CBSE Class 7

These NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.4 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 6.4

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
Answer:
(i) 2 cm, 3 cm, 5 cm
2 cm + 3 cm = 5 cm
third side = 5 cm
∵ Sum of the measures of length of two sides = length of the third side
∴ It is not possible with these sides to form a triangle.

(ii) 3 cm, 6 cm, 7 cm
3 cm + 6 cm = 9 cm
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.

(iii) 6 cm, 3 cm and 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
Thus, a triangle cannot be possible with these sides.

Question 2.
Take any point O in the interior of a triangle PQR. Is
(i) OP + OQ > PQ?
(ii) OQ + OR > QR?
(iii) OR + OP > RP?

Answer:
(i) Yes OP + QQ> PQ
( The sum of lengths of any two sides of a triangle is greater than the length of the third side)
(ii) Yes, OQ + OR > QR
(iii) Yes, OR + OP > RP

Question 3.
AM is a median of a triangle ABC. IsAB + BC + CA > 2 AM?
(Consider the sides of triangles ΔABM and ΔAMC.)
Answer:
Since, the sum of the length of any two sides of a triangle is greater than the length
of the third side
∴ In ΔABM, we have
(AB + BM) > AM …………. (i)
Similarly in ΔACM
(CA + CM) > AM ………. (ii)

Adding (1) and (2), we have
[ (AB + BM) + (CA + CM)] > AM + AM
[ AB + (BM + CM) + CA] > 2 AM
[ AB + BC + CA]> 2AM
Thus, (AB + BC + CA) > 2 AM

Question 4.
ABCD is a quadrilateral.
Is AB + BC + CD + DA > AC + BD?
Answer:
The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

In ABC, we have
AB + BC > AC …………(i)
Similarly, in ΔACD, we have
CD + DA > AC …………(ii)
Adding (i) and (ii), we get
[(AB + BC) + (CD + DA)] > 2 AC …………(iii)
Again In ΔABD, we have
AB + DA > BD …………….(iv)
In ΔBCD, we have
BC + CD > BD …(v)
Adding (iv) and (v), we get
[(AB + DA) + (BC + CD)] > 2 BD …………..(vi)
Now Adding (iii) and (vi), we get
2[(AB + BC) + (CD+ DA)] > 2 (AC + BD) (AB + BC + CD + DA) > (AC + BD)

Question 5.
ABCD is a quadrilateral.
Is AB + BC + CD + DA < 2(AC + BD)?
Answer:
Since the sum of the length of any two sides of a triangle is greater than the length of the third side.

∴ In ΔAOB, we have (OA + OB) > AB …………(i)
Similarly,
In ΔOBC, we have
(OB + OC) > BC …………..(ii)
In ΔOCD, we have
(OC + OD) > CD ………….(iii)
In ΔOAD we have
(OA + OD) > AD …………..(iv)
Adding (i), (ii), (iii) and (iv), we have
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?
Answer:
Since, the sum of the lengths of any two sides of a triangle is greater than the length of the third side.
Sum of 12 cm and 15 cm is greater than the length of the third side,
i.e. (12 cm + 15 cm) > Third side 27 cm > Third side (or) Third side < 27 cm.
Also, the difference of the lengths of any two sides is less than the length of the third side.
(15 cm – 12 cm) < Third side 3 cm < Third side
Thus, we have 3 cm < Third side < 27 cm.
∴ The third side should be of any length between 3 cm and 27 cm.

These NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.3 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 6.3

Question 1.
Find the value of the unknown x in the following diagrams:


Answer:
(i) x + 50° + 60°= 180°
(angle sum property of a triangle)
x + 110° = 180°
x = 180°- 110°
= 70°
∠x = 70°

(ii) x + 90° + 30°= 180°
(angle sum property of a triangle)
x + 120° = 180°
x = 180° – 120°
= 60°
∠x = 60°

(iii) 30° + 110° + x = 180°
(angle sum property of a triangle)
140°+ x = 180°
x = 180° – 140°
= 40°
∠x =40°

(iv) x + x + 50°= 180°
(angle sum property of a triangle)
2x + 50° = 180°
2x = 180° – 50°
2x = 130°
x = \(\frac{130^{\circ}}{2}\) = 65°
∠x = 65°

(v) x + x + x = 180°
(angle sum property of a triangle)
3x = 180°
x = \(\frac{180^{\circ}}{3}\)
x = 60°
∠x = 60°

(vi) x + 2x + 90° = 180°
(angle sum property of a triangle)
3x + 90° = 180°
3x = 180° – 90°
= 90°
x = \(\frac{90^{\circ}}{3}\)
= 30°
∴ ∠x = 30°

Question 2.
Find the values of the unknown x and y in the following diagrams:

Answer:
(i) ∠y + 120°= 180°
(linear pair of angles)
∠y = 180° – 120°
= 60°
∠x + ∠y + 50° = 180°
(using the angle sum property of a triangle)
∠x + 60° + 50° = 180°
∠x + 110° = 180°
∠x = 180°- 110°
= 70°
Thus, ∠x = 70° and ∠y = 60°

(ii) ∠y = 80°
(vertically opposite angles are equal)
∠x + ∠y + ∠50° = 180°
(using the angle sum property of a triangle)
∠x + 80° + 50°= 180°
∠x + 130° = 180°
∠x = 180° – 130°
= 50°
Thus, ∠x = 50° and ∠y = 80°

(iii) 50° + 60° + ∠y = 180°
(using the angle sum property of a triangle)
110° + ∠y – 180°
∠y = 180°- 110°
= 70°
∠x and ∠y form a linear pair.
∠x + ∠y = 180°
∠x + 70° = 180°
∠x = 180° – 70°
= 110°
Thus, ∠x =110° and ∠y = 70°

(iv) ∠x = 60°
(vertically opposite angles)
∠x+ ∠y + 30° = 180°
(angle sum property of a triangle)
60° + ∠y + 30° = 180°
∠y + 90° = 180°
∠y = 180° – 90°
∠y = 90°
Thus, ∠x = 60° and ∠y = 90°

(v) ∠y = 90°
(vertically opposite angles are equal)
∠x + ∠x + ∠y = 180°
(angle sum property of triangle)
2∠x + 90° = 180°
2x = 180° – 90°
2x = 90°
x = \(\frac{90^{\circ}}{2}\) = 45°
Thus, ∠y = 90°,∠x = 45°

(vi) One angle of the triangle = y
Each of the other two angles is equal to their vertically opposite angle ‘x’
∠x + ∠x + ∠y = 180°
(angle sum property of a triangle)
2x + y = 180°
2x + x = 180°
3x = 180°
x = \(\frac{180^{\circ}}{3}\)
x = 60°
y = x
(Vertically opposite angles) y = 60°
Thus, x = 60° and y = 60°

These NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.2 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 6.2

Question 1.
Find the value of the unknown exterior angle x in the following diagrams:


Answer:
(i) x = 50° + 70°
= 120°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)

(ii) x = 45° + 65°
= 110°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)

(iii) x = 30° + 40°
= 70°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)

(iv) x = 60° + 60°
= 120°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)

(v) x = 50° + 50°
= 100°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)

(vi) x = 30° + 60°
= 90°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)

Question 2.
Find the value of the unknown interior angle x in the following figures:

Answer:
(i) x + 50°= 115°
x = 115° – 50°
= 65°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)

(ii) x + 70° = 100°
x = 100° – 70°
= 30°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)

(iii) x + 90° = 125°
x = 125° – 90°
= 35°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)

(iv) x + 60° = 120°
x = 120° – 60°
= 60°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)

(v) x + 30° = 80°
x = 80° – 30°
= 50°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)

(vi) x + 35° = 75°
x = 75° – 35°
= 40°
(The exterior angle of a triangle is equal to the sum of its two interior opposite angles)

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°