CLASS-7 MATHS CHAPTER-6 THE TRIANGLE AND ITS PROPERTIES

 CLASS-7 MATHS 
CHAPTER-6
THE TRIANGLE AND ITS PROPERTIES

                                                                

                                                            EXERCISE-6.1

Question 1: In ΔPQR, D is the mid-point of .

is __________.

PD is __________.

Is QM = MR?

Answer:

(i) Altitude

(ii) Median

(iii) No

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)

(b)

(c)

Here, it can be observed that for ΔXYZ, YL is an altitude drawn exterior to side XZ which is extended up to point L.

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

Answer:

Draw a line segment AD perpendicular to BC. It is an altitude for this triangle. It can be observed that the length of BD and DC is also same. Therefore, AD is also a median of this triangle.

 EXERCISE-6.2

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

Answer:

(i) x = 50° + 70° (Exterior angle theorem)

x = 120°

(ii) x = 65° + 45° (Exterior angle theorem)

= 110°

(iii) x = 40° + 30° (Exterior angle theorem)

= 70°

(iv) x = 60° + 60° (Exterior angle theorem)

= 120°

(v) x = 50° + 50° (Exterior angle theorem)

= 100°

(vi) x = 30° + 60° (Exterior angle theorem)

= 90°

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

Answer:

(i) x + 50° = 115° (Exterior angle theorem)

x = 115° − 50° = 65°

(ii) 70° + x = 100° (Exterior angle theorem)

x = 100° − 70° = 30°

(iii) x + 90° = 125° (Exterior angle theorem)

x = 125° − 90° = 35°

(iv) x + 60° = 120° (Exterior angle theorem)

x = 120° − 60° = 60°

(v) x + 30° = 80° (Exterior angle theorem)

x = 80° − 30° = 50°

(vi) x + 35° = 75° (Exterior angle theorem)

x = 75º − 35º = 40°

 EXERCISE-6.3

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

Answer: The sum of all interior angles of a triangle is 180°. By using this property, these problems can be solved as follows.

(i) x + 50° + 60° = 180°

x + 110° = 180°

x = 180° − 110° = 70°

(ii) x + 90° + 30° = 180°

x + 120° = 180°

x = 180° − 120° = 60°

(iii) x + 30° + 110° = 180°

x + 140° = 180°

x = 180° − 140° = 40°

(iv) 50° + x + x = 180°

50° + 2x = 180°

2x = 180° − 50° = 130°

(v) x + x + x = 180°

3x = 180°

(vi) x + 2x + 90° = 180°

3x = 180° − 90° = 90º

Question 2: Find the value of the unknowns x and y in the following diagrams:

Answer:

(i) y + 120° = 180° (Linear pair)

y = 180° − 120º = 60º

x + y + 50° = 180° (Angle sum property)

x + 60° + 50° = 180°

x + 110° = 180°

x = 180° − 110° = 70°

(ii) y = 80° (Vertically opposite angles)

y + x + 50° = 180° (Angle sum property)

80° + x + 50° = 180°

x + 130º = 180°

x = 180° − 130º = 50°

(iii) y + 50° + 60° = 180° (Angle sum property)

y = 180° − 60° − 50° = 70°

x + y = 180° (Linear pair)

x = 180° − y = 180° − 70° = 110°

(iv) x = 60º (Vertically opposite angles)

30° + x + y = 180°

30° + 60° + y = 180°

y = 180° − 30° − 60° = 90°

(v) y = 90° (Vertically opposite angles)

x + x + y = 180° (Angle sum property)

2x + y = 180°

2x + 90° = 180°

2x = 180° − 90° = 90°

 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

Answer:In a triangle, the sum of the lengths of either two sides is always greater than the third side.

(i) Given that, the sides of the triangle are 2 cm, 3 cm, 5 cm.

It can be observed that,

2 + 3 = 5 cm

However, 5 cm = 5 cm

Hence, this triangle is not possible.

