Andhra Pradesh SCERT Class 7 Maths The Triangle and its Properties Question and Answers Solutions
Andhra Pradesh SCERT 7th Class Maths Solutions Chapter 6 The Triangle and its Properties Question and answers. Students who are searching for Andhra Pradesh Class 7 Maths Chapter 6 can find here Solution of this chapter.
Board |
Andhra Pradesh (AP Board) |
Class |
7th |
Subject |
Maths |
Topic |
Solution |
EXERCISE 6.1
1.) In ∆ PQR, D is the mid-point of QR.
PM is _________________.
PD is _________________.
Is QM = MR?
ANSWER:
Given, in ∆ PQR, D is the mid-point of QR.
We know,
A median connects a vertex of a triangle to the mid-point of the opposite side.
PD is median of a ∆ PQR
We know,
An altitude has one end point at a vertex of the triangle and the other on the line containing the opposite side.
PM is altitude of a ∆ PQR
From figure,
QM is not equal to MR.
2.) Draw rough sketches for the following:
(a) In ∆ABC, BE is a median.
ANSWER:
We have to draw ∆ABC, BE is a median.
(b) In ∆PQR, PQ and PR are altitudes of the triangle.
ANSWER:
We have to draw ∆PQR, PQ and PR are altitudes of the triangle.
3.) Verify by drawing a diagram if the median and altitude of an isosceles triangle can be same.
ANSWER:
We know,
In an isosceles triangle, 2 sides and 2 base angle is equal.
From figure,
The median and altitude of an isosceles triangle is same.
EXERCISE 6.2
1.) Find the value of the unknown exterior angle x in the following diagrams:
ANSWER:
Here, we have to find unknown exterior angle x in the given diagram.
We know,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
From figure, 500,700 are interior opposite angles.
Exterior angle x of a triangle = 500 + 700
Exterior angle x of a triangle = 1200
ANSWER:
Here, we have to find unknown exterior angle x in the given diagram.
We know,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
From figure, 650,450 are interior opposite angles.
Exterior angle x of a triangle = 650 + 450
Exterior angle x of a triangle = 1100
ANSWER:
Here, we have to find unknown exterior angle x in the given diagram.
We know,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
From figure, 300,400 are interior opposite angles.
Exterior angle x of a triangle = 300 + 400
Exterior angle x of a triangle = 700
ANSWER:
Here, we have to find unknown exterior angle x in the given diagram.
We know,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
From figure, 600,600 are interior opposite angles.
Exterior angle x of a triangle = 600 + 600
Exterior angle x of a triangle = 1200
ANSWER:
Here, we have to find unknown exterior angle x in the given diagram.
We know,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
From figure, 500,500 are interior opposite angles.
Exterior angle x of a triangle = 500 + 500
Exterior angle x of a triangle = 1000
ANSWER:
Here, we have to find unknown exterior angle x in the given diagram.
We know,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
From figure, 300,600 are interior opposite angles.
Exterior angle x of a triangle = 300 + 600
Exterior angle x of a triangle = 900
2.) Find the value of the unknown interior angle x in the following figures:
ANSWER:
Here, we have to find unknown interior angle x in the given diagram.
We know,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
From figure, x, 500 are interior opposite angles.
Exterior angle of a triangle = x0 + 500
Exterior angle of a triangle 1150
Interior angle x of a triangle = 1150 – 500
Interior angle x of a triangle = 650
ANSWER:
Here, we have to find unknown interior angle x in the given diagram.
We know,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
From figure, x, 700 are interior opposite angles.
Exterior angle of a triangle = x0 + 700
Exterior angle of a triangle 1000
Interior angle x of a triangle = 1000 – 700
Interior angle x of a triangle = 300
ANSWER:
Here, we have to find unknown interior angle x in the given diagram.
We know,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
From figure, x, 900 are interior opposite angles.
Exterior angle of a triangle = x0 + 900
Exterior angle of a triangle 1250
Interior angle x of a triangle = 1250 – 900
Interior angle x of a triangle = 350
ANSWER:
Here, we have to find unknown interior angle x in the given diagram.
