Author name: Prasanna

These NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Exercise 6.1

Question 1.
In ΔPQR, D is the mid-point of \(\overline{\text { QR }}\).
PM is ……………
PD is ……………
Is QM = MR?

Answer:
\(\overline{\mathrm{PM}}\) is an altitude of ΔPQR.
\(\overline{\mathrm{PD}}\) is a median of ΔPQR.
QM ≠ MR

Question 2.
Draw rough sketches for the following:
(a) In ΔABC, BE is a median.
(b) In ΔPQR, PQ and PR are altitudes of the triangle.
(c) In ΔXYZ, YL is an altitude in the exterior of the triangle.
Answer:
(a) In the ΔABC, \(\overline{\mathrm{BE}}\) is the median.

(b) In the right ΔQPR, \(\overline{\mathrm{PQ}}\) and \(\overline{\mathrm{PR}}\) are altitudes of the triangle.

(c) In the figure, \(\overline{\mathrm{YL}}\) is an altitude of ΔXYZ.

Question 3.
Verify by drawing a diagram if the median and altitude of an isosceles triangle can be same.
Answer:

ABC is an isosceles triangle having
AB = AC …(i)
Draw its median AD.
Consider ΔADB and ΔADC
AB = AC (By (i))
AD = AD (Common)
BD = DC
(Since AD is median)
⇒ ΔADB = ΔADC (SSS)
⇒ ∠ADB = ∠ADC = 90°
(Corresponding angles)
∠ADB = 90°
∵ \(\overline{\mathrm{AD}}\) is a perpendicular to \(\overline{\mathrm{BC}}\).
Thus, \(\overline{\mathrm{AD}}\) is the median as well as the altitude of ΔABC.

These NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry InText Questions

Question 1.
A student attempted to draw a triangle whose rough figure is given here. He drew QR first. Then with Q as centre, he drew an arc of 3 cm and with R as centre, he drew an arc of 2 cm. But he could not get R What is the reason? What property of triangle do you know in connection with this problem?
im
Can such a triangle exist? (Remember the property of triangles ‘The sum of any two sides of a triangle is always greater than the third side’!)
Answer:
Such a triangle does not exist, because a triangle is possible only when the sum of the lengths of any two sides is greater than the third side. But here, 2 cm + 3 cm, i.e. 5 cm < 6 cm.

Question 2.
In ∆ABC if AB = 3 cm, AC = 5 cm and m∠C = 30°. Can we draw this triangle?
Answer:
We may draw AC = 5 cm and ∠C = 30°. CA is one arm of ∠C. But we observe that point B cannot be located uniquely. So, the given data is not sufficient for construction of ∆ABC.

Question 3.
In the example (see Page 201, NCERT Textbook), length of a side and measures of two angles were given. Now study the following problem:
In ∆ABC, if AC = 7 cm, m∠A = 60° and m∠B = 50°, can you draw the triangle? (Angle sum property of a triangle may help you!)
Answer:
We are given the line segment AC. ∠A is given but ∠C is not given. We can determine ∠C using angle-sum property.
∠C = 180° – (∠C + ∠A).
= 180° – (60° + 50°)
= 180° – 110° = 70°
Now, with the help of ∠C, we can construct the triangle.

These NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles InText Questions

NCERT In-text Question Page No. 94
Question 1.
List ten figures around you and identify the acute, obtuse and right angles found in them.
Answer:
It is an activity. Please do it yourself.

NCERT In-text Question Page No. 95
Question 1.
Which pairs of following angles are complementary?


Answer:
(i) Since 70° + 20° = 90°
∴ Angles 70 and 20 are complementary.

(ii) ∵ 75° + 25° = 100° and 100° ≠ 90°
Angles 75° and 25° are not complementary.

(iii) ∵ 48° + 52° = 100° and 100° = 90°
The angles 48° and 52° are not complementary.

(iv) ∵ 35° + 55° = 90°
∵ The angles 35 and 55 are complementary.

Question 2.
What is the measure of the complement of each of the following angles?
(i) 45°
(ii) 65°
(iii) 41°
(iv) 54°
Answer:
(i) Let the complement of 45° be x
∴ x + 45° = 90°
x = 90° – 45°
= 45°
The complement of 45° is 45°.

(ii) Let the complement of 65° be p
∴ p + 65° =90°
p = 90° – 65°
= 25°
The complement of 65° is 25°.

(iii) Let the complement of 41° be m
∴ m + 41° = 90°
m = 90° – 41°
= 49°
The complement of 41° is 49°.

