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Here we are providing Lines and Angles Class 9 Extra Questions Maths Chapter 6 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 Lines and Angles with Answers Solutions

Extra Questions for Class 9 Maths Chapter 6 Lines and Angles with Solutions Answers

Lines and Angles Class 9 Extra Questions Very Short Answer Type

Lines And Angles Class 9 Extra Questions Question 1.
If an angle is half of its complementary angle, then find its degree measure.
Solution:
Let the required angle be x
∴ Its complement = 90° – x
Now, according to given statement, we obtain
x = \(\frac{1}{2}\)(90° – x)
⇒ 2x = 90° – x
⇒ 3x = 90°
⇒ x = 30°
Hence, the required angle is 30°.

Class 9 Lines And Angles Extra Questions Question 2.
The two complementary angles are in the ratio 1 : 5. Find the measures of the angles.
Solution:
Let the two complementary angles be x and 5x.
∴ x + 5x = 90°
⇒ 6x = 90°
⇒ x = 15°
Hence, the two complementary angles are 15° and 5 × 15° i.e., 15° and 75°.

Lines And Angles Extra Questions Question 3.
In the given figure, if PQ || RS, then find the measure of angle m.

Solution:
Here, PQ || RS, PS is a transversal.
⇒ ∠PSR = ∠SPQ = 56°
Also, ∠TRS + m + ∠TSR = 180°
14° + m + 56° = 180°
⇒ m = 180° – 14 – 56 = 110°

Extra Questions On Lines And Angles Class 9 Question 4.
If an angle is 14o more than its complement, then find its measure.
Solution:
Let the required angle be x
∴ Its complement = 90° – x
Now, according to given statement, we obtain
x = 90° – x + 14°
⇒ 2x = 104°
⇒ x = 52°
Hence, the required angle is 52o.

Class 9 Maths Chapter 6 Extra Questions Question 5.
If AB || EF and EF || CD, then find the value of x.

Solution:
Since EF || CD ∴ y + 150° = 180°
⇒ y = 180° – 150° = 30°
Now, ∠BCD = ∠ABC
x + y = 70°
x + 30 = 70
⇒ x = 70° – 30° = 40°
Hence, the value of x is 40°.

Lines And Angles Class 9 Important Questions Question 6.
In the given figure, lines AB and CD intersect at O. Find the value of x.

Solution:
Here, lines AB and CD intersect at O.
∴ ∠AOD and ∠BOD forming a linear pair
⇒ ∠AOD + ∠BOD = 180°
⇒ 7x + 5x = 180°
⇒ 12x = 180°
⇒ x = 15°

Lines And Angles Extra Questions Class 9 Question 7.
In the given figure, PQ || RS and EF || QS. If ∠PQS = 60°, then find the measure of ∠RFE.

Solution:
Since PQ || RS
∴ ∠PQS + ∠QSR = 180°
⇒ 60° + ∠QSR = 180°
⇒ ∠QSR = 120°
Now, EF || QS ⇒ ∠RFE = ∠QSR [corresponding ∠s]
⇒ ∠RFE = 120°

Extra Questions Of Lines And Angles Class 9 Question 8.
In the given figure, if x°, y° and z° are exterior angles of ∆ABC, then find the value of x° + y° + z°.

Solution:
We know that, an exterior angle of a triangle is equal to sum of two opposite interior angles.
⇒ x° = ∠1 + ∠3
⇒ y° = ∠2 + ∠1
⇒ z° = ∠3 + ∠2
Adding all these, we have
x° + y° + z° = 2(∠1 + ∠2 + ∠3)
= 2 × 180°
= 360°

Lines and Angles Class 9 Extra Questions Short Answer Type 1

Class 9 Maths Lines And Angles Extra Questions Question 1.
In the given figure, AB || CD, ∠FAE = 90°, ∠AFE = 40°, find ∠ECD.

Solution:
In AFAE,
ext. ∠FEB = ∠A + F
= 90° + 40° = 130°
Since AB || CD
∴ ∠ECD = FEB = 130°
Hence, ∠ECD = 130°.

Lines And Angles Class 9 Hots Questions Question 2.
In the fig., AD and CE are the angle bisectors of ∠A and ∠C respectively. If ∠ABC = 90°, then find ∠AOC.

Solution:
∵ AD and CE are the bisector of ∠A and ∠C

In ∆AOC,
∠AOC + ∠OAC + ∠OCA = 180°
⇒ ∠AOC + 45o = 180°
⇒ ∠AOC = 180° – 45° = 135°.

Lines And Angles Class 9 Questions Question 3.
In the given figure, prove that m || n.

Solution:
In ∆BCD,
ext. ∠BDM = ∠C + ∠B
= 38° + 25° = 63°
Now, ∠LAD = ∠MDB = 63°
But, these are corresponding angles. Hence,
m || n

Class 9 Maths Chapter 6 Extra Questions With Solutions Question 4.
In the given figure, two straight lines PQ and RS intersect each other at O. If ∠POT = 75°, find the values of a, b, c.

Solution:
Here, 4b + 75° + b = 180° [a straight angle]
5b = 180° – 75° = 105°
b – \(\frac{105^{\circ}}{5}\) = 21°
∴ a = 4b = 4 × 21° = 84° (vertically opp. ∠s]
Again, 2c + a = 180° [a linear pair]
⇒ 2c + 84° = 180°
⇒ 2c = 96°
⇒ c = \(\frac{96^{\circ}}{2}\) = 48°
Hence, the values of a, b and c are a = 84°, b = 21° and c = 48°.

Class 9 Maths Ch 6 Extra Questions Question 5.
In figure, if AB || CD. If ∠ABR = 45° and ∠ROD = 105°, then find ∠ODC.

Solution:

Through O, draw a line ‘l’ parallel to AB.
⇒ line I will also parallel to CD, then
∠1 = 45°[alternate int. angles]
∠1 + ∠2 + 105° = 180° [straight angle]
∠2 = 180° – 105° – 45°
⇒ ∠2 = 30°
Now, ∠ODC = ∠2 [alternate int. angles]
= ∠ODC = 30°

Question 6.
In the figure, ∠X = 72°, ∠XZY = 46°. If YO and ZO are bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OYZ and ∠YOZ.

Solution:
In ∆XYZ, we have
∠X + XY + ∠Z = 180°
⇒ ∠Y + ∠Z = 180° – ∠X
⇒ ∠Y + ∠Z = 180° – 72°
⇒ Y + ∠Z = 108°
⇒ \(\frac{1}{2}\) ∠Y + \(\frac{1}{2}\)∠Z = \(\frac{1}{2}\) × 108°
∠OYZ + ∠OZY = 54°
[∵ YO and ZO are the bisector of ∠XYZ and ∠XZY]
⇒ ∠OYZ + \(\frac{1}{2}\) × 46° = 54°
∠OYZ + 23° = 54°
⇒ ∠OYZ = 549 – 23° = 31°
In ∆YOZ, we have
∠YOZ = 180° – (∠OYZ + ∠OZY)
= 180° – (31° + 23°) 180° – 54° = 126°

Lines and Angles Class 9 Extra Questions Short Answer Type 2

Question 1.
Prove that if two lines intersect each other, then the bisectors of vertically opposite angles are in the same line.

