Author name: Prasanna

HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.1

Haryana State Board HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.1 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 8 Comparing Quantities Exercise 8.1

Question 1.
Find ratio of:
(a) 5 Rs. to 50 paise
(b) 15 kg to 210 gm.
(c) 9 m to 27 cm
(d) 30 days to 36 hours.
Solution:
(a) Ratio between 5 Rs. to 50 paise.
= Ratio between 500 paise to 50 paise = 500 : 50 = 10 : 1.
(6) Ratio between 15 kg to 210 gm.
= 15:210 = 15 : 210 = 1: 14
(c) Ratio between 9 m to 27 cm
= 9 x 100 : 27 = 100 : 3
(d) Ratio between 30 days to 36 hours.
= Ratio between (30 x 24) hours to 36 hours
= 720 : 36 = 20 :1.

HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.1

Question 2.
In a computer dab there are 3 computers for every 6 students. How many computers will be needed for 24 students ?
Solution:
Let x computers are needed for 24 students.
Then the given information can be put in the following tabular form :
HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.1 1
It is a case of direct variation,
\(\frac{3}{6}=\frac{x}{24}\)
or x x 6 = 3 x 24
∴ x = \(\frac{3 \times 24}{6}\) =12
Hence, 12 computers are needed for 24 students.

HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.1

Question 3.
Population of Rajasthan = 570 lakh, Population of UP = 1660 lakhs and area of Rajasthan = 3 lakh sq. km and Area of UP = 2 lakh sq. km.
(i) How many people are there per km2, in both these States ?
(ii) Which state is less populated ?
(iii) How could you say this ?
Solution:
(i) In Rajasthan, No. of people in per sq. km = \(\frac{570}{3}\) = 190
In UP, No. of people in per sq. km = \(\frac{1660}{2}\) = 830 e.g., 190, 830
(ii) Rajasthan is less populated, density of Rajasthan = 190
(iii) Population density
= \(\frac{Total Population}{Total Area(in sq.km}\)
= Total no. of people/sq. km.

HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.1 Read More »

HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.2

Haryana State Board HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.2 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 8 Comparing Quantities Exercise 8.2

Question 1.
Convert the given fractional number to per cents.
(a) \(\frac{1}{8}\)
(b) \(\frac{5}{4}\)
(c) \(\frac{3}{40}\)
(d) \(\frac{2}{7}\)
Solution:
HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.2 1

Question 2.
Convert the given decimal fractions to per cents.
(a) 0.65
(b) 2.1
(c) 0.02
(d) 12.35
Solution:
HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.2 2

HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.2

Question 3.
Estimate what part of the figures is coloured and hence find the per cent which is coloured.
HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.2 3
Solution:
(j) Percentage of coloured part
= \(\frac{1}{4}\) × 100% = 25%
(ii) Percentage of coloured part
= \(\frac{3}{5}\) × 100% = 60%
(iii) Percentage of coloured part
= \(\frac{3}{8}\) × 100% = 37\(\frac{1}{2}\)%

Question 4.
Find :
(а) 15% of 250
(b) 25% of 150
(c) 1% of 1 hour
(d) 20% of Rs. 2500
(e) 75% of 1 kg.
Solution:
HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.2 4

Question 5.
Find the whole quantity if:
(a) 5% of it is 600
(b) 12% of it is Rs. 1080
(c) 40% of it is 500 km.
(d) 70% of it is 14 mins.
Solution:
(a) Hence, 5% of x = 600
\(\frac{5}{100}\) × x — 600
or \(\frac{x}{20}=\frac{600}{1}\)
or, x = 20 × 600
∴ x = 12000.

HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.2

(b) 12% of x = 1080
or, \(\frac{12}{100}\) × x = 1080
or, 12x = 1080 × 100
or, x = \(\frac{1080 \times 100}{12}\)
∴ x = Rs. 9000

(c) 40% of x = 500
or, \(\frac{40}{100}\) × x = 50
or, 2x = 5 × 500
or x = \(\frac{5 \times 500}{2}\)
100 = 500
or, 2x = 5 × 500
or, x = \(\frac{5 \times 500}{2}\)
∴ x = 1250 km.

(d) 70% of x = 14 mins.
or, \(\frac{70}{100}\) × x = 14
or, 7x = 14 × 10
or x = \(\frac{14 \times 10}{7}\)
x = 20 mins.

(e) 8% of × 8 = 40 litres 40
or, \(\frac{8}{100}\) × x = 40
or, 2x = 40 × 25
or, x = \(\frac{40 \times 25}{2}\)
x = 500 litres.

Question 6.
Convert given per cents to decimals and also to fractions in simplest forms :
(a) 25%
(b) 150%
(c) 20%
(d) 5%
Solution:
HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.2 5

HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.2

Question 7.
In a city, 30% are females, 40% are males and remaining are children. What per cent are children ?
Solution:
Percent of children
= [100 – (30 + 40)]%
= (100 – 70)% = 30%

Question 8.
Out of 15,000 voters in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?
Solution:
Percent of vote = 60%
Percent of not vote = (100 – 60)%
= 40%
No. of actually did not vote
= 40% of 15000 = \(\frac{40}{100}\) × 15000 = 6000

Question 9.
Meeta saves Rs. 400 from her salary. If this is 10% of her salary. What is her salary ?
Solution:
Let her salary be Rs. x.
Hence, 10% of x = 400
or \(\frac{10}{100}\) × x = 400
or x = 400 × 10
= Rs. 4000

HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.2

Question 10.
A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win ?
Solution:
No. of total match = 20
percentage of won match = 25%
∴ Total number of winning match
= 20 of 25% = 20 × \(\frac{25}{100}\) = 5

HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.2 Read More »

HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3

Haryana State Board HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 8 Comparing Quantities Exercise 8.3

Question 1.
Tell what is the profit or loss in the following transction. Also find profit % or loss % in each case.
(a) Gardening shears bought for Rs. 250 and sold for Rs. 325.
(b) A fridge bought for Rs. 12,000 arid sold at Rs. 13,500.
(c) A cupboard bought for Rs. 2500 and sold at Rs. 3000.
(d) A skirt bought for Rs. 250 and sold at Rs. 150.
Solution:
(a) C.P. = Rs. 250
S.P. = Rs. 325
Profit or gain = S.P. – C.P.
= 325 – 250 = Rs. 75
Gain % = \(\frac{\text { Gain }}{\text { C.P. }}\) x 100
= \(\frac{75}{250}\) x 100
∴ gain % = 30%
C.P. = Rs. 12,000,
S.P. = Rs. 13,500
Profit = 13,500-12,000
= Rs. 1,500
Profit % = \(\frac{1500}{12000}\) x 100 = 12\(\frac{1}{2}\)%

HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3

(c) C.P. = Rs. 2500 S.P. = Rs. 3000
Profit = 3000 – 2500 = Rs. 500
Profit % = \(\frac{500}{2500}\) x 100 = 20%

(d) C.P. = Rs. 250 S.P. = Rs. 150
Loss = C.P. – S.P.
= 250 – 150 = Rs. 100
Loss % = \(\frac{\text { Loss }}{\text { C.P. }}\) x 100
= \(\frac{100}{250}\) x 100 = 40%

Question 2.
Convert each part of the ratio to percentage:
(a) 3:1
(b) 2:3:5
(c) 1:4
(d) 1:2:5
Solution:
(a) Total part of Ratio = 3 + 1 = 4
= \(\frac{3}{4}=\frac{3}{4}\) x 100% = 75%
= \(\frac{1}{4}=\frac{1}{4}\) x 100% = 25%

(b) Total part of Ratio = 2 + 3 + 5 = 10
HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 1

(c) Total part of Ratio =1 + 4 = 5
HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 2

(d) Total part of ratio = l + 2 + 5 = 8
HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 3

HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3

Question 3.
The population of a city decreased from 25000 to 24500. Find the percent decrease.
Solution:
Percent decrease
HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 4
∴ Percent decrease = 2%.

Question 4.
Arun bought a car for Rs. 3,50,000. The next year, the price went upto Rs. 3,70,000. What was the percent increase ?
Solution:
Percent increase
HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 5

Question 5.
I buy a T.V. for Rs. 10,000 and sell it at a profit of 20%. What do I get for it ?
Solution:
C.P. = Rs. 10,000, S.P. = ?
Profit = 20%
HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 6
∴ S.P. = Rs. 12000
Profit = S.P. – C.P.
= 12000 -10000
∴ Profit = Rs. 2000

Question 6.
Juhi sells a washing machine for Rs. 13,500. She loses 20% in the bargain. What was the price at which she bought it ? Solution:
S.P. = Rs. 13,500, loss = 20%,
C.P. = ?

or 8 C.P. = 13500 x 10
∴ C.P = \(\frac{135000}{8}\) = Rs. 16875.

HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3

Question 7.
(i) Chalk contains calcium, carbon and oxygen in the ratio 10 : 3 : 12. Find percent of carbon in chalk.
(ii) If in a stick of chalk, carbon is 3g, what is the weight of the chalk stick ?
Solution:
(i) Calcium : Carbon : Oxygen
= 10:3 : 12
Total part of Ratio = 10 + 3 + 12 = 25
Percent of calcium in chalk = \(\frac{10}{25}\) x 100% = 40%
Percent of carbon in chalk = \(\frac{3}{25}\) x 100% = 12%
Percent of oxygen in chalk = \(\frac{12}{25}\) x 100% = 48%

(ii) Let total weight of carbon = x
According to question, 12% of x = 3 gm
or, \(\frac{12}{100}\) x x = 3
x = \(\frac{3 x 100}{100}\) = 25
Hence total weight of carbon = 25 gm.

Question 8.
Amina buys a book for Rs. 275 and sells it at a loss of 15%. IIow much does she sell it for ?
Solution:
C.P. = Rs. 275, S.P. = ?
Loss = 15%
S.P = \(\left(\frac{100-\operatorname{Loss} \%}{100}\right)\) x 275 = Rs. 233.75

Question 9.
Find the amount to be paid at end of 3 years in each case :
(a) Principal = Rs. 1200 at 12% p.a.
(b) Principal = Rs. 7500 at 5%p.a.
Solution:
(a) S.I = \(\frac{\mathrm{PRT}}{100}\)
= \(\frac{1200 \times 12 \times 3}{100}\) = Rs. 432
A = P + I = 1200 + 432
= Rs. 1632

(b) S.I. = \(\frac{7500 \times 5 \times 3}{100}\) = Rs. 1125
A = P + I = 7500 + 1125
= Rs. 8625.

HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3

Question 10.
What is the rate of interest which gives an interest of Rs. 280 on a sum of Rs. 56000 for 2 years ?
Solution:
P =Rs. 56000,
S.I. = Rs. 280, T = 2 years,
R =?
HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 8
.’. Rate of interest = 0.25%.

Question 11.
If Meena gives an interest of Rs. 45 for the one year at 9% rate p.a. What is sum she has borrowed ?
Solution:
S.I. = Rs. 45, R = 9%,
T = 1 year, P = ?
P = \(\frac{\text { S.I. } \times 100}{R \times T}=\frac{45 \times 100}{9 \times 1}\)
.’. Sum = Rs. 500.

