Author name: Prasanna

Haryana State Board HBSE 6th Class Maths Solutions Chapter 1 Knowing Our Numbers Ex 1.3 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 1 Knowing Our Numbers Exercise 1.3

Question 1.
Estimate :
(a) 730 + 998
(b) 796-314
(c) 12,904 + 2,888
(d) 28,292 – 21,496.
Solution:
(a) We find 998 > 730.
Round off to hundreds

Actual answer = 730 + 998 = 1728
1728 is rounded to 1700.

(b) To begin with, we round off to hundreds

Actual answer = 796 – 314 = 482
482 is rounded to 500.

(c) We find 12,904 > 2,888
Round off to thousands
12,904 is rounded to 13,000
+2,888 is rounded to 3,000
∴ Estimated sum = 16,000
Actual answer = 12,904 + 2,888 = 15,792

(d) To begin with, we round off to thousands
28,292 is rounded to 28,000
-21,496 is rouned to 21,000
∴ Estimated difference = 7,000
Actual answer = 28,292 – 21496
= 6,796

Question 2.
Give a rough estimate (by round off to nearest hundreds) and also a closer estimate (by’rounding off to nearest tens):
(a) 439 + 334 + 4,317
(b) 1,08,734- 47,599
(c) 8,325 – 491
(d) 4,89,348 – 48,365.
Make four more of such examples.
Solution:
To begin with, we round off to thousands.
439 rounds to 0
334 rounds to 0
and 4,317 rounds to 4,000
Rough estimated sum= 0 + 0 + 4,000
= 4,000
This is not a reasonable estimate.
Actual sum = 439 + 334 + 4,317 = 5,090.
To get a closer estimate, let us try rounding off each number to hundreds.

(b) To begin with, we round off to ten thousands.

Actual difference = 1,08,734 – 47,599 = 61135
To get a closer estimate, let us try rounding off each number to hundreds.

(c) To begin with, we round off to hundreds

Actual difference = 8,325 – 491 = 7,834
To get a closer estimate let us try rounding off each number to tens.

This is a better and more meaningful estimate.

(d) To begin with, we round off to ten thousands

Actual difference = 4,89,348 – 48,365
= 4,40,983
To get a closer estimate, let us try rounding off each number to thousands.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 1 Knowing Our Numbers Ex 1.2 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 1 Knowing Our Numbers Exercise 1.2

Question 1.
A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final days was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.
Solution:

Question 2.
Shekhar is a famous cricket player. He has so far scored 6980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need ?
Solution:

Question 3.
In an election, the successful candidate registered 5,77,500 votes and his nearest rival secured 3,48,700 votes. By what margin did the successful candidate win the election?
Solution:

Question 4.
Kirti bookstore sold books worth Rs. 2,85,891 in the first week of June. The bookstore sold books worth Rs. 4,00,768 in the second week of the month. How much was the sale for the two weeks together ? In which week was the sale greater and by how much ?
Solution:

Question 5.
Find the difference between the greatest and the least numbers that can be written using the digits 6, 2, 7, 4, 3 each only once.
Solution:

Question 6.
A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006 ?
Solution:
Number of screws produced in one day = 2,825
Number of screws produced in 31 days = 2,825 x 31 = 87,575 Ans.

Question 7.
A merchant had Rs. 78,592 with him. He placed an order for purchasing 40 radio sets at Rs. 1,200 each. How much money will remain with him after the purchase ?
Answer:
Cost of one radio set = 1,200
Cost of 39 radio sets = Rs. 1,200 x 40

Question 8.
A student multiplied 7236 by 65 instead of multiplying by 56. How much was his answer greater than the correct answer ?
Solution:

Question 9.
To stitch a shirt 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain ?
Solution:
Cloth needed to stitch one shirt = 2 m 15 cm = 215 cm
Total length of cloth = 40 m = 4,000 cm.
Now,

Hence, 18 shirts can be stitched and 130 cm = 1 m 30 cm cloth will remain.