(iv) Let the complement of 54° be y
∴ y + 54° = 90°
∴ y = 90° – 54°
= 36°
The complement of 54° is 36°.

Question 3.
The difference in the measures of two complementary angles is 12°. Find the measures of the angles.
Answer:
∴ Let one of the angle be x
∴ The difference =12°
∴ The other angle = x + 12°
Sum of the measures of two angles = 90°
Since x and x + 12° are complementary angles
∴ x + x + 12° = 90
2x + 12 = 90°
Transpose 12 to R.H.S
2x = 90 – 12
2x = 78°
Dividing both sides by 2, we get
\(\frac{2 x}{2}=\frac{78}{2}\)
x = 39°
∴ The other angle = 39° +12°
= 51°
The measures of the angle are 39° and 51°.

NCERT In-text Question Page No. 96 & 97
Question 1.
Find the pairs of supplementary angles in following figures:

Answer:
(i) 110° and 50° are not a pair of
supplementary angles as 110° + 50° = 160° ≠ 180°

(ii) 105° and 65° are not a pair of supplementary angles as
105°+ 65° = 170° ≠ 180°

(iii) 50° and 130° are a pair of supplementary angles as
50° + 130° = 180°

(iv) 45° and 45° are not a pair of supplementary angles as
45°+ 45° = 90° ≠ 180°

Question 2.
What will be the measure of the supplement of each one of the following angles?
(i) 100°
(ii) 90°
(iii) 55°
(iv) 125°
Answer:
(i) Let the supplement of 100° be x.
∴ 100°+ x = 180°
or x = 180° – 100° = 80°
∴ The measure of the supplement of 100° is 80°.

(ii) Let the supplement of 90° be x.
∴ x + 90° = 180°
or x = 180°- 90°= 90°
∴ The measure of the supplement of 90° is 90°.

(iii) Let the supplement of 55° be m.
∴ 55° + m = 180°
or m = 180° – 55°
or m = 125°
∴ The supplement of 55° is 125°.

(iv) Let the supplement of 125° be y.
∴ y + 125°= 180°
or y = 180° – 125°
or y = 55°
∴ The supplement of 125° is 55°.

Question 3.
Among two supplementary angles the measure of the larger angle is 44° more than the measure of the smaller. Find their measures.
Answer:
Let the smaller angle be x
∴ The measure of the larger angle = (x + 44°)
Since the two angles are supplementary,
x + (x + 44°) = 180°
2x + 44° = 180°
2x = 180° – 44°
2x = 136°
x = \(\frac{136}{2}\) = 68°
∴ The smaller angle = 68°
Larger angle= 68° + 44° =112°
So, the supplementary angles are 68° and 112°.

NCERT In-text Question Page No. 97 & 98
Question 1.
Are the angles marked 1 and 2 adjacent? If they are not adjacent, say, ‘why’.

Answer:
(i) Yes, ∠1 and ∠2 are adjacent angles.
(ii) ∠1 and ∠2 are adjacent angles.
(iii) ∠1 and ∠2 are not adjacent angles because they have no common vertex.
(iv) No, ∠1 and ∠2 are not adjacent angles because ∠1 is a part of ∠2.
(v) Yes, ∠1 and ∠2 are adjacent angles.

Question 2.
In the given figure, are the following adjacent angles?
(a) ∠AOB and ∠BOC
(b) ∠BOD and ∠BOC

Justify your answer.
Answer:
(a) Yes, ∠AOB and ∠BOC are adjacent angles, because they have common vetex O and their non-common arms (OA and OC) are on either side of the common arm OB.

(b) No, because ∠BOC is a part of ∠BOD.

NCERT In-text Question Page No. 99
Question 1.
Check which of the following pairs of angles form a linear pair.

Answer:
(i) yes,
∵ 140° + 40° = 180°
∴ The given pair of angles forms a linear pair.

(ii) No.
∵ 60° + 90° = 150° and 150° ≠ 180°
∴ The given pair of angles does not form a linear pair.

(iii) No.
∵ 90° + 80° = 170° and 170° ≠ 180°
∴ The given pair of angles does not form a linear pair.

(iv) Yes.
∵ 115° + 65° = 180°
∴ The given pair of angles forms a linear pair.

NCERT In-text Question Page No. 101
Question 1.
In the given figure if ∠1 = 30°, find ∠2 and ∠3

Answer:
∠3 and ∠1 are vertically opposite angles.
.’. ∠3 = ∠1
Since ∠1 = 30°, So ∠3 = 30°
Again ∠3 and ∠2 form a linear pair
.’. ∠3 + ∠2 = 180°
30° + ∠2 = 180°
∠2 = 180° – 30°= 150°
Thus ∠2 = 150° and ∠3 = 30°

Question 2.
Give an example for vertically opposite angles in your surrounding.
Answer:
Please do it yourself.