HBSE 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 Read More »

HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.2

Haryana State Board HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.2 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 7 Congruence of Triangles Exercise 7.2

Question 1.
Which congruence criterion do you use in the following ?
(a) Given:AC = DF
AB = DF
BC = EF
∴ ΔABC ≅ ΔDEF
HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.2 1

HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.2

(b) Given : ZX = RP
RQ = ZY
HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.2 2

(c) Given:
HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.2 3

(d) Given: EB = DB
AE = BC
∠A = ∠C = 90°
∴ ΔABE ≅ ΔCDB
HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.2 4
Solution:
(a) In ΔABC and ΔDEF, we have
AC = DF (given)
AB = DE (given)
BC = EF (given)
Hence, ΔABC ≅ ΔDEF (By SSS congruence Criterion)

HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.2

(b) In ΔPQR and ΔXYZ, we have
ZX = RP (given)
RQ = ZY (given)
∠PQR = ∠XYZ (given)
But included angle
∠PQR ≠ ∠XZY
So, SAS congrunce rule cannot be applied and we cannot conclude that the ΔPQR and ΔXYZ are cogruent.

(c) In ΔLMN and ΔGFH, we have
∠MLN = ∠FGH (given)
∠NML = ∠GFH (given) Included side between ∠MLN, ∠NML and ∠FGH, ∠GFH are
ML = GF
Hence ΔLMN ≅ ΔGFH (By ASA congruence Criterion)
(d) In ΔABE and ΔCDB, we have
EB = DB (given)
AE = BC (given)
∠A = ∠C = 90° (given)
i.e. ∠A =∠C = 90°
Side AE = SideBC hypotenuse EB = hypotenuse DB
Hence, ΔABE ≅ ΔCDB (By RHS Congruence Criterion)

Question 2.
You want to show that ΔART ≅ ΔPEN
(a) If you have to use SSS criterion, then you need to show
(i) AR =
(ii) RT =
(iii) AT =
(b) If it is given that m∠T = m∠N and you are to use SAS criterion, you need to have
HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.2 6

(i) R T =
and
(ii) PN =
(c) If it is given that AT = PN and you are to use ASA criterion, you need to have (i) ? (ii) ?
Solution:
(a) In ΔART and ΔPEN, we have
(i) AR = PE
(ii) RT = EN
(iii) AT = PN
Hence, ΔART ≅ ΔPEN (By SSS congruence rule)

(b) In ΔART and ΔPEN, we have
∠T = ∠N
(i) RT = EN
(ii) AT = PN i.e., PN = AT
Hence, ΔART ≅ ΔPEN (By SAS Congruence Criterion)

(c) In ΔART and ΔPEN, we have
AT = PN
(i) ∠RAT = ∠EPN ⇒ A ⇔  P
(u) ∠RTA = ∠ENP ⇒ T ⇔ T
Hence, ΔART ≅ ΔEPN (By ASA Congruence Criterion)

HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.2

Question 3.
You have to show that ΔAMP ≅ ΔAMQ. In the following proof, supply the missing reasons.
HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.2 5
Solution:
In ΔAMP Δ AMQ
1. PM = QM ⇒ 1. PM = QM (one side of a right-angled triangle are respectively equal to the one side of another)
2. m∠PMA = m∠QMA ⇒ 2. ∠M = ∠M = 90°
3. AM = AM ⇒ 3. AM = AM = Common
4. ΔAMP ≅ ΔAMQ ⇒ 4. ΔAMP ≅  ΔAMQ (By SAS congruence rule)

Question 4.
In ΔABC, ∠A = 30° , ∠B = 40° and ∠C = 110°
In ΔPQR, ∠P = 30°, ∠P = 40° and ∠R = 110°
A student says that ΔABC ≅ ΔPQR by AAA congruence criterion. Is he justified ? Why of why not ?
Solution:
In ΔABC and PQR, we have
∠A = ∠P = 30° (given)
∠B = ∠Q = 40° (given)
∠C = ∠R = 110° (given)
HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.2 7
Using Angle sum property of triangle
∠A + ∠B + ∠C = 30° + 40° + 110°
= 180°
∠P+ ∠Q + ∠R = 30° + 40° + 110°
= 180°
If AB = PQ, BC = QR and AC = PR,
Hence ΔABC ≅ ΔPQR
(By AAA congruence criterion)

HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.2

Question 5.
In the figure, the two triangles are congruent. The corresponding parts are marked. We can write. ΔRAT s ?
HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.2 8
Solution:
In ΔRAT and ΔWON, we have
RA = WO (given)
AT = ON (given)
included ∠RAT = included ∠WON
Hence ΔRAT ≅ ΔWON
(By SAS congruence rule)
or, In ΔRAT and ΔWON, we have
∠R = ∠W
∠A = ∠O
∠T = ∠N and if RT = WN
Hence ΔRAT ≅ ΔWON (By AAA congruence rule)

Question 6.
Complete the congruence statement:
HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.2 9
Solution:
In ΔTBC
BA is the bisector of base TC
Hence, ∠BAC = ∠BAT = 90°
and AC = AT
Now, In ΔBCA and ΔBTA, we have
∠BAC = ∠BAT =90° (given)
and hypotenuse BC = hypotenuse BT (given)
and Side AC = side AT (given)
BA = BA = Common
Hence, ΔBCA ≅ ΔBTA
(By RHS congruence rule)
or, ΔBCA ≅ ΔBTA
(By SSS congruence rule)
and ; In ΔQRS and QPT, we have
QR = TP (given)
SR = QR (given)
SQ = QT (given)
∠R = ∠P (given)
∠RQS = ∠QTP
and included side QR = TP (given)
Hence, ΔQRS ≅ ΔQPT
(By SSS congruence rule)
and, ΔQRS ≅ ΔQTP
(By ASA congruence rule)

HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.2

Question 7.
In a squared sheet, draw two triangles of equal areas such that:
HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.2 10
(i) the triangles are congru¬ent.
(ii) the triangles are not con¬gruent.
What can you say answer their perimeters
Solution:
(i) In ΔABD and ΔCBD,
AD = BC = 4 cm (given)
AB = CD = 4 cm (given)
∠A = ∠C = 90° (given)
Hence, ΔABD ≅ ΔCBD (By SAS congruence rule)

(ii) In squared sheet, we have two triangles and two triangles are congrunet.
Hence, two triangles perimeters are equal.