Question 10.
Medicine is packed in boxes, each such box weighing 4 kg 500 g. How many such boxes can be loaded in a van which cannot carry beyond 800 kg ?
Solution:
800 kg = 800 x 1000 = 8,00,000 g
4 kg 500 g = 4 x 1000 + 500 = 4,500 g

Hence, 177 boxes can be loaded in the van.

Question 11.
The distance between the school and the house of a student is 1 km 875 m. Everyday she walks both ways between her school and home. Find the total distance covered by her in five days.
Solution:
Distance between the school and the house
= 1km 875 m
= 1 x 1000 + 875
= 1,875 m
Both ways distance
= 2x 1,875 m = 3,750 m

∴ Total distance covered by her in five days = 5 x 3,750 m
= 18,750 m
= 18 km 750 m
Now.

Question 12.
A vessel has 4 litres and 500 ml 6f curd. In how many glasses, each of 25 ml capacity, can it be filled ?
Solution:
Total volume of curd = 4l and 500 ml
= 4 x 1000 + 500
= 4,500 ml
Capacity of one glass = 25 ml
Now,

Hence, total number of glasses = 180.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 1 Knowing Our Numbers Ex 1.1 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 1 Knowing Our Numbers Exercise 1.1

Question 1.
Fill in the blanks :
(a) 1 lakh = …………….. thousands.
(b) 1 million = …………. hundred thousands.
(c) 1 crore = …………. lakhs.
(d.) 1 crore = …………. millions.
(e) 1 million = …………. lakh.
Answer:
(a) 10
(b) 10
(c) 10
(d) 10
(e) 10.

Question 2.
Place commas correctly and write the numerals :
(a) Seventy-three lakh seventy-five thousand three hundred seven.
(b) Nine crore five lakh forty-one.
(c) Seven crore fifty-two lakh twenty-one thousand three hundred two.
(d) Fifty-eight million four hundred twenty-three thousand two hundred, two.
(e) Twenty-three lakh thirty thousand ten.
Answer:
(a) 73,75,307;
(b) 9,05,00,041;
(c) 7,52,21,302;
(d) 58,423,202
(e) 23,30,010.

Question 3.
Insert commas suitably and write the names according to Indian System of Numeration :
(a) 87595762
(b) 8546283
(c) 99900046
(d) 98432701.
Answer:
Number — Number Name
(a) 8,75,95,762– Eight crore seventy-five lakh ninety-five thousand seven hundred sixty-two.
(6) 85,46,283 — Eighty-five lakh forty-six thousand two hundred eighty-three.
(c) 9,99,00,046 — Nine crore ninety-nine lakh forty-six.
(d) 9,84,32,701 — Nine crore eighty-four lakh thirty-two thousand seven hundred one.

Question 4.
Insert commas suitably and write the names according to Inter-national system of numeration :
(a) 78921092
(b) 7452283
(c) 99985102
(d) 48049831.
Answer:
Number — Number Name
(a) 78,921,092 — Seventy eight million, nine hundred twenty-one thousand and ninety-two.
(b) 7,452,283 — Seven million, four hundred fifty-two thousand, two hundred and eighty-three.
(c) 99,985,102 — Ninety-nine million, nine hundred eighty-five thousand, one hundred and two.
(d) 48,049,831 — Forty-eight million, forty nine thousand, eight hundred and thirty-one.

Haryana State Board HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles InText Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 7 Congruence of Triangles InText Questions

Discuss, Think & Write (Page No. 137):

Question 1.
When two triangles, say ABC and PQR are given, there are, in all, six possible matchings or correspondences. Two of them are:
(i) ABC ⇔PQR and
(ii) ABC ⇔ QRP
Find the other four correspondences by using two cut outs of triangles. Will all these correspondences lead to congruence ? Think about it.
Solution:
For better understanding of the correspondence let us use a diagram.

(i) The correspondence is ABC ⇔ PQR. This means A ⇔ P ; B ⇔ Q ; and C ⇔ R.
So; (i) \(\overline{\mathrm{QR}} \leftrightarrow \overline{\mathrm{BC}}\) (ii) ∠R ↔ ∠C and
(iii) \(\overline{\mathrm{PQ}} \leftrightarrow \overline{\mathrm{AB}}\)
Similarly we find the all possible matchings or correspondences, e.g.