We know,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
From figure, x, 600 are interior opposite angles.
Exterior angle of a triangle = x0 + 600
Exterior angle of a triangle 1200
Interior angle x of a triangle = 1200 – 600
Interior angle x of a triangle = 600
ANSWER:
Here, we have to find unknown interior angle x in the given diagram.
We know,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
From figure, x, 300 are interior opposite angles.
Exterior angle of a triangle = x0 + 300
Exterior angle of a triangle 800
Interior angle x of a triangle = 800 – 300
Interior angle x of a triangle = 500
ANSWER:
Here, we have to find unknown interior angle x in the given diagram.
We know,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
From figure, x, 300 are interior opposite angles.
Exterior angle of a triangle = x0 + 350
Exterior angle of a triangle 750
Interior angle x of a triangle = 750 – 350
Interior angle x of a triangle = 400
EXERCISE 6.3
1.) Find the value of the unknown x in the following diagrams:
ANSWER:
We have to find unknown x in the diagram.
We know,
The sum of the measures of the three angles of a triangle is 180°
500 + 600 + x = 180°
1100 + x = 180°
x = 180° – 1100
x = 70°
ANSWER:
We have to find unknown x in the diagram.
We know,
The sum of the measures of the three angles of a triangle is 180°
300 + 900 + x = 180°
1200 + x = 180°
x = 180° – 1200
x = 60°
ANSWER:
We have to find unknown x in the diagram.
We know,
The sum of the measures of the three angles of a triangle is 180°
300 + 1100 + x = 180°
1400 + x = 180°
x = 180° – 1400
x = 40°
ANSWER:
We have to find unknown x in the diagram.
We know,
The sum of the measures of the three angles of a triangle is 180°
500 + x + x = 180°
500 + 2x = 180°
2x = 180° – 500
2x = 130°
X = 650
ANSWER:
We have to find unknown x in the diagram.
We know,
The sum of the measures of the three angles of a triangle is 180°
x + x + x = 180°
3x = 180°
x = 180°/3
X = 600
ANSWER:
We have to find unknown x in the diagram.
We know,
The sum of the measures of the three angles of a triangle is 180°
500 + x + 2x = 180°
900 + 3x = 180°
3x = 180° – 900
3x = 90°
X = 300
2.) Find the values of the unknowns x and y in the following diagrams:
ANSWER:
We have to find values of the unknowns x and y in the given diagram.
We know,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
From figure, x, 500 are interior opposite angles.
Exterior angle of a triangle = x0 + 500
Exterior angle of a triangle 1200
Interior angle x of a triangle = 1250 – 500
Interior angle x of a triangle = 700
Now,
We know,
The sum of the measures of the three angles of a triangle is 180°
500 + x + y = 180°
500 + 700 + y = 180°
y = 180° – 120°
y = 60°
ANSWER:
We have to find values of the unknowns x and y in the given diagram.
We know,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
From figure, x, 500 are interior opposite angles.
Y = 800 = opposite angles.
Now,
We know,
The sum of the measures of the three angles of a triangle is 180°
500 + x + 800 = 180°
500 + 800 + x = 180°
x = 180° – 130°
x = 50°
ANSWER:
We know,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
We know,
The sum of the measures of the three angles of a triangle is 180°
500 + 600 + y = 180°
y = 180° – 110°
y = 70°
Now
Exterior angle and y forms angles in linear pair.
x + y = 180°
x + 70°= 180°
x = 110°
ANSWER:
We have to find values of the unknowns x and y in the given diagram.
We know,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
x= 600 = opposite angles.
We know,
The sum of the measures of the three angles of a triangle is 180°
300 + 600 + y = 180°
y = 180° – 90°
y = 90°
ANSWER:
We have to find values of the unknowns x and y in the given diagram.
We know,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
y = 900 = opposite angles.
We know,
The sum of the measures of the three angles of a triangle is 180°
900 + x + x = 180°
2x = 180° – 90°
2x = 90°
X = 45°
EXERCISE 6.4
1.) Is it possible to have a triangle with the following sides?
