Category: CBSE Class 9

Here we are providing Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Heron’s Formula with Answers Solutions

Extra Questions for Class 9 Maths Chapter 12 Heron’s Formula with Solutions Answers

Heron’s Formula Class 9 Extra Questions Very Short Answer Type

Heron’s Formula Class 9 Extra Questions Question 1.
Find the area of an equilateral triangle having side 6 cm.
Solutioin:
Area of an equilateral triangle = \(\frac{\sqrt{3}}{4}\) × (side)² = \(\frac{\sqrt{3}}{4}\) × 6 × 6 = 9√3 cm²

Herons Formula Class 9 Extra Questions Question 2.
If the perimeter of an equilateral triangle is 90 m, then find its area.
Solutioin:

Herons Formula Extra Questions Question 3.
If every side of a triangle is doubled, then find the percent increase in area of triangle so formed.
Solutioin:
Let the sides of the given triangle be, a units, b units and c units.

Hence, percent increase = 300%

Class 9 Herons Formula Extra Questions Question 4.
If the length of a median of an equilateral triangle is x cm, then find its area.
Solutioin:

Let each equal sides of given equilateral triangle be 2y². We know that median is also perpendicular bisector.
∴ y² + x² = 4y²
⇒ x² = 3y²
⇒ x = √3y
or
⇒ y = \(\frac{x}{\sqrt{3}}\)
Now, area of given triangle = \(\frac{1}{2}\)
× 2y × X = y × x = \(\frac{x}{\sqrt{3}}\)× x = \(\frac{x^{2}}{\sqrt{3}}\)

Heron’s Formula Class 9 Extra Questions Short Answer Type 2

Class 9 Maths Chapter 12 Extra Questions Question 1.
Find the area of a triangle whose sides are 11 m, 60 m and 61 m.
Solutioin:
Let a = 11 m, b = 60 m and c = 61 m :

Chapter 12 Maths Class 9 Extra Questions Question 2.
Suman has a piece of land, which is in the shape of a rhombus. She wants her two sons to work on the land and produce different crops. She divides the land in two equal parts by drawing a diagonal. If its perimeter is 400 m and one of the diagonals is of length 120 m, how much area each of them will get for his crops ?
Solutioin:
Here, perimeter of the rhombus is 400 m.
∴ Side of the rhombus = \(\frac{400}{4}\) = 100 m
Let diagonal BD = 120 m and this diagonal divides the rhombus ABCD into two equal parts.

Hence, area of land allotted to two sons for their crops is 4800 m² each.

Extra Questions For Class 9 Maths Chapter 12 With Solution Question 3.
The perimeter of a triangular field is 144 m and its sides are in the ratio 3:4:5. Find the length of the perpendicular from the opposite vertex to the side whose length is 60 m.
Solutioin:
Let the sides of the triangle be 3x, 4x and 5x
∴ The perimeter of the triangular field = 144 m
⇒ 3x + 4x + 5x = 144
⇒ 12x = 144

Extra Questions On Herons Formula Class 9 Question 4.
Find the area of the triangle whose perimeter is 180 cm and two of its sides are of lengths 80 cm and 18 cm. Also, calculate the altitude of the triangle corresponding to the shortest side.
Solutioin:
Perimeter of given triangle = 180 cm
Two sides are 18 cm and 80 cm
∴ Third side = 180 – 18 – 80 = 82 cm

Hence, area of triangle is 720 cm² and altitude of the triangle corresponding to the shortest side is 80 cm.

Heron’s Formula Class 9 Extra Questions Long Answer Type

Extra Questions Of Herons Formula Class 9 Question 1.
Calculate the area of the shaded region.

Solutioin:

= 2 × 2 × 3 × 7 = 84 cm²
Area of shaded region = Area of ∆ABC – Area of ∆AOB
= 84 cm² – 30 cm² = 54 cm²

Class 9 Maths Herons Formula Extra Questions Question 2.
The sides of a triangular park are 8 m, 10 m and 6 m respectively. A small circular area of diameter 2 m is to be left out and the remaining area is to be used for growing roses. How much area is used for growing roses ? (use n = 3.14)
Solutioin:
The sides of the triangular park are 8 m, 10 m and 6 m.

Radius of the circle = \(\frac{2}{2}\) = 1 m
Area of the circle = πr² = 3.14 × 1 × 1 = 3.14 m²
∴ Area to be used for growing roses = Area of the park – area of the circle
= 24 – 3.14 = 20.86 m²

Heron’s Formula Class 9 Extra Questions HOTS

Herons Formula Class 9 Extra Questions With Solutions Pdf Question 1.
OPQR is a rhombus, whose three vertices P, Q and R lie on the circle with centre 0. If the radius of the circle is 12 cm, find the area of the rhombus.
Solutioin:
Since diagonals bisect each other at 90°.
∴ In right ∆QLR, (LR)² + (LQ)² = (QR)²

Class 9 Maths Ch 12 Extra Questions Question 2.
How much paper of each shade is needed to make a kite given in the figure, in which ABCD is a square with diagonal 60 cm?

Solutioin:
Since diagonals of a square are of equal length and bisect each other at right angles, therefore,
Area of ∆AOD = \(\frac{1}{2}\) × 30 × 30 = 450 cm²
Area of ∆AOD = Area of ∆DOC = Area of ∆BOC
= Area of ∆AOB = 450 cm²
[∵ ∆AOD = ∆AOB ≅ ∆BOC ≅ ∆COD, ∵ they² have equal area]
Now, area of ACEF (by Heron’s formula)
Here a = 20 cm, b = 20 cm and c = 30 cm

Now, area of orange shaded paper in kite
= Area of ∆AOD + Area of ∆CEF
= 450 cm² + 198.4 cm²
= 648.4 cm²
Area of blue shaded paper in kite
= Area of ∆AOB + Area of ∆COD
= 450 cm² + 450 cm² = 900 cm²
Area of black shaded paper in kite = Area of ∆BOC = 450 cm².

Heron’s Formula Class 9 Extra Questions Value Based (VBQs)

Heron’s Formula Class 9 Extra Questions With Solutions Question 1.
Sister Nivedita has trapezium shaped plot which she divided into three triangular portion for different purposes. I – for providing free education for orphan children, II – for providing dispensary for the needy villagers and III – for the library for villagers. Find the area of trapezium plot given in the figure. Which qualities of sister Nivedita are being depicted in question ?

Solutioin:
Here, ABCD is the trapezium with AB || DC.
Through C, Draw CF ⊥ AB
For – I: For area of ∆EBC

But, we know, Area of ∆EBC = \(\frac{1}{2}\)(base × height)
⇒ \(\frac{1}{2}\)1 × 30 × CF = 336
15 × CF = 336
⇒ CF = \(\frac{336}{15}\) = 22.4 m
Now, area of trapezium shaped plot = \(\frac{1}{2}\) (20 + 50)(22.4)
= 35 x 22.4 = 784 m²
Caring, kind, social, generous and visionary lady.

Ch 12 Maths Class 9 Extra Questions Question 2.
In an exhibition, an umbrella is made by stiching 10 triangular pieces of cloth with same message written on two triangular pieces. If each piece of cloth measures 60 cm, 60 cm and 20 cm, find how much cloth is required for each message.
Why we should respect women, educate women, save women and empower women ?

Solutioin:
Here, each triangular piece is an isosceles triangle with sides 60 cm, 60 cm and 20 cm.

Now, there are 2 triangular pieces with same message.
∴ Total area of cloth for each message = 2 × 591.61 = 1183.22 cm²
We should respect women, educate women, save women and empower women to serve humanity and give them equal opportunity so far they deprived.

Here we are providing Areas of Parallelograms and Triangles Class 9 Extra Questions Maths Chapter 9 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Areas of Parallelograms and Triangles with Answers Solutions

Extra Questions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles with Solutions Answers

Areas of Parallelograms and Triangles Class 9 Extra Questions Very Short Answer Type

Areas Of Parallelograms And Triangles Class 9 Extra Questions With Answers Question 1.
Two parallelograms are on equal bases and between the same parallels. Find the ratio of their areas.
Solution:
1:1 [∵ Two parallelograms on the equal bases and between the same parallels are equal in
area.]

Class 9 Maths Chapter 9 Extra Questions Question 2.
In ∆XYZ, XA is a median on side YZ. Find ratio of ar(∆XYA) : ar(∆XZA).

Solution:
Here, XA is the median on side YZ.
∴ YA = AZ
Draw XL ⊥ YZ

Parallelogram Questions For Class 9 Question 3.
ABCD is a trapezium with parallel sides AB = a cm and DC = b cm (fig.). E and F are the mid-points of the non parallel sides. Find the ratio of ar(ABFE) and ar(EFCD).

Solution:

Area Of Parallelograms And Triangles Class 9 Extra Questions Question 4.
ABCD is a parallelogram and Q is any point on side AD. If ar(∆QBC) = 10 cm², find ar(∆QAB) + ar(∆QDC).

Solution:
Here, ∆QBC and parallelogram ABCD are on the same base BC and lie between the same parallels BC || AD.
∴ ar(||^gm ABCD) = 2 ar(∆QBC) ar(∆QAB) + ar(∆QDC) + ar(∆QBC) = 2 ar(∆QBC)
ar(∆QAB) + ar(∆QDC) = ar(∆QBC)
Hence, ar(∆QAB) + ar(∆QDC) = 10 cm² [∵ ar(∆QBC) = 10 cm² (given)]

Areas Of Parallelograms And Triangles Class 9 Questions With Answers Question 5.
WXYZ is a parallelogram with XP ⊥ WZ and ZQ ⊥ WX. If WX = 8 cm, XP = 8 cm and ZQ = 2 cm, find YX.

Solution:
ar(||^gm WXYZ) = ar(||^gm WXYZ)
WX × ZQ = WZ × XP
8 × 2 = WZ × 8
⇒ WZ = 2 cm
Now, YX = WZ = 2 cm [∵ opposite sides of parallelogram are equal]

Questions On Areas Of Parallelograms And Triangles Class 9 Question 6.
In figure, TR ⊥ PS, PQ || TR and PS || QR. If QR = 8 cm, PQ = 3 cm and SP = 12 cm, find ar(quad. PQRS).

Solution:
Here,
PS || QR [given]
∴ PQRS is a trapezium
Now, TR ⊥ PS and PQ || TR [given]
⇒ PQ ⊥ PS
∴ PQ = TR = 3 cm [given]
Now, ar(quad. PQRS) = \(\frac{1}{2}\) (PS + QR) × PQ = \(\frac{1}{2}\)(12 + 8) × 3 = 30 cm²

Area Of Parallelogram Class 9 Extra Questions Question 7.
In the given figure, ABCD is a parallelogram and L is the mid-point of DC. If ar(quad. ABCL) is 72 cm, then find ar(∆ADC).

Solution:
In ||^gm ABCD, AC is the diagonal

Areas Of Parallelograms And Triangles Questions With Answers Question 8.
In figure, TR ⊥ PS, PQ || TR and PS || QR. If QR = 8 cm, PQ = 3 cm and SP = 12 cm, find ar (PQRS).

Solution:
Here, PS || QR
∴ PQRS is a trapezium in which PQ = 3 cm, QR = 8 cm and SP = 12 cm
Now, TR I PS and PQ || TR
∴ PQRT is a rectangle [∵ PQ || TR, PT || QR and ∠PTR = 90°]
⇒ PQ = TR = 3 cm
Now, ar(PQRS) = \(\frac{1}{2}\)(PS + QR) × TR = \(\frac{1}{2}\)(12 + 8) × 3 = 30 cm².

Areas of Parallelograms and Triangles Class 9 Extra Questions Short Answer Type 1

Area Of Parallelogram Class 9 Questions Question 1.
ABCD is a parallelogram and O is the point of intersection of its diagonals. If ar(A AOD) = 4 cm\(²\) find area of parallelogram ABCD.

Solution:
Here, ABCD is a parallelogram in which its diagonals AC and BD intersect each other in O.
∴ O is the mid-point of AC as well as BD.
Now, in ∆ADB, AO is its median
∴ ar(∆ADB) = 2 ar(∆AOD)
[∵ median divides a triangle into two triangles of equal areas]
So, ar(∆ADB) = 2 × 4 = 8 cm²
Now, ∆ADB and ||^gm ABCD lie on the same base AB and lie between same parallels AB and CD
∴ ar(ABCD) = 2 ar(∆ADB).
= 2 × 8
= 16cm²

Areas Of Parallelograms And Triangles Question 2.
In the given figure of ∆XYZ, XA is a median and AB || YX. Show that YB is also a median.

Solution:
Here, in ∆XYZ, AB || YX and XA is a median.
∴ A is the mid-point of YZ. Now, AB is a line segment from mid-point of one side (YZ) and parallel to another side (AB || YX), therefore, it bisects the third side XZ.
⇒ B is the mid-point of XZ.
Hence, YB is also a median of ∆XYZ.

Question 3.
ABCD is a trapezium. Diagonals AC and BD intersect each other at O. Find the ratio ar (∆AOD) : ar (∆BOC).
Solution:

Here, ABCD is a trapezium in which diagonals AC and BD intersect each other at O. ∆ADC and ABCD are on the same base DC and between the same ‘parallels i.e., AB || DC.
∴ ar(∆ADC) = ar(∆BCD)
⇒ ar(∆AOD) + ar(∆ODC)
= ar(ABOC) + ar(AODC)
⇒ ar(∆AOD) = ar(∆BOC)
⇒ \(\frac { ar(∆AOD) }{ ar(∆BOC) } \) = 1

Question 4.
ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ (fig.). If AQ intersects DC at P, show that ar(∆BPC) = ar(∆DPQ).

Solution:
In ||^gm ABCD,
ar(∆APC) = ar(∆BCP) …(i)
[∵ triangles on the same base and between the same parallels have equal area]
Similarly, ar(∆ADQ) = ar(∆ADC) i …(ii)
Now, ar(∆ADQ) – ar(∆ADP) = ar(∆ADC) – ar(∆ADP)
ar(∆DPQ) = ar(∆ACP) … (iii)
From (i) and (iii), we have
ar(∆BCP) = ar(∆DPQ)
or ar(∆BPC) = ar(∆DPQ)

Question 5.
In the figure, PQRS is a parallelogram with PQ = 8 cm and ar(∆PXQ) = 32 cm². Find the altitude of gm PQRS and hence its area.

Solution:
Since parallelogram PQRS and APXQ are on the same base PQ and lie between the same
parallels PQ || SR
∴ Altitude of the ∆PXQ and ||^gm PQRS is same.
Now, \(\frac{1}{2}\)PQ × altitude = ar(∆PXQ)
⇒ \(\frac{1}{2}\) × 8 × altitude = 32
altitude = 8 cm
ar(||^gm PQRS) = 2 ar(∆PXQ)
= 2 × 32 = 64 cm²
Hence, the altitude of parallelogram PQRS is 8 cm and its area is 64 cm².