Solution:
Let AB and CD be two intersecting lines intersecting each other in O.
OP and OQ are bisectors of ∠AOD and ∠BOC.
∴ ∠1 = ∠2 and ∠3 = ∠4 …(i)
Now, ∠AOC = ∠BOD [vertically opp. ∠s] ……(ii)
⇒ ∠1 + ∠AOC + ∠3 = ∠2 + ∠BOD + ∠4 [adding (i) and (ii)]
Also, ∠1 + ∠AOC + ∠3 + ∠2 + ∠BOD + ∠4 = 360° (∵ ∠s at a point are 360°]
⇒ ∠1 + ∠AOC + ∠3 + ∠1 + ∠AOC + ∠3 = 360° [using (i), (ii)]
⇒ ∠1 + ∠AOC + ∠3 = 180°
or ∠2 + ∠BOD + ∠4 = 180°
Hence, OP and OQ are in the same line.

Question 2.
In figure, OP bisects ∠BOC and OQ bisects ∠AOC. Prove that ∠POQ = 90°

Solution:
∵ OP bisects ∠BOC
∴ ∠BOP = ∠POC = x (say)
Also, OQ bisects. ∠AOC
∠AOQ = ∠COQ = y (say) .
∵ Ray OC stands on ∠AOB
∴ ∠AOC + ∠BOC = 180° [linear pair]
⇒ ∠AOQ + ∠QOC + ∠COP + ∠POB = 180°
⇒ y + y + x + x = 180°.
⇒ 2x + 2y = 180°
⇒ x + y = 90°
Now, ∠POQ = ∠POC + ∠COQ
= x + y = 90°

Question 3.
In given figure, AB || CD and EF || DG, find ∠GDH, ∠AED and ∠DEF.

Solution:
Since AB || CD and HE is a transversal.
∴ ∠AED = ∠CDH = 40° [corresponding ∠s]
Now, ∠AED + ∠DEF + ∠FEB = 180° [a straight ∠]
40° + CDEF + 45° = 180°
∠DEF = 180° – 45 – 40 = 95°
Again, given that EF || DG and HE is a transversal. .
∴ ∠GDH = ∠DEF = 95° [corresponding ∠s]
Hence, ∠GDH = 95°, ∠AED = 40° and ∠DEF = 95°

Question 4.
In figure, DE || QR and AP and BP are bisectors of ∠EAB and ∠RBA respectively. Find ∠APB.

Solution:

Here, AP and BP are bisectors of ∠EAB and ∠RBA respectively.
⇒ ∠1 = ∠2 and ∠3 = ∠4
Since DE || QR and transversal n intersects DE and QR at A and B respectively.
⇒ ∠EAB + ∠RBA = 180°
[∵ co-interior angles are supplementary]
⇒ (∠1 + ∠2) + (∠3 + ∠4) = 180°
⇒ (∠1 + ∠1) + (∠3 + ∠3) = 180° (using (i)
⇒ 2(∠1 + ∠3) = 180°
⇒ ∠1 + ∠3 = 90°
Now, in ∆ABP, by’angle sum property, we have
∠ABP + ∠BAP + ∠APB = 180°
⇒ ∠3 + ∠1 + ∠APB = 180°
⇒ 90° + ∠APB = 180°
⇒ ∠APB = 90°

Question 5.
In the given figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = \(\frac{1}{2}\) (∠QOS – ∠POS).

Solution:
Given that OR is perpendicular to PQ
⇒ ∠POR = ∠ROQ = 90°
∴ ∠POS + ∠ROS = 90°
⇒ ∠ROS = 90° – ∠POS
Adding ∠ROS to both sides, we have
∠ROS + ∠ROS = (90° + ∠ROS) – ∠POS
⇒ 2∠ROS = ∠QOS – ∠POS
⇒ ∠ROS = \(\frac{1}{2}\) (∠QOS – ∠POS).

Question 6 .
If two parallel lines are intersected by a transversal, then prove that the bisectors of any two corresponding angles are parallel.
Solution:
Given : AB || CD and transversal PQ meet these lines at E and F respectively. EG and FH are
the bisectors of pair of corresponding angles ∠PEB and ∠EFD.
To Prove : EG || FH Proof
∵ EG and FH are bisectors of ∠PEB respectively.

∠PEG = ∠EFH
Which are corresponding angles of EG and FH
∴ EG || FH.

Question 7.
In the given figure, m and n are two plane mirrors perpendicular to each other. Show that incident rays CA is parallel to reflected ray BD.

Solution:
Let normals at A and B meet at P.
As mirrors are perpendicular to each other, therefore, BP || OA and AP || OB.
So, BP ⊥ PA i.e., ∠BPA = 90°
Therefore, ∠3 + ∠2 = 90° [angle sum property] …(i)
Also, ∠1 = ∠2 and ∠4 = ∠3 [Angle of incidence = Angle of reflection]
Therefore, ∠1 + ∠4 = 90° [from (i)) …(ii]
Adding (i) and (ii), we have
∠1 + ∠2 + ∠3 + ∠4 = 180°
i.e., ∠CAB + ∠DBA = 180°
Hence, CA || BD

Lines and Angles Class 9 Extra Questions Long Answer Type

Question 1.
If two parallel lines are intersected by a transversal, prove that the bisectors of two pairs of interior angles form a rectangle.
Solution:
Given : AB || CD and transversal EF cut them at P and Q respectively and the bisectors of
pair of interior angles form a quadrilateral PRQS.

To Prove : PRQS is a rectangle.
Proof : ∵ PS, QR, QS and PR are the bisectors of angles
∠BPQ, ∠CQP, ∠DQP and ∠APQ respectively.
∴∠1 =\(\frac{1}{2}\) ∠BPQ, ∠2 = \(\frac{1}{2}\)∠CQP,
∠3 = \(\frac{1}{2}\)∠DQP and ∠4 = \(\frac{1}{2}\)∠APQ
Now, AB || CD and EF is a transversal
∴ ∠BPQ = ∠CQP
⇒ ∠1 = ∠2 (∵∠1 \(\frac{1}{2}\) ∠BPQ and ∠2 = \(\frac{1}{2}\)∠QP)
But these are pairs of alternate interior angles of PS and QR
∴ PS || QR
Similarly, we can prove ∠3 = ∠4 = QS || PR
∴ PRQS is a parallelogram.
Further ∠1 + ∠3 = \(\frac{1}{2}\)∠BPQ + \(\frac{1}{2}\)∠DQP = \(\frac{1}{2}\) (∠BPQ + ∠DQP)
= \(\frac{1}{2}\) × 180° = 90° (∵ ∠BPQ + ∠DQP = 180°)
∴ In ∆PSQ, we have ∠PSQ = 180° – (∠1 + ∠3) = 180° – 90° = 90°
Thus, PRRS is a parallelogram whose one angle ∠PSQ = 90°.
Hence, PRQS is a rectangle.

Question 2.
If in ∆ABC, the bisectors of ∠B and ∠C intersect each other at O. Prove that ∠BOC = 90° + \(\frac{1}{2}\)∠ A.

Solution:
Let ∠B = 2x and ∠C = 2y
∵OB and OC bisect ∠B and ∠C respectively.
∠OBC = \(\frac{1}{2}\)∠B = \(\frac{1}{2}\) × 2x = x
and ∠OCB = \(\frac{1}{2}\)∠C = \(\frac{1}{2}\) × 2y = y
Now, in ∆BOC, we have
∠BOC + ∠OBC + ∠OCB = 180°
⇒ ∠BOC + x + y = 180°
⇒ ∠BOC = 180° – (x + y)
Now, in ∆ABC, we have
∠A + 2B + C = 180°
⇒ ∠A + 2x + 2y = 180°
⇒ 2(x + y) = \(\frac{1}{2}\)(180° – ∠A)
⇒ x + y = 90° – \(\frac{1}{2}\)∠A …..(ii)
From (i) and (ii), we have
∠BOC = 180° – (90° – \(\frac{1}{2}\)∠A) = 90° + \(\frac{1}{2}\) ∠A

Question 3.
In figure, if I || m and ∠1 = (2x + y)°, ∠4 = (x + 2y)° and ∠6 = (3y + 20)°. Find ∠7 and ∠8.

