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

NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1.

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

Ex 12.1 Class 9 Maths Question 1.
A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side a. Find the area of the signal board, using Heron’s formula.If its perimeter is 180 cm, what will be the area of the signal board?
Solution:
Let each side of the equilateral triangle be a.
Semi-perimeter of the triangle,

Ex 12.1 Class 9 Maths Question 2.
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see figure). The advertisements yield an earning of ₹5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?

Solution:
Let the sides of the triangular will be
a = 122m, b = 12cm, c = 22m
Semi-perimeter, s = a+b+c2
(122+120+224)m = 2642 m = 132m
The area of the triangular side wall

Rent for 1 year (i.e. 12 months) per m2 = Rs. 5000
∴ Rent for 3 months per m2 = Rs. 5000 x 312
= Rent for 3 months for 1320 m2 = Rs. 5000 x 312 x 1320 = Rs. 16,50,000. Ex 12.1 Class 9 Maths Question 3.
There is a slide in a park. One of its side Company hired one of its walls for 3 months.walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN” (see figure). If the sides of the wall are 15 m, 11 m and 6m, find the area painted in colour.

Thus, the required area painted in colour
= 20√2 m2

Ex 12.1 Class 9 Maths Question 4.
Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Solution:
Let the sides of the triangle be a =18 cm, b = 10 cm and c = x cm
Since, perimeter of the triangle = 42 cm
∴ 18cm + 10 cm + xcm = 42
x = [42 – (18 + 10)cm = 14cm
Now, semi-permimeter, s = 422cm = 21 cm
NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula Ex 12.1 Q4
Thus, the required area of the triangle = 2111−−√ cm2 Ex 12.1 Class 9 Maths Question 5.
Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.
Solution:
Let the sides of the triangle be
a = 12x cm, b = 17x cm, c = 25x cm
Perimeter of the triangle = 540 cm
Now, 12x + 17x + 25x = 540
⇒ 54x = 54 ⇒ x = 10
∴ a = (12 x10)cm = 120cm,
b = (17 x 10) cm = 170 cm
and c = (25 x 10)cm = 250 cm
Now, semi-perimeter, s = 5402cm = 270 cm

Ex 12.1 Class 9 Maths Question 6.
An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Solution:
Let the sides of an isosceles triangle be
a = 12cm, b = 12cm,c = x cm
Since, perimeter of the triangle = 30 cm
∴ 12cm + 12cm + x cm = 30 cm
⇒ x = (30 – 24) = 6
Now, semi-perimeter, s = 302cm =15 cm

Thus, the required area of the triangle = 9√15 cm2

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

Ex 12.2 Class 9 Maths Question 1.
A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9m,BC = 12m,CD = 5m and AD = 8 m.
How much area does it occupy?
Solution:
Given, a quadrilateral ABCD with ZC = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m.
Let us join B and D, such that ABCD is a right angled triangle.

Now, to find the area of ∆ABD, we need the length of BD.
In right-angled ∆BCD, by Pythagoras theorem
BD2 = 502 + CD2
⇒ BD2 = 122 + 52
⇒ BD2 = 144 + 25 = 169
⇒ BD = 13 m
Now, for ∆ABD, we have
a = AB = 9 m, b = AD = 8 m, c = BD = 13 m

∴ Area of quadrilateral ABCD = area of ∆BCD + area of ∆ABD = 30 m2 + 35.5 m2
= 65.5 m2 (approx.) Ex 12.2 Class 9 Maths 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:
Given a quadrilateral ABCD with AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

For ∆ABC, a = AB = 3 cm, b = BC = 4 cm and c = AC = 5 cm

Now, area of quadrilateral ABCD = area of ∆ABC + area of ∆ACD
= 6 cm2 + 9.2 cm2 = 15.2 cm2 (approx.) Ex 12.2 Class 9 Maths Question 3.
Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used.

Solution:
For surface I:
It is an isosceles triangle whose sides are a = 5 cm, b = 5 cm, c = 1 cm

= (0.75 x 3.3) cm2
= 2.475 cm2 (approx.) For surface II:
It is a rectangle with length 6.5 cm and breadth 1 cm.
∴ Area of surface II = Length x Breadth
= (6.5 x 1) cm2 = 6.5 cm2

For surface III:
It is a trapezium whose parallel sides are 1 cm and 2 cm as shown in the figure given below:

For surface IV and V:
Surface V is a right-angled triangle with base 6cm arid height 1.5 cm.
Also, area of surface IV = area of surface V
= 12 x base x height
= (12 x 6 x 15) cm2 = 4.5 cm2
Thus, the total area of the paper used = (area of surface I) + (area of surface II) + (area of surface III) + (area of surface IV) + (area of surface V) = [2.475 + 6.5 + 1.3 + 4.5 + 4.5] cm2
= 19.275 cm2
= 19.3 cm2 (approx.) Ex 12.2 Class 9 Maths 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:
For the given triangle, we have a = 28 cm, b = 30 cm, c = 26 cm