Question 6.
In ∆ABC. D and E are points on side BC such that CD = DE = EB. If ar(∆ABC) = 27 cm, find ar(∆ADE)

Solution:
Since in ∆AEC, CD = DE, AD is a median.
∴ ar(∆ACD) = ar(∆ADE)
[∵ median divides a triangle into two triangles of equal areas]
Now, in ∆ABD, DE = EB, AE is a median
ar(∆ADE) = ar(∆AEB)… (ii)
From (i), (ii), we obtain
ar(∆ACD) = ar(∆ADE) = ar(∆AEB)\(\frac{1}{3}\)ar(∆ABC)
∴ ar(∆ADE) = \(\frac{1}{3}\) × 27 = 9 cm²

Areas of Parallelograms and Triangles Class 9 Extra Questions Short Answer Type 2

Question 1.
For the given figure, check whether the following statement is true or false. Also justify your answer. PQRS is a trapezium with PQ || SR, PS || RU and ST || RQ, then ar(PURS) = ar(TQRS)

Solution:
Since ST || RQ and SR || TQ [given]
⇒ STQR is a ||^gm
Similarly, PS || UR and SR || PU [given]
⇒ PSRU is a ||^gm
Also, similarly, ||^gm STQR and ||^gm PSRU lie on same base SR and between same parallels PQ and SR.
∴ ar(||^gm STQR) = ar(||^gm PSRU)
Hence, the given statement is true.

Question 2.
In the given figure, WXYZ is a quadrilateral with a point P on side WX. If ZY || WX, show that :
(i) ar(∆ZPY) = ar(∆ZXY)
(ii) ar(∆WZY) = ar(∆ZPY)
(iii) ar(∆ZWX) = ar(∆XWY)

Solution:
∆ZPY and ∆ZXY lie on same base ZY and between same parallels ZY and WX
∴ ar(∆ZPY) = ar(∆ZXY)
Again, (∆WZY) and (∆ZPY) lie on same base ZY and between same parallels ZY and WX
∴ ar(∆WZY) = ar(∆ZPY)
Also, ∆zwX and ∆XWY lie on same base XW and between same parallels ZY and WX
∴ ar(∆ZWX) = ar(∆XWY)

Question 3.
In ∆ABC ; D, E and F are mid-points of sides BC, AC and AB respectively. A line through C drawn parallel to DE meets FE produced to G. Show that ar(∆FDE) = ar(∆EGC).

Solution:
Here, in ∆ABC; D, E and F are the mid-points of sides BC, AC and AB respectively. A line through C is drawn parallel to DE meets FE produced at G.
Since a line segment drawn through the mid-points of two sides, is parallel to third side and is half of it.
∴ DE || AB, EF || BC and FD || AC
⇒ AEDF, EFDC and EFDB are parallelograms
Also, a diagonal of a parallelogram divides it into two congruent triangles
∴ ∆AFE ≅ ∆DEF
∆DEF ≅ ∆FDB and
∆DEF ≅ ∆EDC
∴ ar(∆FDE) ≅ ar(∆EDC) …(i)
Again, in quad. EGCD, we have
CG || DE and DC || EG [given]
∴ EGCD is a parallelogram .
∴ ∆EDC = ∆CGE
⇒ ar(AEDC) = ar(ACGE) … (ii)
From (i) and (ii), we obtain
ar(∆FDE) = ar(∆EGC)

Question 4.
In ∆PQR, A and B are points on side QR such that they trisect QR. Prove that ar(∆PQB) = 2ar(∆PBR).
Solution:
Here, in ∆PQR, A and B are points on side QR
such that QA = AB = BR.

Through P, draw a line / parallel to QR
Now, APQA, APAB and APBR on the equal bases
and between the same parallels l || QR
⇒ ar(∆PQA) = ar(∆PAB) = ar(∆PBR) …. (i)
Now, ar(∆PQB) = ar(∆PQA) + ar(∆PAB)
= 2ar(∆PQA)[using (i)]
= 2ar(∆PBR) [using (i)]

Question 5.
The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). Show that ar(||^gm ABCD) = ar(||^gm PBQR).

Solution:

Join AC and QP, also it is given that ∆Q || CP
∴ ∆ACQ and ∆APQ are on the same base ∆Q and lie between the same parallels ∆Q || CP.
∴ ar(∆ACQ) = ar(∆APQ)
or ar(∆ABC) + ar(∆ABQ) = ar(ABPQ) + ar(∆ABQ)
ar(∆ABC) = ar(ABPQ)
or \(\frac{1}{2}\)ar(||^gm ABCD) = \(\frac{1}{2}\)ar (||^gm PBQR)
or ar(||^gm ABCD) = ar(||^gm PBQR)

Areas of Parallelograms and Triangles Class 9 Extra Questions Long Answer Type

Question 1.
EFGH is a parallelogram and U and T are points on sides EH and GF respectively. If ar(∆EHT) = 16 cm, find ar(∆GUF).

Solution:
∴ ar(∆EHT) = \(\frac{1}{2}\) ar(||^gm EFGH) …..(i)
Similarly, ∆GUF and parallelogram EFGH are on the same base GF and lie between the same parallels GF and HE
∴ ar(∆GUF) = \(\frac{1}{2}\) ar(||^gm EFGH) …..(ii)
From (i) and (ii), we have
ar(∆GUF) = ar(∆EHT)
= 16 cm² [∵ ar(∆EHT) = 16 cmcm²] [given]

Question 2.
ABCD is a parallelogram and P is any point in its interior. Show that :
ar(∆APB) + ar(∆CPD) = ar(∆BPC) + ar(∆APD)

Solution:
Through P, draw a line LM || DA and EF || AB
Since ∆APB and ||^gm ABFE are on the same base AB and lie between the same parallels AB and EF.
∴ ar(∆APB) = \(\frac{1}{2}\) ar(||^gm ABFE) … (i)

Similarly, ACPD and parallelogram DCFE are on the same base DC and between the same parallels DC and EF.
∴ ar(∆CPD) = \(\frac{1}{2}\) ar(||^gm DCFE) … (ii)
Adding (i) and (ii), we have
ar(∆APB) + ar(∆CPD) = \(\frac{1}{2}\) ar (||^gm ABFE) + ar(||^gm DCFE)
= \(\frac{1}{2}\) ar(|lgm ABCD) … (iii)
Since ∆APD and parallelogram ADLM are on the same base AB and between the same parallels AD and ML
∴ ar(∆APD) = \(\frac{1}{2}\) ar(||^gm ADLM) …..(iv)
Similarly, ar(∆BPC) = \(\frac{1}{2}\) = arc||^gm BCLM) ….(v)
Adding (iv) and (u), we have
ar(∆APD) + ar (∆BPC) = \(\frac{1}{2}\) ar(||^gm ABCD) ….(vi)
From (iii) and (vi), we obtain
ar(∆APB) + ar(∆CPD) = ar(∆APD) + ar(ABPC)

Question 3.
In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP.

Solution:
In ∆PAD, ∠A = 90° and DA = PA = AB
⇒ ∠ADP = ∠APD = \(\frac{90^{\circ}}{2}\) = 45°
Similarly, in ∆QBC, ∠B = 90° and BQ = BC = AB
⇒ ∠BCQ = ∠BQC = \(\frac{90^{\circ}}{2}\) = 45°
In ∆PAD and ∆QBC, we have
PA = BQ [given]
∠A = ∠B [each = 90°]
AD = BC [sides of a square]
⇒ ∠PAD ≅ ∆QBC [by SAS congruence rule]
⇒ PD = QC [c.p.c.t.]
Now, in APDC and ∆QCD
DC = DC [common]
PD = QC [prove above]
∠PDC = ∠QCD [each = 90° + 45° = 135°]
⇒ ∆PDC = ∆QCD [by SAS congruence rule]
⇒ PC = QD or DQ = CP

Areas of Parallelograms and Triangles Class 9 Extra Questions HOTS

Question 1.
In the given figure, PQRS, SRNM and PQNM are parallelograms, Show that :
ar(∆PSM) = ar(∆QRN).

Solution:
Since PQRS is a parallelogram.
∴ PS = QR and PS || QR
Since SRNM is also a parallelogram.
∴ SM = RN and SM || RN
Also, PQNM is a parallelogram
∴ PM || QM and PM = QM
Now, in APSM and ∆QRN
PS = QR
SM = RN
PM = QN
∆PSM ≅ ∆QRN [by SSS congruence axiom]
∴ ar (∆PSM) = ar (∆QRN) [congruent triangles have same areas)

Areas of Parallelograms and Triangles Class 9 Extra Questions Value Based (VBQs)

Question 1.
Naveen was having a plot in the shape of a quadrilateral. He decided to donate some portion of it to construct a home for orphan girls. Further he decided to buy a land in lieu of his donated portion of his plot so as to form a triangle.
(i) Explain how this proposal will be implemented?
(ii) Which mathematical concept is used in it?
(iii) What values are depicted by Naveen?
Solution:

(i) Let ABCD be the plot and Naveen decided to donate some portion to construct a home for orphan girls from one corner say C of plot ABCD. Now, Naveen also purchases equal amount of land in lieu of land CDO, so that he may have triangular form of plot. BD is joined. Draw a line through C parallel to DB to meet AB produced in P.
Join DP to intersect BC at 0.
Now, ABCD and ABPD are on the same base and between same parallels CP || DB.
ar(∆BCD) = ar(∆BPD) ar(∆COD) + ar(∆DBO) = ar(∆BOP) + ar(∆DBO)
ar(ACOD) = ar(ABOP) ar(quad. ABCD)
= ar(quad. ABOD) + ar(∆COD)
= ar(quad. ABOD) + ar(∆BOP)
[∵ ar(ACOD) = ar(ABOP)] (proved above]
= ar(∆APD)
Hence, Naveen purchased the portion ABOP to meet his requirement.
(ii) Two triangles on the same base and between same parallels are equal in area.
(iii) We should help the orphan children.

Question 2.
A flood relief camp was organized by state government for the people affected by the natural calamity near a city. Many school students volunteered to participate in the relief work. In the camp, the food items and first aid centre kits were arranged for the flood victims. The piece of land used for this purpose is shown in the figure.
(a) If EFGH is a parallelogram with P and Q as mid-points of sides GH and EF respectively, then show that area used for first aid is half of the total area.
(b) What can you say about the student volunteers working for the relief work?

Solution:
(a) Here, EFGH is a ||^gm
∴ EF = GH and EF || GH

Hence, area used for first aid is half of the total area.

(b) Students working for the noble cause show compassion towards the affected people. They also realize their social responsibility to work for helping the ones in need.

Here we are providing Quadrilaterals Class 9 Extra Questions Maths Chapter 8 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Quadrilaterals with Answers Solutions

Extra Questions for Class 9 Maths Chapter 8 Quadrilaterals with Solutions Answers

Quadrilaterals Class 9 Extra Questions Very Short Answer Type

Quadrilateral Class 9 Extra Questions Question 1.
If one angle of a parallelogram is twice of its adjacent angle, find the angles of the parallelogram.
Solution:
Let the two adjacent angles be x and 2x.
In a parallelogram, sum of the adjacent angles are 180°
∴ x + 2x = 180°
⇒ 3x = 180°
⇒ x = 60°
Thus, the two adjacent angles are 120° and 60°. Hence, the angles of the parallelogram are 120°, 60°, 120° and 60°.

Class 9 Quadrilaterals Extra Questions Question 2.
If the diagonals of a quadrilateral bisect each other at right angles, then name the
quadrilateral.
Solution:
Rhombus.

Quadrilaterals Class 9 Extra Questions With Solutions Question 3.
Three angles of a quadrilateral are equal and the fourth angle is equal to 144o. Find each of the equal angles of the quadrilateral.
Solution:
Let each equal angle of given quadrilateral be x.
We know that, sum of all interior angles of a quadrilateral is 360°
∴ x + x + x + 144° = 360°
3x = 360° – 144°
3x = 216°
x = 72°
Hence, each equal angle of the quadrilateral is of 72o measures.

Extra Questions On Quadrilaterals Class 9 Question 4.
If ABCD is a parallelogram, then what is the measure of ∠A – ∠C ?
Solution:
∠A – ∠C = 0° (opposite angles of parallelogram are equal]

Quadrilaterals Class 9 Extra Questions Question 5.
PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram.
Solution:

Here, PQ = SR = 12 cm
Let PS = x and PS = QR
∴ x + 12 + x + 12 = Perimeter
2x + 24 = 40
2x = 16
x= 8
Hence, length of each side of the parallelogram is 12 cm, 8 cm, 12 cm and 8 cm.

Class 9 Maths Quadrilaterals Extra Questions Question 6.
Two consecutive angles of a parallelogram are (x + 60)° and (2x + 30)°. What special name can you give to this parallelogram?
Solution:
We know that consecutive interior angles of a parallelogram are supplementary.
∴ (x + 60° + (2x + 30)° = 180°
⇒ 3x° + 90° = 180°
⇒ 3x° = 90°
⇒ x° = 30°
Thus, two consecutive angles are (30 + 60)°, 12 x 30 + 30)”. i.e., 90° and 90°.
Hence, the special name of the given parallelogram is rectangle.

Class 9 Quadrilaterals Extra Questions Pdf Question 7.
ONKA is a square with ∠KON = 45°. Determine ∠KOA.
Solution:
Since ONKA is a square
∴ ∠AON = 90°
We know that diagonal of a square bisects its ∠s
⇒ ∠AOK = ∠KON = 45°
Hence, ∠KOA = 45°

Quadrilaterals Extra Questions Class 9 Question 8.
In quadrilateral PQRS, if ∠P = 60° and ∠Q : ∠R : ∠S = 2 : 3 : 7, then find the measure of ∠S.
Solution:
Let ∠Q = 2x, ∠R = 3x and ∠S = 7x
Now, ∠P + ∠Q + ∠R + ∠S = 360°
⇒ 60° + 2x + 3x + 7x = 360°
⇒ 12x = 300°
x = \(\frac{300^{\circ}}{12}\) = 25°
∠S = 7x = 7 x 25° = 175°

Quadrilaterals Class 9 Extra Questions Short Answer Type 1

Questions On Quadrilaterals For Class 9 Question 1.
ABCD is a parallelogram in which ∠ADC = 75° and side AB is produced to point E as shown in the figure. Find x + y.
Solution:

Here, ∠C and ∠D are adjacent angles of the parallelogram.
∴ ∠C + ∠D = 180°
⇒ x + 75° = 180°
⇒ x = 105°
Also, y = x = 105° [alt. int. angles]
Thus, x + y = 105° + 105° = 210°

Class 9 Maths Chapter 8 Extra Questions With Solutions Question 2.
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Solution:

Given: A parallelogram ABCD, in which AC = BD.
To Prove: ΔBCD is a rectangle.
Proof : In ΔABC and ΔBAD
AB = AB (common]
AC = BD (given]
BC = AD(opp. sides of a ||^gm]
⇒ ΔABC ≅ ΔBAD
[by SSS congruence axiom]
⇒ ∠ABC = ∠BAD (c.p.c.t.)
Also, ∠ABC + ∠BAD = 180° (co-interior angles)
∠ABC + ∠ABC = 180° [ ∵ ∠ABC = ∠BAD ]
2∠ABC = 180°
∠ABC = 1/2 x 180° = 90°
Hence, parallelogram ABCD is a rectangle.