Solution:
Here, ∠1 and ∠4 are forming a linear pair
∠1 + ∠4 = 180°
(2x + y)° + (x + 2y)° = 180°
3(x + y)° = 180°
x + y = 60
Since I || m and n is a transversal
∠4 = ∠6
(x + 2y)° = (3y + 20)°
x – y = 20
Adding (i) and (ii), we have
2x = 80 = x = 40
From (i), we have
40 + y = 60 ⇒ y = 20
Now, ∠1 = (2 x 40 + 20)° = 100°
∠4 = (40 + 2 x 20)° = 80°
∠8 = ∠4 = 80° [corresponding ∠s]
∠1 = ∠3 = 100° [vertically opp. ∠s]
∠7 = ∠3 = 100° [corresponding ∠s]
Hence, ∠7 = 100° and ∠8 = 80°

Question 4.
In the given figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28o and ∠QRT = 65°. Find the values of x, y and z.

Solution:
Here, PQ || SR .
⇒ ∠PQR = ∠QRT
⇒ x + 28° = 65°
⇒ x = 65° – 28° = 37°
Now, in it. ∆SPQ, ∠P = 90°
∴ ∠P + x + y = 180° [angle sum property]
∴ 90° + 37° + y = 180°
⇒ y = 180° – 90° – 37° = 53°
Now, ∠SRQ + ∠QRT = 180° [linear pair]
z + 65° = 180°
z = 180° – 65° = 115°

Question 5.
In figure, AP and DP are bisectors of two adjacent angles A and D of a quadrilateral ABCD. Prove that 2∠APD = ∠B + ∠C.

Solution:
In quadrilateral ABCD, we have
∠A + ∠B.+ ∠C + ∠D = 360°

Question 6.
In figure, PS is bisector of ∠QPR ; PT ⊥ RQ and Q > R. Show that ∠TPS = \(\frac{1}{2}\)(∠Q – ∠R).

Solution:
Since PS is the bisector of ∠QPR
∴∠QPS = ∠RPS = x (say)
In ∆PRT, we have
∠PRT + ∠PTR + ∠RPT = 180°
⇒ ∠PRT + 90° + ∠RPT = 180°
⇒ ∠PRT + ∠RPS + ∠TPS = 90°
⇒ ∠PRT + x + ∠TPS = 90°
⇒ ∠PRT or ∠R = 90° – ∠TPS – x
In ΔΡQT, we have
∠PQT + ∠PTQ + ∠QPT = 180°
⇒ ∠PQT + 90° + ∠QPT = 180°
⇒ ∠PQT + ∠QPS – TPS = 90°
⇒ ∠PQT + x – ∠TPS = 90° [∵∠QPS = x]
⇒ ∠PQT or ∠Q = 90° + ∠TPS – x …(ii)
Subtracting (i) from (ii), we have
⇒ ∠Q – ∠R = (90° + ∠TPS – x) – 190° – ∠TPS – x)
⇒ ∠Q – ∠R = 90° + ∠TPS – X – 90° + ∠TPS + x
⇒ 2∠TPS = 2Q- ∠R
⇒ ∠TPS = \(\frac{1}{2}\)(Q – ∠R)

Lines and Angles Class 9 Extra Questions HOTS

Question 1.
In the given figure, p ll q, find the value of x.

Solution:

Extend line p to meet RT at S.
Such that MS || QT
Now, in ARMS, we have
∠RMS = 180° – ∠PMR (linear pair]
= 180° – 120°
= 60°
∠RMS + ∠MSR + ∠SRM = 180° [by angle sum property of a ∆]
⇒ 60° + ∠MSR + 30o = 180°
⇒ MSR = 90°
Now, PS || QT – ∠MSR = ∠RTQ
⇒ ∠RTQ = x = MSR = 90° (corresponding ∠s]

Question 2.
In a triangle, if ∠A = 2∠B = 6∠C, find the measures of ∠A, ∠B and ∠C and find the value of \(\frac{\angle A+2 \angle B}{3 \angle C}\).
Solution:
Let us consider ∠A = 2∠B = 6∠C = x
∴∠A = x
2∠B = x = ∠B = \(\frac{1}{2}\)

Lines and Angles Class 9 Extra Questions Value Based (VBQs)

Question 1.
Students in a school are preparing banner for a rally to make people aware for saving electricity. What value are they exhibiting by doing so ? Parallel lines I and m are cut by transversal t, if ∠4 = ∠5 and ∠6 = ∠7, what is measure of angle 8 ?

Solution:
Here, given that ∠4 = ∠5 and ∠6 = ∠7
Now, I || m and t is a transversal
∴ ∠4 + ∠5 + ∠6 + ∠7 = 180° [∵ co-interior angles are supplementary]
∠5 + ∠5 + ∠6 + ∠6 = 180° [using (i)]
2(∠5 + ∠6) = 180°
∠5 + ∠6 = 90°
We know that, sum of all interior angles of a triangle is 180°
∴ ∠8 + ∠5 + ∠6 = 180°
⇒ ∠8 + 90° = 180° [using (ii)]
⇒ ∠8 = 180° – 90° = 90°
Save electricity, save energy.

Question 2.
To protect poor people from cold weather, Ram Lal. has given his land to make a shelter home for them. What value is being exhibited by him? In the given figure, ‘sides QP and RQ of ∆PQR are produced to point S and T respectively. If ∠PQT = 110° and ∠SPR = 135°, find ∠PRQ.

Solution:
Here,
∠SPR + ∠QPR = 180° [a linear pair]
135° + ∠QPR = 180° [∵ ∠SPR = 135°]∠
⇒ ∠QPR = 180° – 135° = 45°
In ∆PQR, by exterior angle property, we have
∠QPR + ∠PRQ = ∠PQT
45° + ∠PRQ = 110°
∠PRQ = 110° – 45° = 65°
Helpful nature, service of mankind and helping the needy people.

Question 3.
In an activity of mathematics, a teacher ask students to divide a circular sheet of radius 15 cm into 6 equal parts to write 6 values they like. What is the central angle subtended by each part ? Give the six values of your preference.
Solution:

Draw a circle of radius 15 cm and divide the circular region into six equal parts by constructing an angle of measure 60° at the centre.
Central angle subtended by each part
= \(\frac{360^{\circ}}{6}\) = 60°
The six values are : Honesty, Truth, Regularity, Tolerance, Punctuality and Sincerity.

Here we are providing Coordinate Geometry Class 10 Extra Questions Maths Chapter 7 with Answers Solutions, Extra Questions for Class 10 Maths was designed by subject expert teachers.  https://ncertmcq.com/extra-questions-for-class-10-maths/

Extra Questions for Class 10 Maths Coordinate Geometry with Answers Solutions

Extra Questions for Class 10 Maths Chapter 7 Coordinate Geometry with Solutions Answers

Coordinate Geometry Class 10 Extra Questions Very Short Answer Type

Answer the following questions in one word, one sentence or as per the exact requirement of the question.

