) 15°
Solution:
(i) Steps of Construction:

1. Take a straight line AB.
2. Draw an arc, taking A as centre which intersects AB at C.
3. From C take another arc \(\widehat{\mathrm{CD}}\) such that AB = \(\widehat{\mathrm{CD}}\)
4. From C and D take similar arc which intersects at E.
5. The required ∠FAB = 30°.

(ii) Steps of construction:

1. Follow the instruction of question (2) i.e. ∠FAB = 45°.
2. Take the same arc from points C and K, intersect at G.
3. The required ∠HAB = 22½°

(iii) Step of construction:

1. Follow the instruction of 3:
(i) ∠EAB = 30°.
2. Take a small arc from the points G and C that intersects at F.
3. The required angle ∠FAB = 15°.

Question 4.
Construct the following angles and verify by measuring them by a protractor:
(i) 75°
(ii) 105°
(iii) 135°.
Solution:
(i) Steps of construction:
1. Follow as in question 1. i.e. ∠EAC = 90°.
2. Take the same arc from points D and E which intersect at G.
3. The required ∠KAB = 75°.

(ii) Steps of construction:
1. Follow the instruction as in the previous question upto ∠EAB = 90°.
2. Take the same arc from points D and E intersect at F.
3. Join F to A.
4. The required ∠GAB = 105°.

(iii) Steps of construction:
1. Follow the instruction as ∠EAB = 90°.
2. Take the same arc from the points E and F, Intersect at G.
3. Join G to A.
4. The required ∠HAB = 135°.

Question 5.
Construct an equilateral triangle, given its side, and justify the construction.
Solution:
Steps of construction:

1. Take a straight line XY.
2. By measuring 6 cm on the scale, cut one point C on XY line the reverse the same to get B point. Here, BC = 6 cm
3. From B and C points draw an arc of the same length intersect at A.
4. We find here AB = BC = CA = 6 cm. Thus ABC is an equilateral triangle.

These NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.6

Question 1.
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Solution:
Given: Two circles of centre O and O’ intersect each other at A and B.
To prove that: ∠OAO’ = ∠OBO’
Construction: Join OO’.
Proof: In ∆AOO’ and ∆BOO’
AO = BO (Radii of the same circle)
AO’ = BO’ (Radii of the same circle)
OO’ = OO’ (Common)
So, by S-S-S congruency condition
∆AOO’ ≅ ∆BOO’
∴ ∠OAO’ = ∠OBO’ (By CPCT)

Question 2.
Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
Solution:
Let O be the centre of the given circle and let its radius be r cm.
Draw OM ⊥ AB and ON ⊥ CD.
Since OM ⊥ AB, ON ⊥ CD, and AB || CD.
Therefore point M, O and N are collinear. So MN = 6 cm.
Let OM = x cm, Then ON = 6 – x cm
Join OA and OC, then OA = OC = r.
Since the perpendicular from the centre to a chord of the circle bisect the chord.

∴ AM = MB = \(\frac {5}{2}\) cm
and CN = ND = \(\frac {11}{2}\) cm.
In right triangle OAM,
OA² = OM² + AM²
r² = x² + \(\left(\frac{5}{2}\right)^{2}\) ……(i)
Now, in right triangle OCN
OC² = ON² + CN²
r² = (6 – x)² + \(\left(\frac{11}{2}\right)^{2}\) …….(ii)
From equation (i) and (ii)
\(x^{2}+\left(\frac{5}{2}\right)^{2}=(6-x)^{2}+\left(\frac{11}{2}\right)^{2}\)

Hence, the radius of the circle is \(\frac{5 \sqrt{5}}{2}\) cm.

Question 3.
The length of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance of 4 cm from the centre, what is the distance of the other chord from the centre.
Solution:
Let AB and CD be two parallel chords of a circle with centre O such that AB = 6 cm and CD = 8 cm.
Let the radius of the circle be r cm.
Draw OP ⊥ AB and OQ ⊥ CD.
The length of OP is 4 cm.

