By using Ganita Prakash Class 7 Solutions and Part 1 Chapter 7 A Tale of Three Intersecting Lines Class 7 Question Answer, students can improve their problem-solving skills.
Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines Solutions
Ganita Prakash Class 7 Chapter 7 Solutions
Class 7 Maths Ganita Prakash Chapter 7 Solutions A Tale of Three Intersecting Lines
Page: 146
Question 1.
What happens when the three vertices lie on a straight line?
Solution:
If the three vertices lie on a straight line, then no triangle will be formed and in that case we say the three points to be ‘Collinear’.
7.1. Equilateral Triangles
Question 2.
Construct a triangle in which all the sides are of length 4 cm.
Solution:
To construct a Δ of length 4 cm each, we first draw a line segment PQ of length 4 cm.
At P, we draw an arc of length 4 cm sufficiently long.
At Q, we draw an arc of length 4 cm sufficiently long to cut the already created arc.
Name the point of intersection R and join it with P and Q.
So, we get a △PQR.
Page: 147
Question 3.
How do we make this construction more efficient?
Solution:
We can make construction more efficient by using pair of compasses along with scale.
7.2. Constructing a Triangle when its Sides are Given
Page: 148
Question 4.
Construct a triangle of sidelengths 4 cm, 5 cm and 6 cm.
Solution:
Let us start with the longest side. Draw a line segment XY = 6 cm.
At X, draw an arc of radius 5 cm, sufficiently long.
At Y, draw another arc of radius 4 cm so that it cuts the other arc created,
in the previous step. Name the point of intersection Z.
In this way, we have obtained a Δ XYZ with sides XY = 6 cm, XZ = 5 cm and YZ = 4 cm.
Note: You can start the construction with a different length from amongst the given lengths.
Page: 149
Question 5.
How do we construct this triangle more efficiently?
Solution:
We have used compass and ruler for our construction so it is an efficient way.
Page: 150
Question 6.
Construct triangles having the following sidelengths (all the units are in cm) :
(a) 4, 4, 6
(b) 3, 4, 5
(c) 4, 4, 5
(d) 4, 4, 8
(e) 3.5, 3.5, 3.5
Solution:
(a)
Note: AC = BC = 4 cm.
So, △ABC is isosceles.
(b)
Note: △PQR is a right-angled Δ as ∠PRQ is 90°.
(c)
Note : DE = DF = 5 cm.
So, △DEF is an isosceles triangle.
(d)
Note: AB ≠ BC, BC ≠ AC, AC ≠ AB.
All sides are of different length.
So, △ABC is ‘Scalene’.
(e)
Page: 150-151
Question 7.
Figure it Out
1. Use the points on the circle and/or the centre to form isosceles triangles.
Solution:
Draw a circle of any radius. Give centre name M. Take a point outside the circle. Name N. Draw another circle taking centre N and with the same radius as the previous circle.
Let the two circles intersect at P and Q.
Join P, M, N to get a △MNP with PM = PN.
Note: We can also join M, N, Q to get an isosceles triangle.
2. Use the points on the circles and/or their centres to form isosceles and equilateral triangles. The circles are of the same size.
Solution:
For constructing isosceles triangle, follow the steps described in solution 1, above.
For constructing an equilateral Δ, draw a circle with centre A and of any radius.
Take a point on the circle and draw another circle so that it intersects the already constructed circle at C and D. Name the centre B.
Now join A , B and C to get an equilateral triangle.
Note: We can get an equilateral Δ by joining A, B and D also.
Question 8.
Construct a triangle with sidelengths 3 cm and 8 cm.
Solution:
We can not draw a triangle with given sidelengths as sum of 3 cm+4 cm = 7 cm which is less than the third side which is given as 8 cm.
So, given sidelengths do not satisfy the triangle inequality.
Question 9.
Here is another set of lengths: 2 cm,3 cm , and 6 cm. Check if a triangle is possible for these sidelengths.
Solution:
2 cm, 3 cm and 6 cm, do not satisfy triangle inequality, so, the Δ can not be constructed with the given sidelengths.
Question 10.
Try to find more sets of lengths for which a triangle construction is impossible. See if you can find any pattern in them.