Coordinate Geometry Class 10 Extra Questions Question 1.
What is the area of the triangle formed by the points 0 (0, 0), A (-3, 0) and B (5, 0)?
Solution:
Area of ∆OAB = \(\frac{1}{2}\) [0(0 – 1) – 3(0 – 0) + 5(0 – 0)] = 0
⇒ Given points are collinear

Coordinate Geometry Extra Questions Class 10 Question 2.
If the centroid of triangle formed by points P (a, b), Q (b, c) and R (c, a) is at the origin, what is the value of a + b + c?
Solution:

Coordinate Geometry Class 10 Extra Questions With Solutions Pdf Question 3.
AOBC is a rectangle whose three vertices are A (0, 3), 0 (0, 0) and B (5, 0). Find the length of its diagonal.
Solution:

Class 10 Coordinate Geometry Extra Questions Question 4.
Find the value of a, so that the point (3, a) lie on the line 2x – 3y = 5.
Solution:
Since (3, a) lies on the line 2x – 3y = 5
Then 2(3) – 3(a) = 5
– 3a = 5 – 6
– 3a = -1
⇒ a = \(\frac{1}{3}\)

Coordinate Geometry Class 10 Important Questions Question 5.
Find distance between the points (0, 5) and (-5, 0).
Solution:
Here x₁ = 0, y₁ = 5, x₂ = -5 and y₂ = 0)

Extra Questions Of Coordinate Geometry Class 10 Question 6.
Find the distance of the point (-6,8) from the origin.
Solution:
Here x₁ = -6, y₁ = 8
x₂ = 0, y₂ = 0

Class 10 Maths Chapter 7 Extra Questions With Solutions Question 7.
If the distance between the points (4, k) and (1, 0) is 5, then what can be the possible values of k?
Solution:
Using distance formula

Coordinate Geometry Class 10 Extra Questions With Solutions Question 8.
If the points A (1, 2), B (0, 0) and C (a, b) are collinear, then what is the relation between a and b?
Solution:
Points A, B and C are collinear
⇒ 1(0 – b) + 0 (b – 2) + a(2 – 0) = 0
⇒ -b + 2a = 0 or 2a = b

Extra Questions On Coordinate Geometry Class 10 Question 9.
Find the ratio in which the line segment joining the points (-3, 10) and (6, – 8) is divided by (-1, 6).
Solution:
In Fig. 6.6, let the point P(-1, 6) divides the line joining A(-3, 10) and B (6, -8) in the ratio k : 1

Hence, the point P divides AB in the ratio 2 : 7.

Coordinate Geometry Class 10 Extra Questions Pdf Question 10.
The coordinates of the points P and Q are respectively (4, -3) and (-1, 7). Find the abscissa of a point R on the line segment PQ such that \(\frac{P R}{P Q}\) = \(\frac{3}{5}\).
Solution:

Coordinate Geometry Class 10 Extra Questions Short Answer Type 1

Coordinate Geometry Class 10 Questions With Solutions Question 1.
Write the coordinates of a point on x-axis which is equidistant from the points (-3, 4) and (2, 5).
Solution:
Let the required point be (x, 0).
Since, (x, 0) is equidistant from the points (-3, 4) and (2, 5) .

Class 10 Maths Coordinate Geometry Extra Questions Question 2.
Find the values of x for which the distance between the points P (2, -3) and Q (x, 5) is 10.
Solution:

Coordinate Geometry Questions Class 10 Question 3.
What is the distance between the points (10 cos 30°, 0) and (0, 10 cos 60°)?
Solution:

Extra Questions On Distance Formula Class 10 Question 4.
In Fig. 6.8, if A(-1, 3), B(1, -1) and C (5, 1) are the vertices of a triangle ABC, what is the length of the median through vertex A?
Solution:”

Ch 7 Maths Class 10 Extra Questions Question 5.
Find the ratio in which the line segment joining the points P (3, -6) and Q (5,3) is divided by the x-axis.
Solution:
Let the required ratio be λ : 1

Given that this point lies on the x-axis

Thus, the required ratio is 2 : 1.

Class 10 Maths Ch 7 Extra Questions Question 6.
Point P (5, -3) is one of the two points of trisection of the line segment joining the points A (7, -2) and B (1, -5). State true or false and justify your answer.
Solution:
Points of trisection of line segment AB are given by

∴ Given statement is true.

Coordinate Geometry Extra Questions Question 7.
Show that ∆ABC, where A(-2, 0), B(2, 0), C(0, 2) and APQR where P(-4, 0), Q(4, 0), R(0,4) are similar triangles.
OR
Show that ∆ABC with vertices A(-2, 0), B(0, 2) and C(2, 0) is similar to ∆DEF with vertices D(-4, 0), F(4,0) and E(0, 4).
[∆PQR is replaced by ∆DEF]
Solution:

Chapter 7 Maths Class 10 Extra Questions Question 8.
Point P (0, 2) is the point of intersection of y-axis and perpendicular bisector of line segment joining the points, A (-1, 1) and B (3, 3). State true or false and justify your answer.
Solution:
The point P (0, 2) lies on y-axis

AP ≠ BP
∴ P(0, 2) does not lie on the perpendicular bisector of AB. So, given statement is false.

Coordinate Geometry Class 10 Questions Question 9.
Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Solution:
Let A (5,-2), B (6, 4) and C (7, -2) be the vertices of a triangle

Here, AB = BC
∴ ∆ABC is an isosceles triangle.

Class 10 Coordinate Geometry Important Questions Question 10.
If (1, 2), (4, y), (x, 6) and (3,5) are the vertices of a parallelogram taken in order, find x and y.
Solution:
Let A(1, 2), B(4, y), C(x, 6) and D(3, 5) be the vertices of a parallelogram ABCD.
Since, the diagonals of a parallelogram bisect each other.

Hence, x = 6 and y = 3.

Question 11.
Find the ratio in which y-axis divides the line segment joining the points A(5, -6) and B(-1, 4). Also, find the coordinates of the point of division.
Solution:
Let the point on y-axis be P(0, y) and AP : PB = k : 1

Question 12.
Let P and Q be the points of trisection of the line segment joining the points A(2, -2) and B(-7, 4) such that P is nearer to A. Find the coordinates of P and Q.
Solution:
∵ P divides AB in the ratio 1 : 2.

Question 13.
Find the ratio in which the point (-3, k) divides the line-segment joining the points (-5, 4) and (-2, 3). Also find the value of k.
Solution:
Let Q divide AB in the ratio of p : 1

Question 14.
The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q(2, -5) and R(-3, 6), find the coordinates of P.
Solution:
Let the point P be (2y, y)

Hence, coordinates of point P are (16, 8).

Question 15.
If two adjacent vertices of a parallelogram are (3, 2) and (-1, 0) and the diagonals intersect at (2, -5), then find the coordinates of the other two vertices.
Solution:
Let other two coordinates are (x, y) and (x’, y’)
O is mid point of AC and BD

Hence, co-ordinates are (1, -12) and (5, -10).

Coordinate Geometry Class 10 Extra Questions Short Answer Type 2

Question 1.
Determine, if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Solution:
Let A (1, 5), B (2, 3) and C (-2, -11) be the given points. Then we have

Clearly, AB + BC ≠ AC
∴ A, B, C are not collinear.