(ii) ABC ⇔ QRP : ∠A ⇔ ∠Q, ∠B ⇔ ∠R,
∠C ⇔ ∠P
\(\overline{\mathrm{AB}} \leftrightarrow \overline{\mathrm{QR}}, \overline{\mathrm{BC}} \leftrightarrow \overline{\mathrm{RP}}, \overline{\mathrm{AC}} \leftrightarrow \overline{\mathrm{QR}}\)

(iii) ABC ⇔ XYZ,
(iv) ABC ⇔ YZX
(v) ABC ⇔ ZXY,
(vi) ABC ⇔ DEF.

Try These (Page No. 140-141):

Question 1.
In Fig. lengths of the sides of the triangles are indicated. By applying the SSS congruence rule, state which pairs of triangles are congruent. In case of congruent triangles, write the result in symbolic form:

Solution:
From fig(i);
In triangle ABC and PQR
AB = PQ = 1.5
BC = QR = 2.5
and AC = PR = 2.2
This shows that the three sides of one triangle are equal to the three sides of the other triangle. So, by SSS congruence rule, the two triangles are congruent. From the above three equality relations, it can be easily seen that A ⇔ P.B ⇔ Q and C ⇔ R.
So, we have ΔABC ≅ ΔPQR .
From fig. (ii);
In triangles DEF and NLM,
DE = MN = 3.2
EF = LM = 3
FD = NL = 3.5
This shows that the three sides of one triangle are equal to the three sides of the other triangle. So, by SSS congruence rule, the two triangles are congruent.
From the above three equality relations, it can be easily seen that
D ⇔ N, E ⇔ M and F ⇔ L.
So, we have ΔDEF ≅ ΔNLM.
From Fig. (iii);
In triangles ABC and PQR AB = 2, PQ = 5 ∴ AB ≠ PQ
AC = 5, PR = 4 ∴ AC ≠ PR
BC = 4, QR = 2.5 ∴ BC ≠ QR
Hence, ΔABC and ΔPQR are not congruent. From Fig. (iv);
In triangles ABD and ADC,
AB = AC = 3.5
BD = DC = 2.5
AD = AD = Common
This shows that the three sides of one triangle are equal to the three sides of the other triangle. So by SSS congruence rule, the two triangles are congruent.
From the above three equlity relations, it can be easily seen that
A ⇔ A,B ⇔ C and D ⇔ D.
So, we have ΔABC ≅ ΔADC.

Question 2.
In Fig., AB = AC and D is the mid-point of \(\overline{\mathbf{B C}}\)
(r) State the three pairs of equal parts in ΔADB and ΔADC.
(ii) Is ΔADB ≅ ΔADC ? Give reasons.
(iii) Is ∠B = ∠C Why ?

Solution:
From Fig. ;
In triangles ADB and ADC,
(i) AB = AC
AD = AD = Common BD = DC
(ii) This shows that the three sides of one triangle are equal to the three sides of the other triangle. So, by SSS congruence rule, the two triangles are congruent.
From the above three equality relations, it can be easily seen that
A ⇔ A, B ⇔ C and D ⇔ D
So, we have ΔADB ≅ ΔADC.
(iii) By corresponding parts of congruent triangles (C.P.C.T) are equals
hence, ∠B = ∠C.

Question 3.
In Fig., AC = BD and AD = BC. Which of the following statements is meaningfully written ?
(i) ΔABC ≅ ΔABD.
(ii) ΔABC ≅ ΔBAD.

Solution:
From Fig.
(i) In triangles ABC and ABD
AD = BC
BD = AC
AB = AB = Common
This shows that the three sides of one triangle are equal to the three sides of the other triangle. So, by SSS congruence rule, the two angles are congruent.

From the above three equality relations, it can be easily seen that
A ⇔ B, D ⇔ C and ∠CAB = ∠DBA.
So, we have AABC s AABD
(ii) Similarly, above three equality relations ∠ADB = ∠ACB, ∠ABD = ∠BAC and ∠DAB = ∠ABC.
So, we have ΔABC ≅ ΔBAD.