(i) 2 cm, 3 cm, 5 cm
ANSWER:
We have to find that it is possible to have a triangle with the given sides.
We know,
The sum of the lengths of any two sides of atriangle is greater than the third side.
2 cm + 3 cm = 5 cm
Which is not greater than the third side.
Triangle having sides 2 cm, 3 cm, 5 cm is not possible.
(ii) 3 cm, 6 cm, 7 cm
ANSWER:
We have to find that it is possible to have a triangle with the given sides.
We know,
The sum of the lengths of any two sides of a triangle is greater than the third side.
3 cm + 6 cm = 9 cm
Which is greater than the third side.
Triangle having sides 3 cm, 6 cm, 7 cm is possible.
(iii) 6 cm, 3 cm, 2 cm
ANSWER:
We have to find that it is possible to have a triangle with the given sides.
We know,
The sum of the lengths of any two sides of a triangle is greater than the third side.
2 cm + 3 cm = 5 cm
Which is not greater than the third side.
Triangle having sides 6 cm, 3 cm, 2 cm is not possible.
- Take any point O in the interior of a triangle PQR. Is
(i) OP + OQ > PQ?
ANSWER:
We join OP and OQ
From figure, a new triangle is form.
We know,
The sum of the lengths of any two sides of a triangle is greater than the third side.
OP + OQ > PQ
(ii) OQ + OR > QR?
ANSWER:
From figure, a new triangle is form.
We know,
The sum of the lengths of any two sides of a triangle is greater than the third side.
OQ + OR > QR
(iii) OR + OP > RP?
ANSWER:
From figure, a new triangle is form.
We know,
The sum of the lengths of any two sides of a triangle is greater than the third side.
OR + OP > RP
3.) AM is a median of a triangle ABC.
Is AB + BC + CA > 2 AM?
(Consider the sides of triangles∆ABM and ∆AMC.)
ANSWER:
Given, AM is a median of a triangle ABC.
BM = MC
In triangle AMB
AM + BM > AB ———(i)
In triangle AMC
AM + MC > AC
But BM = MC
AM + MC > AC
We put BM instead of MC
AM + BM > AC —————— (ii)
In triangle ABC,
AB + AC > BC ————– (iii)
From equation (i),(ii) and (iii)
AB + BC + CA > 2 AM
4.) ABCD is a quadrilateral.
Is AB + BC + CD + DA > AC + BD?
ANSWER:
ABCD is a quadrilateral.
In triangle ABC,
AB + BC > AC
In triangle BCD,
BC + CD > BD
In triangle ABD,
AB + AD > BD
In triangle ACD,
AD + CD> AC
Adding all equations we get,
AB + BC + CD + DA > AC + BD
5.) ABCD is quadrilateral. Is
AB + BC + CD + DA < 2 (AC + BD)?
ANSWER:
ABCD is a quadrilateral.
In triangle ABC,
AB + BC > AC
In triangle BCD,
BC + CD > BD
In triangle ABD,
AB + AD > BD
In triangle ACD,
AD + CD > AC
Adding all equations we get,
AB + BC + CD + DA < 2 (AC + BD)
EXERCISE 6.5
1.) PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR.
ANSWER:
Given that,
PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm
We have to find QR
By Pythagoras theorem,
The square on the hypotenuse = sum of the squares on the legs
QR2= PQ2 + PR2
QR2= 102 + 242
QR2= 100 + 576
QR2 = 676
QR = 26 CM.
2.) ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.
ANSWER:
Given that,
ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm
We have to find BC.
By Pythagoras theorem,
The square on the hypotenuse = sum of the squares on the legs
AB2 = AC2 + BC2
252 = 72 + BC2
BC2 = 625 – 49
BC2 = 576
BC = 24 cm.
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:
Given that,
A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a.