Solution:
Some of the sets of sidelengths with which we can not draw a triangle are :
(i) 10 cm, 6 cm, 4 cm
(ii) 2 cm, 3 cm, 8 cm
(iii) 12 cm, 14 cm, 30 cm
(iv) 6 cm, 4 cm, 1 cm
Page: 152-153
Question 11.
Can this understanding be used to tell something about the existence of a triangle having sidelengths 10 cm, 15 cm and 30 cm ?
Solution:
The triangle with sides 10 cm, 15 cm and 30 cm, can not be drawn.
Question 12.
Can we say anything about the existence of a triangle having sidelengths 3 cm, 3 cm and 7 cm ? Verify your answer by construction.
Solution:
A triangle can not be constructed with the sidelengths 3 cm, 3 cm and 7 cm , as these lengths do not satisfy ‘Triangle inequality’.
Clearly from the above construction, we see that the arcs at A and B both of 3 cm radius do not intersect.
Therefore, the triangle can not be drawn.
Question 13.
“In the rough diagram in Fig. 7.4, is it possible assign lengths in a different order such that the direct paths are always coming out to be shorter than the roundabout paths? If this possible, then a triangle might exist.”
Solution:
No, even after rearranging the sides a triangle can not be drawn.
Question 14.
Is such rearrangement of lengths possible in the triangle?
Solution:
Yes, such rearrangement of lengths are possible in a triangle but triangle can be made only with the sidelengths which satisfy triangle inequality in all possible arrangements and rearrangements.
Page: 154
Question 15.
Figure it Out
1. We checked by construction that there are no triangles having sidelengths 3 cm, 4 cm and 8 cm; and 2 cm, 3 cm and 6 cm. Check if you could have found this without trying to construct the triangle.
Solution:
For sidelengths 3 cm, 4 cm and 8 cm we can see that 3 + 4 = 7 < 8.
So, these sidelengths do not satisfy triangle inequality.
Hence, we can not construct triangle with these sidelengths.
Exactly on the same basis as above, we can conclude that with sidelengths 2 cm, 3 cm and 6 cm, we can not draw a triangle.
2. Can we say anything about the existence of a triangle for each of the following sets of lengths?
(a) 10 km, 10 km and 25 km
(b) 5 mm, 10 mm and 20 mm
(c) 12 cm, 20 cm and 40 cm
Solution:
Here,
(a)10 km + 10 km = 20 km < 25 km
(b)5 mm + 10 km = 15 mm < 20 mm
(c)12 cm + 20 cm = 32 cm < 40 cm
So, in all the above cases sum of two smaller sides is smaller than the third side. Thus, Δ s can not be drawn in any of the above cases.
3. For each set of lengths seen so far, you might have noticed that in at least two of the comparisons, the direct length was less than the sum of the other two (if not, check again!). For example, for the set of lengths 10 cm, 15 cm and 30 cm , there are two comparisons where this happens :
10 < 15 + 30
15 < 10 + 30 But this does not happen for the third length: 30 > 10 + 15.
Solution:
Yes, first two comparisons are correctly set up but the third comparison in not a triangle inequality.
So, triangle can not be constructed in this case.
Question 16.
Will this always happen? That is, for any set of lengths, will there be at least two comparisons where the direct length is less than sum of the other two? Explore for different sets of lengths.
Solution:
Consider the lengths :
7, 12, 23
we have
7 + 12 < 23 12 + 23 > 7
7 + 23 > 12
So, we find that the direct length is less than the sum of the other two.
Let us consider another set of sidelengths :
9,21,40: 9 + 21 < 40 we have 21 + 40 > 9
9 + 40 > 21
Here again we see the same conclusion.
Question 17.
Further, for a given set of lengths, is it possible to identify which lengths will immediately be less than the sum of the other two, without calculations?
Solution:
We can arrange the given sidelengths in ascending order or descending order.
In this way we can choose the smallest value which will be less than the sum of the other two.
Question 18.
Given three sidelengths, what do we need to compare to check for the existence of a triangle?
Solution:
For the existence of a triangle we need to check that the sum of any two sides is greater than the third side. (Triangle Inequality).
For example: Consider △ABC, if
AB + BC > AC
AC + BC > AB and
AB + AC > BC.