Area of the given parallelogram = Area of the given triangle
∴ Area of the parallelogram = 336 cm2
⇒ base x height = 336
⇒ 28 x h = 336, where ‘h’ be the height of the parallelogram.
⇒ h = 33628 = 12
Thus, the required height of the parallelogram = 12 cm Ex 12.2 Class 9 Maths 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 area of grass field will each cow be getting?
Solution:
Here, each side of the rhombus = 30 m.
Let ABCD be the given rhombus and the diagonal, BD = 48 m

Sides ∆ABC are a = AB = 30m, b = AD = 30m, c = BD = 48m

Since, a diagonal divides the rhombus into two congruent triangles.
∴ Area of triangle II = 432 m2
Now, total area of the rhombus = Area of triangle I + Area of triangle II
= 432 m2 + 432 m2= 864 m2
Area of grass for 18 cows to graze = 864 m2
⇒ Area of grass for 1 cow to graze = 86418 m2
= 48 m2 Ex 12.2 Class 9 Maths Question 6.
An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

Solution:
Let the sides of each triangular piece be
a = 20 cm, b = 50 cm, c = 50 cm

Ex 12.2 Class 9 Maths Question 7.
A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. How much paper of each shade has been used in it?

Solution:
Each shade of paper is divided into 3 triangles i.e., I, II, III

8 cm
For triangle I:
ABCD is a square [Given]
∵ Diagonals of a square are equal and bisect each other.
∴ AC = BD = 32 cm
Height of AABD = OA = (12 x 32 )cm
= 16 cm
Area of triangle I = (12 x 32 x 16 ) cm2
= 256cm2

For triangle II:
Since, diagonal of a square divides it into two congruent triangles.
So, area of triangle II = area of triangle I
∴ Area of triangle II = 256 cm2 For triangle III:
The sides are given as a = 8 cm, b = 6 cm and c = 6 cm

Thus, the area of different shades are:
Area of shade I = 256 cm2
Area of shade II = 256 cm2
and area of shade III = 17.92 cm2 Ex 12.2 Class 9 Maths 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 figure). Find the cost of polishing the tiles at the rate of 50 paise per cm .

Solution:
Let the sides of the triangle be a = 9 cm, b = 28 cm, c = 35 cm

Total area of all the 16 triangles = (16 x 88.2) cm2 = 1411.2 cm2 (approx.)
Cost of polishing the tiles = Rs. 0.5 per cm2
∴ Cost of polishing all the tiles = Rs. (0.5 x 1411.2) = Rs. 705.60 (approx.) Ex 12.2 Class 9 Maths 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:
Let the given field is in the form of a trapezium ABCD such that parallel sides are AB = 10 m and DC = 25 m
Non-parallel sides are AD = 13 m and BC = 14 m.
We draw BE || AD, such that BE = 13 m.

The given field is divided into two shapes (i) ∆BCE, (ii) parallelogram ABED For ∆BCE:
Sides of the triangle are a = 13 m, b = 14 m, c = 15 m

(ii) For parallelogram ABED:
Let the height of the ∆BCE corresponding to the side EC be h m.
Area of a triangle = 12 x base x height
∴ 12 x 15 x h = 84
⇒ (10 + 82×215 = 565
Now, area of a parallelogram = base x height
= (10 x 565) = (2 x 56) m2 = 112 m2
So, area of the field
= area of ∆BCE + area of parallelogram ABED
= 84 m2 + 112 m2 = 196 m2

NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula (हीरोन सूत्र) (Hindi Medium) Ex 12.1

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1.

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1

Ex 6.1 Class 9 Maths Question 1
In figure, lines AB and CD intersect at 0. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Solution:
Since AB is a straight line,
∴ ∠AOC + ∠COE + ∠EOB = 180°
or (∠AOC + ∠BOE) + ∠COE = 180° or 70° + ∠COE = 180° [ ∵∠AOC + ∠BOE = 70° (Given)]
or ∠COE = 180° – 70° = 110°
∴ Reflex ∠COE = 360° – 110° = 250°
Also, AB and CD intersect at O.
∴∠COA = ∠BOD [Vertically opposite angles]
But ∠BOD = 40° [Given]
∴ ∠COA = 40°
Also, ∠AOC + ∠BOE = 70°
∴ 40° + ∠BOE = 70° or ∠BOE = 70° -40° = 30°
Thus, ∠BOE = 30° and reflex ∠COE = 250°. Ex 6.1 Class 9 Maths Question 2.
In figure, lines XY and MN intersect at 0. If ∠POY = 90° , and a : b = 2 : 3. find c.