Question 2.
Find the distance between the following pairs of points:
(i) (-5, 7), (-1, 3)
(ii) (a, b), (-a, -b)
Solution:
(i) Let two given points be A (-5, 7) and B (-1, 3).
Thus, we have x₁ = -5 and x₂ = -1
y₁ = 7 and y₂ = 3

(ii) Let two given points be A (a,b) and B(-a, -b)
Here, x₁ = a and x₂ = -a; y₁ = b and y₂ = -b

Question 3.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for
your answer: (i) (-1, -2), (1, 0), (- 1, 2), (-3, 0) (ii) (4, 5), (7, 6), (4, 3), (1, 2)
Solution:
(i) Let A(-1, -2), B(1, 0), C(-1, 2) and D(-3, 0) be the four given points.
Then, using distance formula, we have,

Hence, four sides of quadrilateral are equal and diagonals AC and BD are also equal.
∴ Quadrilateral ABCD is a square.

(ii) Let A (4, 5), B (7, 6), C (4, 3) and D (1, 2) be the given points. Then,

∴ ABCD is a parallelogram.

Question 4.
Find the value of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
OR
A line segment is of length 10 units. If the coordinates of its one end are (2, -3) and the abscissa of the other end is 10, find its ordinate.
Solution:
We have, PQ = 10

Squaring both sides, we have
⇒ (8)² + (y + 3)² = 100
⇒ (y + 3)² = 100 – 64
⇒ (y + 3)² = 36 or y + 3 ± 16
⇒ y + 3 = 6, y + 3 = -6 or y = 3, y = -9
Hence, values of y are – 9 and 3.

Question 5.
If Q(0, 1) is equidistant from P(5,-3) and R(x, 6) find the value of x. Also, find the distances of QR and PR.
Solution:
Since, point Q(0, 1) is equidistant from P(5, -3) and R(x, 6).
Therefore, QP = QR
Squaring both sides, we have, Qp² = QR²
⇒ (5 – 0)² + (-3 -1)² = (x – 0)² + (6 – 1)²
25 + 16 = x² + 25
x² = 16
∴ x = ±4
Thus, R is (4, 6) or (-4, 6).

Question 6.
Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
OR
Find the coordinates of a point on the x-axis which is equidistant from the points A(2, -5) and B(-2, 9).
Solution:
Let P(x, 0) be any point on x-axis.
Now, P(x, 0) is equidistant from point A(2,-5) and B(-2, 9)
∴ AP = BP

Squaring both sides, we have
(x – 2)² + 25 = (x + 2)² + 81
⇒ x² + 4 – 4x + 25 = x² + 4 + 4x + 81
⇒ -8x = 56
∴ x = \(\frac{56}{-8}\)
∴ The point on the x-axis equidistant from given points is (-7,0).

Question 7.
Find the relation between x and y, if the points (x, y), (1, 2) and (7, 0) are collinear.
Solution:
Given points are A(x, y), B(1, 2) and C(7, 0)
These points will be collinear if the area of the triangle formed by them is zero.
Now, ar(∆ABC) = \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
⇒ 0 = \(\frac{1}{2}\) [x(2 – 0) + 1(0 – y) + 7(y – 2)]
⇒ 0 = \(\frac{1}{2}\) (2x – y + 7y – 14)
⇒ 2x + by – 14 = 0
⇒ x + 3y = 7, which is the required relation between x and y.

Question 8.
Find a relation between x and y such that the point (x, y) is equidistant from the point (3,6) and (-3, 4).
Solution:
Let P(x, y) be equidistant from the points A(3, 6) and B(-3, 4)
i.e., PA = PB
Squaring both sides, we get
AP² = BP²
⇒ (x – 3)² + (y – 6)² = (x + 3)² + (1 – 4)²
⇒ x² – 6x + 9 + y² – 12y + 36 = x² + 6x + 9 +y² – 8y + 16
⇒ -12x – 4y + 20 = 0
⇒ 3x + y – 5 = 0, which is the required relation.

Question 9.
Find the coordinates of the point which divides the line joining of (- 1, 7) and (4, – 3) in the ratio 2 : 3.
Solution:
Let P(x, y) be the required point. Thus, we have

So, the coordinates of P are (1, 3).

Question 10.
Find the coordinates of the points of trisection of the line segment joining (4, – 1) and (-2,-3).
Solution:
Let the given points be A(4, -1) and B(-2,-3) and points of trisection be P and Q.

Lęt AP = PQ = QB = k
PB = PQ + QB = k + k = 2k
AP : PB = k : 2k = 1 : 2.
Therefore, coordinates of P are

Question 11.
Find the ratio in which the line segment joining A(1, -5) and B(-4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
Solution:
Let the required ratio bek: 1. Then, the coordinates of the point of division is

Question 12.
The points A(4, -2), B(7, 2), C(0, 9) and D(-3, 5) form a parallelogram. Find the length of the altitude of the parallelogram on the base AB.
Solution:
Let, DE = h be the height of parallelogram ABCD w.r.t. base AB.

Question 13.
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4).
Solution:
Let the coordinates of A be (x, y)
Now, C is the centre of circle therefore, the coordinates of

Hence, coordinates of A are (3, -10).

Question 14.
If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AB = 5 AB and P lies on the line segment AB.
Solution:

Question 15.
Find the coordinates of the points which divide the line segment joining A (-2, 2) and B (2, 8) into four equal parts (Fig. 6.21).
Solution:
Let P, Q, R be the points that divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.

Since, Q divides the line segment AB into two equal parts, i.e., Q is the mid-point of AB.

Question 16.
Find the area of a rhombus if its vertices (3, 0), (4, 5), (- 1, 4) and (-2, – 1) are taken in order.
Solution:
Let A(3, 0), B(4, 5), C(-1, 4) and D(-2, -1) be the vertices of a rhombus.
Therefore, its diagonals

Question 17.
Find the area of the triangle whose vertices are: (-5, -1), (3, -5), (5, 2)
Solution:
Let A(x₁, y₁) = (-5, -1), B(x₂, y₂) = (3 – 5), C(x₃, y₃) = (5, 2)
∴ area of ∆ABC = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\)(-5(-5 – 2) + 3 (2 + 1) + 5(-1 + 5)]
= \(\frac{1}{2}\)(35 + 9 + 20) = \(\frac{1}{2}\) × 64 = 32 sq units.

Question 18.
If the point A (0, 2) is equidistant from the points B(3, p) and C(p, 5), find p. Also find the length of AB.
Solution:
Given that A (0, 2) is equidistant from B (3, p) and C (p, 5)
∴ AB = AC
or
AB² = AC²
(3 – 0)² + (0 – 2)² = (p – 0)² + (5 – 2)²
3² + p² + 4 – 4p = p² + 9
= 4 – 4p = 0
⇒ 4p = 4
⇒ p = 1

Question 19.
If the points A (-2, 1), B (a, b) and C (4, -1) are collinear and a – b = 1, find the values of a and b.
Solution:
Since the given points are collinear, then area of ∆ABC = 0
⇒ \(\frac{3}{5}\)[x1 ( y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 0
Given, x₁ = -2, y₁ = 1, x₂ = 0, y₂ = b, x₃ = 4, y₃ = -1
Putting the values,
\(\frac{1}{2}\)[-2(b + 1) + a(-1 – 1) + 4(1 – b)] = 0
⇒ -26 – 2 – 2a + 4- 4b = 0
⇒ 2a + 65 = 2
a + 3b = 1 ….. (i)
Given, a – b = 1 … (ii)
Subtracting (i) from (ii), we have
– 4b = 0
⇒ b = 0
Substituting the value of b in (ii), we have a = 1.