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: If O is a point in the interior of a given triangle, then three triangles ΔOPQ, ΔOQR, and ΔORP can be constructed. In a triangle, the sum of the lengths of either two sides is always greater than the third side.

(i) Yes, as ΔOPQ is a triangle with sides OP, OQ, and PQ.

OP + OQ > PQ

(ii) Yes, as ΔOQR is a triangle with sides OR, OQ, and QR.

OQ + OR > QR

(iii) Yes, as ΔORP is a triangle with sides OR, OP, and PR.

OR + OP > PR

Question 3:AM is a median of a triangle ABC. Is AB + BC + CA > 2 AM?

(Consider the sides of triangles ΔABM and ΔAMC.)

Answer: In a triangle, the sum of the lengths of either two sides is always greater than the third side.

In ΔABM,

AB + BM > AM (i)

Similarly, in ΔACM,

AC + CM > AM (ii)

Adding equation (i) and (ii),

AB + BM + MC + AC > AM + AM

AB + BC + AC > 2AM

Yes, the given expression is true.

Question 4:ABCD is quadrilateral. Is AB + BC + CD + DA >AC + BD?

Answer: In a triangle, the sum of the lengths of either two sides is always greater than the third side.

Considering ΔABC,

AB + BC > CA (i)

In ΔBCD,

BC + CD > DB (ii)

In ΔCDA,

CD + DA > AC (iii)

In ΔDAB,

DA + AB > DB (iv)

Adding equations (i), (ii), (iii), and (iv), we obtain

AB + BC + BC + CD + CD + DA + DA + AB > AC + BD + AC + BD

2AB + 2BC + 2CD +2DA > 2AC + 2BD

2(AB + BC + CD + DA) > 2(AC + BD)

(AB + BC + CD + DA) > (AC + BD)

Yes, the given expression is true.

 EXERCISE-6.5

Question 1:PQR is a triangle right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

Answer:

By applying Pythagoras theorem in ΔPQR,

(PQ)² + (PR)² = (RQ)²

(10)² + (24)² = RQ²

100 + 576 = (QR)²

676 = (QR)²

QR = 26 cm

Question 2:ABC is a triangle right angled at C. If AB = 25 cm and AC = 7 cm, find BC.

Answer:

By applying Pythagoras theorem in ΔABC,

(AC)² + (BC)² = (AB)²

(BC)² = (AB)² − (AC)²

(BC)² = (25)² − (7)²

(BC)² = 625 − 49 = 576

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 a. Find the distance of the foot of the ladder from the wall.

Answer:

By applying Pythagoras theorem,

(15)² = (12)² + a²

225 = 144 + a²

a² = 225 − 144 = 81

a = 9 m

Therefore, the distance of the foot of the ladder from the wall is 9 m.

Question 4: Which of the following can be the sides of a right triangle?

(i) 2.5 cm, 6.5 cm, 6 cm

Answer: (i) 2.5 cm, 6.5 cm, 6 cm

(2.5)² = 6.25

(6.5)² = 42.25

(6)² = 36

It can be observed that,

36 + 6.25 = 42.25

(6)² + (2.5)² = (6.5)²

The square of the length of one side is the sum of the squares of the lengths of the remaining two sides. Hence, these are the sides of a right-angled triangle. Right angle will be in front of the side of 6.5 cm measure.

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.

Answer:

In the given figure, BC represents the unbroken part of the tree. Point C represents the point where the tree broke and CA represents the broken part of the tree. Triangle ABC, thus formed, is right-angled at B.

Applying Pythagoras theorem in ΔABC,

AC² = BC² + AB²

AC² = (5 m)² + (12 m)²

AC² = 25 m² + 144 m² = 169 m²

AC = 13 m

Thus, original height of the tree = AC + CB = 13 m + 5 m = 18 m

Question 6: Angles Q and R of a ΔPQR are 25° and 65°.