Quadrilateral Extra Questions Class 9 Question 3.
In the figure, ABCD is a rhombus, whose diagonals meet at O. Find the values of x and y.

Solution:
Since diagonals of a rhombus bisect each other at right angle.
In ∴ ΔAOB, we have
∠OAB + ∠x + 90° = 180°
∠x = 180° – 90° – 35°
= 55°
Also,
∠DAO = ∠BAO = 35°
∠y + ∠DAO + ∠BAO + ∠x = 180°
⇒ ∠y + 35° + 35° + 55° = 180°
⇒ ∠y = 180° – 125o = 55°
Hence, the values of x amd y are x = 55°, y = 55°.

Extra Questions For Class 9 Maths Quadrilaterals With Solutions Question 4.
ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see fig.). Show that :
(i) ΔAPB = ΔCQD
(ii) AP = CQ
Solution:

Given : In ||gm ABCD, AP and CQ are perpendiculars from the
vertices A and C on the diagonal BD.
To Prove: (i) ΔAPB ≅ ΔCQD
(ii) AP = CQ
Proof : (i) In ΔAPB and ΔCQD
AB = DC (opp. sides of a ||gm ABCD]
∠APB = ∠DQC (each = 90°)
∠ABP = ∠CDQ (alt. int. ∠s]
⇒ ΔAPB ≅ ΔCQD[by AAS congruence axiom]
(ii) ⇒ AP = CQ [c.p.c.t.]

Quadrilaterals Class 9 Extra Questions Short Answer Type 2

Extra Questions Of Quadrilaterals Class 9 Question 1.
The diagonals of a quadrilateral ABCD are perpendicular to each other. Show that the quadrilateral formed by joining the mid-points of its sides is a rectangle.

Solution:
Given: A quadrilateral ABCD whose diagonals AC and BD are perpendicular to each other at O. P, Q, R and S are mid-points of side AB, BC, CD and DA respectively are joined are formed quadrilateral PQRS.
To Prove: PQRS is a rectangle.
Proof : In ∆ABC, P and Q are mid-points of AB and BC respectively.
∴ PQ || AC and PQ = \(\frac{1}{2}\) AC … (i) (mid-point theorem]
Further, in SACD, R and S are mid-points of CD and DA respectively.
SR || AC and SR = \(\frac{1}{2}\) AC … (ii) (mid-point theorem]
From (i) and (ii), we have PQ || SR and PQ = SR
Thus, one pair of opposite sides of quadrilateral PQRS are parallel and equal.
∴ PQRS is a parallelogram.
Since PQ|| AC PM || NO
In ∆ABD, P and S are mid-points of AB and AD respectively.
PS || BD (mid-point theorem]
⇒ PN || MO
∴ Opposite sides of quadrilateral PMON are parallel.
∴ PMON is a parallelogram.
∠ MPN = ∠ MON (opposite angles of ||gm are equal]
But ∠MON = 90° [given]
∴ ∠MPN = 90° ⇒ ∠QPS = 90°
Thus, PQRS is a parallelogram whose one angle is 90°
∴ PQRS is a rectangle.

Class 9 Maths Chapter 8 Extra Questions Question 2.
In the fig., D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. Prove that DEF is also an equilateral triangle.

Solution:
Since line segment joining the mid-points of two sides of a triangle is half of the third side.
Therefore, D and E are mid-points of BC and AC respectively.
⇒ DE = \(\frac{1}{2}\)AB …(i)
E and F are the mid-points of AC and AB respectively.
∴ EF = \(\frac{1}{2}\)BC … (ii)
F and D are the mid-points of AB and BC respectively.
∴ FD = \(\frac{1}{2}\) AC … (iii)
Now, SABC is an equilateral triangle.
⇒ AB = BC = CA
⇒ \(\frac{1}{2}\)AB = \(\frac{1}{2}\)BC = \(\frac{1}{2}\)CA
⇒ DE = EF = FD (using (i), (ii) and (iii)]
Hence, DEF is an equilateral triangle.

Quadrilateral Class 9 Questions Question 3.
In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY is a parallelogram. Show that ABCD is a parallelogram.

Solution:
Since BXDY is a parallelogram.
XO = YO
DO = BO
[∵ diagonals of a parallelogram bisect each other]
But AX = CY …. (iii) (given]
Adding (i) and (iii), we have
XO + AX = YO + CY
⇒ AO = CO …. (iv)
From (ii) and (iv), we have
AO = CO and DO = BO
Thus, ABCD is a parallelogram, because diagonals AC and BD bisect each other at O.

Class 9 Maths Ch 8 Extra Questions Question 4.
ABCD is a quadrilateral in which the bisectors of ∠A and ∠C meet DC produced at Y and BA produced at X respectively. Prove that
∠X +∠Y = \(\frac{1}{2}\)(∠A + ∠C)
Solution:

Here, ∠1 = ∠2 and ∠3 = ∠4
In ΔXBC, we have
∠X + ∠B + ∠4 = 180°
∠X + ∠B + \(\frac{1}{2}\)∠C = 180
In ΔADY, we have
∠2 + ∠D + ∠Y= 180°
\(\frac{1}{2}\) ∠A + ∠D + ∠Y = 180°
Adding (i) and (ii), we have
∠X + ∠Y + ∠B + ∠D + \(\frac{1}{2}\) ∠C + \(\frac{1}{2}\) ∠A = 360°
Also, in quadrilateral ABCD
∠A + ∠B + ∠C + ∠D = 360°
∠X + ∠Y + ∠B + ∠D + \(\frac{1}{2}\) ∠C + \(\frac{1}{2}\) ∠A = ∠A + ∠B + ∠C + ∠D
∠X + ∠Y = ∠A – \(\frac{1}{2}\) ∠A + \(\frac{1}{2}\) ∠C – \(\frac{1}{2}\) ∠C
∠X+ ∠Y = \(\frac{1}{2}\) (∠A + ∠C)

Quadrilaterals Class 9 Extra Questions Long Answer Type

Quadrilateral Questions For Class 9 Question 1.
In the figure, P, Q and R are the mid-points of the sides BC, AC and AB of ΔABC. If BQ and PR intersect at X and CR and PQ intersect at Y, then show that XY = \(\frac{1}{4}\) BC.

Solution:
Here, in ΔABC, R and Q are the mid-points of AB and AC respectively.
∴ By using mid-point theorem, we have
RQ || BC and RQ = \(\frac{1}{2}\) BC
∴ RQ = BP = PC [∵ P is the mid-point of BC]
∴ RQ || BP and RQ || PC
In quadrilateral BPQR
RQ || BP, RQ = BP (proved above]
∴ BPQR is a parallelogram. [∵ one pair of opp. sides is parallel as well as equal]
∴ X is the mid-point of PR. [∵ diagonals of a ||gm bisect each other]
Now, in quadrilateral PCQR
RQ || PC and RQ = PC [proved above)
∴ PCQR is a parallelogram [∵ one pair of opp. sides is parallel as well as equal]
∴ Y is the mid-point of PQ [∵ diagonals of a ||gm bisect each other]
In ΔPQR
∴ X and Y are mid-points of PR and PQ respectively.

Class 9 Quadrilaterals Important Questions Question 2.
In the given figure, AE = DE and BC || AD. Prove that the points A, B, C and D are concyclic. Also, prove that the diagonals of the quadrilateral ABCD are equal.

Solution:
Since AE = DE
∠D = ∠A …. (i) [∵ ∠s opp. to equal sides of a Δ]
Again, BC || AD
∠EBC = ∠A …. (ii) (corresponding ∠s]
From (i) and (ii), we have
∠D = ∠EBC …. (iii)
But ∠EBC + ∠ABC = 180° (a linear pair]
∠D + ∠ABC = 180° (using (iii)]
Now, a pair of opposite angles of quadrilateral ABCD is supplementary
Thus, ABCD is a cyclic quadrilateral i.e., A, B, C and D’are concyclic. In ΔABD and ΔDCA
∠ABD = ∠ACD [∠s in the same segment for cyclic quad. ABCD]
∠BAD = ∠CDA [using (i)]
AD = AD (common]
So, by using AAS congruence axiom, we have
ΔABD ≅ ΔDCA
Hence, BD = CA [c.p.c.t.]

Question 3.
In ΔABC, AB = 8 cm, BC = 9 cm and AC = 10 cm. X, Y and Z are mid-points of AO, BO and CO respectively as shown in the figure. Find the lengths of the sides of ΔXYZ.

Solution:
Here, in ΔABC, AB = 8 cm, BC = 9 cm, AC = 10 cm.
In ΔAOB, X and Y are the mid-points of AO and BO.
∴ By using mid-point theorem, we have
XY = \(\frac{1}{2}\) AB = \(\frac{1}{2}\) x 8 cm = 4 cm
Similarly, in Δ𝜏BOC, Y and Z are the mid-points of BO and CO.
∴ By using mid-point theorem, we have
YZ = \(\frac{1}{2}\) BC = \(\frac{1}{2}\) x 9cm = 4.5 cm
And, in Δ𝜏COA, Z and X are the mid-points of CO and AO.
∴ ZX = \(\frac{1}{2}\) AC = \(\frac{1}{2}\) x 10 cm = 5 cm
Hence, the lengths of the sides of ΔXYZ are XY = 4 cm, YZ = 4.5 cm and ZX = 5 cm.

Question 4.
PQRS is a square and ∠ABC = 90° as shown in the figure. If AP = BQ = CR, then prove that ∠BAC = 45°

Solution:
Since PQRS is a square.
∴ PQ = QR … (I) [∵ sides of a square are equal]
Also, BQ = CR … (ii) [given]
Subtracting (ii) from (i), we obtain
PQ – BQ = QR – CR
⇒ PB = QC … (iii)
In Δ𝜏APB and Δ𝜏BQC
AP = BQ[given
∠APB = ∠BQC = 90°](each angle of a square is 90°)
PB = QC (using (iii)]

So, by using SAS congruence axiom, we have
ΔAPB ≅ ΔBQC
∴ AB = BC [c.p.c.t.]
Now, in ΔABC
AB = BC [proved above]
∴ ∠ACB = ∠BAC = x° (say) [∠s opp. to equal sides]
Also, ∠B + ∠ACB + ∠BAC = 180°
⇒ 90° + x + x = 180°
⇒ 2x° = 90°
x° = 45°
Hence, ∠BAC = 45°

Question 5.
ABCD is a parallelogram. If the bisectors DP and CP of angles D and C meet at P on side AB, then show that P is the mid-point of side AB.
Solution:

Since DP and CP are angle bisectors of ∠D and ∠C respectively.
: ∠1 = ∠2 and ∠3 = ∠4
Now, AB || DC and CP is a transversal
∴ ∠5 = ∠1 [alt. int. ∠s]
But ∠1 = ∠2 [given]
∴ ∠5 = ∠2

Now, in ABCP, ∠5 = ∠2
⇒ BC = BP … (I) [sides opp. to equal ∠s of a A]
Again, AB || DC and DP is a transversal.
∴ ∠6= ∠3 (alt. int. Δs]
But ∠4 = ∠3 [given]
∴ ∠6 = ∠4
Now, in ΔADP, ∠6 = ∠4
⇒ DA = AP …. (ii) (sides opp. to equal ∠s of a A]
Also, BC = DA… (iii) (opp. sides of parallelogram)
From (i), (ii) and (iii), we have
BP = AP
Hence, P is the mid-point of side AB.

Question 6.
In the figure, ΔBCD is a trapezium in which AB || DC. E and F are the mid-points of AD and BC respectively. DF and AB are produced to meet at G. Also, AC and EF intersect at the point O. Show that :
(i) EO || AB
(ii) AO = CO

Solution:
Here, E and F are the mid-points of AD and BC respectively.
In ΔBFG and ΔCFD
BF = CF [given]
∠BFG = ∠CFD (vert. opp. ∠s]
∠BGF = ∠CDF (alt. int. ∠s, as AB || DC)
So, by using AAS congruence axiom, we have
ΔBFG ≅ ΔCFD
⇒ DF = FG [c.p.c.t.)
Now, in ΔAGD, E and F are the mid-points of AD and GD.
∴ By mid-point theorem, we have
EF || AG
or EO || AB
Also, in ΔADC, EO || DC
∴ EO is a line segment from mid-point of one side parallel to another side.
Thus, it bisects the third side.
Hence, AO = CO

Here we are providing Number Systems Class 9 Extra Questions Maths Chapter 1 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Number Systems with Answers Solutions

Extra Questions for Class 9 Maths Chapter 1 Number Systems with Solutions Answers

Number Systems Class 9 Extra Questions Very Short Answer Type

Number System Class 9 Extra Questions With Solutions Question 1.
Simplify: (√5 + √2)².
Solution:
Here, (√5 + √2² = (√5² + 2√5√2 + (√2)²
= 5 + 2√10 + 2 = 7 + 2√10

Question 2.
Find the value of √(3)^-2.
Solution:

Number System Class 9 Extra Questions Question 3.
Identify a rational number among the following numbers :
2 + √2, 2√2, 0 and π
Solution:
O is a rational number.

Class 9 Maths Chapter 1 Extra Questions Question 4.
Express 1.8181… in the form \(\frac{p}{q}\) where p and q are integers and q ≠ 0.
Solution:
Let x =1.8181… …(i)
100x = 181.8181… …(ii) [multiplying eqn. (i) by 100]
99x = 180 [subtracting (i) from (ii)]
x = \(\frac{180}{99}\)
Hence, 1.8181… = \(\frac{180}{99}\) = \(\frac{20}{11}\)

Class 9 Number System Extra Questions Question 5.
Simplify : √45 – 3√20 + 4√5
Solution:
√45 – 3√20 + 4√5 = 3√5 – 6√5 + 4√5 = √5.