Think, Discuss & Write (Page No. 141):

Question 1.
ABC is an isosceles triangle with AB = AC (Fig. 7.15)
Take a trace copy of ΔABC and also name it as ΔABC.
(i) State the three pairs of equal parts in ΔABC and ΔACB.
(ii) Is ΔABC ≅ ΔACB ? Why or why not ?
(iii) Is ∠B = ∠C ? Why or why not ?

Solution:
in Triangle ΔABC and ΔACB
AB = AC, AC = AB and BC = CB

But in isosceles triangle two sides are equal and base angles are equal.
Hence, ∠B = ∠C and ∠C = ∠B
From the above three equality rela-tions, it can be easily seen that, A ⇔ A, B ⇔ C and C ⇔ B
So, we have ΔABC ≅ ΔACB. (by SAS congruence rule)

(i) AB = C, AC = AB, BC = CB
(ii) Yes, ΔABC ≅ ΔACB
(iii) Yes, ∠B = ∠C. (by CPCT and by isosceles triangle).

Try These (Page No. 143-144):

Question 1.
Which angle is included between the sides DE and EF of ΔDEF ?
Solution:
∠DEF or ∠E is included between the sides DE and EF of ΔDEF.

Question 2.
By applying SAS congruence rule, you want to establish that ΔPQR s ΔFED. It is given that PQ – EF and RP = DE What additional information (pair of equal parts) is needed to establish the congruence ?
Solution:
In ΔPQR and ΔDEF,
PQ = EF =3 and RP = DF = 3.5
If ∠F = ∠P then the ΔPQR ≅ ΔFED
Since, ∠F = 40° but ∠P is not given.

Question 3.
In Fig., measures of some parts of the tirangles are indicated, by applying SAS congruence rule, state the pairs of congruent triangles, if any, in each case. In case of congruent triangles, write them in symbolic form.

Solution:
From Fig. 7.17, ΔACB ≅ ΔRPQ;
ΔPRS ≅ ΔRPQ
Now, In ΔACB and ΔRPQ
AC = RP =2.5,
BC = PQ = 3
and included ∠C = included
∠P =35°
Also, A ⇔ R , B ⇔ Q and C ⇔ P

Therefore, ΔACB = ΔRPQ (By SAS congruence rule)
Now, in a ΔPRS and ΔRPQ
PR = RP = Common
RS = PQ =3.5
and included ∠R = included ∠P = 30°
Also, R ⇔ P.S ⇔ Q and P ⇔ R
Thererefore, ΔPRS ≅ ΔRPQ (By SAS congruence rule)

Question 4.
In Fig. \(\overline{\mathbf{A B}}\) and \(\overline{\mathrm{CD}}\) bisect each other at O
(i) State the three pairs of equal parts in two triangles AOC and BOD.
(ii) Which of the following statements are true ? (Fig.)
(a) ΔAOC ≅ ΔDOB
(b) ΔAOC ≅ ΔBOD

Solution:
(i) From Fig.
AB and CD bisect each other at O.
Therefore, AO = OB, CO = OD,
AC = DB
and ∠AOC = ∠DOB = Vertically opposite angles
= 30°
∠ACO = ∠BDO = alternate interior angle
= 70°
Hence, ∠CAO = ∠DBO = alternate interior angle

ii) (a) In ΔAOC and ΔDOB
AC = DB = 3,
CO = OD

and included ∠C = included
∠D =70°
Also, A ⇔ B,C ⇔ D and O ⇔ O
Therefore, ΔAOC ≅ ΔDOB
(6) In AAOC s ABOD
AO = OB,
CO = OD and
Included angle ∠AOC = included angle ∠BOD = 30°
Also, A ⇔ B, C ⇔ D and O ⇔ O
Therefore, ΔAOC ≅ ΔBOD
(By SAS congruence rule)

Try These (Page No. 145-146):

Question 1.
What is the side included between the angles M and N of MNP

Solution:
Side MN = MP
included angle = ∠NMP
and side NM = NP
included angle = ∠MNP