Write which of the following is true:

(i) PQ² + QR²= RP²

(ii) PQ² + RP²= QR²

(iii) RP² + QR²= PQ²

Answer:  The sum of the measures of all interior angles of a triangle is 180º.

∠PQR + ∠PRQ + ∠QPR = 180º

25º + 65º + ∠QPR = 180º

90º + ∠QPR = 180º

∠QPR = 180º − 90º = 90º

Therefore, Δ PQR is right-angled at point P.

Hence, (PR)² + (PQ)² = (QR)²

Thus, (ii) is true.

Question 7: Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

Answer:

In a rectangle, all interior angles are of 90º measure. Therefore, Pythagoras theorem can be applied here.

(41)² = (40)² + x²

1681 = 1600 + x²

x² = 1681 − 1600 = 81

x = 9 cm

Perimeter = 2(Length + Breadth)

= 2(x + 40)

= 2 (9 + 40)

= 98 cm

CLASS-6 MATHS CHAPTER-4 BASIC GEOMETRICAL IDEAS

                                                                         Chapter-4

Basic geometrical ideas

                                                                                       Exercise-4.1

Question 1:Use the figure to name:

(a) Five points

(b) A line

(c) Four rays

(d) Five line segments

Answer:(a) The five points are D, E, O, B, and C.

(b) 

(c) 

(d) 

Question 2:Name the line given in all possible (twelve) ways, choosing only two letters at a time from the four given.

Answer:

Exercise-4.2

Question 3:Use the figure to name:

(a) Line containing point E.

(b) Line passing through A.

(c) Line on which O lies

(d) Two pairs of intersecting lines.

Answer:

(a) 

(b) 

(c) 

(d) , 

Question 4:How many lines can pass through (a) one given point? (b) Two given points?

Answer:(a) Infinite number of lines can pass through a single point.

(b) Only one line can pass through two given points.

Question 5:

Draw a rough figure and label suitably in each of the following cases:

(a) Point P lies on.

(b) and intersect at M.

(c) Line l contains E and F but not D.

(d) and meet at O.

Answer:

(a)

(b)

(c)

(d)

Question 6: Consider the following figure of line. Say whether following statements are true or false in context of the given figure.

(a) Q, M, O, N, P are points on the line.

(b) M, O, N are points on a line segment.

(c) M and N are end points of line segment.

(d) O and N are end points of line segment.

(e) M is one of the end points of line segment.

(f) M is point on ray.

(g) Ray is different from ray.

(h) Ray is same as ray.

(i) Ray is not opposite to ray.

(j) O is not an initial point of.

(k) N is the initial point of and.

Answer:

(a) True

(b) True

(c) True

(d) False

(e) False

(f) False

(g) True

(h) False

(i) False

(j) False

(k) True

Exercise-4.3

Question 1:Classify the following curves as (i) Open or (ii) Closed.

Answer:

(a) Open

(b) Closed

(c) Open

(d) Closed

(e) Closed

Question 2:Draw rough diagrams to illustrate the following:

(a) Open curve (b) Closed curve.

Answer:

(a) Open curve

(b) Closed curve

Question 3:

Draw any polygon and shade its interior.

Answer:

Question 4:Consider the given figure and answer the questions:

(a) Is it a curve? (b) Is it closed?

Answer:

(a) Yes

(b) Yes

Question 5:Illustrate, if possible, each one of the following with a rough diagram:

(a) A closed curve that is not a polygon.

(b) An open curve made up entirely of line segments.

(c) A polygon with two sides.

Answer:

(a)

(b)

(c) This is not possible as the polygon having the least number of sides is a triangle, which has three sides in it.

Exercise-4.4

Question 1:Name the angles in the given figure.

Answer:

∠BAD, ∠ADC, ∠DCB, ∠CBA

Question 2:In the given diagram, name the point (s)

(a) In the interior of ∠DOE

(b) In the exterior of ∠EOF

(c) On ∠EOF

Answer:

(a) A

(b) C, A, D

(c) B, E, O, F