Category: Class 7

MP Board Class 7th Maths Solutions Chapter 12 Algebraic Expressions Ex 12.2

Question 1.
Simplify by combining like terms:
(i) 21b – 32 +7b – 20b
(ii) -z² + 13z² – 5z + 7z³ – 15z
(iii) p – (p – q) – q – (q – p)
(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
(v) 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² – 3y²
(vi) (3y² + 5y – 4) – (8y – y² – 4)
Solution:
(i) 21b – 32 + 7b – 20b
= 21b + 7b – 20b – 32
= (21 + 7 – 20)b – 32
= 8b – 32

(ii) -z² + 13z² – 5z + 7z³ – 15z
= 7z³ + (-1 + 13) z² + (-5 -15) z
= 7z³ + 12z² – 20z

(iii) p – (p – q) – q – (q – p)
= p – p + q – q – q + p
= p – q

(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
= 3a – 2b – ab – a + b – ab + 3 ab + b – a
= (3 – 1 – 1)a + (-2 + 1 + 1)b + (-1 – 1 + 3)ab
= a + ab

(v) 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² – 3y²
= (5 + 3) x²y + (-5 + 1) x² + (- 3 – 1 – 3 )y² + 8xy²
= 8x²y – 4x² – 7y² + 8xy²

(vi) (3y² + 5y – 4) – (8y – y² – 4)
= 3y² + 5y – 4 – 8y + y² + 4
= (3 + 1) y² + (5 – 8) y + 4 – 4
= 4y² – 3y

Question 2.
Add:
(i) 3mn, -5mn, 8mn, -4mn
(ii) t – 8tz, 3tz – z, z – 1
(iii) -7mn + 5, 12mn + 2, 9mn – 8, -2mn – 3
(iv) a + b – 3, b – a + 3, a – b + 3
(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
(vii) 4x²y, -3xy², 5xy², 5x²y
(viii) 3 p²q² – 4pq + 5, -10p²q², 15 + 9pq + 7p²q²
(ix) ab – 4a, 4b – ab, 4a – 4b
(x) x² – y² – 1, y² – 1 – x², 1 – x² – y².
Solution:
(i) 3 mn + (-5 mn) + 8 mn + (-4 mn)
= (3 – 5 + 8 – 4 )mn = 2 mn

(ii) (t – 8tz) + (3tz – z) + (z – t)
= t – 8tz + 3tz – z + z – t
= (1 – 1)t + (- 8 + 3)tz + (-1 + 1)z
= -5 tz

(iii) (-7mn + 5) + (12mn + 2) + (9mn – 8) + (-2mn – 3)
=-7mn + 5 + 12mn + 2 + 9mn – 8 – 2mn – 3
= (-7 + 12 + 9 – 2)mn + (5 + 2 – 8 – 3)
= 12mn – 4

(iv) (a + b – 3) + (b – a + 3) + (a – b + 3)
= a + b – 3 + b – a + 3 + a – b + 3
= (1 -1 + 1)a + (1 + 1 – 1)b + (- 3 + 3 + 3)
= a + b + 3

(v) (14x + 10y – 12xy – 13) + (18 – 7x – 10y + 8xy) + 4xy
= 14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy
= (14 – 7)x + (10 – 10)y + (12 + 8 + 4 )xy + (-13 + 18)
= 7x + 5

(vi) (5m – 7n) + (3n – 4m + 2) + (2m – 3mn – 5)
= 5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5
= (5 – 4 + 2)m + (- 7 + 3)n – 3mn + (2 – 5)
= 3m – 4n – 3mn – 3

(vii) (4x²y) + (-3xy²) + (-5xy²) + (5x²y)
= 4x²y – 3xy² – 5xy² + 5x²y
= (4 + 5) x²y + (-3 – 5) xy²
= 9x²y – 8xy²

(viii) (3p²q² – 4pq + 5) + (-10p²q²) + (15 + 9pq +7p²q²)
= 3p²q² – 4pq + 5 – 10p²q² + 15 + 9pq + 7p²q²
= (3 – 10 + 7) p²q² + (-4 + 9)pq + (5 + 15)
= 5pq + 20

(ix) (ab – 4a) +(4 b – ab) + (4a- 4b)
= ab – 4a + 4b – ab + 4a – 4b
= (1 – 1)ab + (-4 + 4 )a + (4 – 4)b = 0