If all the above inequalities are satisfied, then the
triangle can be constructed.
Page: 155
Question 19.
Why do we not need to check the other two sides?
Solution:
We need not to check the other two sides because if sum of two smaller sides is greater than the largest side then the other two will automatically follow.
Question 20.
Now, suppose that a circle of radius 5 cm is constructed, centred at B. Can you draw a rough diagram of the resulting figure?
Solution:
Page: 156
Question 21.
Figure it Out
-Which of the following lengths can be the sidelengths of a triangle? Explain your answers. Note that for each set, the three lengths have the same unit of measure.
(b) 2, 2, 5
(b) 3,4,6
(c) 2, 4, 8
(d) 5, 5, 8
(e) 10, 20, 25
(f) 10, 20, 35
(g) 24, 26, 28
Solution:
(a) Here, 2 + 2 < 5 ∴ These can not be the sidelengths of a triangle. (b) Here, 3 + 4 > 6
4 + 6 > 3
3 + 6 > 4
∴ These can be the sidelengths of a triangle.
(c) Here, 2 + 4 < 8 ∴ These can not be the sidelengths of a triangle. (d) Here, 5 + 8 > 5, 5 + 5 > 8
∴ These can be the sidelengths of a triangle.
(e) Here, 10 + 20 > 25, 20 + 25 > 10,
25 + 20 > 10
∴ These can be the sidelengths of a triangle.
(f) Here, 10 + 20 < 35 ∴ These can not be the sidelengths of a triangle. (g) Here, 24 + 26 > 28,
28 + 26 > 24,
24 + 28 > 26.
So, These can be the sidelengths of a triangle.
Question 22.
Will triangles always exist when a set of lengths satisfies the triangle inequality? How can we be sure?
Solution:
Yes. Whenever a set of lengths satisfy triangle inequality a triangle can be made.
This can be explained as follows :
We get a △ABC.
Page: 159
Question 23.
How will the two circles turn out for a set of lengths that do not satisfy the triangle inequality? Find 3 examples of sets of lengths for which the circles :
(a) touch each other at a point,
(b) do not intersect.
Solution:
(a)
(i)
XY = 5 cm, XP = 3 cm, PY = 2 cm.
(ii)
XY = 6 cm, XP = 3 cm and YP = 3 cm.
(iii)
XY = 6 cm, XP = 2 cm and PY = 4 cm.
(b)
(i)
XP = 3 cm, QY = 3 cm, XY = 7 cm.
(ii)
XY = 9 cm
XP = 3.5 cm
QY = 2 cm.
(iii)
XY = 4 cm
XP = 2 cm, QY = 1 cm.
Question 24.
Frame a complete procedure that can be used to check the existence of a triangle.
Solution:
To check whether the three given lengths will form a triangle or not, we do the following :
(i) Take the three lengths as AB, BC and AC.
(ii) Check whether AB + BC > AC; if it is Yes, then.
(iii) Check whether BC + AC > AB; if it is Yes, then.
(iv) Check whether AC + AB > BC; if Yes, then our answer will be Yes, the triangle with these measurements exists.
If answer of (ii), (iii), or (iv) is no, then we will stop only at that step, and our answer should be ‘No such triangle exists.’
Question 25.
Figure it Out
1. Check if a triangle exists for each of the following set of lengths :
(a) 1, 100, 100
(b) 3, 6, 9
(c) 1, 1, 5
(d) 5, 10, 12
Solution:
(a) Since 1 + 100 > 100
100 + 100 > 1
∴, Δ is possible.
(b) Since 3 + 6 = 9
That is, the sum of two sides is equal to the third side.
So, the triangle is not possible.
(c) Since, 1 + 1 < 5 So, triangle is not possible.
(d) Since 5 + 10 > 12
12 + 10 > 5
5 + 12 > 10
So, these lengths satisfy the triangle inequality.
Hence, Δ is possible
2. Does there exist an equilateral triangle with sides 50, 50, 50? In general, does there exist an equilateral triangle of any sidelength?
Solution:
Yes, there does exist an equilateral triangle with sides 50, 50, 50.
In general, there does exist an equilateral triangle of any sidelength.