In right triangle OAP,
r² = OP² + AP²
⇒ r² = (4)² + (3)² [∵ AP = \(\frac {1}{2}\) AB = 3 cm]
⇒ r² = 16 + 9 = 25
⇒ r = 5 cm ……(i)
Now, in right triangle OQC,
r² = (OQ)² + (CQ)²
⇒ (5)² = (OQ)² + (4)²
[∵ r = 5 cm prove above and CQ = \(\frac {1}{2}\) CD = 4 cm]
⇒ 25 = (OQ)² + 16
⇒ (OQ)² = 25 – 16 = 9
⇒ OQ = 3 cm
Therefore the distance of the chord CD from the centre is 3 cm.

Question 4.
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with a circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Solution:
Given: Vertex of an angle ABC be located outside a circle, and AD = CE.
Let ∠AOC = ∠x, ∠AOD = ∠z and ∠DOE = ∠y.

To prove that: ∠ABC = \(\frac {1}{2}\) (x – y)
Proof: AD = CE (given)
∴ ∠AOD = ∠COE (Angle made by equal chord at the centre are equal)
Therefore,
∠x + ∠y + ∠z + ∠z = 360°
⇒ ∠x + ∠y + 2∠z = 360°
⇒ 2∠z = 360° – ∠x – ∠y
⇒ ∠z = 180 – \(\frac {1}{2}\) ∠x – \(\frac {1}{2}\) ∠y ……(A)
Now, ∠ODB = ∠OAD + ∠DOA
(Exterior angle is equal to sum of opposite interior angles)
∠ODB = ∠OAD + ∠z ……(i)
Again, ∠OAD + ∠z + ∠ODA = 180°
(Sum of all three angles of triangles)
or, ∠OAD + ∠z + ∠OAD = 180°
(Angle opposite to equal sides are equal and OA = OD radii of circle)
or, 2∠OAD = 180° – ∠z
or, ∠AOD = \(\frac {1}{2}\) (180° – ∠z) = 90° – \(\frac {1}{2}\) ∠z
Putting the value of ∠AOD in equation (i)
∴ ∠ODB = 90° – \(\frac {1}{2}\) ∠z + ∠z
(∵ ∠OAD = 90° – \(\frac {1}{2}\) ∠z)
or, ∠ODB = 90° + \(\frac {1}{2}\) ∠z ……..(ii)
Similarly, ∠OEB = 90° + \(\frac {1}{2}\) ∠z ……(iii)
Now, in ODBE
∠ODB + ∠B + ∠OEB + ∠y = 360°
(Sum of all angle of ∆ is 360°)
90 + \(\frac {1}{2}\) ∠z + ∠B + 90 + \(\frac {1}{2}\) ∠z + ∠y = 360°
(Use equation (ii) and (iii))
or, 180 + ∠z + ∠B + ∠y = 360
or, ∠B = 180 – ∠y – ∠z …..(iv)
⇒ ∠ABC = \(\frac {1}{2}\) [(Angle subtended by the chorde DE at the centre) – (Angle subtended by the chord AC at the centre)]
⇒ ∠ABC = \(\frac {1}{2}\) [(Difference of the angles subtended by the chorde DE and AC at the centre)]

Question 5.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Solution:
We have a rhombus ABCD such that its diagonals AC and BD intersect at O.
P, Q, R and S are the mid-points of DC, AB, AD and BC respectively
∴ \(\frac {1}{2}\) AD = \(\frac {1}{2}\) BC
⇒ RA = SB
⇒ RA = OQ ……(ii)
[∴ PQ is drawn parallel to AD and AD = BC]
⇒ \(\frac {1}{2}\) AB = \(\frac {1}{2}\) AD
⇒ AQ = AR ……(iii)
From (i), (ii) and (iii), we have
AQ = QB = OQ
i.e., A circle drawn with Q as centre, will pass through A, B and O.
Thus, the circle passes through the intersection ‘O’ of the diagonals rhombus ABCD.

Question 6.
ABCD is a parallelogram. The circle through A, B, and C intersect CD (produced if necessary) at E. Prove that AE = AD.
Solution:
We have a circle passing through A, B, and C is drawn such that it intersects CD at E.
∴ ∠AEC + ∠B = 180° ……(i)
[Opposite angles of a cyclic quadrilateral are supplementary]
But ABCD is a prallelogram (Given)
∴ ∠D = ∠B ……(ii)
[Opposite angles of a parallelogram are equal]
From (i) and (ii), we have
∠AEC + ∠D = 180° …….(iii)
But ∠AEC + ∠AED = 180° (Linear pair) ……(iv)