Question 2.
You want to establish ΔDEF = ΔMNP, using the ASA congruence rule. Your are given that ∠D = ∠M and ∠F = ∠P. What information is needed to establish the congruence ? (Draw a rough figure and then try !)
Solution:
In ΔDEF and ΔMNP,

∠D = ∠M, ∠F = ∠P and DF = MP i.e. D ⇔ M, F ⇔ P and DF ⇔MP
For ASA congruence rule, we need the two sides which the two angles ∠D = ∠M and ∠F = ∠P are included, so the addtional information are as follows:
DF = MP
Hence ΔDEF ≅ ΔMNP.

Question 3.
In Fig., measures of some parts are indicated. By applying ASA congruence rule, state which parts of triangles are congruent. In case of congruence, write the result in symbolic form.

Solution:
(i) In ΔABC and ΔEFD, we have
∠BAC=∠EFD = 40° (given)
∠ABD = ∠DEF = 60° (given)
AB = EF = 3.5 (given)
Hence ΔABC ≅ ΔEFD (By ASA congruence rule)

(ii) In ΔPQR and ΔDEF,
∠PQR = ∠EDF = 90° (given)
∠PRQ = ∠DEF = 50° (given)
PR = EF = 3.5 (given)
Hence ΔPQR ≅ ΔDEF (By ASA congruence rule)

(Hi) In ΔRQP and ΔLNM,
∠PRQ = ∠MLN = 60° (given)
∠PQR = ∠LMN = 30° (given)
RQ = LN = 6 (given)
Hence ΔRQP ≅ ΔLNM (By ASA congruence rule)

(iv) In ΔDAB and ΔCBA,
∠DAB = ∠CBA = (45°+ 30°)
= 75° (given)
∠ABD = ∠BAC = 30° (given)
AB = AR = common
Hence ΔDAB ≅ ΔCBA (By ASA congruence rule)

Question 4.
Given below are measurements of some parts of two triangles. Examine whether the two triangles are congruent or not, by ASA congruence rule. In case of congruence, write it in symbolic form.

Solution:

If under a correspondence
By applying ASA congruence rule, ∠D = ∠Q = 60° (given)
∠F = ∠R = 80° (given)
DF = QR = 5 cm (given)
Hence ADEF ≅ APQR

(ii) In ADEF and APQR
∠D = ∠Q = 60° (given)
∠F = ∠R = 80° (given)
DF = QR = 5 cm (given)
Hence ΔDEF ≅ ΔPQR

(iii) In ΔDEF and ΔPQR
∠D = ∠Q = 60° (given)
∠F = ∠R = 80° (given)

But included side DF ≠ QR
Hence ASA congruence rule not satisfied.
So, ASA congruence rule cannot be applied and we cannot conclude that two triangles are congruent.

(iii) In ΔDEF and ΔPQR
∠E = ∠P = 80° (given)
∠F = ∠R = 30° (given)
But included side EF ≠ PR

So, ASA congruence rule cannot be applied and we cannot conclude that two triangles are congruent.

Question 5.
In Fig., ray AZ bisects ∠DAB as well as ∠DCB.

(i) State the three pairs of equal parts triangles BAC and DAC.
(ii) Is ΔBAC ~ ΔDAC ? Give reasons.
(iii) Is AB = AD ? Justify your answer.
(iv) Is CD = CB ? Give reasons.
Solution:
Since ray AZ bisects ∠DAB as well as∠DCB.
Hence ∠DAC = ∠BAC and ∠DCA = ∠BCA

(i) In ΔBAC and ΔDAC,
∠BAC = ∠DAC (given)
∠BCA = ∠DCA (given)
AC = AC = Common
Hence ΔBAC = ΔDAC (by ASA congruence rule)

(ii) From solution (i), by ASA congruence rule, we have
ΔBAC s ΔDAC.

(iii) Since ΔBAC = ΔDAC
hence by CPCT (corresponding part of congruent triangle)
AD = AB
DC = CB
and ∠D ↔ ∠B
(iv) By CPCT, CD = CB
because ΔBAC = ΔDAC.