Question 8.
If ΔABC and ΔPQR are to be congruent name one additional pair of corresponding parts. What criterion did you see ?
HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.2 11
Solution:
Since ΔABC ≅ ΔPQR
Hence, ∠ABC = ∠RQP = 90°
Hypotenuse AC = Hypotenuse PR
Side AB = Side PQ
So, ΔABC ≅ ΔPQR
(By RHS congruence criterion)
or ∠B = ∠Q = 90
∠C = ∠R
Included side BC = included side QR
Hence ΔABC ≅ ΔPQR
(By ASA congruence criterion)

HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.2

Question 9.
Explain, why ΔABC ΔFED
HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.2 12
Solution:
∵ ∠B = ∠E (each 90°)
∠A = ∠F (given)
∠C = ∠D
(third angles are equal) Also BC = ED (given)
Two angles (∠B and ∠C) and included side BC of ΔABC are respectively equal to two angles (∠E and ∠D) and the included side ED of ΔDEF.
∴ ΔABC ≅ ΔFED.

HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.2 Read More »

HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.1

Haryana State Board HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.1 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 7 Congruence of Triangles Exercise 7.1

Question 1.
Complete the following statements:
(a) Two line segments are congruent if
(b) Among two congruent angles one has a measure 70°; the measure of the other angle is .
(c) When we write ∠A ≅ ∠B we actually mean
Solution:
(a) their length are same.
(b) 70° (c) ∠A ≅ ∠B.

Question 2.
Give any two real-life examples for congruent shapes.
Solution:
(i) Ears
(ii) Eyes

Question 3.
If ∠ABC = ∠FED under the correspondence ABC ⇔ FED, write all the corresponding congruent parts of the triangles.
HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.1 1
Solution:
If ΔABC ≅ ΔFED
then AB = FE
BC = ED
AC = FD
and ∠A = ∠F,
∠B =∠E and ∠C = ∠D.

HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.1

Question 4.
If ΔDEF ≅ ΔBCA, write the part (s) of ABCA that correspond to
(i) ∠E
(ii) \(\overline{E F}\)
(iii) ∠F
(iv) \(\overline{D F}\)
HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.1 2
(i) ∠E ⇔ ∠F
(ii) \(\overline{E F}\) ⇔ \(\overline{C A}\)
(iii) ∠F ⇔∠A
(iv) \(\overline{D F}\) ⇔ \(\overline{B A}\)

HBSE 7th Class Maths Solutions Chapter 7 Congruence of Triangles Ex 7.1 Read More »

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions

Haryana State Board HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions

Try These (Page No. 113-114):

Question 1.
Write the six elements (i.e. the 3 sides and the 3 angles) of AABC.
Solution:
Six elements of ΔABC are : ∠A, ∠B, ∠C, \(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}\) and \(\overline{\mathrm{CA}}\).
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 1

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions

Question 2.
Write the :
(i) Side opposite to
the vertex Q of B ΔPQR.
(ii) Angle opposite to the side LM of ΔLMN
(iii) Vertex oppostie to the side RT of ΔRST.
Solution:
A triangle is simple closed curve mode of three line segments. It has three vertices, three sides and three angles.

(i) Side opposite to the vertex Q of ΔPQR is PR.
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 2

(ii) Angle opposite to side LM of ΔLMN is ∠LMN
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 3

(iii) Vertex opposite to the side RT of ΔRST is S.
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 4

Question 3.
Look at the Fig. and classify each of the triangles according to its (a) Sides (b) Angles
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 6
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 6
Solution:
(a) Sides : We may classify the triangles according to their sides as follows:
I. A triangle having no two sides equal is called a scalene triangle.
II. A triangle having two sides equal is called an isosceles triangle.
III. A triangle having all sides equal is called an equilateral triangle.
Hence, (ii) ΔPQR is scalene triangle.
(i) ΔABC, (iii) ΔLMN, (v) ΔABC and
(vi) ΔPQR are isosceles triangle, and;
(iv) ARST is equilateral triangle.

(b) Angles : According to angles :
I. A triangle, all of whose angles are acute, is called an acute angle triangle.
II. A triangle, one of whose angles is right angle, is called right angle triangle.
III. A triangle one of whose angles is obtuse, is called an obtuse angle triangle.
Hence, (i) ΔABC, (iv) ΔRST are acute angle triangle.
(ii) ΔPQR, (vi) ΔPQR arc right angle triangle.
(iii) ΔLMN, (v) ΔABC are obtuse angle triangle.

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions

Think, Discuss & Write (Page No. 114):

Question 1.
How many medians can a triangle have? .
Solution:
The line segment, joining the mid point to its opposite vertex are called a median of the triangle. In a triangle has three vertex. Hence in a triangle has three median. (Fig. 6.19)

Question 2.
Does a median lie wholly in the interior of the triangle ? (If you think that this is not true, draw a figure to show such a case).
Solution:
In a triangle ABC three median AL, BM, CN meet the point O. Hence a median lie wholly in the interior of the triangle.