Solution:
Since XOY is a straight line.
∴ b+a+∠POY= 180°
But ∠POY = 90° [Given]
∴ b + a = 180° – 90° = 90° …(i)
Also a : b = 2 : 3 ⇒ b = 3a2 …(ii)
Now from (i) and (ii), we get
3a2 + A = 90°
⇒ 5a2 = 90°
⇒ a = 90∘5×2=36∘ = 36°
From (ii), we get
b = 32 x 36° = 54°
Since XY and MN interstect at O,
∴ c = [a + ∠POY] [Vertically opposite angles]
or c = 36° + 90° = 126°
Thus, the required measure of c = 126°. Ex 6.1 Class 9 Maths Question 3.
In figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Solution:
ST is a straight line.
∴ ∠PQR + ∠PQS = 180° …(1) [Linear pair]
Similarly, ∠PRT + ∠PRQ = 180° …(2) [Linear Pair]
From (1) and (2), we have
∠PQS + ∠PQR = ∠PRT + ∠PRQ
But ∠PQR = ∠PRQ [Given]
∴ ∠PQS = ∠PRT Ex 6.1 Class 9 Maths Question 4.
In figure, if x + y = w + ⇒, then prove that AOB is a line.

Solution:
Sum of all the angles at a point = 360°
∴ x + y + ⇒ + w = 360° or, (x + y) + (⇒ + w) = 360°
But (x + y) = (⇒ + w) [Given]
∴ (x + y) + (x + y) = 360° or,
2(x + y) = 360°
or, (x + y) = 360∘2 = 180°
∴ AOB is a straight line. Ex 6.1 Class 9 Maths Question 5.
In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that

Solution:
rara POQ is a straight line. [Given]
∴ ∠POS + ∠ROS + ∠ROQ = 180°
But OR ⊥ PQ
∴ ∠ROQ = 90°
⇒ ∠POS + ∠ROS + 90° = 180°
⇒ ∠POS + ∠ROS = 90°
⇒ ∠ROS = 90° – ∠POS … (1)
Now, we have ∠ROS + ∠ROQ = ∠QOS
⇒ ∠ROS + 90° = ∠QOS
⇒ ∠ROS = ∠QOS – 90° ……(2)
Adding (1) and (2), we have
2 ∠ROS = (∠QOS – ∠POS)
∴ ∠ROS = 12(∠QOS−∠POS) Ex 6.1 Class 9 Maths Question 6.
It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.
Solution:
XYP is a straight line.

∴ ∠XYZ + ∠ZYQ + ∠QYP = 180°
⇒ 64° + ∠ZYQ + ∠QYP = 180°
[∵ ∠XYZ = 64° (given)]
⇒ 64° + 2∠QYP = 180°
[YQ bisects ∠ZYP so, ∠QYP = ∠ZYQ]
⇒ 2∠QYP = 180° – 64° = 116°
⇒ ∠QYP = 116∘2 = 58°
∴ Reflex ∠QYP = 360° – 58° = 302°
Since ∠XYQ = ∠XYZ + ∠ZYQ
⇒ ∠XYQ = 64° + ∠QYP [∵∠XYZ = 64°(Given) and ∠ZYQ = ∠QYP]
⇒ ∠XYQ = 64° + 58° = 122° [∠QYP = 58°]
Thus, ∠XYQ = 122° and reflex ∠QYP = 302°.

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (रेखाएँ और कोण) (Hindi Medium) Ex 6.1

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2

Ex 6.2 Class 9 Maths Question 1.
In figure, find the values of x and y and then show that AB || CD.

Solution:
In the figure, we have CD and PQ intersect at F.

∴ y = 130° …(1)
[Vertically opposite angles]
Again, PQ is a straight line and EA stands on it.
∠AEP + ∠AEQ = 180° [Linear pair]
or 50° + x = 180°
⇒ x = 180° – 50° = 130° …(2)
From (1) and (2), x = y
As they are pair of alternate interior angles.
∴ AB || CD Ex 6.2 Class 9 Maths Question 2.
In figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Solution:
AB || CD, and CD || EF [Given]
∴ AB || EF
∴ x = z [Alternate interior angles] ….(1)
Again, AB || CD
⇒ x + y = 180° [Co-interior angles]
⇒ z + y = 180° … (2) [By (1)]
But y : z = 3 : 7
z = 73 y = 73(180°- z) [By (2)]
⇒ 10z = 7 x 180°
⇒ z = 7 x 180° /10 = 126°
From (1) and (3), we have
x = 126°. Ex 6.2 Class 9 Maths Question 3.
In figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