Try These (Page No. 148):

Question 1.
In Fig., measures of some parts of triangles are given. By applying RHS congruence rule, state which pairs of triangles are congruent. In case of congruent triangles, write the result in symbolic form.


Solution:
If under a correspondence the hypotenuse and one side of a right-angled triangle are respectively equal to the hypotenuse and one side of another right angled triangle, then the triangles are congruent.

(i) In ΔPQR and ΔDEF,
∠Q = ∠E = 90°
Hypotenuse PR = Hypotenuse DF = 6
Side PQ ≠ side DE
So, the triangles are not congruent.

(ii) In ΔCAB and ΔDAB,
∠C = ∠D = 90°
Hypotenuse AB = Hypotenuse
AB = Common
Side CA = side DB = 2
So, ΔCAB = ΔDAB (By RHS congruence rule)

(iii) In ΔBCA and DCA,
∠B = ∠D = 90°
Hypotenuse CA = Hypotenuse
CA = Common
Side BA = side DA = 3.6
So, ΔBCA = ΔDCA (By RHS congruence rule)

(iv) In ΔPQS and ΔPRS,
∠PSQ = ∠PSR = 90°
Hypotenuse PQ = Hypotenuse PR = 3
Side PS = side PS = Common
So, ΔPQS ≅ ΔPRS.

Question 2.
Is to be established by RHS congruence rule that ΔABC S ΔRPQ. What additional information is needed, if it is given that ∠B = ∠P = 90° and AB = RP ?

Solution:
In ΔABC and ΔRPQ
∠B = ∠P = 90°
Side AB = Side RP
The additional information are as follows :
If hypotenuse AC = hypote-nuse RQ
So, ΔABC ≅ ΔRPQ.

Question 3.
In Fig. , BD and CE are altitudes of ΔABC such that BD = CE.

(i) State the three pairs of equal parts in ΔCBD and ΔBCE.
(ii) Is ΔCBD ≅ ΔBCE ? Why or why not ?
(iii) Is ∠DCB ≅ ∠EBC ? Why or why not ?
Solution:
BD = CE,
(i) ∠BEC = ∠BDC = 90° (given)
BD = EC
BE = DC
i. e., ΔCBD = ΔBCE (SAS congruence)
(ii) Yes, because SAS congruency.
(tit) By CPCT, ∠DCB = ∠EBC. (Fig. 7.30)
hence ΔCBD ≅ ΔBCE.

Question 4.
ABC is an isosceles triangle with AB = AC and AD is one of its altitudes (Fig. 7.31)
(i) State the three pairs of equal parts in ΔADB and ΔADC.
(ii) Is ΔADB ≅ ΔADC ? Why or why not ?
(iii) Is ∠B = ∠C ? Why or why not ?
(iv) Is BD = CD ? Why or why not ?
Sol. Since, ABC is an isosceles triangle.
Hence, AB = AC and AD is altitude of BC.

(i) In ΔADB and ΔADC
∠ADB = ∠ADC = 90° (given)
hypotenuse AB = hypotenuse AC (given)
SideBD = Side DC (given)

(ii) So, ΔADB ≅ ΔADC (By RHS congruence rule)

(iii) By CPCT, ∠B = ∠C
(Isosceles’s base angles are equal)

(iv) By CPCT, BD = CD (AD is one of its altutude of side BC, hence BD = CD)

Haryana State Board HBSE 7th Class Maths Solutions Chapter 15 Visualising Solid Shapes InText Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 15 Visualising Solid Shapes InText Questions

Try These (Page 277)

Question 1.
Match the shape with the name:



Solution:
(i) (b) Cylinder
(ii) (d) Sphere
(iii) (a) Cuboid
(iv) (c) Cube
(v) (f) Cone
(vi) (e) Pyramid

Try These (Page 278)

Question 1.
Match the 2 dimensional figure with the names:


Solution:
(i) Rectangle
(ii) Circle
(iii) Triangle
(iv) Square
(v) Quadrilateral

Try These (Page 279)

Question 1.
Do this : Complete the following table :

Try These (Page 281)

Question 1.
Here you find four nets. There are two correct nets among them to make a tetrahedron. See if you can work out which nets will make a tetrahedron.