Think, Discuss & Write (Page No. 115):

Question 1.
How many altitudes can a triangle have ?
Solution:
An altitude has one end point at a vertex of the triangle and the other on the line containing the opposite side. In a triangle has three vertex. So it has three altitude.
e.g. AD, BE, CF are altitude of triangle.
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 7

Question 2.
Draw rough sketches of altit udes from A to \(\overline{\mathrm{BC}}\) for the given triangles :
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 8
Solution:
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 9

Question 3.
Will an altitude always lie in the interior of triangle ? If you think that this need not be true, draw a rough sketch to show such a case.
Solution:
Yes.

Question 4.
Can you think of a triangle in which two altitudes of the triangle are two of its sides ?
Solution:
A triangle has three vertices, and for each vertex there is an altitude. Thus, a triangle has three altitudes. Thus, a triangle in which two altitude of the triangle and two of its sides.

Question 5.
Can the altitude and median be same for a triangle ?
(Hint: For Q. No. 4 and 5, investigate by drawing the altitudes for every type of triangle.)
Solution:
No, the altitude and median arc not same for a tringle.

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions

Think, Discuss & Write (Page No. 118):

Question 1.
Exterior angles can be formed for a triangle in many ways. Three of them are shown here (fig)
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 10
There are three more ways of getting exterior angles. Try to produce those rough sketches.
Solution:
We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.
i.e. ∠1 + ∠2 = ∠3
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 11
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 12
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 13

Question 2.
Are the exterior angles formed at each vertex of a triangle equal?
Solution:
(See Fig. ) ∠4 =∠5 (vertically opposite angles )
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 14

Question 3.
What can you say about the sum of an exterior angle of a triangle and its adjacent interior angle ?
Solution:
∠ACX is called an exterior angle of the ΔABC atC. (Fig. 6.28)
∠ACB is not adjacent ot ∠ACX.
∠ACB is the interior adjacent angle.
∠A and ∠B are not the interior adjacent angles. They are called the interior angles. They are called the interior opposite angles corresponding to exterior angles ∠ACX.
i.e. Sum of an exterior angles and at vertex -C vertically other angles are equal to 360°.

Think, Discuss & Write (Page No. 118):

Question 1.
What can you say about each of the interior opposite angles, when the exterior angles is
(i) a right angle? (ii) an obtuse angle ? (iii) an acute angle.
Solution:
An exterior angle of a triangle is equal to the sum of its interior oppsite singles, (i) a right angle, i.e. (ii) an obtuse angle.

Question 2.
Can the exterior angle of a triangle be a straight angle ?
Solution:
No; because a straight angle = 180°.

Try These (Page No. 118):

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions

Question 1.
An exterior angle of a triangle is of measure 70° and one of its interior opposite angles is of measure 25°. find the measure of the other interior oppsite angle.
Solution:
Let us take m to be the number of
parmit’s marbles other interior opposite angle
= exterior angle — first interior
= oppsite angle
= 70° – 25° = 45°

Question 2.
The two interior opposite angles of an exterior angle of a triangle are 60° and 80°. Find the measure of the exterior angle.
Solution:
Exterior angles = 60° + 80° = 140°.

Question 3.
Is something wrong in this diagram (Fig.) ? Comment.
Solution:
We know that,
The sum of interior opposite angles = exterior angles
but from figure,
50 ≠ 50°
Hence this diagram ‘ is wrong.
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 15

Try These (Page No. 122):

Question 1.
Two angles of a triangle are 30° and 80°.
Find the third angle.
Solution:
By angle – sum property of a triangle
30° + 80° + x° = 180°
110° +x° =180°
∴ x° = 180° – 110° = 70°
∴ third angle = x° = 70°
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 16

Question 2.
One of the angles of a triangle is 80° and the other two anlges are equal. Find the measure of each of the equal angles.
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 17
Solution:
By angle – sum property of a triangle
x° + x° + 80° = 180°
or, 2x°+ 80° = 180°
or, 2x° = 180° -80°
= 100°
= \(\frac{100^{\circ}}{2}\) = 50°
angles are 50°, 50°, 80°

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions

Question 3.
The three angles of a triangle are in the ratio 1:2:1. Find all the angles of the triangle. Classify the triangle in two different ways.
Solution:
Let the ratio be x :2x:x By angle – sum property of a triangle
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 18
x + 2r + x = 180°
or 4x = 180°
or x = \(\frac{180^{\circ}}{4}\) = 45°
∴ angles are 45°, 90°, 45°
Thus the triangles are Right angled and isosceles triangles.

Think, Discuss & Write (Page No. 122):

Question 1.
Can you have a triangle with two right angles ?
Solution:
Since, the three angles of a triangle have a total measure of 180°.
Hence we cannot have a triangle with two right angles.

Question 2.
Can you have a triangle with two obtuse angles ?
Solution:
No, Because obtuse angle > 90°.

Question 3.
Can you have a triangle with two acute angles ?
Solution:
Yes, Because acute angle < 90°.

Question 4.
Can you have a triangle with all the three angles greater than 60° ?
Solution:
No, Because the three angles of a triangle have a total measure of 180°.

Question 5.
Can you have a triangle with all the three angles equal to 60° ?
Solution:
Yes, Because 60° + 60° + 60° = 180°.

Question 6.
Can you have a triangle with all the three angles less than 60° ?
Solution:
No, Because the three angles of a triangle have a total measure of 180°.