Class 9 Maths Number System Extra Questions With Solutions Question 6.
Find the value of’

Solution:

Class 9 Maths Ch 1 Extra Questions Question 7.
Find the value of

Solution:

Number Systems Class 9 Extra Questions Short Answer Type 1

Extra Questions For Class 9 Maths Chapter 1 With Solution Pdf Question 1.
Evaluate : (√5 + √2² + (√8 – √5)²
Solution:
(√5 + √2)² + (√8 – √5² = 5 + 2 + 2√10 + 8 + 5 – 2√40
= 20 + 2√10 – 4√10 = 20 – 2√10

Extra Questions On Number System Class 9 Question 2.
Express \(23 . \overline{43}\) in \(\frac{p}{q}\) form, where p, q are integers and q ≠ 0.
Solution:
Let x = \(23 . \overline{43}\)
or x = 23.4343…         ….(i)
100x = 2343.4343…    …(ii) [Multiplying eqn. (i) by 100]
99x = 2320 [Subtracting (i) from (ii)
⇒ x = \(\frac{2320}{99}\)
Hence, \(23 . \overline{43}\) = \(\frac{2320}{99}\)

Number System Extra Questions Question 3.
Let ‘a’ be a non-zero rational number and ‘b’ be an irrational number. Is ‘ab’ necessarily an irrational ? Justify your answer with example.
Solution:
Yes, ‘ab’ is necessarily an irrational.
For example, let a = 2 (a rational number) and b = √2 (an irrational number)
If possible let ab = 2√2 is a rational number.
Now, \(\frac{ab}{a}\) = \(\frac{2 \sqrt{2}}{2}\) = √2 is a rational number.
[∵ The quotient of two non-zero rational number is a rational]
But this contradicts the fact that √2 is an irrational number.
Thus, our supposition is wrong.
Hence, ab is an irrational number.

Chapter 1 Maths Class 9 Extra Questions Question 4.
Let x and y be a rational and irrational numbers. Is x + y necessarily an irrational number? Give an example in support of your answer.
Solution:
Yes, x + y is necessarily an irrational number.
For example, let x = 3 (a rational number) and y = √5 (an irrational number)
If possible let x + y = 3 + √5 be a rational number.
Consider \(\frac{p}{q}\) = 3 + √5, where p, q ∈ Z and q ≠ 0.
Squaring both sides, we have

∵ \(\frac{p}{q}\) is a rational
⇒ √5 is a rational
But this contradicts the fact that √5 is an irrational number.
Thus, our supposition is wrong.
Hence, x + y is an irrational number.

Number Systems Class 9 Extra Questions Short Answer Type 2

Ch 1 Maths Class 9 Extra Questions Question 1.
Represent √3 on the number line.
Solution:


On the number line, take OA = 1 unit. Draw AB = 1 unit perpendicular to OA. Join OB.
Again, on OB, draw BC = 1 unit perpendicular to OB. Join OC.
By Pythagoras Theorem, we obtain OC = √3. Using
compasses, with centre O and radius OC, draw an arc, which intersects the number line at point
D. Thus, OD = √3 and D corresponds to √3.

Class 9 Chapter 1 Maths Extra Questions Question 2.
Represent √3.2 on the number line.
Solution:
First of all draw a line of length 3.2 units such that AB = 3.2 units. Now, from point B, mark a distance of 1 unit. Let this point be ‘C’. Let ‘O’ be the mid-point of the distance AC. Now, draw a semicircle with centre ‘O’ and radius OC. Let us draw a line perpendicular to AC passing through the point ‘B’ and intersecting the semicircle at point ‘D’.
∴ The distance BD = √3.2

Now, to represent √3.2 on the number line. Let us take the line BC as number line and point ‘B’ as zero, point ‘C’ as ‘1’ and so on. Draw an arc with centre B and radius BD, which intersects the number line at point ‘E’.
Then, the point ‘E’ represents √3.2.

Class 9 Ch 1 Maths Extra Questions Question 3.
Express 1.32 + 0.35 as a fraction in the simplest form.
Solution:
Let . x = 1.32 = 1.3222…..(i)

Multiplying eq. (i) by 10, we have
10x = 13.222…
Again, multiplying eq. (i) by 100, we have
100x = 132.222… …(iii)
Subtracting eq. (ii) from (iii), we have
100x – 10x = (132.222…) – (13.222…)
90x = 119
⇒ x = \(\frac{119}{90}\)
Again, y = 0.35 = 0.353535……
Multiply (iv) by 100, we have …(iv)
100y = 35.353535… (v)
Subtracting (iv) from (u), we have
100y – y = (35.353535…) – (0.353535…)
99y = 35
y = \(\frac{35}{99}\)

Extra Questions For Class 9 Maths Ch 1 Question 4.
Find the square root of 10 + √24 + √60 + √40.
Solution:

Question 5.
If x = 9 + 4√5, find the value of √x – \(\frac{1}{\sqrt{x}}\).
Solution:
Here,
x = 9 + 4√5
x = 5 + 4 + 2 x 2√5
x = (√5² + (2² + 2 x 2x √5).
x = (√5 + 2)²
√x = √5 + 2
Now, \(\frac{1}{\sqrt{x}}\) = \(\frac{1}{\sqrt{5}+2}\)

Question 6.
If x = \(\frac{1}{\sqrt{5}-2}\) , find the value of x³ – 3² – 5x + 3
Solution:

∴ x – 2 = √5
Squaring both sides, we have
x² – 4x + 4 = 5
x² – 4x – 1 = 0 …(i)
Now, x³ – 3² – 5x + 3 = (x² – 4x – 1) (x + 1) + 4
= 0 (x + 1) + 4 = 4 [using (i)]

Question 7.
Find ‘x’, if 2^x-7 × 5^x-4 = 1250.
Solution:
We have 2^x-7 × 5^x-4 = 1250
⇒ 2^x-7 × 5^x-4 = 2 5 × 5 × 5 × 5
⇒ 2^x-7 × 5^x-4 = 21 × 54
Equating the powers of 2 and 5 from both sides, we have
⇒ x – 7 = 1 and x – 4 = 4
⇒ x = 8 and x = 8
Hence, x = 8 is the required value.

Question 8.
Evaluate:

Solution:

Number Systems Class 9 Extra Questions Long Answer Type

`Question 1.
If x = \(\frac { \sqrt { p+q } +\sqrt { p-q } }{ \sqrt{p+q}-\sqrt{p-q} }\), then prove that q² – 2px + 9 = 0.
Solution:


Squaring both sides, we have
⇒ q²x² + p² – 2pqx = p² – q²
⇒ q²x² – 2pqx + q² = 0
⇒ q(q² – 2px + q) = 0
⇒ qx² – 2px + q = 0 (∵ q ≠ 0)

Question 2.
If a = \(\frac{1}{3-\sqrt{11}}\) and b = \(\frac{1}{a}\), then find a² – b²
Solution:

Question 3.
Simplify:

Solution:

Question 4.
Prove that:

Solution:

Question 5.
Find a and b, if

Solution:

Number Systems Class 9 Extra Questions HOTS

Question 1.
If x^a = y, y^b = z and z^c = x, then prove that abc = 1.
Solution:
We have x^a = y, y^b = z and z^c = x

Question 2.
Prove that:

Solution:
Taking L.H.S., we have

Question 3.
Show that:

Solution:

Number Systems Class 9 Extra Questions Value Based (VBQs)

Question 1.
Sudhir and Ashok participated in a long jump competition along a straight line marked as a number line. Both start the jumps one by one but in opposite directions. From ‘O’ Ashok jumps one unit towards the positive side while Sudhir jumps double in units as Ashok jumps, along negative side. After jumping 4 jumps each, at which point Ashok and Sudhir reached. What is the distance between their final positions ? Ashok argue that he is the winner since Sudhir is at negative side. Who do you think is winner and why? What is the value of the competition ?

Solution:
After jumping four jumps each, Ashok reached at 4 in positive direction and Sudhir reached at -8 i.e., in negative direction. Distance between their final positions is 12 units. Here, distance covered by Sudhir is 8 units and distance covered by Ashok is 4 units. Thus, Sudhir is the winner. Competition inculcate spirit of performance.

Question 2.
Manu went to his mathematics teacher and asked him “Sir, I want some chocolates to distribute among my classmates for my birthday but I have no money. Can you provide me some chocolates”. Teacher told Manu, I am giving you two numbers \(\frac{1}{3+2 \sqrt{2}}\) and \(\frac{1}{3-2 \sqrt{2}}\) and if you can find the value of sum of their squares, then I will provide you as many chocolates as the resulting value of sum of squares of given numbers. Find the number of chocolates. What value is depicted from this action?
Solution:

= (3 – 2√2)² + (3 + 2√2²
= 9 + 8 – 2 × 3 × (2√2) + 9 + 8 + 2 × 3 × 2√2 = 34.
Hence, resulting value of sum of squares of numbers = number of chocolates = 34. By doing this, teacher motivates the students to use their knowledge and apply it in day to day life with caring and kindness.

Here we are providing Coordinate Geometry Class 9 Extra Questions Maths Chapter 3 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Coordinate Geometry with Answers Solutions

Extra Questions for Class 9 Maths Chapter 3 Coordinate Geometry with Solutions Answers

Coordinate Geometry Class 9 Extra Questions Very Short Answer Type

Coordinate Geometry Class 9 Extra Questions Question 1.
Write the signs convention of the coordinates of a point in the second quadrant.
Solution:
(-ve, +ve)

Class 9 Coordinate Geometry Extra Questions Question 2.
Write the value of ordinate of all the points lie on x-axis.
Solution:
0

Coordinate Geometry Extra Questions Class 9 Question 3.
Write the value of abscissa of all the points lie on y-axis.
Solution:
0

Class 9 Maths Chapter 3 Extra Questions Question 4.
If in coordinates of a point B(3, -2), signs of both coordinates are interchanged, then it will lie in which quadrant ?
Solution:
When signs of both coordinates of B(3, -2) are interchanged, then coordinates of new point are B'(-3, 2) and it will lie in second quadrant.

Extra Questions On Coordinate Geometry Class 9 Question 5.
Find distances of points C(-3, -2) and D(5, 2) from x-axis and y-axis.
Solution:
Distances of point C(-3, -2) from x-axis is 2 units in the negative direction and from y-axis is 3 units in the negative direction. Distances of point D(5, 2) from x-axis is 2 units and from y-axis is 5 units.

Extra Questions Of Coordinate Geometry Class 9 Question 6.
Find the values of x and y, if two ordered pairs (x – 3, – 6) and (4, x + y) are equal.
Solution:
Here, two ordered pairs are equal.
⇒ Their first components are equal and their second components are separately equal.
⇒ x – 3 = 4 and x + y = -6
⇒ x = 7 and 7 + y = -6 ⇒ y = – 13
Hence, x = 7 and y = – 13.

Questions On Coordinate Geometry Class 9 Question 7.
In which quadrant does the point (-1, 2) lie ?
Solution:
(-1, 2) lie in second quadrant.

Coordinate Geometry Class 9 Important Questions Question 8.
Find the distance of the point (0, -5) from the origin.
Solution:
5 units.

Class 9 Maths Coordinate Geometry Extra Questions Question 9.
Write the shape of the quadrilateral formed by joining (1, 1), (6, 1), (4, 5) and (3, 5) on graph paper.
Solution:
Trapezium.

Coordinate Geometry Class 9 Extra Questions Short Answer Type 1

Coordinate Geometry Class 9 Questions With Solutions Question 1.
In the given figure, ABCD is a rectangle with length 6 cm and breadth 3 cm. O is the mid-point of AB. Find the coordinates of A, B, C and D.

Solution:

We have taken 1 cm = 1 unit and origin O is the mid-point of AB
∴ OA = OB = 3 cm
and BC = AD = 3 cm
Thus, the coordinates of A are (-3, 0)
the coordinates of B are (3, 0)
the coordinates of C are (3, 3)
the coordinates of D are (-3, 3)

Ch 3 Maths Class 9 Extra Questions Question 2.
Write the coordinates of A, B, C and D from the figure given alongside.

Solution:
Coordinates of the point A are (5, 0)
Coordinates of the point B are (5, 3)
Coordinates of the point C are (-2, 4)
Coordinates of the point D are (0, -2)

Class 9 Maths Ch 3 Extra Questions Question 3.
A point lies on x-axis at a distance of 9 units from y-axis. What are its coordinates? What will be the coordinates of a point, if it lies on y-axis at a distance of -9 units from x-axis ?
Solution:
As shown in graph, the coordinates of a point which lies on x-axis at a distance of 9 units from y-axis are (9, 0) and the coordinates of a point which lies at a distance of -9 units from x-axis are (0, -9).

Coordinate Geometry Class 9 Hots Questions Question 4.
Plot the point P(2, -6) on a graph paper and from it draw PM and PN perpendiculars to x-axis and y-axis respectively. Write the coordinates of the points M and N.
Solution:

As shown in graph, coordinates of M are (2, 0) and coordinates of N are (0, -6).

Coordinate Geometry Class 9 Extra Questions Short Answer Type 2

Chapter 3 Maths Class 9 Extra Questions Question 1.
Without plotting the points indicate the quadrant in which they lie, if :
(i) ordinate is 5 and abscissa is – 3
(ii) abscissa is -5 and ordinate is – 3
(iii) abscissa is – 5 and ordinate is 3
(iv) ordinate is 5 and abscissa is 3
Solution:
(i) Clearly, point (-3, 5) lies in 2nd quadrant.
(ii) Clearly, point (-5, – 3) lies in 3rd quadrant.
(ii) Clearly, point (-5, 3) lies in 2nd quadrant.
(iv) Clearly, point (3, 5) lies in 1st quadrant.

Coordinate Geometry Class 9 Extra Questions With Solutions Question 2.
Plot the points A(1, 4), B(-2, 1) and C(4, 1). Name the figure so obtained on joining them in order and also, find its area.
Solution:

Triangle.
Area of ∆ABC = \(\frac{1}{2}\) × BC × Height
= \(\frac{1}{2}\) × 6 × 3
= 9 sq. units

Coordinate Geometry Questions Class 9 Question 3.
Plot the following points, join them in order and identify the figure thus formed : A(1, 3), B(1, -1), C(7, -1) and D(7, 3)
Write the coordinates of the point of intersection of the diagonals.
Solution:

ABCD is a rectangle.
Point of intersection of the diagonals AC and BD is (4, 1).

Class 9 Chapter 3 Maths Extra Questions Question 4.
Plot the points A(2, 5), B(8,5) and C(5, -3) and join AB, BC and CA. What figure do you obtain ?
Solution:

We obtain an isosceles triangle in which AC = BC.

Extra Questions For Class 9 Maths Chapter 3 With Solution Question 5.
(i) Plot the points M(4, 3), N(4, 0), 0(0, 0), P(0, 3).
(ii) Name the figure obtained by joining MNOP.
(iii) Find the perimeter of the figure.
Solution:
(i)

(ii) As shown in graph, the figure obtained by joining MNOP is rectangle.
(iii) Perimeter of rectangle MNOP = 2 (ON + OP) = 2 (4 + 3) = 2 × 7 = 14 units.