Solution:
AB || CD and GE is a transversal.
∴ ∠AGE = ∠GED [Alternate interior angles]
But ∠GED = 126° [Given]
∴∠AGE = 126°
Also, ∠GEF + ∠FED = ∠GED
or ∠GEF + 90° = 126° [∵ EF ⊥ CD (given)]
x = z [Alternate interior angles]… (1) Again, AB || CD
⇒ x + y = 180° [Co-interior angles]
∠GEF = 126° -90° = 36°
Now, AB || CD and GE is a transversal.
∴ ∠FGE + ∠GED = 180° [Co-interior angles]
or ∠FGE + 126° = 180°
or ∠FGE = 180° – 126° = 54°
Thus, ∠AGE = 126°, ∠GEF=36° and ∠FGE = 54°. Ex 6.2 Class 9 Maths Question 4.
In figure, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠QRS.

Solution:
Draw a line EF parallel to ST through R.

Since PQ || ST [Given]
and EF || ST [Construction]
∴ PQ || EF and QR is a transversal
⇒ ∠PQR = ∠QRF [Alternate interior angles] But ∠PQR = 110° [Given]
∴∠QRF = ∠QRS + ∠SRF = 110° …(1)
Again ST || EF and RS is a transversal
∴ ∠RST + ∠SRF = 180° [Co-interior angles] or 130° + ∠SRF = 180°
⇒ ∠SRF = 180° – 130° = 50°
Now, from (1), we have ∠QRS + 50° = 110°
⇒ ∠QRS = 110° – 50° = 60°
Thus, ∠QRS = 60°. Ex 6.2 Class 9 Maths Question 5.
In figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

Solution:
We have AB || CD and PQ is a transversal.
∴ ∠APQ = ∠PQR
[Alternate interior angles]
⇒ 50° = x [ ∵ ∠APQ = 50° (given)]
Again, AB || CD and PR is a transversal.
∴ ∠APR = ∠PRD [Alternate interior angles]
⇒ ∠APR = 127° [ ∵ ∠PRD = 127° (given)]
⇒ ∠APQ + ∠QPR = 127°
⇒ 50° + y = 127° [ ∵ ∠APQ = 50° (given)]
⇒ y = 127°- 50° = 77°
Thus, x = 50° and y = 77°. Ex 6.2 Class 9 Maths Question 6.
In figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.

Solution:
Draw ray BL ⊥PQ and CM ⊥ RS

∵ PQ || RS ⇒ BL || CM
[∵ BL || PQ and CM || RS]
Now, BL || CM and BC is a transversal.
∴ ∠LBC = ∠MCB …(1) [Alternate interior angles]
Since, angle of incidence = Angle of reflection
∠ABL = ∠LBC and ∠MCB = ∠MCD
⇒ ∠ABL = ∠MCD …(2) [By (1)]
Adding (1) and (2), we get
∠LBC + ∠ABL = ∠MCB + ∠MCD
⇒ ∠ABC = ∠BCD
i. e., a pair of alternate interior angles are equal.
∴ AB || CD.

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3

Ex 6.3 Class 9 Maths Question 1.
In figure, sides QP and RQ of ∆PQR are produced to points S and T, respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Solution:
We have, ∠TQP + ∠PQR = 180°
[Linear pair]
⇒ 110° + ∠PQR = 180°
⇒ ∠PQR = 180° – 110° = 70°
Since, the side QP of ∆PQR is produced to S.
⇒ ∠PQR + ∠PRQ = 135°
[Exterior angle property of a triangle]
⇒ 70° + ∠PRQ = 135° [∠PQR = 70°]
⇒ ∠PRQ = 135° – 70° ⇒ ∠PRQ = 65° Ex 6.3 Class 9 Maths Question 2.
In figure, ∠X = 62°, ∠XYZ = 54°, if YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OZY and ∠YOZ.

Solution:
In ∆XYZ, we have ∠XYZ + ∠YZX + ∠ZXY = 180°
[Angle sum property of a triangle]
But ∠XYZ = 54° and ∠ZXY = 62°
∴ 54° + ∠YZX + 62° = 180°
⇒ ∠YZX = 180° – 54° – 62° = 64°
YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively.
∴ ∠OYZ = 12∠XYZ = 12(54°) = 27°
and ∠OZY = 12∠YZX = 12(64°) = 32°
Now, in ∆OYZ, we have
∠YOZ + ∠OYZ + ∠OZY = 180°
[Angle sum property of a triangle]
⇒ ∠YOZ + 27° + 32° = 180°
⇒ ∠YOZ = 180° -27° – 32° = 121°
Thus, ∠OZY = 32° and ∠YOZ = 121° Ex 6.3 Class 9 Maths Question 3.
In figure, if AB || DE, ∠BAC = 35° and ∠CDE = 539 , find ∠DCE.