We have to find the distance of the foot of the ladder from the wall I.e. a
By Pythagoras theorem,
The square on the hypotenuse = sum of the squares on the legs
Here hypotenuse = Ladder
(Ladder) 2 = (12)2 + a2
(15) 2 = (12)2 + a2
225 – 144 = a2
a2= 81
a = 9 m.
The distance of the foot of the ladder from the wall is 9m.
4.) Which of the following can be the sides of a right triangle?
(i) 2.5 cm, 6.5 cm, 6 cm.
ANSWER:
We have to find the given sides are of right triangle or not.
We know,
If the Pythagoras property holds, the triangle must be right-angled
By Pythagoras theorem,
6.52 = 2.52 + 62
42.25 = 6.25 + 36
42.25 = 42.25
2.5 cm, 6.5 cm, 6 cm are sides of right triangle.
(ii) 2 cm, 2 cm, 5 cm.
ANSWER:
We have to find the given sides are of right triangle or not.
We know,
If the Pythagoras property holds, the triangle must be right-angled
By Pythagoras theorem,
52 = 22 + 22
25 = 4 + 4
25 is not equal to8
2 cm, 2 cm, 5 cm are not sides of right triangle.
(iii) 1.5 cm, 2cm, 2.5 cm.
ANSWER:
We have to find the given sides are of right triangle or not.
We know,
If the Pythagoras property holds, the triangle must be right-angled
By Pythagoras theorem,
2.52 = 1.52 + 22
6.25 = 2.25 + 4
6.25 = 6.25
1.5 cm, 2cm, 2.5 cm are sides of right triangle.
6.) Angles Q and R of a ∆PQR are 25º and 65º.
Write which of the following is true:
(i) PQ2 + QR2 = RP2
ANSWER:
Given that,
Angles Q and R of a ∆PQR are 25º and 65º
We know,
The sum of the measures of the three angles of a triangle is 180°
250 + 650 + P = 180°
P = 180° – 90°
P = 90°
1 angle 90° then given triangle is right angled triangle.
Side opposite to 90°is hypotaneous.
QR is hypotaneous.
PQ2 + QR2 = RP2is not possible.
(ii) PQ2 + RP2 = QR2
ANSWER:
Given that,
Angles Q and R of a ∆PQR are 25º and 65º
We know,
The sum of the measures of the three angles of a triangle is 180°
250 + 650 + P = 180°
P = 180° – 90°
P = 90°
1 angle 90° then given triangle is right angled triangle.
Side opposite to 90° is hypotaneous.
QR is hypotaneous.
PQ2 + RP2 = QR2this is possible.
(iii) RP2 + QR2 = PQ2
ANSWER:
Given that,
Angles Q and R of a ∆PQR are 25º and 65º
We know,
The sum of the measures of the three angles of a triangle is 180°
250 + 650 + P = 180°
P = 180° – 90°
P = 90°
1 angle 90° then given triangle is right angled triangle.
Side opposite to 90° is hypotaneous.
QR is hypotaneous.
RP2 + QR2 = PQ2is not possible.
7.) Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
ANSWER:
Given that,
The rectangle whose length is 40 cm and a diagonal is 41 cm
We have to find the perimeter of the rectangle.
By Pythagoras theorem,
412 = 402 + (Breadth of rectangle)2
1681 – 1600 = (Breadth of rectangle)2
(Breadth of rectangle)2 = 81
(Breadth of rectangle)= 9 cm
Now,
The perimeter of the rectangle = 2 x (Length + Breadth)
The perimeter of the rectangle = 2 x (40 + 9)
The perimeter of the rectangle = 98 cm.
8.) The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
ANSWER:
Given that,
The diagonals of a rhombus measure 16 cm and 30 cm
We have to find its perimeter.
Diagonals are bisect in 2 equal parts.
By using Pythagoras theorem,
(Side of rhombus)2 = 82 + 152
(Side of rhombus)2 = 64 + 225
(Side of rhombus)2 = 289
(Side of rhombus) = 17 cm.
Now,
Perimeter of rhombus = 4 x Side of rhombus
Perimeter of rhombus = 4 x 17
Perimeter of rhombus = 68 cm