(x) (x² – y² – 1) + (y² – 1 – x²) + (1 – x² – y²)
= x² – y² – 1 + y² – 1 – x² + 1 – x² – y²
= (1 – 1 – 1)x² + (-1 + 1 – 1)y² + (-1 – 1 + 1)
= -x² – y² – 1

Question 3.
Subtract:
(i) -5y² from y²
(ii) 6xy from -12xy
(iii) (a – b) from (a + b)
(iv) a(b – 5) from b (5 – a)
(v) – m2 + 5 mn from 4mi2 – 3mn + 8
(vi) -x² + 10x – 5 from 5x – 10
(vii) 5a² – 7ab + 5b² from 3ab – 2a² – 2b²
(viii) 4pq – 5q² – 3p² from 5p² + 3q² – pq
Solution:
(i) y² – (-5y²) = y² + 5y² = 6y²
(ii) -12xy – (6xy) = -12xy – 6xy = -18xy
(iii) (a + b) – (a – b) = a + b – a + b = 2b

(iv) b(5 – a) – a(b – 5)
= 5b – ab – ab + 5a
= 5a + 5b – 2ab

(v) (4m² – 3mn + 8) – (- m² + 5mn)
= 4m² – 3mn + 8 + m² – 5mn
= (4 + 1)m² + (- 3 – 5 )mn + 8
= 5m² – 8mn + 8

(vi) (5x – 10) – (-x² + 10x – 5)
= 5x – 10 + x² – 10x + 5
= x² + (5 – 10)x + (-10 + 5)
= x² – 5x – 5

(vii) (3ab – 2a² – 2b²) – (5a² – 7ab + 5b²)
= 3ab – 2a² – 2b² – 5a² + 7ab – 5b²
= (3 + 7)ab + (- 2 – 5)a² + (- 2 – 5 )b²
= 10ab – 7a² – 7b²

(viii) (5p² + 3q² – pq) – (4pq – 5a² – 3p²)
= 5p² + 3q² – pq – 4pq + 5q² + 3p²
= (5 + 3 )p² + (3 + 5 )q² + (-1 – 4 )pq
= 8p² + 8q² – 5pq

Question 4.
(a) What should be added to x² + xy + y² to obtain 2x² + 3xy?
(b) What should be subtracted from 2a + 8b + 10 to get – 3a + 7b + 16?
Solution:
(a) Let a be the required term.
∴ a + (x² + y² + xy) = 2x² + 3xy
⇒ a = 2x² + 3xy – (x² + y² + xy)
= 2x² + 3xy – x² – y² – xy
= (2 – 1) x² – y² + (3 – 1)xy
= x² – y² + 2xy

(b) Let p be the required term.
∴ (2a + 8b + 10) -p = -3a + 7b + 16
⇒ p = 2a + 8b + 10 – (- 3a + 7b + 16)
= 2a + 8b + 10 + 3a – 7b – 16
= (2 + 3)a + (8 – 7)b + (10 – 16)
= 5a + b – 6

Question 5.
What should be taken away from 3x² – 4y² + 5xy + 20 to obtain -x² – y² + 6xy + 20 ?
Solution:
Required term
= (3x² – 4y² + 5xy + 20) – (-x² – y² + 6xy + 20)
= 3x² – 4y² + 5xy + 20 + x² + y² – 6xy – 20
= (3 + 1)x² + (- 4 + 1) y² + (5 – 6)xy + (20 – 20)
= 4x² – 3y² – xy

Question 6.
(a) From the sum of 3x – y + 11 and -y – 11, subtract 3x – y – 11.
(b) From the sum of 4 + 3x and 5 – 4x + 2x², subtract the sum of 3x² – 5x and -x² + 2x + 5.
Solution:
(a) Sum of 3x – y + 11 and – y – 11
= (3x – y + 11) + (-y – 11)
= 3x – y + 11 – y – 11
= 3x + (- 1 – 1) y + (11 – 11)
= 3x – 2y
Now, required difference
= (3x – 2y) – (3x – y – 11)
= 3x – 2y – 3x + y + 11
= (3 – 3)x + (- 2 + 1) y + 11
= -y + 11

(b) Sum of 4 + 3x and 5 – 4x + 2x²
= (4 + 3x) + (5 – 4x + 2x²)
= 4 + 3x + 5 – 4x + 2x²
= (3 – 4)x + 2x² + 4 + 5
= – x + 2x² + 9
Now, sum of 3x² – 5x and -x² + 2x + 5
= (3x² – 5x) + (-x² + 2x + 5)
= 3x² – 5x – x² + 2x + 5
= (3 – 1) x² + (- 5 + 2)x + 5
= 2x² – 3x + 5
Required difference
= (- x + 2x² + 9) – (2x² – 3x + 5)
= -x + 2x² + 9 – 2x² + 3x – 5
= (-1 + 3)x + (2 – 2) x² + (9 – 5)
= 2x + 4