Try These (Page No. 123- 124):

Question 1.
Find angles x in each Fig.
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 19
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 20
Solution:
In an isosceles triangle
(a) two sides have same length
(b) base angles opposite to the equal sides are equal.
Hence, (i) x° = 40°
(ii) x° + 45° + 45° = 180°
or, x° + 90° = 180°
.’. x° = 180°-90° = 90°

(iii) x° = 50°
(iv) x°+x° + 100° = 180°
2x° = 180°-100° = 80°
x° = \(\frac{80^{\circ}}{2}\) = 40°

(v) x° + x° + 90° = 180°
or, 2x° = 180°-90° = 90°
x° = \(\frac{90^{\circ}}{2}\) = 45°

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions

(vi) x° + x° + 40° = 180°
or, 2x° = 180°-40° = 140°
x° = \(\frac{140^{\circ}}{2}\) = 70°

(vii) x° + 120° = 180° = Linear pair
or, x° = 180°-120°
.. x° = 60°. (Fig. )
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 21

(viii) y° + 110° = Linear pair = 180°
or, y° =180° —110°
y° =70°
Now, x° + x° + y° = 180°
or, 2x° + 70° =180°
2x° =180°-70° = 110°
.’. x = \(\frac{110}{2}\) = 55°
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 22

(ix) 30° =30°
= vertically opposite angle
x° =30°,
Because base angles opposite to the equal sides are equal. (Fig.)
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 23

Question 2.
Find angles x andy each Fig.
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 24
In isosceles triangle base angled are equal. Hence y° = z°
Solution:
(i) z° + 120° = Linear pair = 180°
∴ z° =180° -120°
∴ z° = 60°
or, x°+y° + 2° = 180° (.-. y° = z°)
x° + 60° + 60° = 180°
or, x° = 180°-120° = 60°
∴ x° = 60°,
y° = 60°
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 25

(ii) x° + x° + 90° = 180°
or, 2x° = 180° – 90° = 90°
∴ x° = \(\frac{90}{2}\) = 45°
or, x° + y° = 180° = linear pair
45° +y° = 180°
∴ y° = 180°-45° = 135°
∴ x° = 45°,
y° = 135°

(iii) 92°+ 92° = vertically opposite angle
x° + x° + 92° = 180°
or, 2x° = 180° – 92° = 88°
∴ x = \(\frac{88}{2}\) = 44°
or, x° + y° = linear pair = 180°
44° + y = 180°
y = 180°-44° = 136°
∴ x° = 44°,
y° = 136°.

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions

Think, Discuss & Write (Page No. 127):

Question 1.
Is the sum of any two angles of a triangle always greater than the third angle ?
Solution:
In a triangle, the sum of any two sides of a triangle is always greater than the third side.

Think, Discuss & Write (Page No. 130):

Question 1.
Find the unknown length x in the following figures :
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 26
Solution:
By Pythagoras property,
i.e. h2 = p2 + b2
Hence, (i) x2 = 32 + 42
x2 = 9 + 16
x = \(\sqrt{25}\)
∴ x = 5

(ii) x2 = (12)2 + 52
∴ x = \(\sqrt{169}\) = 13
∴ x = 13

(iii) x2 = 82 + 152 = 64 + 225 = 289
∴ x = \(\sqrt{289}\) = 17
∴ x = 17

(iv) x2 = 72 + 242 = 49 + 576 = 625
∴ x = \(\sqrt{625}\)
∴ x = 25

(v) h2 = p2 + b2
In ΔABD, b2 = h2 – p2 = (37)2 – (12)2
= 1369-144 = 1225
b = \(\sqrt{1225}\) = 35
= 35 + 35 = 70

(vi) (12)2 = 32 + x2
144 – 9 = x2
or, 135 = x2,
∴ x = \(\sqrt{135}\)

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 27

Think, Discuss & Write (Page No. 131):

Question 1.
Which is the longest side in the triangle PQR right angled at P ?
Solution:
Longest side = QR.

Question 2.
Which is the longest side in the triangle ABC right angled at B ?
Solution:
Longest side = AC.

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions

Question 3.
Which is the longest side of a right triangle?
Solution:
Hypotenuse is the longest side of a right triangle.

Question 4.
“The diagonal of a rectangle produce by itself the same area as produced by its length and breadth”-This is Baudhayan theorem and Compare it with the Pythagorean property.
Solution:
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions 28

Pythagoras, a Greek philospher of sixth century B.C. is A B said to have found a very important and useful property of right angled triangles. The Indian Mathemetician Baudhayan has also given an equivalent form of this property.
By Pythagoras property, In a right angle triangle,
The square on the hypotenuse = Sum of the squares on the legs.
i.e. h2 = p2 + b2
From above figure,
(BD)2 = (BC)2 + (CD)2
i.e. h2 – p2 + b2
This is Baudhayan Theorem.

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions Read More »

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.5

Haryana State Board HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.5 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Exercise 6.5

Question 1.
PQR is a triangle right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.
Solution:
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.5 1
(PR)2 = (PQ)2 + (QR)2
⇒ (QR)2 =(PR)2 – (PQ)2
= (24)2-(10)2
= 576 – 100 = 476
∴ QR = \(\sqrt{476}\)
= \(\sqrt{2 \times 2 \times 7 \times 17}=\sqrt[2]{119}\)

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.5

Question 2.
ABC is a triangle right angled at C. If AB = 25 cm and AC = 7 cm, find BC.
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.5 2
Solution:
(AB)2 = (AC)2 + (BC)2
(25)2 = 72 + (BC)2
625-49 = (BC)2
576 = (BC)2
\(\sqrt{576}\) =, BC
24 = BC
∴  BC = 24 cm.

Question 3.
A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance (Fig. 6.56). Find the distance of the foot of the ladder from the wall.
Solution:
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.5 3
Let ? = x
(15)2 = (12)2 + x2
225 -144 = x2
81 = x2
\(\sqrt{81}\) = x
9 = x
∴  x = 9
The distance of the foot of the ladder from the wall = 9mm.