Question 6.
Plot D(-2, -3) on the graph paper. Also, plot reflections of D in x-axis and y-axis.
Solution:

Reflection of D(-2, -3) in x-axis is D'(-2, 3) and reflection of D(-2, -3) in y-axis is D'(2, -3).

Question 7.
If the coordinates of a point M are (-2, 9) which can also be expressed as (1 + x, y) and y > 0, then find in which quadrant do the following points lie : Ply, x), Q(2, x), R(x, y − 1), S(2x, -3y).
Solution:
Here, M(-2, 9) can also be expressed as (1 + x, y²)
∴ (1 + x, y²) ⇔ (-2, 9) and y > 0
⇒ 1 + x = -2 and y² = 9
⇒ x = -3 and y = √9 = 3 (∵ y > 0)
P (y, x) = P(3, -3), it lies in IV quadrant
Q(2, x) = Q(2, -3), also lies in IV quadrant
R(x², y – 1) = R((-3)², 3 – 1) = R(9, 2), it lies in I quadrant
S(2x, -3y) = S(2 × (-3), -3 × 3) = S(-6, -9), it lies in III quadrant

Question 8.
In the given figure, PQR is an equilateral triangle with coordinates of Q and R as (-2, 0) and (2, 0) respectively. Find the coordinates of the vertex P.
Solution:

Here,
QR = OQ + OR.
= 2 + 2 = 4 units
∴ ∆PQR is an equilateral triangle.
∴ PQ = PR = QR = 4 units
In right-angled ∆OPQ, ∠POQ = 90°
∴ We have PQ² = OP² + OQ²
⇒ OP² = PQ² – OQ²
= 4² – 2² = 16 – 4 = 12
⇒ OP = √12 = 2√3 units
∴ The coordinates of P are (0, 2√3).

Coordinate Geometry Class 9 Extra Questions HOTS

Question 1.
Plot the points A(3, 2), B(-2, 2), C(-2, -2) and D(3, -2) in the cartesian plane. Join these points and name the figure so formed.
Solution:

Figure so formed is ABCD a rectangle.

Question 2.
Write the coordinates of two points on X-axis and two points on Y-axis which are at equal distances from the origin. Connect all these points and make them as vertices of quadrilateral. Name the quadrilateral thus formed.
Solution:
Let a be the equal distance from origin on both axes. Now, the coordinates of two points on equal distance ‘a’on x-axis are Pla, 0) and R(-a, 0). Also, the coordinates of two points on equal distance ‘a’ on Y-axis are Q(0, a) and S(0, -a). Join all the four points on the graph. Now, PQRS, thus formed is a square.

Coordinate Geometry Class 9 Extra Questions Value Based (VBQs)

Question 1.
On environment day, class-9 students got five plants of mango, silver oak, orange, banyan and amla from soil department. Students planted the plants and noted their locations as (x, y).

Plot the points (x, y) in the graph and join them in the given order. Name the figure you get. Which social act is being done by students of class-9 ?
Solution:

The given trees (points) are Mango (2, 0), Silver Oak (3, 4), Orange (0,7), Banyan (-3, 4) and Amla (-2, 0). The location of these trees are Orange (0,7) shown in the graph.
On joining the points of mango, silver oak, orange, banyan and amla in order, the figure so formed is a regular pentagon.
Planting more trees helpful in reducing pollution and make the environment clean and green for the coming generations.

Here we are providing Linear Equations for Two Variables Class 9 Extra Questions Maths Chapter 4 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Linear Equations for Two Variables with Answers Solutions

Extra Questions for Class 9 Maths Chapter 4 Linear Equations for Two Variables with Solutions Answers

Linear Equations for Two Variables Class 9 Extra Questions Very Short Answer Type

Linear Equations In Two Variables Class 9 Extra Questions Question 1.
Linear equation x – 2 = 0 is parallel to which axis ?
Solution:
Here, linear equation is x – 2 ⇒ 0 x = 2
Thus, it is parallel to the y-axis.

Linear Equations Class 9 Extra Questions Question 2.
Express x in term of y: \(\frac{x}{7}\) + 2y = 6
Solution:
Given equation is
\(\frac{x}{7}\) + 2y = 6
⇒ \(\frac{x}{7}\) = 6 – 2y
Thus, x = 7(6 – 2y).

Linear Equations In Two Variables Extra Questions Class 9 Question 3.
If we multiply or divide both sides of a linear equation with a non-zero number, then what will happen to the solution of the linear equation ?
Solution:
Solution remains the same.

Class 9 Maths Chapter 4 Extra Questions Question 4.
Find the value of k for which x = 0, y = 8 is a solution of 3x – 6y = k.
Solution:
Since x = 0 and y = 8 is a solution of given equation
3x – 6y = k
3(0) – 6(8) = k
⇒ k = – 48

Class 9 Linear Equations In Two Variables Extra Questions Question 5.
Write the equation of a line which is parallel to x-axis and is at a distance of 2 units from the origin.
Solution:
Here, required line is parallel to x-axis and at a distance of 2 units from the origin.
∴ Its equation is
y + 2 = 0
or y – 2 = 0

Extra Questions For Class 9 Maths Chapter 4 With Solution Question 6.
Find ‘a’, if linear equation 3x – ay = 6 has one solution as (4, 3).
Solution:
Since (4, 3) is a solution of given equation.
∴ 3(4) – a(3) = 6
⇒ 12 – 3a = 6
⇒ a = \(\frac{-6}{-3}\)
Hence, a = 2

Linear Equations In Two Variables Class 9 Extra Questions With Solutions Question 7.
Cost of a pen is two and half times the cost of a pencil. Express this situation as a linear equation in two variables.
Solution:
Let cost of a pen be ₹ x and cost of a pencil be ₹ y.
According to statement of the question, we have
x = 2\(\frac{1}{2}\) y
⇒ 2x = 5y or 2x – 5y = 0

Extra Questions On Linear Equations In Two Variables Class 9 Question 8.
In an one day international cricket match, Raina and Dhoni together scored 198 runs. Express the statement as a linear equation in two variables.
Solution:
Let runs scored by Raina be x and runs scored by Dhoni be y.
According to statement of the question, we have
x + y = 198
x + y – 198 = 0

Extra Questions For Class 9 Maths Linear Equations In Two Variables Question 9.
The cost of a table is 100 more than half the cost of a chair. Write this statement as a linear equation in two variables.
Solution:
Let the cost price of a table be ₹ x and that of a chair be ₹ y.
Since the cost price of a table is 100 more than half the cost price of a chair.
∴x = \(\frac{1}{2}\)y + 100
⇒ 2x = y + 200 or 2x – y – 200 = 0.

Linear Equations for Two Variables Class 9 Extra Questions Short Answer Type 1

Class 9 Maths Ch 4 Extra Questions Question 1.
Write linear equation representing a line which is parallel to y-axis and is at a distance of 2 units on the left side of y-axis.
Solution:
Here, required equation is parallel to y-axis at a distance of 2 units on the left side of y-axis.
x = -2 or x + 2 = 0

Chapter 4 Maths Class 9 Extra Questions Question 2.
In some countries temperature is measured in Fahrenheit, whereas in countries like India it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius :
F = \(\left(\frac{9}{5}\right)\)C + 32°
If the temperature is – 40°C, then what is the temperature in Fahrenheit?
Solution:
Given linear equation is
F = \(\left(\frac{9}{5}\right)\)C + 32°
Put C = -40°, we have
F = \(\frac{9}{5}\)(-40°) + 32°
F = – 72° + 32°
F= -40°

Ch 4 Maths Class 9 Extra Questions Question 3.
Give equation of two lines on same plane which are intersecting at the point (2, 3).
Solution:
Since there are infinite lines passing through the point (2, 3).
Let, first equation is x + y = 5 and second equation is 2x + 3y = 13.
Clearly, the lines represented by both equations intersect at the point (2, 3).

Class 9 Maths Linear Equations In Two Variables Extra Questions Question 4.
If ax + 3y = 25, write y in terms of x and also, find the two solutions of this equation.
Solution:

Linear Equations In 2 Variables Class 9 Extra Questions Question 5.
Find the value of k, if (1, -1) is a solution of the equation 3x – ky = 8. Also, find the coordinates of another point lying on its graph.
Solution:
Since (1, -1) is a solution of the equation 3x – ky = 8
∴ 3(1) – k(-1) = 8
⇒ k = 8 – 3 = 5
Thus, the given equation is
3x – 5y = 8

Hence, the coordinates of another point lying on the graph of 3x – 5y = 8 is (6, 2).

Linear Equations In Two Variables Class 9 Important Questions Question 6.
Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is the value of y, when x = 5 ?
Solution:
Given y varies directly as x implies y = kx
But y = 12 for x = 4
⇒ 4k = 12 = k = 3
Put k = 3 in y = kx, we have
y = 3x
Now, when x = 5, y = 3 x 5 = y = 15 …(i)

Linear Equations for Two Variables Class 9 Extra Questions Short Answer Type 2

Linear Equations In Two Variables Class 9 Extra Questions Pdf Question 1.
A fraction becomes , when 2 is subtracted from the numerator and 3 is added to the denominator. Represent this situation as a linear equation in two variables. Also, find two solutions for this.
Solution:
Let numerator and denominator of the given fraction be respectively x and y. According to the statement, we obtain
\(\frac{x-2}{y+3}\) = \(\frac{1}{4}\)
⇒ 4x – 8 = y + 3
⇒ 4x – y – 11 = 0
Which is the required linear equation. When y = 1, then x = 3. When y = 5, then x = 4. Hence, the two solutions are (3, 1) and (4, 5).

Extra Questions For Class 9 Maths Chapter 4 Question 2.
The path of an aeroplane is given by the equation 3x – 4y = 1 2: Represent the path graphically. Also, show that the point (-4,-6) lies on the graph.
Solution:
Given equation is 3x – 4y = 12
∴ y = \(\frac{3 x-12}{4}\).

When x = 0, then y = -3
When x = 4, then y = 0
Table of solutions is

Plot the points (0, – 3), (4, 0) on the graph and join them to get the required graph. From the graph, we see, when
x = -4, then y = -6.
Therefore, (-4,- 6) lies on the graph of given equation.

Class 9 Chapter 4 Maths Extra Questions Question 3.
Express y in terms of x for the equation 3x – 4y + 7 = 0. Check whether the points (23, 4) and \(\left(0, \frac{7}{4}\right)\) lie on the graph of this equation or not.
Solution:
Given equation is 3x – 4y + 7 = 0
∴y = \(\frac{3 x+7}{4}\)
When x = -1, then y = 1
When x = 3, then y = 4
When x = 7, then y = 7
Table of solutions is :


Plot the points (-1, 1), (3, 4), (7, 7). and join them to get the given graph. From the graph, we see that, when x = 0, then y \(\frac{7}{4}\) = 5 and when y = 4,
then x = 3. Thus, (0, \(\frac{7}{4}\)) lies on the graph, whereas (23, 4) does not lie on the graph.

Extra Questions Of Chapter 4 Class 9 Maths Question 4.
The following observed values of x and y are thought to satisfy a linear equation. Write the linear equation :

Draw the graph using the values of x, y as given in the above table. At what points the graph of the linear equation :
(i) cuts the x-axis (ii) cuts the y-axis
Solution:
Let ax + by + c = 0 ……(i)
be the linear equation in two variables. From the table, we have two points A(6, -2) and B(-6, 6) which lie on the graph of the linear equation
⇒ 6a – 2b + c = 0
and -6a + 6b + c = 0
Adding the above two equations, we obtain

Which is the required linear equation in two variables. Plots the points A(6, -2) and B(-6, 6) on the graph. Join them to get line AB.
From the graph, we see that the graph cuts the x-axis at (3, 0) and the y-axis at (0, 2).

Linear Equations for Two Variables Class 9 Extra Questions Long Answer Type

Question 1.
Write the equations of the lines drawn in following graph :

Also, find the area enclosed between these lines.
Solution:
Equations of the lines drawn in the graph are as :
x = -1 or x + 1 = 0,
x = 2 or x – 2 = 0,
y = 1 or y – 1 = 0 and
y = 3 or y – 3 = 0
Figure formed by these lines is a rectangle of dimensions 3 units by 2 units.
Hence, the area enclosed between given lines = 6 sq. units.

Question 2.
If (2, 3) and (4, 0) lie on the graph of equation ax + by = 1. Find the value of a and b. Plot the graph of equation obtained.
Solution:
(2, 3) and (4, 0) lie on the graph of equation
ax + by = 1 …(i)
∴ We have 2a + 3b = 1 … (ii)
and 4a + 0 = 1

Which is required linear equation.
Put x= 0 in eq. (iii), we have
⇒ 3(0) + 2y = 12
⇒ 2y = 12
⇒ y = 6
Put x = 2 in eq. (iii), we have
⇒ 3(2) + 2y = 12
⇒ 2y = 6
⇒ y = 3
Put x = 4 in eq. (iii), we have
⇒ 3(4) + 2y = 12
⇒ 2y = 0
⇒ y = 0
We have the following table :

By plotting the points (0, 6), (2, 3) and (4, 0). Joining them, we obtained the graph of 3x + 2y = 12.

Question 3.
Draw the graphs of the following equations on the same graph sheet :
x = 4, x = 2, y = 1 and y – 3 = 0.
Solution:
Graphs of the given equations are drawn on graph sheet.

Question 4.
Cost of 1 pen is ₹ x and that of 1 pencil is ₹ y. Cost of 2 pens and 3 pencils together is ₹ 18. Write a linear equation which satisfies this data. Draw the graph for the same.
Solution:
Here, cost of 1 pen is ₹x and that
of 1 pencil is ₹ y According to the statement of the question, we have
2x + 3y = 18
⇒ x = \(\frac{18-3 y}{2}\)
When y = 0, x = 9
When y = 4, x= 3.
When y = 6, x = 0
Table of solutions is :

Plot the points (0, 6), (3, 4) and (9, 0). Join them in pairs to get the required line.

Question 5.
Sum of two numbers is 8. Write this in the form of a linear equation in two variables. Also, draw the line given by this equation. Find graphically the numbers, if difference between them is 2.
Solution:
Let the two numbers be x and y.
It is given that sum of two numbers is 8.
∴ x + y = 8
y = 8 – x
When x = 0,
When x = 4, y = 4
When x = 8, y = 0

Table of solutions is :

Plot the points (0, 8), (4, 4), (8, 0) and join them in pairs, we get the required graph.
When difference between two number is 2, then
x – y = 2, x > y
⇒ x = y + 2

When x = 0, y = -2
When x = 2, y = 0
When x = 4, y = 2
Table of solutions is :

Plot these poitns (0, -2), (2, 0), (4, 2) and join them to get the required line.
Graphically, the numbers are : (-2; 4), (-1, -3), (0, -2), (1, -1), (2, 0), (3, 1), (4, 2), (5, 3), (6, 4), (7,5) etc.