This is possible because, if let us suppose the sidelengths be given as a, a, and a, then it will satisfy the triangle inequality because- a+a>a for any value of ‘a’.
3. For each of the following, give at least 5 possible values for the third length so there exists a triangle having these as sidelengths (decimal values could also be chosen) :
(a) 1, 100
(b) 5,5
(c) 3, 7
Solution:
(a) The third length will be greater than 100 – 1 = 99 and less than 100 + 1 = 101.
So, we take 5 values as
99.4, 99.6 99.8, 100 and 100.5 for the third side so that the triangle is possible.
(b) The third length must lie between 0 and 10.
So, we can choose 3, 4, 5, 6 or 7 as the third length which will make the triangle possible.
(c) The third length must lie between 7 – 3 = 4 and 3 + 7 = 10
So, we can choose 5 values as
5, 5.6, 6, 6.8, 7.5, so, that the Δ becomes possible.
7.3. Construction of Triangles when some Sides and Angles are Given
Page: 161
Question 26.
Figure it Out
Construct triangles for the following measurements where the angle is included between the sides :
(a) 3 cm, 75°, 7 cm
(b) 6 cm, 25°, 3 cm
(c) 3 cm, 120°, 8 cm
Solution:
(a)
(i) Draw a line segment YZ = 7 cm.
(ii) At Y , make an angle 75° with the help of protector.
(iii) Mark a point X on the other arm of the angle drawn in the last step, equal to 3 cm.
Join XZ.
Hence, we have obtained a △XYZ based on the given measurements.
(b)
(i) Draw a line segment BC = 6 cm.
(ii) At B draw an angle equal to 25°.
(iii) Mark a point A on the other arm of the angle such that BA = 3 cm.
Join AC.
Hence, we have obtained a △ABC.
(c)
(i) Draw QR = 8 cm.
(ii) At Q draw an angle equal to 120°.
(iii) On the other arm mark P such that PQ = 3 cm.
Join PR.
Hence, we obtain a △PQR.
Question 27.
We have seen that triangles do not exist for all sets of sidelengths. Is there a-combination of measurements in the case of two sides and the included angle where a triangle is not possible? Justify your answer using what you observe during construction.
Solution:
No, If we have given two sidelengths and an included angle we can draw a triangle in any case.
The angle sum property will take care of the other two angles !!! (c)
Page: 162
Question 28.
Figure it Out
Construct triangles for the following measurements :
(a) 75°, 5 cm, 75°
(b) 25°, 3 cm, 60°
(c) 120°, 6 cm, 30°
Solution:
(a)
(i) Draw BC = 5 cm.
(ii) At B and C draw angles 75° each.
(iii) Let the free arms meet at A. So, Δ ABC is the required △.
(b)
(i) Draw VW = 3 cm.
(ii) At V draw an 25° and at W draw an angle 60°.
(iii) Let the free arms meet at U.
So, we get the required △UVW.
(c)
Question 29.
Do triangles exist for every combination of two angles and their included side? Explore.
Solution:
No. The triangle will be formed only if the sum of the two given angles is strictly less than 180°. This is because we must have some value left for the third angle so that angle sum property of Δ is satisfied.
Question 30.
Find examples of measurements of two angles with the included side where a triangle is not possible.
Solution:
If we take angles 70° and 120° and length of the included side 7 cm, then triangle will not be possible. As an other example, if we take angles as 75° and 110° and included length 4 cm, then also, we will not get a Δ.
Note: There can be many such examples. Make sure that the two angles must sum more than or equal to 180°.
Question 31.
It is clear that if the line from B is “inclined” sufficiently to the right then it will not meet the line l.
(a) Try to find a possible ∠B (marked in the figure) for this to happen.
(b) What could be smallest value of ∠B for the lines to not meet?
Solution:
(a) Any angle greater than or equal to 140° drawn at B.
(b) smallest value of ∠B = 140°.
Page: 163
Question 32.
Figure it Out
1. For each of the following angles, find another angle for which a triangle is (a) possible, (b) not possible. Find at least two different angles for each category.