Solution:

Try These (Page 286)

Question 1.
Try to guess the number of cubes in the following arrangements (Fig.):

Solution:
(i) Numerofcubes = 24
Let the edge of cube = 2 cm
Height = 4 cm
Breadth = 6 cm
Length = 8cm
(ii) Number of cubes = 8
Let the edge of cube = 2 cm
Length = 6 cm
Height = 4cm
Breadth = 4 cm
(iii) Number of cubes = 9
Lettheedgeofcube = 2cm
Length = 6cm
Breadth = 6 cm
Height = 4 cm

Try These (Page 287)

Question 1.
Two dice are placed side by side as shown; Can you say what the total would be on the face opposite to
(а) 5 + 6
(b) 4 + 3
(Remember that in a dice sum of numbers on opposite faces is 7)

Solution:
(i) (a) 5 + 6 → The face opposite to 5 is 2 and 6 is 1.
So, 2 + 1 = 3
(b) 4 + 3 → The face opposite to 4 is 3 and 3 is 4.
So, 3 + 4 = 7.

Question 2.
Three cubes each with 2 cm edge are placed side by side to form a cuboid. Try to make an oblique sketch and say what could be its length, breadth and height.
Sol. (Fig.)

Length = 6 cm
Breadth = 2 cm
Height = 2 cm.

Try These (Page 291)

Question 1.
For each solid, the three views (1), (2), (3) are given. Identify for each solid the corresponding top, front and side views.
Solid In views


Solution:

Question 2.
Draw a view of each solid as seen from the direction indicated by the arrow.

Solution:

Haryana Board HBSE 7th Class Sanskrit Solutions रुचिरा भाग 2

• Chapter 1 सुभाषितानि
• Chapter 2 दुर्बुद्धिः विनश्यति
• Chapter 3 स्वावलम्बनम्
• Chapter 4 हास्यबालकविसम्मेलनम्
• Chapter 5 पण्डिता रमाबाई
• Chapter 6 सदाचारः
• Chapter 7 सड.कल्पः सिद्धिदायकः
• Chapter 8 त्रिवर्णः ध्वजः
• Chapter 9 अहमपि विद्यालयं गमिष्यामि
• Chapter 10 विश्वबंधुत्वम्
• Chapter 11 समवायो हि दुर्जयः
• Chapter 12 विद्याधनम्
• Chapter 13 अमृतं संस्कृतम्
• Chapter 14 अनारिकायाः जिज्ञासा
• Chapter 15 लालनगीतम्

Haryana State Board HBSE 7th Class Maths Solutions Chapter 15 Visualising Solid Shapes Ex 15.1 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 15 Visualising Solid Shapes Exercise 15.1

Question 1.
Identify the nets which can be used to make cubes (cut out copies of the nets and try it) :

Solution:

Question 2.
Dice are cubes with dots on each face.
Opposite faces of a die always total to seven dots on them. Fig.

Here are two nets to make dice (cubes); the numbers inserted in each square indicate the number of dots in that box.

Insert suitable numbers in the blanks, remembering that the number on the opposite faces should total to 7.
Solution:

Question 3.
Can this be a net for a die ? Explain your answer. (Shown Fig. 15.11.)

Solution:
(Shown in Fig. 15.12). No, the net is not for die, because the sum of the opposite laces is not 7.

Question 4.
Here is an incomplete net for making a cube. Complete it in at least two different ways. Remember that a cube has six faces. How many are there in the net here ? (Give two separate diagrams. If you like, you may use a squared sheet for easy manipulation.)

Solution:

Question 5.
Match the nets with appropriate solids :

Solution:
(a) → (ii)
(b) → (iii)
(c) → (iv)
(d) → (i)

Haryana State Board HBSE 7th Class Maths Solutions Chapter 15 Visualising Solid Shapes Ex 15.2 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 15 Visualising Solid Shapes Exercise 15.2

Question 1.
Use isometric dot paper and make an isometric sketch for each one of the given shapes:
Solution:
Let us take m to be the number of pannit’s marbles.Fig. 15.16.