Linear Equations for Two Variables Class 9 Extra Questions HOTS

Question 1.
Solve for y :

Solution:

Question 2.
A and B are friends. A is elder to B by 5 years. B’s sister C is half the age of B while A’s father D is 8 years older than twice the age of B. If the present age of D is 48 years, find the present ages of A, B and C.
Solution:
Let the age of B’s sister i.e., C be x years.
∴Age of B be 2x years, age of A be (2x + 5) years.
And age of A’s father i.e., D be 2(2x) + 8 = 4x + 8 years
According to the statement of the question, we have
4x + 8 = 48
⇒ 4x = 48 – 8 = 40
⇒ x = \(\frac{40}{4}\) = 10
Age of A = (2x + 5) years i.e., (2 × 10 + 5) years = 25 years
Age of B = 2x years = 2 × 10 years = 20 years
and Age of C = \(\frac{1}{2}\) of B’s age = \(\frac{1}{2}\) × 20 years = 10 years

Linear Equations for Two Variables Class 9 Extra Questions Value Based (VBQs)

Question 1.
Mrs Sharma lost her purse containing 50 rupee and 100 rupee notes amount to 1500 in a shop. Next day shopkeeper found the purse during dusting. He immediately went to Mrs Sharma’s house and returned the purse and rupees. Mrs Sharma appreciates the shopkeeper for his act.
(i) Represent the situation as an equation and draw the graph.
(ii) What value do you learn from shopkeeper’s act ?
Solution:
(i) Let the number of 50 rupee notes be x
And the number of 100 rupee notes be y
∴ We have 50x + 100y= 1500
⇒ x + 2y = 30
(dividing each term by 50)
2y = 30 – x

We have the following table :

By plotting the points (8, 11), (10, 10) and (12, 9) on the graph and by joining them, we obtain the straight line represented by equation
(i) as shown in graph.
(ii) We should be always honest to feel good.

Question 2.
In an election, a good candidate may lose because 40% of voters do not cast their votes due to various reasons. Form an equation and draw the graph with data. From the graph, find :
(i) The total number of voters, if 720 voters cast their votes.
(ii) The number of votes cast, if the total number of voters are 1000.
(iii) What message did you get from above information ?
Solution:
(i) We have, total number of voters who do not cast their votes = 40%
⇒ Total number of voters who cast their votes = 60%
Let the total number of voters be x and number of voters cast their votes be y

Thus, we have the following table :

By plotting the points (100, 60), (200, 120), (300, 180) on the graph and by joining them, we get the graph of equation (i) as shown in figure.

From the graph, we see that:
(i) When total votes polled = 720
i.e., y = 720, the total number of voters i.e., x = 1200
Hence, total number of voters = 1200.
(ii) When total number of voters x = 1000
Number of votes cast is 600.
(iii) Every voters should cast vote to elect an honest candidate.

Here we are providing Introduction to Euclid’s Geometry Class 9 Extra Questions Maths Chapter 5 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Introduction to Euclid’s Geometry with Answers Solutions

Extra Questions for Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry with Solutions Answers

Introduction to Euclid’s Geometry Class 9 Extra Questions Very Short Answer Type

Class 9 Maths Chapter 5 Extra Questions With Solutions Question 1.
Give a definition of parallel lines. Are there other terms that need to be defined first ? What are they and how might you define them?
Solution:
Two coplanar lines in a plane) which are not intersecting are called parallel lines. The other term intersecting is undefined.

Introduction To Euclidean Geometry Class 9 Extra Questions Question 2.
Give a definition of perpendicular lines. Are there other terms that need to be defined first ? What are they and how might you define them?
Solution:
Two coplanar (in a plane) lines are perpendicular if the angle between them at the point
of intersection is one right angle. The other terms point of intersection and one right angle are undefined.

Euclid’s Geometry Class 9 Extra Questions Question 3.
Give a definition of line segment. Are there other terms that need to be defined first ? What are they and how might you define them ?
Solution:
A line segment PQ of a line ‘l is the continuous part of the line I with end points P and Q.

Here, continuous part of the line ‘l is undefined.

Euclid’s Geometry Class 9 Extra Questions With Solutions Question 4.
Solve the equation a – 15 = 25 and state which axiom do you use here.
Solution:
a – 15 = 25
On adding 15 to both sides, we obtain
a – 15 + 15 = 25 + 15 [using Euclid’s second axiom]
a = 40

Questions On Euclid Geometry Class 9 Question 5.
Ram and Ravi have the same weight. If they each gain weight by 2 kg, how will their new weights be compared ?
Solution:
Let x kg be the weight each of Ram and Ravi.
On adding 2 kg,
Weight of Ram and Ravi will be (x + 2) kg each.
According to Euclid’s second axiom, when equals are added to equals, the wholes are equal.
So, weight of Ram and Ravi are again equal.

Chapter 5 Maths Class 9 Extra Questions Question 6.
If a point C be the mid-point of a line segment AB, then write the relation among AC, BC and AB.
Solution:
Here, C is the mid-point of AB
⇒ AC = BC
⇒ AC = BC = \(\frac{1}{2}\)AB

Euclid Geometry Class 9 Questions And Answers Question 7.
If a point P be the mid-point of MN and C is the mid-point of MP, then write the relation between MC and MN.
Solution:
Here, P is the mid-point of MN and C is the mid-point of MP.

∴ MC = \(\frac{1}{4}\)MN

Extra Questions For Class 9 Maths Chapter 5 Question 8.
How many lines does pass through two distinct points ?
Solution:
One and only one.

Euclid Geometry Class 9 Questions Question 9.
In the given figure, if AB = CD, then prove that AC = BD. Also, write the Euclid’s axiom used for proving it.

Solution:
Here, given that
AB = CD
By using Euclid’s axiom 2, if equals are added to equals, then the wholes are equal, we have
AB + BC = CD + BC
⇒ AC = BD

Introduction to Euclid’s Geometry Class 9 Extra Questions Short Answer Type 2

Euclids Geometry Class 9 Extra Questions Question 1.
Define :
(a) a square (b) perpendicular lines.
Solution:
(a) A square : A square is a rectangle having same length and breadth. Here, undefined terms are length, breadth and rectangle.
(b) Perpendicular lines : Two coplanar (in a plane) lines are perpendicular, if the angle between them at the point of intersection is one right angle. Here, the term one right angle is undefined.

Class 9 Maths Chapter 5 Important Questions Question 2.
In the given figure, name the following :
(i) Four collinear points
(ii) Five rays
(iii) Five line segments
(iv) Two-pairs of non-intersecting line segments.

Solution:
(i) Four collinear points are D, E, F, G and H, I, J, K
(ii) Five rays are DG, EG, FG, HK, IK.
(iii) Five line segments are DH, EI, FJ; DG, HK.
(iv) Two-pairs of non-intersecting line segments are (DH, EI) and (DG, HK).

Class 9 Maths Chapter 5 Extra Questions Question 3.
In the given figure, AC = DC and CB = CE. Show that AB = DE. Write the Euclid’s axiom to support this.

Solution:
We have
AC = DC
CB = CE
By using Euclid’s axiom 2, if equals are added to equals, then wholes are equal.
⇒ AC + CB = DC + CE
⇒ AB = DE.

Question 4.
In figure, it is given that AD=BC. By which Euclid’s axiom it can be proved that AC = BD?

Solution:
We can prove it by Euclid’s axiom 3. “If equals are subtracted from equals, the remainders are equal.”
We have AD = BC
⇒ AD – CD = BC – CD
⇒ AC = BD

Question 4.
If A, B and C are any three points on a line and B lies between A and C (see figure), then prove that AB + BC = AC.

Solution:
In the given figure, AC coincides with AB + BC. Also, Euclid’s axiom 4, states that things
which coincide with one another are equal to one another. So, it is evident that:
AB + BC = AC.

Question 5.
In the given figure, AB = BC, BX = BY, show that
AX = CY.

Solution:
Given that AB = BC
and BX = BY
By using Euclid’s axiom 3, equals subtracted from equals, then the remainders are equal, we have
AB – BX = BC – BY
AX = CY

Question 6.

In the above figure, if AB = PQ, PQ = XY, then AB = XY. State True or False. Justify your answer.
Solution:
True. ∵ By Euclid’s first axiom “Things which are equal to the same thing are equal to one another”.
∴ AB = PQ and XY = PQ ⇒ AB = XY

Question 7.
In the given figure, if ∠1 = ∠3, ∠2 = ∠4 and ∠3 = ∠4, write the relation between ∠1 and ∠2, using an Euclid’s axiom.

Solution:
Here, ∠3 = ∠4, ∠1 = ∠3 and ∠2 = ∠4. Euclid’s first axiom says, the things which are equal to equal thing are equal to one another. So ∠1 = ∠2.,

Question 8.
In the given figure, we have ∠1 = ∠2, ∠3 = ∠4. Show that ∠ABC = ∠DBC. State the Euclid’s Axiom used.
Solution:
Here, we have 1 = ∠2 and ∠3 = ∠4. By using Euclid’s Axiom 2. If equals are added to
equals, then the wholes are equal..
∠1 + ∠3 = ∠2 + ∠4
∠ABC = ∠DBC.

Question 9.
In the figure, we have BX and \(\frac{1}{2}\) AB =\(\frac{1}{2}\) BC. Show that BX = BY.

Solution:
Here, BX = \(\frac{1}{2}\) AB and BY = \(\frac{1}{2}\) BC …(i) [given]
Also, AB = BC [given]
⇒ \(\frac{1}{2}\)AB = \(\frac{1}{2}\)BC …(ii)
[∵ Euclid’s seventh axiom says, things which are halves of the same thing are equal to one another]
From (i) and (ii), we have BX = BY

Question 10.
In the given figure, AC = XD, C is mid-point of AB and D is mid-point of XY. Using an Euclid’s axiom, show that AB = XY.

Solution:
∵ C is the mid-point of AB
AB = 2AC
Also, D is the mid-point of XY
XY = 2XD
By Euclid’s sixth axiom “Things which are double of same things are equal to one another.”
∴ AC = XD = 2AC = 2XD ⇒ AB = XY

Introduction to Euclid’s Geometry Class 9 Extra Questions HOTS

Question 1.
For given four distinct points in a plane, find the number of lines that can be drawn through :
(i) When all four points are collinear.
(ii) When three of the four points are collinear.
(iii) When no three of the four points are collinear.

Solution:

(i) Consider the points given are A, B, C and D.
When all the four points are collinear :
One line \(\overrightarrow{\mathrm{AD}}\).

(ii) When three of the four points are collinear :
4 lines
Here, we have four lines \(\stackrel{\leftrightarrow}{A B}, \stackrel{\leftrightarrow}{B C}, \stackrel{\leftrightarrow}{B D}, \stackrel{\leftrightarrow}{A D}\) (four).

(iii) When no three of the four points are collinear :
6 lines Here, we have
\(\stackrel{\leftrightarrow}{A B}, \stackrel{\leftrightarrow}{B C}, \stackrel{\leftrightarrow}{A C}, \stackrel{\leftrightarrow}{A D}, \stackrel{\leftrightarrow}{B D}, \stackrel{\leftrightarrow}{C D}\) (six).

Question 2.
Show that : length AH > sum of lengths of AB + BC + CD.

Solution:
We have
AH = AB + BC + CD + DE + EF + FG + GH
Clearly, AB + BC + CD is a part of AH.
⇒ AH > AB + BC + CD
Hence, length AH > sum of lengths AB + BC + CD.

Introduction to Euclid’s Geometry Class 9 Extra Questions Value Based (VBQs)

Question 1.
Rohan’s maid has two children of same age. Both of them have equal number of dresses. Rohan on his birthday plans to give both of them same number of dresses. What can you say about the number of dresses each one of them will have after Rohan’s birthday? Which Euclid’s axiom is used to answer this question? What value is Rohan depicting by doing so ? Write one more Euclid’s axiom.
Solution:
Here, Rohan’s maid has two children of same age group and both of them have equal number of dresses. Rohan on his birthday plans to give both of them same number of dresses.
∴ By using Euclid’s Axiom 2, if equals are added to equals, then the whole are equal. Thus, again both of them have equal number of dresses. Value depicted by Rohan are caring and other social values. According to Euclid’s Axiom 3, if equals are subtracted from equals, then the remainders
are equal.

Question 2.
Three lighthouse towers are located at points A, B and C on the section of a national forest to protect animals from hunters by the forest department as shown in figure. Which value is department exhibiting by locating extra towers ? How many straight lines can be drawn from A to C? State the Euclid Axiom which states the required result. Give one more. Postulate.

Solution:
One and only one line can be drawn from A to C. According to Euclid’s Postulate, “A straight line may be drawn from any point to any other point:” An other postulate : “A circle may be described with any centre and any radius.” Wildlife is a part of our environment and conservation of each of its element is important for ecological balance.

Get Chapter Wise MCQ Questions for Class 9 English with Answers of Beehive, Moments, English Language and Literature PDF Free Download prepared here according to the latest CBSE syllabus and NCERT curriculum https://ncert.nic.in/. Students can practice CBSE Class 9 English MCQs Multiple Choice Questions with Answers of Beehive, Moments to score good marks in the examination.

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Class 9 English MCQs Multiple Choice Questions with Answers Beehive, Moments

Practicing these CBSE NCERT Objective MCQ Questions of Class 9 English with Answers Pdf of Beehive, Moments, English Language and Literature will guide students to do a quick revision for all the concepts present in each chapter and prepare for final exams.