NCERT In-text Question Page No. 104
Question 1.
Find examples from your surrounding where lines intersect at right angles.
Answer:
Please do it yourself.

Question 2.
Find the measure of the angles made by the intersecting lines at the vertices of an equilateral triangle.
Answer:
Points of intersection are A, B and C.
Measure of ∠A = 60°
Measure of ∠B = 60°
Measure of ∠C = 60°

Question 3.
Draw any rectangle and find the measures of angles at the four vertices made by the intersecting lines.
Answer:
Measure of ∠A = 90°
Measure of ∠B = 90°
Measure of ∠C = 90°
Measure of ∠D = 90°

Question 4.
If two lines intersect, do they always intersect at right angles?
Answer:
No.

NCERT In-text Question Page No. 105
Question 1.
Suppose two lines are given. How many transversals can you draw for these lines?
Answer:
We can draw an infinite number of transversals to two given lines.

Question 2.
If a line is a transversal to three lines, how many points of intersections are there?
Answer:
As shown in the adjoining figure, there are 3 distinct points of intersection.

Question 3.
Try to identify a few transversals in your surroundings.
Answer:
Please do it yourself.

NCERT In-text Question Page No. 106
Question 1.
Name the pairs of angles in each figure:


Answer:
(i) ∠1 and ∠2 are a pair of corresponding angles.
(ii) ∠3 and ∠4 are a pair of alternate interior angles.
(iii) ∠5 and ∠6 are a pair of interior angles on the same side of the transversal.
(iv) ∠7 and ∠8 are a pair of corresponding angles.
(v) ∠9 and Z10 are a pair of alternate interior angles.
(vi) ∠11 and ∠12 are linear pair of angles.

NCERT In-text Question Page No. 109
Question 1.
Find the missing values.
(i) Lines l || m; t is a transversal ∠x = ?

(ii) Lines a || b; c is a transversal ∠y = ?

(iii) 1₁ , 1₂ be two lines t is a transversal Is ∠1 = ∠2?

(iv) Lines l || m; t is a transversal ∠z =?

(v) Lines l || m; t is a transversal ∠x =?

(vi) Lines l || m, p || q; Find a, b, c, d.

Answer:
(i) x = 60°
(x and 60° are alternate interior angles)

(ii) y = 55°
[∴ y and 55° are alternate interior angles]

(iii) No, ∠1 and ∠2 are not equal.
(1₁ and 1₂ are not parallel).

(iv) 60° + z = 180°
[z and 60° are interior angles on the same side of the transversal]
z = 180° – 60° = 120°

(v) x = 120°
[x and 120° are corresponding angles]

(vi) a + 60° = 180°
(Interior angles on same side of transversal)
a = 180 – 60° = 120° a = b = 120°
(alternate exterior angles)
b + d = 180°(linear pair)
d + 120 = 180°
d = 180° – 120° = 60°
⇒ c = b = 120°
(Vertically opposite angles)

NCERT In-text Question Page No. 110
Question 1.
(i) Is l || m? Why?

(ii) Is 1 || m ? Why?

(iii) If || m, what is ∠x?

Answer:
(i) If a transversal intersects two given lines such that alternate angles are equal, then the given lines are parallel.
Since 50° = 50°(alternate angles)
∴ l || m

(ii) l || m as a pair of interior angles on the same side of the transversal are supplementary
Here
x + 130° = 180° (Linear Pair)
x = 50°
y = x = 50°
⇒ l || m (as alternate angles are equal)

(iii) l and m are parallel and t is a transversal
∴ The sum of interior angles on the same side of the transversal is 180°.
x + 70° = 180°
x = 180°- 70°
x = 110°

These NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Exercise 6.4

Question 1.
Find the square root of each of the following numbers by division method.
(i) 2304
(ii) 4489
(iii) 3481
(iv) 529
(v) 3249
(vi) 1369
(vii) 5776
(viii) 7921
(ix) 576
(x) 1024
(xi) 3136
(xii) 900
Solution:
(i) 2304
∴ √2304 = 48