From (iii) and (iv),
We have ∠D = ∠AED
i.e., The base angles of ∆ADE are euqal.
∴ AE = AD

Question 7.
AC and BD are chords of a circle which bisects each other. Prove that
(i) AC and BD are diameters
(ii) ABCD is a rectangle.
Solution:
Given: A circle in which two chords AC and BD are such that they bisect each other.
Let their point of intersection be O.
To Prove: (i) AC and BD are diameters.
(ii) ABCD is a rectangle.
Construction: Join AB, BC, CD, and DA.
Proof: (i) In ∆AOB and ∆COD, we have

(∵ ∠ADC = ∠ABC, opposite angles of || gm ABCD)
From equation (i) and (ii)
∠AED + ∠ABC = ∠ADE + ∠ABC
or, ∠AED = ∠ADE
Thus, in ∆AED, we have
∠AED = ∠ADE
∴ AD = AE. (Side opposite to equal angles are equal)

Question 8.
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angle of the triangle DEF are 90° – \(\frac {1}{2}\) ∠A, 90° – \(\frac {1}{2}\) ∠B and 90° – \(\frac {1}{2}\) ∠C.
Solution:
Given: In ∆ABC, AD, BE and CF is the angle bisector of ∠A, ∠B, and ∠C respectively.
Where D, E, and F lie on the circumcircle of ∆ABC.

To prove that:
(i) ∠FDE = 90° – \(\frac {1}{2}\) ∠A
(ii) ∠DEF = 90° – \(\frac {1}{2}\) ∠B
(iii) ∠EFD = 90° – \(\frac {1}{2}\) ∠C
Proof:
(i) ∠ADF = ∠ACF ……(i) (Angles in the same segments of a circle are equal)
Again, ∠ADE = ∠ABE ……(ii) (Angles in the same segment of atcircle are equal)
Adding (i) and (ii)
∠APF + ∠ACF = ∠ACF + ∠ABE
or, ∠FDE = \(\frac {1}{2}\) ∠C + \(\frac {1}{2}\) ∠B ……(iii)
(∵ CF and BE are the bisector of ∠B and ∠C respectively)
Now, In ∆ABC
∠A + ∠B + ∠C = 180° (Sum of all angles of a ∆ is 180°)
or, \(\frac {1}{2}\) (∠A + ∠B + ∠C) = 90°
or, \(\frac {1}{2}\) ∠B + \(\frac {1}{2}\) ∠C = 90 – \(\frac {1}{2}\) ∠A ……(iv)
From equation (iii) and (iv)
∠FDE = 90° – \(\frac {1}{2}\) ∠A
Similarly we can prove
(ii) ∠DEF = 90 – \(\frac {1}{2}\) ∠B
(iii) ∠EFD = 90 – \(\frac {1}{2}\) ∠C.

Question 9.
Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q, lie on the two circles. Prove that BP = BQ.
Solution:
Let C(0, r) and C(O’, r) be two congruent circles.

Since AB is a common chord of two congruent circles. Therefore,
arc ACB = arc ADB
∴ ∠BPA = ∠BQA
Thus, in ∆BPQ, we have ∠BPA = ∠BQA
∴ BP = BC (Sides opposite to equal angles of a triangle are equal)

Question 10.
In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
Solution:
Given: ABC is a triangle in which the bisector of ∠A intersects the circumcircle of ∆ABC at D.
To prove that: Perpendicular bisector of side BC intersects the angle bisector of ∠A at D.

Construction: Join BD and CD.
Proof: ∠BCD = ∠BAD ……(i)
(Angles in the same segment of a circle are equal)
Again, ∠DBC = ∠DAC …….(ii)
(Angles in the same segment of a circle are equal)
But, ∠BAD = ∠DAC …..(iii)
(∵ AD is the bisector of ∠A)
From equation (i), (ii) and (iii)
∠BCD = ∠DBC
or, BD = CD (Side opposite to equal angles of a ∆ are equal)
So, D must be lying on the perpendicular bisector of BC.
Therefore, the perpendicular bisector of side BC intersects the angle bisector of ∠A).

These NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.5

Question 1.
In Fig. 10.36, A, B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D In a point on the circle other than the arc ABC, find ∠ADC.