MCQ Questions for Class 9 English Language and Literature

MCQ Questions for Class 9 English with Answers Beehive

MCQ Questions for Class 9 English with Answers Beehive Prose

1. The Fun They Had Class 9 MCQ Questions
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4. A Truly Beautiful Mind Class 9 MCQ Questions
5. The Snake and the Mirror Class 9 MCQ Questions
6. My Childhood Class 9 MCQ Questions
7. Packing Class 9 MCQ Questions
8. Reach for the Top Class 9 MCQ Questions
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11. If I Were You Class 9 MCQ Questions

MCQ Questions for Class 9 English with Answers Beehive Poems

1. The Road Not Taken Class 9 MCQ Questions
2. Wind Class 9 MCQ Questions
3. Rain on the Roof Class 9 MCQ Questions
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5. A Legend of the Northland Class 9 MCQ Questions
6. No Men are Foreign Class 9 MCQ Questions
7. The Duck and the Kangaroo Class 9 MCQ Questions
8. On Killing a Tree Class 9 MCQ Questions
9. The Snake Trying Class 9 MCQ Questions
10. A Slumber did my Spirit Seal Class 9 MCQ Questions

MCQ Questions for Class 9 English with Answers Moments

1. The Lost Child Class 9 MCQ Questions
2. The Adventures of Toto Class 9 MCQ Questions
3. Iswaran the Storyteller Class 9 MCQ Questions
4. In the Kingdom of Fools Class 9 MCQ Questions
5. The Happy Prince Class 9 MCQ Questions
6. Weathering the Storm in Ersama Class 9 MCQ Questions
7. The Last Leaf Class 9 MCQ Questions
8. A House is not a Home Class 9 MCQ Questions
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10. The Beggar Class 9 MCQ Questions

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Here we are providing Oh, I Wish I’d Looked After Me Teeth Extra Questions and Answers Class 9 English Literature Reader, Extra Questions for Class 9 English was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-english/

Oh, I Wish I’d Looked After Me Teeth Extra Questions and Answers Class 9 English Literature

Oh, I Wish I’d Looked After Me Teeth Extra Questions and Answers Short Answer Type

Answer the following questions briefly.

Oh, I Wish I Looked After Me Teeth Question Answers Cbse Question 1.
What were the “perils” that the narrator spotted in her teeth? How had they been caused?
Answer:
The “perils” refer to the cavities and tooth decay that she is suffering from. They have been caused by her eating too many sweets as a child.

Oh I Wish I Looked After Me Teeth Question Answers Cbse Question 2.
When did the narrator have “more teeth than fillin’”? What does this tell us about her present condition?
Answer:
As a child, she had more teeth but now she had lost most of her teeth and had to have fillings on her remaining teeth.

Oh I Wish I’d Looked After Me Teeth Question 3.
What does the narrator mean when she says, “My conscience gets horribly pricked”? Why does she feel like this?
Answer:
It means she is feeling very guilty because she realises that she, herself, is responsible for her tooth decay. She also feels guilty at the thought of the number of sweets she has had in the past.

Oh I Wish I Looked After My Teeth Question 4.
How do we know that the narrator had been careless about taking care of her teeth?
Answer:
We know the narrator had been careless about taking care of her teeth as she has mentioned that she had bashed her teeth lightly because she had thought brushing teeth was a waste of time.

Question 5.
Why has the narrator described a filling as a “murder”?
Answer:
The narrator has done this to express the pain that she feels every time that she has to undergo a filling.

Question 6.
Why does the narrator have to look up the dentist’s nose?
Answer:
The narrator does this because she has to lie in the dentist’s chair while he works on her teeth.

Question 7.
What are molars? What is the word that the narrator uses in the poem to describe her teeth? What does the dentist do to them?
Answer:
Molars are teeth that are used for grinding food. She has also called them “choppers”. The dentist drills holes in the teeth and reconstructs them.

Question 8.
What is the mood of the poem?
Answer:
The poem is written in a humorous style, making light of a painful experience, that is, a visit to the dentist.

Question 9.
Why does the narrator say “it was a time of reckonin’” for her now? Why is it ironical?
Answer:
The narrator feels that it was time for her to face the consequences of teasing her mother in the past. It is ironical because in the past she had made fun of her mother’s false teeth and now, she would soon have to get them for herself, too.

Question 10.
What is the rhyme scheme of the poem? Is the title appropriate? Justify your answer.
Answer:
The rhyme scheme is aabba. Yes, the title is appropriate because it expresses the feeling of remorse that the narrator experiences when she has to visit a dentist to treat her decaying teeth. It also highlights her guilt at not having taken proper care of them.

Oh, I Wish I’d Looked After Me Teeth Long Answer Type

Question 1.
Read the following statement where “I” refers to “you” “I can’t afford to, after what Jack’s done to his teeth.” What is it, you think you can’t afford and why? Write a diary entry of not less than 125 words.
Answer:
24 August 20xx
Jack, my friend had not come to school today, so I just dropped in to see him after school. He was in bed with the left side of his face all swollen and in pain. He had a toothache. I never thought tooth aches could be so painful!
He had gone to the dentist who had extracted one of his teeth on the lower jaw because it had cavities. He told Jack it was because he ate so many toffees and did not brush his teeth properly. He also told him that he would have to take out two more teeth after the swelling came down!

I am frightened! I love eating sweets as much as Jack does. But I suppose I can’t afford to like them so much anymore, not after seeing the pain that Jack is suffering. I will have to resist the temptation and cut down on the number of chocolates and toffees I eat. Also, I will have to brush my teeth with greater care if I don’t want cavities in my teeth!

Question 2.
In line 35, the poet has misspelled the word “amalgum”. Why do you think she has done that? Discuss.
Answer:
The word has been misspelt deliberately to create a pun with the word “gum”. On one hand, the word “amalgam” refers to the mixture of mercury and filling used by the dentist to make fillings while the word “gum” refers to the tissues in the jaw area in which the teeth lie embedded.

Oh, I Wish I’d Looked After Me Teeth Reference to Context

Read the extracts given below and answer the following questions.

Question 1.
“Oh, I wish I’d looked after me teeth,
And spotted the perils beneath.
All the toffees I chewed,
And the sweet sticky food,
Oh, I wish I’d looked after me teeth.”

(a) What does the narrator wish for in the first line?
Answer:
The narrator wishes she had taken more care of her teeth.

(b) What “perils” did the narrator face?
Answer:
The “perils” that the narrator faced was the threat of tooth decay and cavities.

(c) What had given rise to these perils?
Answer:
Eating too many sweets had given rise to these perils.

Question 2.
“I wish I’d been that much more willin’
When I had more tooth there than fillin’
To pass up gobstoppers.
From respect to me choppers,
And to buy something else with me shillin”

(a) Explain: “When I had more tooth there than fillin”
Answer:
The narrator talks about a time when the narrator did not have so many cavities in her teeth and did not require so much filling.

(b) What are gobstoppers?
Answer:
Gobstoppers are a type of hard sweet or toffee which is usually round.

(c) Why should she have given up gobstoppers?
Answer:
She should have given up gobstoppers to protect her teeth.

Question 3.
“I wish I’d been that much more willin’
When I had more tooth there than fillin’
To pass up gobstoppers.
From respect to me choppers,
And to buy something else with me shillin”

(a) What is a shillin’?
Answer:
A shilling is a coin that was used in United Kingdom earlier.

(b) Explain “To pass up gobstoppers”?
Answer:
To stop buying or eating gobstoppers.

(c) What is the feeling expressed by the narrator in these lines?
Answer:
The feeling expressed by the narrator is the given lines, is that of regret, guilt and remorse.

Question 4.
“When I think of the lollies I licked, 
And the liquorice all- sorts I picked,
Sherbet dabs, big and little,
All that hard peanut brittle,
My conscience gets horribly pricked.”

(a) What does the line—“My conscience gets horribly pricked” signify?
Answer:
The given line signifies that the narrator is feeling guilty and remorseful.

(b) Why has the narrator listed the sweets she ate?
Answer:
The narrator listed all the sweets she ate to show that she had eaten all kinds of sweets.

(c) What has been the result of the narrator’s fondness for sweets?
Answer:
The result of the narrator’s fondness for sweets is that now she is suffering from tooth decay and cavities and her mouth is full of fillings.

Question 5.
“Oh I showed them the toothpaste all right,
I flashed it about late at night,
But up-and-down brushin’ And pokin’ and fussin’
Didn’t seem worth the time—I could bite!”

(a) Explain “Showed them the toothpaste”
Answer:
She bloodshed her teeth.

(b) Explain “pokin’ and fussin”
Answer:
The terms indicate brushing carefully.

Question 6.
“If I’d known, I was paving the way
To cavities, caps and decay,
The murder of fillin’s Injections and drillin’s,
I’d have thrown all me sherbet away”

(a) What are the narrator’s feelings regarding her visits to the dentist?
Answer:
The narrator finds the visits uncomfortable.

(b) What was it that the narrator did not realise when she ate those sweets?
Answer:
The narrator says she did not know about the damage the sweets would cause her teeth.

(c) Explain ‘paving the way’?
Answer:
Paving the way means making way or in this context it means that all the sweets the narrator was eating was making way for cavities and decay to set in.

Question 7.
“So I lay in the old dentist’s chair, ‘
And I gaze up his nose in despair,
And his drill it do whine,
In these molars of mine.
‘Two amalgum,’ he’ll say, ‘for in there’.”

(a) Why i.s the narrator lying in the “old dentist’s chair”?
Answer:
The narrator is lying in the “old dentist’s chair” as she is getting treatment for her tooth decay.

(b) Explain “drill it do whine”.
Answer:
The dentist’s drill makes a loud whining sound.

(c) What is the dentist doing to her molars?
Answer:
The dentist is drilling her tooth and filling it with tooth filling, he is extracting a rotten tooth and planting a tooth in its place.

Question 8.
“How I laughed at my mother’s false teeth,
As they foamed in the waters beneath.
But now comes the reckonin’
It’s me they are beckonin’
Oh, I wish I’d looked after me teeth.”

(a) How had the narrator behaved when her mother had lost all her teeth?
Answer:
The narrator laughed and made fun of her false teeth.

(b) Explain “As they foamed in the waters beneath”.
Answer:
The narrator’s mother had her false teeth preserved in the water.

(c) “But now comes the reckonin’/It’s me they are beckonin’”. Explain.
Answer:
The narrator feels she will also have to wear false teeth like her mother.

Here we are providing The Man Who Knew Too Much Extra Questions and Answers Class 9 English Literature Reader, Extra Questions for Class 9 English was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-english/

The Man Who Knew Too Much Extra Questions and Answers Class 9 English Literature

The Man Who Knew Too Much Extra Questions and Answers Short Answer Type

Answer the following questions briefly.

The Man Who Knew Too Much Question Answer English Class 9 Question 1.
Why did his fellow trainees dislike Private Quelch?
Answer:
Private Quelch’s fellow trainees disliked him because each time one of them made a mistake he would publicly correct him. Whenever one of them shone in their work, he outshone them. He had a very patronising and condescending attitude towards them.

The Man Who Knew Too Much Question Answers Class 9 English Question 2.
“We used to pride ourselves on aircraft recognition.” How was their pride shattered?
Answer:
The pride of the author and his fellow trainees was shattered when Private Quelch announced that the plane was a North American Harvard Trainer even without looking up at it, adding that it could be identified by the harsh sound of the engine which was caused by the high speed of the airscrews.

The Man Who Knew Too Much Extra Questions and Answers Long Answer Type

The Man Who Knew Too Much Class 9 Question And Answers Pdf Question 1.
“At first, Private Quelch was a hero in the eyes of his fellow soldiers.” Support this observation with suitable examples from the story in about 80-100 words.
Answer:
It is true that at first the narrator and the others at the training camp were in awe of the amount of knowledge Quelch .had about everything under the sun and that is why they had nicknamed him “Professor”. Therefore, the narrator says that when he was able to answer all the questions the Sergeant asked him about rifles accurately, it “enhanced” his glory in the eyes of his colleagues. At another place the narrator mentions, “He had brains. He was sure to get a commission before long.” Again, commenting on his hard working nature he writes, “He worked hard. We had to give him credit for that”, and again, “He was not only miraculously tireless but infuriated us all with his heartiness.” And finally he writes, “At first we had certain respect for him but soon we lived in terror of his approach.”

The Man Who Knew Too Much Class 9 English Question 2.
Private Quelch knew “too much”. Give reasons to prove that he was unable to win the admiration of his superior officers or his colleagues.
Answer:
Though everyone agreed that Quelch knew too much, he soon lost all their respect because of his habit of correcting his colleagues publicly whenever they made a mistake. If anyone shone at his work, he made sure to outshine them. He was always trying to patronise them and show off his knowledge to them. After a while they just couldn’t take his condescending ways any longer and steered clear of him.

The Man Who Knew Too Much Chapter Class 9 Question 3.
Attempt a character sketch of Private Quelch.
Answer:
Private Quelch was a dedicated, committed, and focused man. He had set his heart on becoming an army • officer and getting a stripe and he left no stone unturned to achieve his goal. He borrowed the traihing manual and read it thoroughly, even staying up late at night to prepare himself for the classes to be held the next day. He was very keen to acquaint himself with every aspect of army life and would badger his instructors with all sorts of questions till he got his answers. He worked very hard and during the long marches he never appeared tired or exhausted.

But with all his good qualities, he was not popular because of his habit of showing off and behaving in a condescending manner. If anyone made a mistake he was sure to correct them with no thought to the fact that he was offending their feelings. In fact, he did not even spare his teachers. The Sergeant and the Corporal were highly offended at his interruptions of their lessons and that is why he was sent to the cook house as a punishment by the Corporal. However, the punishment did not seem to have had any effect on his behaviour because he was heard lecturing the cooks on the correct method of peeling potatoes.

The Man Who Knew Too Much English Question 4.
You are the “Professor”. Write a diary entry after your first day at the cookhouse, describing the events that led to this assignment, also express your thoughts and feelings about the events of the day in about 175 words.
Answer:
21 January 20xx
Today was my first day at the cook house! It wasn’t as bad as I had thought it would be! But I was appalled to see how little the cooks know about cooking. Today I had to lecture them on the correct way of peeling potatoes. They have been peeling them so thickly that a lot of the vitamins are getting lost! By the time I complete my term here I will have taught them a thing or two.

After all, I was chosen by Corporal Turnbull for this task. I think he was greatly impressed by my knowledge of grenades. He let me give the whole lecture by myself. Even the Sergeant had been impressed by my knowledge of rifles. I have no doubt they think I am the best trainee in the camp! I know that I march the best and my hut is the cleanest. I simply love reading the training manual. I know the others call me “Professor” behind my back because of all my knowledge. I feel so good. I am eagerly waiting for my stripe. Only then will I fulfil my dream of becoming an army officer. I wonder whether I will get the Best Trainee Award

The Man Who Knew Too Much Extra Questions and Answers Reference to Context

Read the extracts given below and answer the questions that follow.

The Man Who Knew Too Little Class 9 English Question 1.
“I first met Private Quelch at the training depot. A man is liable to acquire in his first week of Army life together with his uniform, rifle, and equipment—a nickname. ”

(a) What was the nickname given to Private Quelch? Why did he get his nickname?
Answer:
The nickname given to Private Quelch was Professor because of his habit of sermonising.

(b) Where did the narrator meet him?
Answer:
The narrator met him at an army training camp.

(c) What do you think does the word “liable” mean in this context?
Answer:
The word ‘liable’ means ‘likely’ in this context.

The Man Who Knew Too Much Private Quelch Class 9 Question 2.
“When he hadfinished, he put questions to us and perhaps in the hope of revenge, he turned with his questions again and again to the Professor. ”

(a) Whom does “he” refer to and why was “he” looking for revenge?
Answer:
In the extracted line, “he” refers to the Sergeant. He was looking for revenge because the Professor had disturbed him during his lecture.