(ii) 4489
∴ √4489 = 67

(iii) 3481
∴ √3481 = 59

(iv) 529
∴ √529 = 23

(v) 3249
∴ √3249 = 57

(vi) 1369
∴ √1369 = 37

(vii) 5776
∴ √5776 = 76

(viii) 7921
∴ √7921 = 89

(ix) 576
∴ √576 = 24

(x) 1024
∴ √1024 = 32

(xi) 3136
∴ √3136 = 56

(xii) 900
∴ √900 = 30

Question 2.
Find the number of digits in the square root of each of the following numbers. (without any calculation).
(i) 64
(ii) 144
(iii) 4489
(iv) 27225
(v) 390625
Solution:
If n stands fnr nnmher of dibits in the given number then
(i) For 64
Here, n = 2 (even number)
Number of digits in the square root = \(\frac{n}{2}=\frac{2}{2}=1\)

(ii) For 144
Here, n = 3 (odd number)
Number of digits in the, square root = \(\frac{n+1}{2}=\frac{3+4}{2}=\frac{4}{2}=2\)

(iii) For 4489
Here, n = 4 (even number)
Number of digits in the square root = \(\frac{\mathrm{n}}{2}=\frac{4}{2}=2\)

(iv) For 27225
Here, n = 5 (odd number)
Number of digits in the square root = \(\frac{\mathrm{n}+1}{2}=\frac{5+1}{2}=\frac{6}{2}=3\)

(v) For 390625
Here, n = 6 (even number)
Number of digits in the square root = \(\frac{n}{2}=\frac{6}{2}=3\)

Question 3.
Find the square root of the following decimal numbers.
(i) 2.56
(ii) 7.29
(iii) 51.84
(iv) 42.25
(v) 31.36
Solution:
(i) 2.56
∴ √2.56 = 1.6

(ii) 7.29
∴ √7.29 = 2.7

(iii) 51.84
∴ √51.84 = 7.2

(iv) 42.25
∴ √42.25 = 6.5

(v) 31.36
∴ √31.36 = 5.6

Question 4.
Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also, find the square root of the perfect square so obtained.
(i) 402
(ii) 1989
(iii) 3250
(iv) 825
(v) 4000
Solution:
(i) 402
The required least number to be subtracted from 402 is 2.
∴ 402 – 2 = 400
∴ √400 = 20

(ii) 1989
The remainder is 53
∴ The required least numbers to be subtracted from the given number is 53.
1989 – 53 = 1936
∴ √1936 = 44

(iii) 3250
The remainder is 1
∴ The required least number to be subtracted from the given number is 1.
3250 – 1 = 3249
∴ √3249 = 57

(iv) 825
The remainder is 41
The required least number to be subtracted from the given number is 41.
825 – 41 = 784
∴ √784 = 28

(v) 4000
The remainder is 31
The required least number to be subtracted from the given number is 31.
4000 – 31 = 3969
∴ √3969 = 63

Question 5.
Find the least number which must be added to each of the following numbers so as to get a perfect square. Also, find the square root of the perfect square, so obtained.
(i) 525
(ii) 1750
(iii) 252
(iv) 1825
(v) 6412
Solution:
(i) 525
The remainder is 41.
i.e. 525 > 22² and the next perfect square number is 23² = 529
The required number to be added = 23² – 525
= 529 – 525
= 4
Hence, 4 should be added to 525 to get a perfect square
number.
525 + 4 = 529
∴ √529 = 23

(ii) 1750
The remainder is 69.
1750 > 41² and the next perfect square number is 42² = 1764
The required number to be added = 42² – 1750
= 1764 – 1750
= 14
The next perfect square number = 1750 + 14 = 1764
∴ √1764 = 42

(iii) 252
The remainder is 27.
This shows that 15² < 252
The next perfect square number is 16² = 256.
The required number to be added = 16² – 252
= 256 – 252
= 4
The perfect square number = 252 + 4 = 256
∴ √256 = 16

(iv) 1825
The remainder is 61.
This shows that 42² < 1825
The next perfect square is 43² = 1849
Hence, the required number to be added = 43² – 1825
= 1849 – 1825
= 24
The perfect square number so obtained 1825 + 24 = 1849
Hence, √1849 = 43

(v) 6412
The remainder is 12.
This shows that 80² < 6412
The next perfect square number is 81² = 6561
The required number to be added = 6561 – 6412 = 149
The perfect square number = 6412 + 149 = 6561
Hence, √6561 = 81

Question 6.
Find the length of the side of a square where area is 441m².
Solution:
Let x be the side of a square.
∴ Area of square = x²
Given, Area of square = 441
x² = 441
⇒ x = √441 = 21
So, length of the side of the square = 21 m.