Question 2.
The dimensions of a cuboid are 5 cm, 3 cm and 2 cm. Draw three different isometric sketches of this cuboid.
Solution:

Question 3.
Three cubes each with 2 cm edge are placed side by side to form a cuboid. Sketch an oblique or isometric sketch of this cuboid.
Solution:

Question 4.
Make an oblique sketch for each one of the given isometric shpaes:
Solution:

Question 5.
Give (i) an oblique sketch and (ii) an isometric sketch for each of the following :
(a) A cuboid of dimensions 5 cm, 3 cm and 2 cm. (Is your sketch unique ?)
(b) A cube with an edge 4 cm long.
An Isometric sheet is attached at the end of the book. You could try to make on it some cubes or cuboids of dimensions specified by your friend.
Solution:

Haryana State Board HBSE 7th Class Maths Solutions Chapter 15 Visualising Solid Shapes Ex 15.3 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 15 Visualising Solid Shapes Exercise 15.3

Question 1.
What cross-sections do you get when you give a
(i) vertical cut
(ii) horizontal cut to the following solids ?
(a) Abrick
(b) Around apple
(c) A die
(d) A circular pipe
(e) An ice cream cone
Solution:
(a) (i) Vertical cut: Here is a brick. When you give the vertical cut to brick you get rectangle. So the cross-section of brick is nearly rectangle in this case.

(ii) Horizontal cut :
When you give the horizontal cut to brick you get rectangle.
So the cross section of brick is nearly rectangle in this case.

(b) (i) Vertical cut : Here is a round apple. When you give the vertical cut to round apple you get nearly circle. So, the cross section is nearly circle in this case.

(ii) Horizontal cut: When you give the horizontal cut to round apple you get nearly circle.
So, the cross-section of round apple is nearly circle.

(c) (i) Vertical cut: Here is a dice. It is like a cube.
When you give a vertical cut, you get a square face. So the cross-section of dice is nearly square in this case.
(ii) Horizontal cut :
When you give the horizontal cut to the dice you get the rectangle. So the cross-section of the dice is nearly rectangle in this case.

(d) (i) Vertical cut: Here is a circular pipe. When you give the horizontal. Cut to circular pipe you get semi cylinder. So, the cross-section of circular pipe is semi cylinder in this case.

(ii) Horizontal cut: When you give the horizontal cut to circular pipe, its cross-section remains same.

(e) (i) Vertical cut: Here is an ice cream cone when you give the vertical cut to an ice cream cone you get nearly triangle. So the cross section of an ice cream cone is nearly triangle.
(ii) Horizontal cone: When you give the horizontal cut to an ice cream cone you get nearly triangle. So the cross section of an ice cream cone is nearly triangle in this case.

Haryana State Board HBSE 7th Class Maths Solutions Chapter 15 Visualising Solid Shapes Ex 15.4 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 15 Visualising Solid Shapes Exercise 15.4

Question 1.
A bulb is kept burning just right above the following solids. Name the shape of the shadows obtained in each case. Attempt to give a rough sketch of the shadow. (You may try to experiment first and then answer these questions).

Solution:
A Ball : The shape of the shadow obtained in case of ball is nearly semi-circle.

Cylindrical pipe : The shape of the shadow obtained in case of cylindrical Pipe is nearly rectangular.
A book: The shape of the shadow obtained in case a book in nearly rectangular.

Question 2.
Here are the shadows of some 3-D objects, when seen under the lamp of an Overhead projector. Identify the solid(s) that match each shadow. (There may be multiple answers for these !)

Solution:
(i) Sphere, iii) Cube, cuboid, (iii) Pyramid, cone, (iv) Cylinder.

Question 3.
Examine if the following are true statements :
(i) The cube can cast a shadow in the shape of a rectangle.
(ii) The cube can cast a shadow in the shape of a hexagon.
Solution:
(i) True, (ii) False.