Question 20.
If the point P (k-1, 2) is equidistant from the points A (3, k) and B (k, 5), find the values of k.
Solution:
Let the given line segment be divided by point Q. Since P is equidistant from A and B,
AP = BP or Ap² = BP²
[3 – (k – 1)]² + (k – 2)² = [k – (k – 1)]² + (5 – 2)²
(3 – k + 1)² + (k – 2)² = (k – k + 1)² + (3)²
(4 – k)² + (k – 2)² = (1)² + (3)²
⇒ 16 + k² – 8k + k² + 4 – 4k = 1 + 9
⇒ 2k² – 12k + 20 = 10
⇒ k² – 6k + 10 = 5
⇒ ķ² – 6k + 5 = 0
⇒ k² – 5k – k + 5 = 0
⇒ k (k – 5) -1(k – 5) = 0
⇒ k = 1 or k = 5

Question 21.
Find the ratio in which the line segment joining the points A(3, -3) and B(-2, 7) is divided by x-axis. Also find the coordinates of the point of division.
Solution:

Here, point Q is on x-axis so its ordinate is O.
Let ratio be k : 1 and coordinate of point Q be (x, 0)

We are given that A(3, -3) and B(-2, 7)

Question 22.
Find the values of k if the points Ask + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear.
Solution:
Points Ask + 1, 2k), B (3k, 2k + 3) and C (5k – 1, 5k) are collinear
∴ Area of ∆ABC = 0

Question 23.
If the point P(x, y) is equidistant from the points A(a + b, b – a) and B(a – b, a + b). Prove that bx = ay.
Solution:
Given, PA = PB or (PA)² = (PB)²
(a + b – x)² + (b – a – y)² = (a – b – x)² + (a + b – y)²
⇒ (a + b)² + x² – 2ax – 25x + (b – a)² + y² – 2by + 2ay
⇒ (a – b)² + x² – 2ax + 2bx + (a + b)² + y² – 2ay – 2by
⇒ 4ay = 4bx or bx = ay
Hence proved.

Question 24.
If the point C(-1, 2) divides internally the line segment joining the points A(2, 5) and B(x, y) in the ratio of 3 : 4, find the value of x² + y².
Solution:

Question 25.
In the given figure, ∆ABC is an equilateral triangle of side 3 units. Find the coordinates of the other two vertices.
Solution:

Since AB = BC = AC = 3 units
∴ Co-ordinates of B are (5, 0)
Let co-ordinates of C be (x, y)
AC² = BC²
[∵ ∆ABC is an equilateral triangle]
Using distance formula
⇒ (x – 2)² + (y – 0)² = (x – 5)² + (y – 0)²
⇒ x²+ 4 – 4x + y = x²+ 25 – 10x + y² 6x = 21
⇒ x = \(\frac{21}{6}\) = \(\frac{7}{2}\)
Again (x – 2)² + (y – 0)² = 9

Coordinate Geometry Class 10 Extra Questions Long Answer Type

Question 1.
Find the value of ‘k”, for which the points are collinear: (7, -2), (5, 1), (3, k).
Solution:
Let the given points be
A (x₁, y₁) = (7, -2), B (x₂, Y₂) = (5, 1) and C (x₃, y₃) = (3, k)
Since these points are collinear therefore area (∆ABC) = 0
⇒ \(\frac{1}{2}\) [x₁(y₂ – Y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 0
⇒ x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂) = 0
⇒ 7(1 – k) + 5(k + 2) + 3(-2 -1) = 0
⇒ 7 – 7k + 5k + 10 – 9 = 0
⇒ -2k + 8 = 0
⇒ 2k = 8
⇒ k = 4
Hence, given points are collinear for k = 4.

Question 2.
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Solution:
Let A (x₁, y₁) = (0, -1), B (x₂, y₂) = (2, 1), C (x₃, y₃) = (0, 3) be the vertices of ∆ABC.
Now, let P, Q, R be the mid-points of BC, CA and AB, respectively.
So, coordinates of P, Q, R are

Ratio of ar (∆PQR) to the ar (∆ABC) = 1 : 4.

Question 3.
Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).
Solution:

Let A(4, -2), B(-3, -5), C(3, -2) and D(2, 3) be the vertices of the quadrilateral ABCD.
Now, area of quadrilateral ABCD
= area of ∆ABC + area of ∆ADC

Question 4.
A median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ABC whose vertices are A (4,-6), B (3, -2) and C (5, 2).
Solution:
Since AD is the median of ∆ABC, therefore, D is the mid-point of BC.

Hence, the median divides it into two triangles of equal areas.

Question 5.
Find the ratio in which the point P (x, 2), divides the line segment joining the points A (12, 5) and B (4, -3). Also find the value of x.
Solution:

The ratio in which p divides the line segment is \(\frac{3}{5}\), i.e., 3 : 5.

Question 6.
If A (4, 2), B (7, 6) and C (1, 4) are the vertices of a ∆ABC and AD is its median, prove that the median AD divides into two triangles of equal areas.
Solution:
Given: AD is the median on BC.
⇒ BD = DC
The coordinates of midpoint D are given by.

Hence, AD divides ∆ABC into two equal areas.

Question 7.
If the point A (2, -4) is equidistant from P (3, 8) and Q (-10, y), find the values of y. Also find distance PQ.
Solution:
Given points are A(2, 4), P(3, 8) and Q(-10, y)
According to the question,
PA = QA

Question 8.
The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of point Care (0, -3). The origin is the mid-point of the base. Find the coordinates of the points A and B. Also find the coordinates of another point D such that BACD is a rhombus.
Solution:
∵ O is the mid-point of the base BC.
∴ Coordinates of point B are (0, 3). So,
BC = 6 units Let the coordinates of point A be (x, 0).
Using distance formula,

∴ Coordinates of point A = (x, 0) = (3√3, 0)
Since BACD is a rhombus.
∴ AB = AC = CD = DB
∴ Coordinates of point D = (-3√3, 0).

Question 9.
Prove that the area of a triangle with vertices (t, t-2), (t + 2, t + 2) and (t + 3, t) is independent of t.
Solution:
Area of a triangle = \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
Area of the triangle = \(\frac{1}{2}\)[t + 2 – t) + (t + 2) (t – t + 2) + (t + 3) (t – 2 – t – 2)]
= \(\frac{1}{2}\) [2t + 2t + 4 – 4t – 12 ]
= 4 sq. units
which is independent of t.
Hence proved.

Question 10.
The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, -2). If the third vertex is (\(\frac{7}{2}\), y), find the value of y.
Solution:
Given: ar(∆ABC) = 5 sq. units

Question 11.
The coordinates of the points A, B and Care (6, 3), (-3,5) and (4,-2) respectively. P(x, y) is any point in the plane. Show that \(\frac { ar(∆PBC) }{ ar(∆ABC) } \) = \(\frac{x+y-2}{7}\)
Solution:
P(x, y), B(-3, 5), C(4, -2), A(6, 3)

Question 12.
In Fig. 6.32, the vertices of ∆ABC are A(4, 6), B(1, 5) and C(7, 2). A line-segment DE is drawn to intersect the sides AB and AC at D and E respectively such that \(\frac{A D}{A B}=\frac{A E}{A C}=\frac{1}{3}\) Calculate the area of ∆ADE and compare it with area of ∆ABC.
Solution:

\

Question 13.
If a = b = 0, prove that the points (a, a²), (b, b²) (0, 0) will not be collinear.
Solution:
∵ We know that three points are collinear if area of triangle = 0
∴ Area of triangle with vertices (a, a²), (b, b²) and (0, 0)

∵ Area of ∆ ≠ 0
∴ Given points are not collinear.