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.5

Question 4.
Which of the following can be the sides of a right triangle ?
(i) 2.5 cm, 6.5 cm, 6 cm.
(ii) 2 cm, 2 cm, 5 cm.
(iii) 1.5 cm, 2 cm, 2.5 cm.
In the case of right angled triangles, identify the right angles.
Solution:
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.5 4
In a right triangle
h2 = p2 + b2
(iii) (1.5 cm, 2 cm, 2.5 cm) is the right angle
h2 = p2 + b2
(2.5)2 = (1.5)2 + (2)2
6.25 = 2.25 + 4
6.25 = 6.25
This is satisfied.

Question 5.
A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.
Solution:
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.5 5
Let the original height of tree = x
∴  x2 = (5)2 + (12)2
= 25 + 144 = 169
∴  x = \(\sqrt{169}\)
∴ The original height of the tree = 13 m.

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.5

Question 6.
Angle Q and R of a APQR are 25° and 65°. Write which of the following is true:
(i) PQ2 + QR2 = RP2
(ii) PQ2 + RP2 = QR2
(iii) RP2 + QR2 = PQ2
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.5 6
Solution:
∠QPR = 180° – (25° + 65°)
= 180°-90°
∴  ∠QPR = 90
By Pythagoras Property,
(QR)2 = (PR)2 + (PQ)2
i.e. h2 = p2 + b2
hence, QR2 = (PQ)2 + (RP)2 i.e. (ii) is true.

Question 7.
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.5 7
Solultion:
From Fig 6.88
(BC) = (BD) – (CD)
= (41) – (40)
= 1681 – 1600 = 81
∴  BC = \(\) = 9 cm.
Thus the perimeter of the rectangle
= 2 ( l x b) = 2(40 + 9)
= 2 x 49 = 98 cm.

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.5

Question 8.
The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
Solution:
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.5 8
Area of Rhombus
= \(\frac{1}{2}\) x d1 x d2 Sq.units.

(where d1 and d2 are diagonals of rhombus) Side of Rhombus (a) = \(\frac{1}{2} \sqrt{\left(d_{1}\right)^{2}+\left(d_{2}\right)^{2}}\)
Perimeter of Rhombus = 4 x a
Here,
d1 = 16 cm
d2 = 30 cm
∴  Side of Rhombus (a)
= \(\frac{1}{2} \times \sqrt{(16)^{2}+(30)^{2}}=\frac{1}{2} \times \sqrt{256 \times 900}\)
= \(\frac{1}{2}\) x \(\sqrt{(16)^{2}+(30)^{2}}\) = \(\frac{1}{2}\) x 34 =17 cm z z
∴ Perimeter of Rhombuse
= 4 x a = 4 x 17= 68
∴ Perimeter of Rhombus = 68 cm.

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.5 Read More »

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.4

Haryana State Board HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.4 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Exercise 6.4

Question 1.
Is it possible to have a triangle with the following sides.
(i) 2 cm, 3 cm, 5 cm
(ii) 3 cm, 6 cm, 7 cm
(iii) 6 cm, 3 cm, 2 cm
Solution:
(i) 2 cm, 3 cm, 5 cm
∵ 2 cm + 3 cm = 5 cm
and third side = 5 cm
∴ Sum of the measure of lengths of two sides = length of third side which is not possible.
∴ A triangle cannot be possible with these sides.

(ii) 3 cm, 6 cm, 7 cm
∵ 3 cm + 6 cm = 9 cm and 9 cm > 7 cm
3 cm + 7 cm = 10 cm and 10 cm > 6 cm
6 cm + 7 cm = 13 cm and 13 cm > 3 cm
Thus, a triangle can be possible with these sides.

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.4

(iii) 6 cm, 3 cm, 2 cm
∵ 6 cm + 3 cm = 9 cm and 9 cm > 2 cm
3 cm + 2 cm = 5 cm and 5 cm < 6 cm
2 cm + 6 cm = 8 cm and 8 cm > 5 cm
∴ A triangle cannot be possible with these sides.

Question 2.
Take any point O in the interior of a triangle PQR. Is
(i) OP + PQ > OQ ?
(ii) OQ + OR > QR ?
(iii) OR + OP > RP ?
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.4 1
Solution:
Sum of the lengths of any two sides of a triangle is greater than the length of than the length of the third side. (Fig. 6.47)
(i) Yes (ii) Yes (iii) Yes

Question 3.
AM is a median of a triangle ABC. Is AB + BC + CA + > 2 AM ?
Solution:
Since the sum of the lengths of any two sides of a triangle is greater than the length of the third side.
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.4 2
∴ In ΔABM, we have AB + BM > AM ..,(i)
Similarly in AACM, CA + CM > AM …(ii)
Adding (i) and (ii), we have (AB + BM) + (CA + CM) > AM + AM
⇒ [AB + (BM + CM) + CA] > 2AM
⇒ [AB + BC + CA] > 2AM

Question 4.
ΔBCD is quadrialteral. Is AB + BC + CD + DA > AC + BD ?
Solution:
In ΔABC, AB + BC > AC …(i)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side.
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.4 3
In ΔACD, CD + DA > AC …(ii)
Adding (i) and (ii), we have
AB + BC + CD + DA > AC + AC …(iii)
In ΔABD, AB + DA > BD …(to)
In ΔBCD, BC + CD > BD …(v)
Adding (iv) and (v), we get
AB + BC + CD + DA > BD + BD …(vi)
Adding (iii) and (vi)
2 (AB + BC + CD + DA) > 2(AC + BD)
AB + BC + CD + DA > AC + BD.