Solution:
AB || DE and AE is a transversal.
So, ∠BAC = ∠AED
[Alternate interior angles]
and ∠BAC = 35° [Given]
∴ ∠AED = 35°
Now, in ∆CDE, we have ∠CDE + ∠DEC + ∠DCE = 180°
{Angle sum property of a triangle]
∴ 53° + 35° + ∠DCE =180°
[∵ ∠DEC = ∠AED = 35° and∠CDE = 53° (Given)]
⇒ ∠DCE = 180° – 53° – 35° = 92°
Thus, ∠DCE = 92° Ex 6.3 Class 9 Maths Question 4.
In figure, if lines PQ and RS intersect at point T, such that ∠ PRT = 40°, ∠ RPT = 95° and ∠TSQ = 75°, find ∠ SQT.

Solution:
In ∆PRT, we have ∠P + ∠R + ∠PTR = 180°
[Angle sum property of a triangle]
⇒ 95° + 40° + ∠PTR =180°
[ ∵ ∠P = 95°, ∠R = 40° (given)]
⇒ ∠PTR = 180° – 95° – 40° = 45°
But PQ and RS intersect at T.
∴ ∠PTR = ∠QTS
[Vertically opposite angles]
∴ ∠QTS = 45° [ ∵ ∠PTR = 45°]
Now, in ∆ TQS, we have ∠TSQ + ∠STQ + ∠SQT = 180°
[Angle sum property of a triangle]
∴ 75° + 45° + ∠SQT = 180° [ ∵ ∠TSQ = 75° and ∠STQ = 45°]
⇒ ∠SQT = 180° – 75° – 45° = 60°
Thus, ∠SQT = 60° Ex 6.3 Class 9 Maths Question 5.
In figure, if PQ ⊥ PS, PQ||SR, ∠SQR = 2S° and ∠QRT = 65°, then find the values of x and y

Solution:
In ∆ QRS, the side SR is produced to T.
∴ ∠QRT = ∠RQS + ∠RSQ
[Exterior angle property of a triangle]
But ∠RQS = 28° and ∠QRT = 65°
So, 28° + ∠RSQ = 65°
⇒ ∠RSQ = 65° – 28° = 37°
Since, PQ || SR and QS is a transversal.
∴ ∠PQS = ∠RSQ = 37°
[Alternate interior angles]
⇒ x = 37°
Again, PQ ⊥ PS ⇒ AP = 90°
Now, in ∆PQS,
we have ∠P + ∠PQS + ∠PSQ = 180°
[Angle sum property of a triangle]
⇒ 90° + 37° + y = 180°
⇒ y = 180° – 90° – 37° = 53°
Thus, x = 37° and y = 53° Ex 6.3 Class 9 Maths Question 6.
In figure, the side QR of ∆PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that

Solution:
In ∆PQR, side QR is produced to S, so by exterior angle property,
∠PRS = ∠P + ∠PQR
⇒ 12∠PRS = 12∠P + 12∠PQR
⇒ ∠TRS = 12∠P + ∠TQR …(1)
[∵ QT and RT are bisectors of ∠PQR and ∠PRS respectively.]
Now, in ∆QRT, we have
∠TRS = ∠TQR + ∠T …(2)
[Exterior angle property of a triangle]
From (1) and (2),
we have ∠TQR + 12∠P = ∠TQR + ∠T
⇒ 12∠P = ∠T
⇒ 12∠QPR = ∠QTR or ∠QTR = 12∠QPR

Indian Coast Guard rescues 12 crew of MV Kanchan in distress off Umargam, Gujarat

  • MV Kanchan stranded because of no power amid inclement weather.
  • ICG’s MV Hermeez rescues all the 12 crew in a swift operation.
  • More ICG vessels deployed for assistance.

Indian Coast Guard rescued all the 12 crew of Motor Vessel (MV) Kanchan stranded off Umargam, Gujarat on July 21, 2021. Maritime Rescue Coordination Centre (MRCC) Mumbai had received information from DG Communication center, Mumbai in the afternoon of July 21, 2021 that MV Kanchan is stranded due to contamination in fuel thereby rendering the engine non-operational and no electrical power onboard, amid inclement weather. Later in evening, the master of the vessel intimated that MV Kanchan, carrying steel coils as cargo, had dropped anchor and was tilting towards Starboard (right) side.