Question 7.
In a right triangle ABC, ∠B = 90°.
(a) If AB = 6 cm, BC = 8 cm, find AC.
(b) If AC = 13 cm, BC = 5 cm, find AB.
Solution:
(a) In the right triangle ABC, ∠B = 90°

By Pythagoras Theorem,
AC² = AB² + BC²
⇒ AC² = 6² + 8²
⇒ AC² = 36 + 64
⇒ AC² = 100
⇒ AC = √100 = 10
∴ Length of AC = 10 cm

(b) In the right triangle ABC, ∠B = 90°

By Pythagoras theorem,
AC² = AB² + BC²
⇒ AB² = AC² – BC²
⇒ AB² = 13² – 5²
⇒ AB² = 169 – 25
⇒ AB² = 144
⇒ AB = √144 = 12
∴ Length of AB = 12 cm

Question 8.
A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this.
Solution:
Let the number of rows be ‘x’
The number of columns is x
The number of plants = x × x = x²
which is a perfect square
Let us find out the square root of 1000 by division method
The remainder is 39
This shows that 31² < 1000
The next perfect square number is 32² = 1024
Minimum number of plants he needs besides 100 plants = 32² – 1000
= 1024 – 1000
= 24

Question 9.
There are 500 children in a school. For a P.T. drill, they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement.
Solution:
Let the number of rows be x and the number of columns is x
∴ Number of children = x × x = x²
which is a perfect square
∴ 500 = x²
The remainder is 16.
Number of children left out in this arrangement = 16.

These NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercise 10.5

Question 1.
Construct the right-angled ∆PQR, where m∠Q = 90° QR = 8 cm and PR = 10 cm. Ans: Steps of Construction
(i) Draw a rough diagram with measures marked on it.
(ii) Draw a line segment QR = 8 cm.
(iii) At Q, Construct ∠RQX = 90° .

(iv) With centre at R and radius 10 cm, draw an arc to cut QX at P.
(v) Join PR.
Thus, PQR is the required right-angled triangle.

Question 2.
Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.
Answer:
Steps of Construction


(i) Draw a rough diagram with measures marked on it.
(ii) Draw a line segment QR = 4 cm.
(iii) At Q, construct ∠RQX = 90
(iv) With R as centre and radius 6 cm draw an arc to cut QX at R
(v) Join PR.
Thus, PQR is the required right angled triangle.

Question 3.
Construct an isosceles right-angled triangle ABC, where m∠ACB = 90° and AC = 6 cm.
Answer:
Steps of Construction

(i) Draw a rough diagram with measures marked on it.
(ii) Draw a line segment AC = 6 cm.
(iii) At ‘C’, construct ∠ACX – 90°.
(iv) With centre ‘C’ and radius 6 cm, draw an arc to cut CX at B.
(v) Join AB.
Thus, ∆ABC is the required isosceles right-angled triangle.

These NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercise 10.4

Question 1.
Construct ∆ABC given m∠A = 60° , m ∠B = 30 and AB = 5.8 cm.
Answer:
Steps of Construction
(i) Draw a rough diagram with measures marked on it.
(ii) Draw a line segment AB = 5.8 cm.
(iii) Construct ∠BAX = 60° at A.
(iv) At B, construct ∠ABY = 30° .

(v) Let the rays AX and BY intersect at C. Thus, ∆ABC is the required triangle.

Question 2.
Construct ∆PQR if PQ = 5 cm, m∠PQR = 105° and m ∠QRP = 40°.
(Hint : Recall angle-sum property of a triangle)
Answer:
Here the line segment PQ is given. We can construct the triangle, if the measure of ∠QPR is known.
∠QPR = 180° – [∠PQR + ∠QRP]
= 180°-(105°+ 40°)
= 180° – 145°
∠QPR = 35°
Steps of Construction


(i) Draw a rough diagram with measures marked on it.
(ii) Draw a line, segment PQ = 5 cm.
(iii) At P, construct ∠QPX = 35 .
(iv) At Q, construct ∠PQY = 105 .
(v) Let the rays \(\overline{\mathrm{PX}}\) and \(\overline{\mathrm{QY}}\) intersect at R.
Thus, ∆PQR is the required triangle.

Question 3.
Examine whether you can construct ∆DEF such that EF = 7.2 cm, m∠E = 110 and m∠F = 80°. Justify your answer.
Answer:
Here m∠E =110 and m∠F = 80°
i. e. m∠E + m∠F= 110°+ 80°= 190°>180°
Since, sum of the three angles of a triangle is 180°.
∆DEF is not possible, having sum of two angles > 180°
Thus, ∆DEF cannot be constructed.