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.4

Question 5.
ABCD is quadrilateral. Is AB + BC + CD + DA + < 2 (AC + BD) ?
Solution:
In ΔOAB
OA + OB > AB …(i)
[Sum of the lengths of any two sides of a triangle is greater than the length of the third side]
In ΔOBC, OB + OC > BC …(ii)
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.4 4
In ΔOCD, OC + OD > DC …(Hi)
In ΔOAD, OA + OD > AD …(iv)
Adding (i), (ii), (iii) and (iv)
2(OA + OB + OC + OD) > AB + BC + CD + DA
⇒ AB + BC + CD + DA < 2(OA + OB + OC + OD)
⇒ AB + BC + CD + DA < 2(OA + OC + OB + OD)
AB + BC + CD + DA < 2(AC + BD).

Question 6.
The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall ?
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.4 5
Solution:
If x cm be the length of third side, we should have
12 + 15 > x ⇒ 27 > x
x + 12 > 15 ⇒ x > 3
and x + 15 > 12 ⇒ x > – 3
The numbers between 3 and 27 satisfy these.
∴ The length of the third side could be any length between 3 cm and 27 cm.

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.4 Read More »

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.3

Haryana State Board HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.3 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Exercise 6.3

Question 1.
Find the value of the unknown x in the following diagrams :
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.3 1
Solution:
By angle – sum’property of triangle,
(i) ∠A + ∠B + ∠C = 180°
or, x° + 50° + 60° = 180°
or, x° + 110 = 180°
or, x° = 180°-110°
∴ x° = 70°.

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.3

(ii) ∠P + ∠Q + ∠R = 180°
or, 90° + 30° + x° = 1800
or, 120 + x° = 180°.
∴ x° = 180° -120° = 60°.

(iii) ∠X + ∠Y + ∠Z = 180°
or, 30° + 110° + x° = 180°
or, 140 + x° = 180°.
∴ x° = 180°-140° = 40°

(iv) y x° +x° + 50° = 180°
or, 2x°+ 50° = 180°
or, 2x° = 180° – 50° = 130°
∴ x° = \(\frac{130}{2}\) = 65°

(v) x° + x° + x° = 180°
or, 3x° = 180°
∴ x° = \(\frac{180}{3}\) = 60°.

(vi) x° + 2x° + 90° 180°
or, 3x° = 180°-90° = 90°
∴ x° = \(\frac{90}{3}\) = 30°. 3

Question 2.
Find the values of the unknowns x and y in the following diagrams:
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.3 2
Solution:
By angle – sum property of a triangle the three angles of a triangle have a total measure of 180°.
The external angle of a triangle is equal to the sum of of its interior opposite angles and the linear pair = 180°.

(i) x° + 50° = 120°
x° = 120°-50° = 70°
Now, x° + y° + 30° = 180°
or, 70° + y° + 30° = 180°
or y° = 180°-70° = 30°
= 180° -100° = 80°.
x° = 70° , y° = 80°.

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.3

ii) y° = 80° vertically opposite angles
Now, x° + y° + 50° = 180°
or, x° + 80° + 50° = 180°
or, x° + 130° = 180°
∴ x° = 180° -130° = 50°
∴ x° = 50°, y° = 80°.

(iii) x° = 50° + 60° = 110°
or, y° + 50° + 60° = 180°
or, y° + 110° = 180°
∴ y°= 180° -110° = 70°
∴ x° = 110°, y° = 70°.

(iv) x° = 60° = vertically opposite angles
Now, x°+y° + 30° = 180°
60° +y° + 30° = 180°
y° + 90° = 180°
∴ y° = 180°-90° = 90°
∴ x° = 60°, y° = 90°.

(v) y° = 90° = = vertically opposite angle
Now, x° + x° +y° = 180°
or, 2x°+ 90° = 180°
or, 2x° = 180°-90° = 90°
∴ x° = \(\frac{90}{2}\) = 45°
∴ angle are, 45°, 45°, 90°.

(vi) x° = x° = vertically opposite angles
Now, x° = y° = vertically opposite angles
∴ x° + x° + x° = 180°

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.3 3
or, 3x° = 180°
∴ x° = \(\frac{180}{3}\) = 60°
∴ x° = 60°,
∴ x° = 60°,
y° = 60°.

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.3 Read More »

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.2

Haryana State Board HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.2 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Exercise 6.2

Question 1.
Find the value of the unknown exterior angle x in the following diagrams:
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.2 1
Solution:
Since the external angle of a traingle is equal to the sum of the sum of its interior opposite angles. Hence
(i) x° = ∠50° + ∠70° = ∠120°
(ii) x° = ∠65° + ∠45° = 110°
(iii) x° = ∠30° + ∠40° = 70°
(iv) x° = ∠60° + ∠60° = 120°
(v) x° = ∠50° + ∠50° = 100°
(vi) x° = ∠30° + ∠60° = 90°.

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.2

Question 2.
find the value of the unknown interior angle x in the following diagrams:
HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.2 2
Solution:
external angle = sum of interior angle
(i) Hence 115° = angle x° + 50°
or, 115°-50° = x°
65° = x°
∴ x° = 65°.

(ii) 30° + x° = 100°
or, x° = 100°-30
∴ x° = 70°.

(iii) x° + 90° = 125°
or, x° = 125°- 90′
x° = 35°.

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.2

(iv) 60° + x° = 120°
or, x° = 120°-60°
∴ x° = 60°.

(v) 30° + x° = 80°
x° = 80° – 30°
∴ x° = 50°.

(vi) x° + 35° = 75°
or, x° = 75° – 35°
∴ x° = 40°

HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.2 Read More »