The MRCC Mumbai instantly activated the International Safety Net (ISN) and MV Hermeez was immediately diverted towards the distressed vessel. Braving rough seas, MV Hermeez safely evacuated all the 12 crew of MV Kanchan in a swift night operation.

Emergency Towing Vessel (ETV) Water Lily has also been deployed by DG Shipping, Mumbai for assisting the stranded vessel. In addition, two tugs have been deployed by the vessel owners to render assistance to the vessel.https://platform.twitter.com/embed/Tweet.html?dnt=false&embedId=twitter-widget-0&features=eyJ0ZndfZXhwZXJpbWVudHNfY29va2llX2V4cGlyYXRpb24iOnsiYnVja2V0IjoxMjA5NjAwLCJ2ZXJzaW9uIjpudWxsfSwidGZ3X2hvcml6b25fdHdlZXRfZW1iZWRfOTU1NSI6eyJidWNrZXQiOiJodGUiLCJ2ZXJzaW9uIjpudWxsfX0%3D&frame=false&hideCard=false&hideThread=false&id=1418026148978515972&lang=en&origin=https%3A%2F%2Fwww.pib.gov.in%2FPressReleasePage.aspx%3FPRID%3D1737631&sessionId=2691f6638f3c000931bde891ca176e2ad2fb583c&theme=light&widgetsVersion=82e1070%3A1619632193066&width=550px

#Savinglives In a swiftly coordinated operation by MRCC #Mumbai, MV Hermeez, braving rough seas, rescued 12 crew of distressed vessel MV Kanchan stranded off Umargam, #Maharashtra on 21 Jul night.Well done Hermeez. @DefenceMinIndia @shipmin_india pic.twitter.com/TiZaofyO9i— Indian Coast Guard (@IndiaCoastGuard) July 22, 2021

India vs Sri Lanka: Shikhar Dhawan, Rahul Dravid, Bhuvneshwar Kumar enjoy dinner ‘with amazing company’

Bhuvneshwar Kumar had played a big role with the bat in pulling India from the jaws of defeat in the second ODI.

HIGHLIGHTS

  • Bhuvneshwar played a big role with the bat in pulling India out of the jaws of defeat in the second ODI
  • Thanks to Suryakumar Yadav’s quickfire 53 from 44 balls, Chahar and Bhuvneshwar did not feel the scoreboard pressure
  • The third ODI will be played on Friday and India will look to complete a series whitewash

Shikhar Dhawan, India’s captain for their limited overs tour of Sri Lanka, posted a photo on his Instagram page of him enjoying a dine-out with vice-captain Bhuvneshwar Kumar and head coach Rahul Dravid a day after India took an unassailable 2-0 lead in the ODI series in dramatic fashion.

Vice President of India addresses the World Universities Summit yesterday

Universities should become thought leaders in finding solutions to global challenges such as climate change and poverty – Vice President

MERU to open up new opportunities for India’s youth by promoting multi-disciplinary research – Education and Skill Development Minister Shri Dharmendra Pradhan

Study in India -Stay in India program will make India a global destination in education – Shri Dharmendra Pradhan

Synergizing education with skill development will open new avenues of socio-economic empowerment – Shri Dharmendra Pradhan

Vice President of India Shri M. Venkaiah Naidu addressed the World Universities Summit organized by O.P. Jindal Global University as a Chief Guest. Union Education and Skill Development Minister Shri Dharmendra Pradhan also addressed the summit. The theme of the summit was “Universities of the Future: Building Institutional Resilience, Social Responsibility and Community Impact”.

Vice President asked universities to become thought leaders in finding solutions to global challenges such as climate change, poverty and pollution. He also wanted the universities to discuss various socio-economic and political issues facing the world and come up with ideas that can be implemented by the governments as per their needs and suitability.

Referring to the benefits of learning in the mother tongue, the Vice President said it enhances one’s grasping and comprehension levels. “To understand a subject in another language, one has to first learn and master that language, which needs a lot of effort. However, this is not the case while learning in one’s mother tongue,” he added.

Highlighting our country’s rich linguistic and cultural heritage, the Vice President said that India is home to hundreds of languages and thousands of dialects. He said, “Our linguistic diversity is one of the cornerstones of our rich cultural heritage.” Emphasizing on the significance of mother language, Shri Naidu said, “Our mother language or our native language is very special to us, as we share an umbilical cord relationship with it.”

Addressing the participants, Union Education and Skill Development Minister Shri Dharmendra Pradhan emphasized on the Government’s commitment to transform India’s education sector bringing it at par with global standards, encouraging research and innovation and on developing well-rounded responsible citizens, who are also global citizens- Vishwa Manav.