(a) 30°
(b) 70°
(c) 54°
(d) 144°
Solution:
| Given angles | Possible | Not possible |
| (a) 30° | (i) 60° (ii) 100° |
(i) 150°(ii) 160° |
| (b) 70° | (i) 30°(ii) 90° | (i) 110°(ii) 115° |
| (c) 54° | (i) 50°(ii) 60° | (i) 130°(ii) 140° |
| (d) 144° | (i) 20°(ii) 15° | (i) 40° (ii) 50° |
2. Determine which of the following pairs can be the angles of a triangle and which can not :
(a) 35°, 150°
(b) 70°, 30°
(c) 90°, 85°
(d) 55°, 150°
Solution:
(a) Can not be angles of a △
(b) Can be angles of a △
(c) Can be angles of a △
(d) Not possible; cannot be angles of a Δ
Question 33.
Like the triangle inequality, can you form a rule that describes the two angles for which a triangle is possible?
Solution:
Yes. The rule is :
The sum of two angles must be strictly less than 180°.
Yes, the sum of the two angles is used for framing this rule.
Page: 164
Let us take two angles, say 60° and 70°, whose sum is less than 180°. Let the included side be 5 cm.
Question 34.
What could the measure of the third angle be? Does this measure change if the length is changed to some other value, say 7 cm ? Construct and find out.
Solution:
The measure of the third angle
= 180° – (60°+70°)
= 180° – (130°)
= 50°
Base length will not effect the measure of the third angle as depicted by the following constructions :
In all the above three constructions we have used different base lengths.
We have got the third angle equal to 50° in each case.
So, the change in the included length will not effect the third angle.
Question 35.
In general, once the two angles are fixed, does the third angle depend on the included side length? Try with different pairs of angles and lengths.
Solution:
No. The third angle does not depend on the included side length, as depicted by the following construction.
Question 36.
Try experimenting with different triangles to see if there is a relation between any two angles and the third one. To find this relation, what data will you keep track of and how will you organise the data you collect?
Solution:
The emphasis must be given to the sum of angles.
Here you will notice that sum of angles of a triangle is 180°.
Page: 165
Question 37.
Figure it Out
1. Find the third angle of a triangle (using a parallel line) when two of the angles are :
(a) 36°, 72
(b) 150°, 15
(c) 90°, 30
(d) 75°, 45
Solution:
(a)
Now three angles at P must sum to 180°.
∴, ∠XPQ + ∠QPR + ∠YPR = 180°.
arrow 36° + ∠QPR + 72° = 180°
arrow 108 °+ ∠QPR = 180°
arrow ∠QPR + 180° – 108° = 72°
∴ the third angle = 72°.
(b)
Here, ∠XPQ = 150 (Alternate interior angles)
∠YPR = 15° (Alternate interior angles)
∴, ∠XPQ + ∠QPR + ∠YPR = 180°.
arrow ∠QPR
= 180° – ∠XPQ – ∠YPR.
∠QPR = 180° – (150°+15°)
= 180° – (165°)
= 15°
Thus, the third angle is 15°.
(c)
Here, ∠XPQ + ∠QPR + ∠YPR = 180°
90° + ∠QPR + 30° = 180° (Alternate interior angles are equal)
arrow ∠QPR = 180° – 90° – 30°
= 60°
So, the third angle = 60°
(d)
Since, XY || BC
∴ ∠XAB = ∠ABC (Alternate interior ∠s)
arrow ∠XAB = 75°
∠YAC = ∠ACB (Alternate interior ∠s)
arrow ∠YAC = 45°
Since, AB and AC are two ways standing on a line XY at A.
Hence, ∠XAB +∠BAC + ∠YAC
arrow 75° + ∠BAC + 45° = 180°
arrow ∠BAC = 180° – 120° = 60°
So, the third angle is 60°
Question 38.
Can you construct a triangle all of whose angles are equal to 70°? If two of the angles are 70° what would the third angle be? If all the angles in a triangle have to be equal, then what must its measure be? Explore and find out.
Solution:
No. We can not construct a triangle with each angle equal to 70°.
If two angles in a Δ are 70° each then the measure of the third angle will be 180°-70°
– 70° = 40°
If all angles of a triangle have to be equal, then each must be equal to 60°.
In this case only we will have sum of angles equal to 180°, which is a must for the construction of a Δ.