Coordinate Geometry Class 10 Extra Questions HOTS

Question 1.
The line joining the points (2, 1) and (5, -8) is trisected by the points P and Q. If the point P lies on the line 2x – y + k = 0, find the value of k.
Solution:

As line segment AB is trisected by the points P and Q.
Therefore,
Case I: When AP : PB = 1 : 2.

⇒ P (3, -2)
Since the point P (3,-2) lies on the line
2x – y + k = 0 =
⇒ 2 × 3-(-2) + k = 0
⇒ k = -8

Case II: When AP : PB = 2 : 1.

Coordinates of point P are

Since the point P(4, -5) lies on the line
2x – y + k = 0
∴ 2 × 4-(-5) + k = 0
∴ k = -13

Question 2.
Prove that the diagonals of a rectangle bisect each other and are equal.
Solution:

Let OACB be a rectangle such that OA is along x-axis and OB is along y-axis. Let OA = a and OB = b.
Then, the coordinates of A and B are (a,0) and (0, b) respectively.
Since, OACB is a rectangle. Therefore,
AC = OB
⇒ AC = b
Also, OA = a
⇒ BC = a
So, the coordinates of Care (a, b).

∴ OC = AB.

Question 3.
In what ratio does the y-axis divide the line segment joining the point P (4, 5) and Q (3, -7)?
Also, find the coordinates of the point of intersection.
Solution:
Suppose y-axis divides PQ in the ratio k : 1. Then, the coordinates of the point of division are|

Question 4.
Find the centre of a circle passing through the points (6, -6), (3, -7) and (3, 3).
Solution:
Let O(x, y) be the centre of circle. Given points are A(6, -6), B(3, -7) and C(3, 3).

Question 5.
If the coordinates of the mid-points of the sides of a triangle are (1, 1), (2, – 3) and (3, 4). Find its centroid.
Solution:

Let P(1, 1), Q(2, -3), R(3, 4) be the mid-points of sides AB, BC and CA respectively, of triangle ABC. Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of triangle ABC. Then, P is the mid-point of AB.

Question 6.
Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, -2) and B(3, 7).
Solution:
Let P(x₁, y₁) be common point of both lines and divide the line segment joining A(2, -2) and B(3, 7) in ratio k : 1.

Question 7.
Show that ∆ABC with vertices A (-2, 0), B (2, 0) and C (0, 2) is similar to ∆DEF with vertices
D(4, 0) E (4, 0) and F (0, 4).
Solution:
Given vertices of ∆ABC and ∆DEF are

A(-2, 0), B(2, 0), C(0, 2), D(-4, 0), E(4, 0) and F(0, 4)


Here, we see that sides of ∆DEF are twice the sides of a ∆ABC.
Hence, both triangles are similar.

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Online Education for The Making of a Scientist Extra Questions and Answers Class 10 English Footprints Without Feet

The Making of a Scientist Extra Questions and Answers Short Answer Type

The Making Of A Scientist Extra Questions And Answers Question 1.
Why did viceroy butterflies copy monarchs?
Answer:
Viceroy butterflies copied monarchs because monarchs do not taste good to birds. Viceroy butterflies on the other hand taste good to birds. So, the more they look similar to monarchs, the less likely they are to become a bird’s prey. Thus they protect themselves.

Making Of A Scientist Extra Questions Question 2.
Why did Richard Ebright give up tagging butterflies?
Answer:
Richard Ebright lost interest in tagging butterflies as it was tedious and there was not much feedback. He could recapture only two butterflies in all the time he did it and they were not more than seventy five miles away from where he lived.

Making Of A Scientist Class 10 Extra Questions Question 3.
What are the ingredients in the making of a scientist?
Answer:
The author gave examples from Richard Ebright’s life to show the ingredients to make a scientist. Start with a first rate mind, add curiosity, and mix in the will to win for the right reasons.

The Making Of A Scientist Class 10 Extra Questions And Answers Question 4.
What was the common belief about the twelve tiny gold spots on a monarch pupa? What is the actual purpose of these tiny gold spots?
Answer:
These twelve tiny gold spots were believed to be ornamental only. The actual purpose of these tiny gold spots is to produce a hormone necessary for the butterfly’s full development.

The Making Of A Scientist Extra Questions Question 5.
“But there was one thing I could do-collect things”. What collection did Ebright make? When did he start making collection?
Answer:
Ebright began collecting butterflies, rocks, fossils and coins. He began as early as when he was in kindergarten. He collected with same determination that had marked all his activities.

The Making Of Scientist Extra Questions Question 6.
What other interests besides science did Richard Ebright pursue?
Answer:
Richard Ebright was a champion debater and public speaker. He was a good canoeist and all-around outdoor person. He was also an expert photographer, particularly of natural and scientific exhibits.

Making Of Scientist Extra Questions Question 7.
How did Richard Ebright’s mother help him to become a scientist?
Answer:
Ebright’s mother was his only companion. She used to encourage the child to learn whatever he wanted to learn. She took him on trips, brought him telescopes, microscopes, cameras, mounting materials and other such equipments.

Extra Questions Of The Making Of A Scientist Question 8.
Which book did Ebright mother get for him? How did it change his life?
Answer:
Ebright’s mother got a children’s book called The Travel of Monarch X’ for him. The book invited readers to help study butterfly migrations and actively participate in tagging butterflies to help in the research being conducted by Dr Frederick A. Urquhart. Ebright then went on to raise an entire flock of butterflies in the basement of his home. In this way the book managed to keep his enthusiasm in the study of butterflies alive for several years and opened the world of science to the young collector who never lost his scientific curiosity. ,

The Making Of A Scientist Questions And Answers Question 9.
Why did Richard Ebright raise a flock of butterflies?
Answer:
At the end of the book, “The travels of Monarch X’, readers were invited to help study butterfly migrations. They were asked to tag butterflies for research by Dr Frederick A. Urquhart. The butterfly collecting season around reading lasts six weeks in late summery. If Ebright went to chase them one by one, he could not catch very many. So he decided to raise a flock of butterflies.

The Making Of A Scientist Important Questions Question 10.
Mention any two Ebright contributions to the world of science.
Answer:
Ebright made valuable contributions to the world of science. He discovered an unknown insect hormone and also determined how the cell could read the blueprint of its DNA.

Making Of Scientist Class 10 Extra Questions Question 11.
What lesson did Ebright learn when he did not win anything at the science fair?
Answer:
When Ebright did not win anything at the science fair, he learnt a lesson that he needed to do real experiments, not simply make a neat display. His entry was slides of frog tissues which he showed under a microscope.

The Making Of Scientist Class 10 Extra Questions Question 12.
What lesson does Ebright learn when he does not win anything at a science fair?
Answer:
Ebright realizes that were display of his collection does not mean science. To win at a science fair he will have to do real experiments and prove his worth.

Question 13.
What experiments and projects does he then undertake?
Answer:
He then undertakes the projects and experiments to find out what actually causes the viral disease that kills nearly all Monarch caterpillars. He then works on a project to test the theory that viceroy butterflies copy monarchs to survive.

Question 14.
What are the qualities that go into the making of a scientist?
Answer:
The author mentions three qualities that go into the making of a scientist—a first-rate mind, curiosity, and the will to win for the right reasons. Richard Ebright was a very intelligent student. He was also a champion debater, a public speaker, a good canoeist and an expert photographer. He always tried to put that extra effort in his work. He was competitive, but for the right reasons. From the very beginning, he had a driving curiosity along with a bright mind; and it was this curiosity that ultimately led him to his theory about cell life.