Shri Pradhan said that the New National Education Policy -2020 has heralded a new imagination for the Indian higher education system. It outlines the vision of the Prime Minister Shri Narendra Modi to make an Aatmanirbhar Bharat. Quality, equity, accessibility and affordability are the four pillars of the new education policy on which a new India will emerge, he added.

Minister Pradhan stated that with the vision of Study In India—Stay In India’, India will move towards becoming a global destination in education. Shri Pradhan highlighted the efforts made by the Government to make education holistic, innovative, linguistically diverse, and multi-disciplinary. No student should suffer due to language limitations or regional linguistic constraints, he added.

He said that the multidisciplinary education and research university (MERU) will open up new opportunities for India’s youth. MERU will promote inter-disciplinary research and make India a global hub of Research and Development, he further added.

Minister Pradhan stressed that synergizing education with skill development will open new avenues of socio-economic empowerment. The NEP will facilitate integration of education with skills and enable India to reap the demographic dividend.

Education and Skill Development Minister Shri Pradhan said that Covid-19 pandemic necessitated the adoption of online learning and use of digital technologies to ensure that learning continues. This mode is going to stay giving way to hybrid methods of learning and knowledge dissemination. Our future planning, therefore, needs to fill a digital divide, he said.

He also congratulated the organizers for event and wished them success in all their endeavors.

Professor (Dr.) D.P. Singh, Chairman, University Grants Commission, Mr. Naveen Jindal, Chancellor, O.P. Jindal Global University, Professor (Dr.) C. Raj Kumar, Founding Vice-Chancellor; Professor Dabiru Sridhar Patnaik, Registrar and other distinguished university leaders from India and overseas, faculty also attended the event.

DRDO successfully flight-tests indigenously developed MPATGM for minimum range

  • Low weight, fire and forget Man Portable Anti-tank Guided Missile
  •   Miniaturised Infrared Imaging seeker
  •   Major boost to Army and AatmaNirbhar Bharat
  •   Raksha Mantri congratulates DRDO 

In a major boost towards AatmaNirbhar Bharat and strengthening of Indian Army, Defence Research and Development Organisation (DRDO) successfully flight-tested indigenously developed low weight, fire and forget Man Portable Antitank Guided Missile (MPATGM) on July 21, 2021. The missile was launched from a man portable launcher integrated with thermal site and the target was mimicking a tank. The missile hit the target in direct attack mode and destroyed it with precision. The test has validated the minimum range successfully. All the mission objectives were met. The missile has already been successfully flight tested for the maximum range. 

The missile is incorporated with state-of-the-art Miniaturized Infrared Imaging Seeker along with advanced avionics. The test brings the development of indigenous third generation man portable Anti-Tank Guided Missile close to completion. 

Raksha Mantri Shri Rajnath Singh has congratulated DRDO and the Industry for the successful test. Secretary Department of Defence R&D and Chairman DRDO Dr G Satheesh Reddy congratulated the team for the successful test.

DRDO successfully flight-tests surface-to-air missile Akash-NG

  • New Generation surface-to-air Missile
  • High manoeuvrability to neutralise aerial threats
  • Boost to Air Defence capabilities of Indian Air Force
  • Raksha Mantri congratulates DRDO 

Defence Research & Development Organisation (DRDO) successfully flight-tested the New Generation Akash Missile (Akash-NG), a surface-to-air Missile from Integrated Test Range (ITR) off the coast of Odisha on July 21, 2021. The flight trial was conducted at around 12:45 PM from a land-based platform with all weapon system elements such as Multifunction Radar, Command, Control & Communication System and launcher participating in deployment configuration. 

The missile system has been developed by Defence Research & Development Laboratory (DRDL), Hyderabad in collaboration with other DRDO laboratories. The launch was witnessed by the representatives of Indian Air Force. In order to capture flight data, ITR deployed a number of Range stations like, Electro Optical Tracking System, Radar and Telemetry. The flawless performance of the entire weapon system has been confirmed by complete flight data captured by these systems. During the test, the missile demonstrated high manoeuvrability required for neutralising fast and agile aerial threats. 

Once deployed, the Akash-NG weapon system will prove to be a force multiplier for the air defence capability of the Indian Air Force. Production agencies Bharat Electronics Limited (BEL) and Bharat Dynamics Limited (BDL) also participated in the trials. 

Raksha Mantri Shri Rajnath Singh has congratulated DRDO, BDL, BEL, Indian Air Force and the Industry for the successful test. Secretary Department of Defence R&D and Chairman DRDO applauded the efforts of the team and said the missile will strengthen the Indian Air Force.

Japanese PM Yoshihide Suga says world should see safe Olympics staged

Tens of thousands of athletes, officials, games staff and media are arriving in Japan amid a local state of emergency and widespread opposition from the general public.