Question 39.
Here is a triangle in which we know ∠B = ∠C and ∠A = 50°. Can you find ∠B and ∠C?
Solution:
By using angle sum property of a Δ we write
∠A + ∠B + ∠C = 180°
arrow 50° + 2 ∠B = 180°
(∵ ∠A = 50° and ∠B = ∠C).
arrow 2 ∠B = 130° arrow ∠B = \(\frac{130}°{2}\) = 65°
Therefore, ∠B and ∠C both will be equal to 65° each.
Question 40. What can we say about the sum of the angles of any triangle?
Solution: The sum of the angles of any triangle is 180°.
7.5. Types of Triangles
Page: 170-171
Question 41.
What are the other types of triangles based on angle measures?
Solution:
On the basis of angle measurements, following are the types of Δ :
(i) Acute angle triangle.
(ii) Right angle triangle.
(iii) Obtuse angle triangle.
Question 42.
What could an acute-angled triangle be? Can we define it as a triangle with one acute angle? Why not?
Solution:
An acute angled triangle be defined as a triangle in which all the angles are an acute angle that is whose measure is less than 90°. We can not define an acute angle triangle which has one angle on acute angle as there will be an acute angle in an obtuse angle triangle also.
Question 43.
Figure it Out
1. Construct a triangle ABC with B C = 5 cm, AB = 6 cm, CA = 5 cm. Construct an altitude from A to BC.
Solution:
Steps of Construction :
Step 1: Draw a line segment AB = 6 cm.
Step 2: Draw an arc of 5 cm at A and B sufficiently long so that the two arcs intersect.
Step 3: Let the two arcs intersect at C.
Step 4: Join A, C and B, C.
In this way we have obtained a Δ ABC.
2. Construct a triangle TRY with RY = 4 cm, TR = 7 cm, ∠R = 140°. Construct an altitude from T to RY.
Solution:
Steps of Construction :
Step 1: Draw a line segment RY = 4 cm.
Step 2: At R draw an angle 140°.
Step 3: On the free arm of the angle cut an arc of length 7 cm from R and name it T,
Step 4: Join T, Y.
In this way we have obtained the required △TRY.
Step 5: To construct altitude from T to RY, extend YR to the left.
Step 6: Place the set square on the line segment (extended) such that one of the edges of the right angle touches the line.
Step 7: Slide the set square along the line YR extended till the vertical edge of the set square touches the vertex T.
Step 8: Draw the altitude to RY from T and name it TM.
3. Construct a right-angled triangle △ABC with ∠B = 90°, AC = 5 cm. How many different triangles exist with these measurements?
Solution:
Steps of Construction :
Step 1: Draw BC = 4 cm.
Step 2 : At B make an angle 90°.
Step 3 : Taking C as centre and radius = 5 cm cut an arc on the free arm of the angle 90°.
Step 4 : Join A, C.
In this way we have obtained a Δ ABC in which ∠B = 90° and AC = 5 cm.
We can construct infinitely many Δ s with these measurements.
4. Through construction, explore if it is possible to construct an equilateral triangle that is (i) right-angled (ii) obtuse-angled. Also construct an isosceles triangle that is (i) right-angled (ii) obtuse-angled.
Solution:
(i)
Draw AB = 4 cm, At A draw ∠A = 90°, cut arcs AC = 4 cm and BD = 4 cm.
We can see it is impossible to make an equilateral triangle in which one angle is a right angle.
(ii)
Draw PQ = 4 cm, At P draw ∠P = 120°, cut arc PR = 4 cm.
Join RQ.
Clearly RQ is not equal to 4 cm.
So, it is impossible to draw a Δ which is equilateral and one of its angles is greater than 90° that is an obtuse angle.
(i)We can construct an isosceles which is right-angled :
Draw BC = 5 cm, At B draw ∠B = 90°, on the free arm cut an arc of length 5 cm.
Join AC.
We have obtained a right-angled Δ which is an isosceles Δ.
(ii) Draw AB = 7 cm, At A draw ∠A = 100°, on the free arm cut on arc = 7 cm, name it C.
Join B and C.
So, Δ ABC is an obtuse isosceles △.