Solution:
We have given,
∠BOC = 30° and ∠AOB = 60°
∠AQC = 30° + 60°
∴ ∠AOC = 90° …….(i)
Again, ∠AOC and ∠ADC subtended by same arc AC.
We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∠AOC = 2∠ADC
⇒ 90° = 2∠ADC (From (i))
⇒ ∠ADC = 45°

Question 2.
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Solution:
In circle whose centre is O. AB is a chord equal to its radius.
∴ OAB is a an equilateral triangle.
We know that each angle of an equilateral triangle is 60°.
∴ AOB = 60°
Now, ∠AOB = 2∠AQB
(The angle subtended by an arc at the centre is double the angle subtanded by it at any point on the remaining part of the circle)
⇒ 60° = 2∠AQB
⇒ ∠AQB = 30°
Again, QAPB is a cyclic quadrilateral.
∴ ∠AQB + ∠APB = 180° (Sum of opposite angles, of a cyclic quadrilateral is 180°)
⇒ 30 + ∠APB = 180°
⇒ ∠APB = 180° – 30° = 150°
Therefore, angle subtended by the chord at a point on the minor arc is 150° and at a point on major arc is 30°.

Question 3.
In Fig. 10.37, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

Solution:
We have given ∠PQR = 100°.
Since, angle subtended by an arc at the centre is double tire angle subtended by it at any point on the remaining part of the circle.
∴ ∠1 = 2∠PQR
⇒ ∠1 = 2 × 100°
⇒ ∠1 = 200°
Again, ∠1 + ∠2 = 360°
⇒ 200° + ∠2 = 360° (∵ ∠1 = 200°)
⇒ ∠2 = 160°
Now, In ∆POR
OP = OR (Radii of circle)
∴ ∠OPR = ∠ORP (Angle opposite to equal sides of ∆ are equal)
Therefore,
∠OPR + ∠ORP + ∠2 = 180 (Angle sum property of triangle)
⇒ ∠OPR + ∠OPR + 160 = 180
⇒ 2∠OPR = 20
⇒ ∠OPR = 10°

Question 4.
In Fig. 10.38, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.

Solution:
In ∆ABC,
∠ABC = 69° and ∠ACB = 31°
∴ ∠ABC + ∠ACB + ∠BAC = 180° (Angle sum property of ∆)
⇒ 69°+ 31° + ∠BAC = 180°
⇒ ∠BAC = 80°
We know that angles in the same segment of a circle are equal.
∴ ∠BAC = ∠BDC
∠BDC = 80° (∵ ∠BAC = 80°)

Question 5.
In Fig. 10.39, 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.

Solution:
We have, ∠BEC = 130° and ∠ECD = 20°
Again, ∠BEC + ∠CED = 180° (linear pair)
⇒ 130° + ∠CED = 180°
⇒ ∠CED = 50°
Now, In triangle CDE
∠CED + ∠ECD + ∠CDE = 180° (Sum of all angles of a ∆)
⇒ 50° + 20° + ∠CDE = 180
⇒ ∠CDE = 110°
We know that angles on the same segments of a circle are equal.
∴ ∠CDB = ∠BAC
or, ∠BAC = 110° (∵ ∠CDB = 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.
Solution:
We know that angles on the same segment of a circle are equal.

∴ ∠DAC = ∠DBC
or, ∠DAC = 70° (∵ ∠DBC = 70°)
Again, ABCD is a cyclic quadrilateral.
∴ ∠DAB + ∠BCD = 180°
(Sum of opposite angles of a cyclic quadrilateral is 180°)
100° + ∠BCD = 180° (∵∠DAB = ∠DAC + ∠BAC)
⇒ ∠BCD = 80°
Now, in triangle ABC,
AB = BC (Given)
∴ ∠BAC = ∠BCA (Angle opposite to equal sides are equal)
or, ∠BCA = 30°
Again, ∠BCD = ∠BCA + ∠ACD
⇒ 80° = 30° + ∠ACD
⇒ ∠ACD = 50°
∴ ∠ECD = 50°

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

Solution:
Given: ABCD is a cyclic quadrilateral in which diagonal AC and BD are diameter of circle.
To prove that: ABCD is a rectangle.
Proof: In ABCD,
AC = BD (Diameter of circle are equal)
Again, ∠ABC = 90° (Angle in a semicircle is a right angle)
Similarly, ∠BAD = ∠ADC = ∠BCD = 90°
Now, we know that, if a quadrilateral has both diagonals are equal and each angle is 90° then the quadrilateral is a rectangle.
∴ ABCD is a rectangle.