The Making of a Scientist Extra Questions and Answers Long Answer Type

Question 1.
Although Richard does not win anything at the science fair but it was a stepping stone for his success. With reference to the story ‘The Making of a Scientist’ of the above statement, give your comments whether competitions are for winning sake or to give your best at work.
Answer:
It is true, no one can deny the fact that every person wants to be a winner. Each has basic wish to reach ‘ at the top. For that many competitions are organised at various levels. But still we must accept that all cannot be winners. Participation is more important than winning. The participant should work hard to reach their level best. Failures should not make us disheartened and best way to overcome failure is to learn through our mistakes. We cannot deny that experimentation and learning are stepping stones to our success. So we should try to give our best.

Question 2.
Besides curiosity a number of other values are required to become a successful scientist. Explain with reference to the chapter, ‘The Making of a Scientist’.
Answer:
From very young age Richard Ebright was competitive and put in extra effort with curiosity for the right reason to win. But his mother was always very dedicated and made his spirits rise high. He did not lose heart even after losing when he was in seventh grade. To him people around were very encouraging.

His mother’s encouragement was really an eye-opener. She took him on trips, bought him telescopes, microscopes, cameras, mounting materials and other equipment. Thus constant support of each other opened a new world. This helps us to conclude that hard work, parental guidance and keen observation are the qualities which help one to excel.

Question 3.
Ebright’s mother played a pivotal role in enabling him to become a successful scientist. This is true for most of our lives. Our parents help us a lot in our education. Their guidance is very important in what we become in later life. Based on your reading of the story how did Ebright’s mother help him in becoming a scientist?
Answer:
Parents play a pivotal role in determining not just our behaviour but also our future in most cases. They are the ones who teach us wrong from right. For most of us, parents are our role models. Ebright’s mother supported her son in becoming a scientist. She gave him an intensive training. She took him on trips, bought him telescopes, microscopes, cameras, mounting materials and other equipments.

She used to keep her son busy. If she found him sitting idle, she would find work for him-not physical work, but learning things. His mother was very supportive. She wrote to Dr Urquhart also so that her son could be busy in research activity. She helped her son a lot. She inspired him to explore new things and instilled a sense of discovery into her child.

Question 4.
To participate in the competition is more necessary than to win a prize. Explain this statement in the light of Ebright’s participation at the country science fair.
Answer:
We know very well to win is a human nature. Everyone wants to get a winning place everywhere. Our life is full of different competitions at different levels. In these, competitions everyone of us wants to become a winner. But it is always not possible. When we enter any competition, we feel a great zeal. We try our best to get the top position there. But if we don’t get or achieve our goal, we feel disappointed.

Great thinkers have said that participation is more important than winning. The same is proved in Ebright success. Richards Ebright participated in the country science fair, but he lost. There he showed slides of frog tissues. He realised that he should have done some real experiments to be a winner. If he did not participate in that competition, the result might have been different.

Question 5.
How can one become a scientist, an economist, a historian…? Does it simply involve reading t many books on the subject? Does it involve observing, thinking and doing experiments?
Answer:
Reading books is just one aspect of learning. This is an exercise in information gathering. It is how your brain processes the information that affects the degree of learning. The first and the foremost criteria to become a genius in one’s chosen field is to have great curiosity and unending hunger to discover more, Next criteria is a sense of closely observing the things, which further helps you to correlate your findings with what you see or experience in the real world. Experiments are must to test your findings against possible variables and in real life situations. And last but not the least, it is an urge, a strong desire to work really hard on your area of interest.

Question 6.
You must have read about cells and DNA in your science books. Discuss Richard Ebright’s work in light of what you have studied. If you get an opportunity to work like Richard Ebright on projects and experiments, which field would you like to work on and why?
Answer:
Ebright’s work is directly related to Biology. Discovery of a cell’s structure has helped the scientific community ! to understand how the organisms function and grow. This has also helped other scientists to discover how disease causing organisms attack us and grow inside our; body. DNA fingerprints help police to pinpoint to the real culprit. This could not have been possible until DNA was discovered. Monarch I butterflies present an amazing example of a tiny creature migrating thousands of miles from North America to the rainforests of Amazon. Some day we can be in a position to develop as a sturdy and reliable navigation system as that of the Monarch butterflies.

Question 7.
Children everywhere wonder about the world around them. The questions they ask are the beginning of scientific inquiry. Given below are some questions that children in India have
asked professor Yash Pal and Dr Rahul Pal as reported in their book, Discovered Questions?
(i) What is DNA fingerprinting? What are its uses?
(ii) How do honeybees identify their own honeycombs?
(iii) What does rainfall in drops?
Can you answer these questions? You will find professor Yash Pal’s and Dr Rahul Pal’s answers
(as given in Discovered Questions) on Page 75.
Answer:
Classroom activities and self attempt. You may try to find answers to these questions. However Prof. Yash Pal’s answers are given below.

(i) DNA exists as strands of bases that carry genetic information specific to each living thing. The sequence of bases of DNA in each of our cells is the same, but differs from that of any differences make the DNA break at different places when certain protein called enzymes are added to it, resulting in smaller DNA fragments of different sizes. These fragments migrate at different rates in an electric field, resulting in a unique pattern: This pattern is referred to as a DNA fingerprint.

Our DNA is inherited from our parents. Some parts come from the father and some from the mother. DNA fingerprinting can help identify percentage, since a son or a daughter would always exhibit a pattern identifiable as coming from both parents. DNA fingerprinting analysis is very useful in forensic science: from a single hair or tiny spot of blood. It is possible to prove the innocence or guilt of a murder suspect.

Similarly, it is also possible to identify human remains after violent accidents have caused disfigurement. It has been suggested that in the not-so-distant future, a DNA fingerprinting profile of the individual will have to accompany applications for an ID card, a bank account and a driving license. Human right groups say this type of “genetic profiling” constitutes an invasion of privacy. As with a lot of new technology, DNA fingerprinting also has a potential for abuse.

(ii) Honeybees are very sophisticated at position, location and navigation. It is known that they use the sun as a guide. They also appear to have a good memory. They convey the information of finding of food to the hive through an amazingly clever dance language. The dance indicates the direction and distance of the food source with respect to the direction of the sun in the sky! If it is dark inside the hive and a light bulb is switched on the dance is modified to include the light bulb as a new reference direction! Since bees have pictorial memory of some sort, a direction-finding mechanism and a way of reckoning distance, they are probably better equipped for getting back home than any of us!

(iii) Rain is the result of condensation of vapour when the air is cooled below the dew point. All the vapour in a cloud cannot condense at the same time and turn into a large pool of water. Pockets of air move up independently and slowly cool till condensation begins and water droplets form. It is believed that most raindrops start out as tiny ice crystals so tiny that they float down, slowly accretion of more moisture on the way; at lower altitudes, the crystals melt into water droplets. In colder climates, the crystals reach the ground as snowflakes.

Question 8.
You also must have wondered about certain things around you. Share these questions with your class, and try and answer them.
Answer:
Classroom activities and self-attempt
Yes, I also must have wondered about certain things around me. These questions are:

• How does a fruit ripe?
• How do trees give us oxygen etc?
• How does the sky colour change as soon as the sunsets.
  (to answer these question, you may discuss with your science teacher).