(b) What had “he” been lecturing the trainees on?
Answer:
The Sergeant had been lecturing the trainees on the mechanism of a rifle.

(c) Can you think of another word that has the same meaning as revenge?
Answer:
avenge, exact retribution, make retaliation for

Question 3.
“No Sergeant. It’s all a matter of intelligent reading. ”
(a) Who is the speaker?
Answer:
Private Quelch is the speaker.

(b) What had the sergeant wanted to know?
Answer:
The sergeant wanted to know whether he had been trained previously.

(c) What did the speaker mean by “intelligent reading”?
Answer:
By “intelligent reading” the speaker meant reading up the training manual the day before.

Question 4.
“That was our introduction to him. ”
(a) Who is the person being talked about?
Answer:
The person being talked about is Private Quelch.

(b) Where did they meet him for the first time?
Answer:
They met him for the first time at rifle training.

(c) What do you mean by “introduction” in this context?
Answer:
introduction means the action of introducing . In this extract, it talks about how the narrator and his batch mates got acquainted with private Quelch.

Question 5.
“In pursuit of his ambition, he worked hard. ”
(a) Who is being referred to as “he”? What was “his” ambition?
Answer:
Private Quelch is being referred to as “he”. His ambition was to become an army officer and to get a stripe before his peers did.

(b) Can you name some other word with the same meaning as ambition?
Answer:
aspiration, desire, goal

(c) How did “he” work hard?
Answer:
Private Quelch worked hard by reading up training manuals, questioning his instructors incessantly and drilling enthusiastically. He was also tireless on route marches and would march to the canteen like a guardsman when officers were in sight.

Question 6.
“At first we had a certain respect for him, but soon we lived in terror of his approach. ”

(a) Whom does the word “him” refer to?
Answer:
Here, “him” refers to Private Quelch.

(b) Why did the speaker respect him?
Answer:
The speaker respected him for his knowledge, intelligence and hard work.

(c) Why was the narrator terrified at his approach?
Answer:
The narrator as well as the other trainees, was terrified at his approach because Private Quelch lectured ‘ everyone on everything under the sun.

Question 7.
“He was not a man to be trifled with.,.He was our hero and we used to tell each other that he was so tough that you could hammer nails into him without his noticing it. ”
(a) Whom does “He” refer to?
Answer:
In this extract “He” refers to Corporal Turnbull.

(b) What does a man “not to be trifled with” mean?
Answer:
It means that he was not a man one could act frivolously with.

(c) Explain: One could hammer nails into him without his noticing it.
Answer:
This exaggerated claim meant that Corporal Turnbull was an immensely strong and sturdy man.

Question 8.
“The squad listened in cowed, horrified kind of silence ”
(a) What was the squad doing at this time?
Answer:
The squad was learning about a grenade from Corporal Turnbull at this time

(b) Why were they horrified?
Answer:
The lecture had been interrupted by the Professor and so the squad were scared of the reaction of the Corporal.

(c) Give the meaning of the word “cowed”?
Answer:
“Cowed” means quieted or subdued.

Question 9.
“Through the open door we could see the three cooks standing against the wall as if at bay. ”
(a) Give the meaning of the phrase “keeping at bay”?
Answer:
“Keeping at bay” means keeping a safe distance from someone or something.

(b) Why were the cooks standing against the wall?
Answer:
The cooks were standing against the wall because they were listening to Private Quelch’s lecture in their kitchen.

(c) Who were the cooks reacting against?
Answer:
The cooks were reacting against Private Quelch.

Question 10.
“Most of us could not help glancing at Private Quelch who stood rigidly to attention and stared straight in front of him with an expression of self-conscious innocence. ”
(a) Why did everyone glance at the Professor?
Answer:
Everyone glanced at the Professor to check his reaction as they were expecting Turnbull to give him some tough punishment.

(b) What did they expect would happen?
Answer:
They expected Private Quelch to be punished for trying to show off his knowledge to the Corporal.

(c) What does the Professor’s “self-conscious innocence” show?
Answer:
The Professor’s self-conscious innocence shows that he did not expect to be punished.

Check the below NCERT MCQ Questions for Class 9 History Chapter 6 Extra Questions and Answers Peasants and Farmers with Answers Pdf free download. https://ncertmcq.com/extra-questions-for-class-9-social-science/

Peasants and Farmers Class 9 Extra Questions History Chapter 6

Class 11 History Chapter 6 Extra Questions And Answers Question 1.
When were the laws of cricket first written?
Answer:
In 1744.

Class 9 History Chapter 6 Extra Questions Question 2.
When was the world’s first cricket club formed and where?
Answer:
At Hambledon, and in 1760s.

Class 9 History Chapter 6 MCQ With Answers Question 3.
When was the Marylebone Cricket Club (MCC) founded?
Answer:
1787.

History Class 9 Chapter 6 Question Answer Question 4.
Where was cricket originally played?
Answer:
Cricket was originally played on country commons. As such it has been a village sport.

Class 9th History Chapter 6 Question Answer Question 5.
What led to the introduction of protective equipment in cricket?
Answer:
The invention of vulcanised rubber led to the introduction of pads in 1848 and gloves afterwards.

Peasants And Farmers Class 9 Extra Questions And Answers Question 6.
Who was Len Hutton?
Answer:
A professional batsman who led the English team in 1930s.

Peasants And Farmers Class 9 Questions And Answers Question 7.
Name the founder of the modern public school system.
Answer:
Thomas Arnold, the headmaster of the famous Rugby School.

Ncert History Class 9 Chapter 6 Question Answer Question 8.
Which game was considered the game for the girls during late 19th century?
Answer:
Croquet, and not cricket. It was a slow-paced elegant game, suitable for women.

Class 9 History Chapter 6 Questions And Answers Question 9.
Where is cricket usually played?
Answer:
Cricket is played in Commonwealth countries.

Class 9 History Chapter 6 Question Answer Question 10.
Which and When was the first cricket club established in India?
Answer:
The Calcutta Cricket Club, in 1792.

Class 8 History Chapter 6 Extra Questions And Answers Question 11.
Mention the place where the origins of the Indian Cricket are found in India?
Answer:
Bombay. (Now Mumbai).

Class 6 History Chapter 9 Extra Questions Question 12.
Name the first Indian community which started playing the game of cricket in the country.
Answer:
Parsis, the small community called the Zoroastrians.

Class 6 History Chapter 6 Extra Questions And Answers Question 13.
Who was India’s captain when she played against England in 1932?
Answer:
C. K. Nayudu.

Class 9 Chapter 6 Extra Questions Question 14.
When was the Imperial Cricket Conference renamed to International Cricket Conference and why?
Answer:
In 1965, to scrap the colonial influence.

Question 15.
Who saw in cricket the money-making potential?
Answer:
Kerry Packer, an Australian television 1 tycoon.

Question 16.
When was the first one-day international played and between whom?
Answer:
In 1971 between England and Australia.

Question 17.
When was the first leg before wicket (LBW) published?
Answer:
In 1799.

Question 18.
State two grounds on which Gandhjji was averse to cricket.
Answer:

1. It was a game for the privileged.
2. It showed the colonial mindset.

Question 19.
Mention any four laws of cricket as were written down in 1744.
Answer:

1. Two umpires to decide all disputes when the game was played.
2. The stamps were to be 22 inches high.
3. The ball was to be between 5 and. 6 ounces.
4. Two stumps were to be, 22 yards apart.

Question 20.
Why was ball allowed to pitch through the air rather titan rolling it along the ground in the 18th century Britain.
Answer:
During the 1760s and 1770s it became common to pitch the ball through the air, rather than roll; it along the ground. This change, gave bowlers the options of length, deception through the air, plus increased pace. It also opened new possibilities for spin and swing. It also helped the batsman to master timing and sort selection. It was at his time that the curved bat came to be replaced by straight one.

Question 21.
Describe the changes in cricket rule which occurred during the 10th century.
Answer:
The 19th-century saw important changes in the cricket history.
These changes include:

• the fule about wide balls;
• The exact circumference of the ball was specified;
• protective equipment like pads and gloves were introduced;
• boundaries came to be known where previously, all shots had to be run;
• overarm bowling became legal.

Question 22.
How do you explain that cricket remained a colonial game?
Answer:
While some English team games like hockey and football became international games, played all over the world, cricket remained a colonial game, limited to countries that had once been part of the British empire. The pre-industrial oddness of cricket made it a hard game to export. It took roots only in countries that the British conquered and ruled.

In these colonies, cricket was. established as a popular sport either by white settlers (as in Soth Africa, Zimbabwe, Australia, New Zealand, the West Indies and Kenya) or by local elites who wanted to copy the habits of their colonial masters as in India.

Question 23.
Explain the fact that the origins of the Indian cricket are to be found in Bombay (Mumbai).
Answer:
The origins of Indian cricket, that is cricket played by Indians are to be found in Bombay and the first Indian community to start playing the game was the small community of Zoroastrians, the Parsis. Brought into close contact with the British because of their interest in trade and the first Indian community to westernize, the Parsis founded the first Indian cricket club, the Oriental Cricket Club in Bombay in 1848. Parsi clubs were founded and sponsored by Parsi businessmen like the Tatas and the Wadias.

Question 24.
Why did Mahatma Gandhi condemn Pentangular as a communally divisive competition?
Answer:
Mahatma Gandhi condemned the quadrangular or the pentangular tournament a communally divisive competition because cricket had been organised in India on communal and racial lines. The teams that played colonial India’s greatest and most famous first-class cricket tournament did not represent regions, as teams in today’s Ranjit Trophy currently do, but religious communities.

Tournament was initially called the Quadrangular because it was played by four teams: the Europeans, the Parsis, the Hindus and the Muslims. It later became the Pentangular when a fifth team was added, namely, the Rest, which comprised all the communities leftover such as the Indian Christians. For example, Vijay Hazare, a Christian, played for the Rest.

Question 25.
State briefly Gandhiji’s views on the colonial port such as cricket and football.
Answer:
Mahatma Gandhiji believed that sport was essential for creating a balance between body and the mind. However, he often emphasized that games like cricket and hockey were imported into India by the British and were replacing traditional games. Such games as cricket, hockey, football and tennis were for the privileged, he believed. They showed a colonial mindset and were a less effective education than the simple exercise of those worked on the land.

Question 26.
How has television coverage changed’ cricket? Explain.
Answer:
Television coverage changed cricket. It expanded the audience for the game by beaming cricket into small towns and villages. It also broadened cricket social base. Children who had never previously had the chance to watch international cricket because they lived outside the big cities, where top-level cricket was played, could now watch and learn by imitating their heroes.

Question 27.
Who was Kerry Packer? What were his innovations in cricket?
Answer:
Kerry Packer, an Australian television tycoon who saw the money-making potential of cricket as a televised sport, signed up fifty- one of the world’s leading cricketers against the wishes of the national cricket boards and for about two years staged unofficial ‘Tests’ and One-day internationals under the name of World Series Cricket. The innovations he introduced during its time to make cricket more attractive to television audiences endured and changed the nature of the game.

Question 28.
Why is 1970s decade significant in the history of cricket?
Answer:
The decade 1970s has been significant in the history of cricket because of the following reasons:

• The year 1970 was notable for the exclusion of South Africa from international cricket.
• The year 1971 was remarkable because the first one-day international cricket was played between England and Australia in Melbourne. ,
• The year 1977 celebrated the 100 years of test matches in cricket.

Question 29.
Write a brief essay on Cricket in the Victorian England.
Answer:
The organisation of cricket in England’reflected the nature of English society; The rich who could afford to play it for pleasure were called amateurs and the poor who played, it for a living were called professionals. The rich were amateurs for two reasons. One, they considered sport a kind of leisure.

To play for the pleasure of playing and not for money was an aristocratic value. Two, there was not enough money in the game for the rich to be interested. The wages’ of professionals were paid by patronage or subscription or gate money. The game was seasonal and did not offer employment the year-round. Most professionals worked as miners or in other, forms of working-class employment in winter, the off-seasons.

The social superiority of amateurs was built into the custom: of cricket. Amateurs were called Gentlemen while professionals had to be content with being described as Players. They even entered the ground from different entrances. The social superiority of the amateur was also the reason that the captain of a cricket team was traditionally a. batsman; not because batsmen were naturally better captains but because they were generally the Gentlemen.

Question 30.
Why did cricket become popular in the Caribbean countries?
Answer:
Despite the exclusiveness of the white cricket elite in the West Indies/the game became hugely popular in the Caribbean. Success at cricket became a measure of racial equality and political progress. At the time of their independence, many of the. political leaders of Caribbean countries like Forbes Burnham and Eric Williams saw in the game a chance for self-respect and international standing.

When the West Indies won its first Test series against England in 1950, it was celebrated as a national achievement, as a way of demonstrating that West Indians were the equals of white Englishmen. There were two ironies to this great victory. One, the West Indian team that won was captained by a white player. The first time a black player led the West Indies Test team was in 1960 when Frank Worrell was named captain. And two, the West Indies cricket team represented not one nation but several dominions that later became independent countries.

Objective Type Questions

1. Fill in the blanks with words given in the bracket:

(i) Palwankar Baloo and Vithal were……………………. . (brothers, cousins)
Answer:
brothers

(ii) The Oriental Cricket Club was ……………………. .(Bombay, New Delhi)
Answer:
Bombay

(iii) The ……………………. were the first to start cricket in India. (Hindus, Parsis)
Answer:
Parsis

(iv) Vijay Hazare was an Indian ……………………. . (European, Christan)
Answer:
Christian.

2. Mark right (✓) or wrong (✗) in the following:

(i) The cricket was first played in rural America.
Answer:
(✗)

(ii) The Laws of Cricket were written in1844.
Answer:
(✗)

(iii) MCC stands for Melbourne Cricket Club.
Answer:
(✗)

(iv) Vijay Hazare was an Indian Christian.
Answer:
(✓)

3. Choose the correct answer from among the four alternatives given below:

(i) The two sets of stumps are apart:
(a) 20 yards
(b) 21 yards
(c) 22 yards
(d) 23 yards.
Answer:
(c) 22 yards

(ii) The world’s first cricket club was formed in:
(a) London
(b) Hambledon
(c) Washington
(d) Melbourne.
Answer:
(b) Hambledon

(iii) Dennis Lillee, the cricket player, belongs to:
(a) England
(b) New Zealand
(c) Australia
(d) South Africa
Answer:
(c) Australia

(iv) C. K. Nayudu was India’s Test Captain in:
(a) 1930
(b) 1931
(c) 1932
(d) 1933
Answer:
(c) 1932.

Extra Questions for Class 9 Social Science