Question 8.
If the non parallel sides of a trapezium are equal, prove that it is cyclic.
Solution:
Given: A trapezium ABCD in which AB || CD and AD = BC.
To prove that: ABCD is a cyclic trapezium.
Construction: Draw DE ⊥ AB and CF ⊥ AB.
Proof: In order to prove that ABCD is a cyclic trapezium, it is sufficient to show that ∠B + ∠D = 180°.

In triangles DEA and CFB, we have
AD = BC (Given)
∠DEA = ∠CFB (Each equal to 90°)
DE = CF (Distance between two parallel lines is always same)
So, by S- A-S congruency condition
∆DEA ≅ ∆CFB
∴ ∠A = ∠B and ∠ADE = ∠BCF (By CPCT)
⇒ 90° + ∠ADE = 90° + ∠BCF (Add 90° both side)
⇒ ∠EDC + ∠ADE = ∠FCD + ∠BCF (∵ ∠EDC = 90° and ∠CD = 90°)
⇒ ∠ADC = ∠BCD
⇒ ∠D = ∠C
Thus, ∠A = ∠B and ∠C = ∠D
∴ ∠A + ∠B + ∠C + ∠D = 360° (Sum of the angles of a equal is 360°)
⇒ 2∠B + 2∠D = 360°
⇒ ∠B + ∠D = 180°
Hence ABCD is a cyclic quadrilateral.

Question 9.
Two circles intersect at two points B and C. Through B, two line segment ABD and PBQ are drawn to intersect the circle at A, D and P, Q respectively (see Fig. 10.40). Prove that ∠ACP = ∠QCD.

Solution:
We know that angles in the same segment of a circle are equal.
∠D = ∠Q and ∠P = ∠A
Again, In ∆PQC,
∠P + ∠Q + ∠PCQ = 180° ……(i)
(Sum of all three angles of the triangle is 180°)
Now, In ∆ADC,
∠A + ∠D + ∠ACD = 180° …….(ii)
(Sum of all three angles of a triangle is 180°)
From equation (i) and (ii)
∠P + ∠Q + ∠PCQ = ∠A + ∠D + ∠ACD
⇒ ∠P + ∠Q + ∠PCQ = ∠P + ∠Q + ∠ACD (∵ ∠A = ∠P and ∠D = ∠Q prove above)
∴ ∠PCQ = ∠ACD
⇒ ∠PCQ – ∠PCD = ∠ACD – ∠PCD (Subtract ∠PCD both side)
⇒ ∠QCD = ∠ACP
⇒ ∠ACP = ∠QCD

Question 10.
If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Solution:
Join AD
Since angle in a semicircle is right angle.
Therefore,

∠ADB = 90° and ∠ADC = 90°
⇒ ∠ADB + ∠ADC = 90° + 90°
⇒ ∠ADB + ∠ADC = 180°
⇒ ∠BDC = 180°
Therefore, BDC is a straight line or, D must be lie on BC.

Question 11.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
Solution:
Given: ABC and DBC are two right angle triangles on same base AC.

To prove that: ∠CAD = ∠CBD
Construction: Take AC as a diameter and draw a circle and join BD.
Proof: We know that angle in a semicircle is a right angle.
Therefore, vertex B of ∆ABC and vertex D of ∆ADC must be lie on the circumference of the circle.
Again we know that angles in the same segment of a circle are equal.
∴ ∠CAD = ∠CBD.

Question 12.
Prove that a cyclic parallelogram is a rectangle.
Solution:
Let ABCD be acyclic parallelogram. In order to prove that it is a rectangle, it is sufficient to show that one of the angles of parallelogram ABCD is a right angle.

Now ABCD is a parallelogram.
∠B = ∠D ……(i) [∵ opposite angle of a || gm are equal]
Also, ABCD is a cyclic quadrilateral
∴ ∠B + ∠D = 180° ……(ii) (Sum of opposite angles of cyclic quadrilateral)
⇒ ∠B + ∠B = 180° [∵ ∠B = ∠D from (i)]
⇒ 2∠B = 180°
⇒ ∠B = 90°
Hence ABCD is a rectangle.