The world needs to see that Japan can stage a safe Olympics, the country’s prime minister told sports officials Tuesday ahead of the Tokyo Games.

Tens of thousands of athletes, officials, games staff and media are arriving in Japan amid a local state of emergency and widespread opposition from the general public.

Events start Wednesday — in softball and women’s soccer — two days ahead of the formal opening ceremony of an Olympics already postponed a year because of the coronavirus pandemic.

“The world is faced with great difficulties,” Japanese Prime Minister Yoshihide Suga told International Olympic Committee members in a closed-door meeting at a five-star hotel in Tokyo, adding “we can bring success to the delivery of the Games.”

“Such fact has to be communicated from Japan to the rest of the world,” Suga said through an interpreter. “We will protect the health and security of the Japanese public.”

He acknowledged Japan’s path through the pandemic toward the Olympics had gone “sometimes backward at times.”

“But vaccination has started and after a long tunnel an exit is now in our sight,” Suga said.

The prime minister’s office said Monday more than 21% of Japan’s 126 million population has been inoculated.

Health experts in Japan have questioned allowing so many international visitors for the games, which end on Aug. 8. There will be no local or foreign fans at events. The Paralympics will follow in late August.

Praising vaccine manufacturers for working on a dedicated Olympic rollout, IOC President Thomas Bach singled out Pfizer BioNTech for “a truly essential contribution.”

This cooperation meant “85% of Olympic Village residents and 100% of IOC members present here have been either vaccinated or are immune” to COVID-19, Bach said.

About 75 of the 101 IOC members were in the room for their first in-person meeting since January 2020. Their previous two meetings, including to re-elect Bach in March, were held remotely.

The IOC declined to say if any members who are not vaccinated had been asked to stay away. One member missing the meeting, Ryu Seung-min of South Korea, tested positive for COVID-19 after arriving on a flight Saturday.

Bach has been met with anti-Olympic chants from protesters on visits in Japan since arriving two weeks ago, including at a state welcome party with Suga on Sunday.

The IOC leader praised his hosts Tuesday, saying “billions of people around the world will follow and appreciate the Olympic Games.”

“They will admire the Japanese people for what they achieved,” Bach said, insisting the games will send a message of peace, solidarity and resilience.

Canceling the Olympics was never an option, Bach said, because “the IOC never abandons the athletes.”

Staging the games will also secure more than $3 billion in revenue from broadcasters worldwide. It helps fund the Switzerland-based IOC, which shares hundreds of millions of dollars among the 206 national teams and also with governing bodies of Olympic sports.

Bach said the IOC is contributing $1.7 billion to Tokyo organizers of the Olympics and the Paralympics.

IOC decisions taken Tuesday, rubber-stamping proposals sent from the Bach-chaired executive board:

— The sport of ski mountaineering was added to the program for the 2026 Winter Games in Milan and Cortina d’Ampezzo. It involves skiing and hiking up and down mountain terrain. Five medal events should be created in sprint and individual races for men and women, and a mixed gender relay.

— The Olympic motto “Faster Higher Stronger” was updated to include the word “Together.” The formal Latin motto will now be “Citius, Altius, Fortius — Communis.”

— The IOC formally recognized the governing bodies of six sports: lacrosse, cheerleading, kickboxing, muay Thai, sambo and ice stock sport.

— Spending by the IOC was $55 million more than its revenue in 2020, when most income from the postponed Tokyo Olympics could not be declared. A “strong, solid” financial position was reported with the IOC’s fund balances — of assets exceeding liabilities — at almost $2.5 billion.

Punjab schools to reopen for classes 10, 11 and 12 from THIS date.

The Punjab government on Tuesday allowed the reopening of schools for Classes 10, 11 and 12 from July 26 onwards, as the Covid situation improves in the state.

The Punjab government on Tuesday allowed the reopening of schools for Classes 10, 11 and 12 from July 26 onwards, as the Covid situation improves in the state.

However, the government has put conditions that staff and teachers should be fully vaccinated and parents’ consent is taken.

Also, the government has raised the number of people in indoor gatherings 150 and outdoors to 300, subject to a cap of 50 per cent capacity.

Not only Punjab, schools in Odisha will also reopen for offline classes from July 26. With slight improvement of the corona situation, the Odisha government on Saturday decided to reopen schools for classes 10 and 12 from July 26, a senior official said.

School and Mass Education Department Principal Secretary Satyabrata Sahu said that offline Matric Exams (class 10) will be held from July 30 to August 5 in the state adhering to Covid protocols.

The decision to resume classroom teaching for the two classes was made keeping in mind the problems faced by students in online mode due to poor internet connectivity, Sahu said.