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 12 Algebraic Expressions Ex 12.1

Question 1.
Get the algebraic expressions in the following cases using variables, constants and arithmetic operations.
(i) Subtraction of z from y.
(ii) One-half of the sum of numbers x and y.
(iii) The number z multiplied by itself.
(iv) One-fourth of the product of numbers p and Q.
(v) Numbers x and y both squared and added.
(vi) Number 5 added to three times the product of numbers m and n.
(vii) Product of numbers y and z subtracted from 10.
(viii) Sum of numbers a and b subtracted from their product.
Solution:
(i) y – z
(ii) \(\frac{1}{2}\)(x + y)
(iii) z²
(iv) \(\frac{1}{4}\)(pq)
(v) x² + y²
(vi) 5 + 3 (mn)
(viii) ab – (a + b)

Question 2.
(i) Identify the terms and their factors in the following expressions. Show the terms and factors by tree diagrams.
(a) x – 3
(b) 1 + x + x²
(c) y – y³
(d) 5xy² + 7x²y
(e) -ab + 2b² – 3a²

(ii) Identify terms and factors in the expressions given below:
(a) -4x + 5
(b) -4x + 5y
(c) 5y + 3y²
(d) xy + 2x²y²
(e) pq + q
(f) 1.2ab – 2.4b + 3.6a
(g) \(\frac{3}{4}\)x + \(\frac{1}{4}\)
(h) 0.1p² + 0.2q²
Solution:
(i)

Question 3.
Identify the numerical coefficients of terms (other than constants) in the following expressions:
(i) 5 – 3t²
(ii) 1 + t + t² + t³
(iii) x + 2xy + 3y
(iv) 100m + 1000n
(v) -p²q² + 7pq
(vi) 1.2a + 0.8b
(vii) 3.14 r²
(viii) 2(l + b)
(ix) 0.1y + 0.01y²
Solution:

Question 4.
(a) Identify terms which contains x and give the coefficient of x.
(i) y²x + y
(ii) 13y² – 8yx
(iii) x + y + 2
(iv) 5 + z + zx
(v) 1 + x + xy
(vi) 12xy² + 25
(vii) 7x + xy²
(b) identify terms which contains y² and give the coefficient of y².
(i) 8 – xy²
(ii) 5y² + 7x
(iii) 2x²y – 15xy² + 7y²
Solution:
(a)

Question 5.
Classify into trinomials.
(i) 4y – 7z
(ii) y²
(iii) x + y – xy
(iv) 100
(v) ab – a – b
(vi) 5 – 3t
(vii) 4p²q – 4pq²
(viii) 7mn
(ix) z² – 3z + 8
(x) a² + b²
(xi) z² + z
(xii) 1 + x + x²
Solution:
The monomials, binomials and trinomials have 1, 2 and 3 unlike terms in it respectively.
(i) 4y – 7z Binomial
(ii) y² Monomial
(iii) x + y – xy Trinomial
(iv) 100 Monomial
(v) ab – a – b Trinomial
(vi) 5 – 3t Binomial
(vii) 4p²q – 4pq² Binomial
(viii) 7mn Monomial
(ix) z² – 3z + 8 Trinomial
(x) a² + b² Binomial
(xi) z² + z Binomial
(xii) 1 + x + x² Trinomial

Question 6.
State whether a given pair of terms is of like or unlike terms.
(i) 1,100
(ii) -7x, \(\frac{5}{2}\)x
(iii) -29x, -29y
(iv) 14xy, 42yx
(v) 4m²p, 4mp²
(vi) 12xz, 12x²z²
Solution:
The terms which have same algebraic factors are called like terms. However, when terms have different algebraic factors, these are called unlike terms.
(i) 1,100 Like
(ii) -7x, \(\frac{5}{2}\)x Like
(iii) -29x, -29y Unlike
(iv) 14xy, 42yx Like
(v) 4m²p, 4mp² Unlike
(vi) 12xz, 12x²z² Unlike

Question 7.
Identify like terms in the following:
(a) -xy², -4yx², 8x², 2xy², 7y, -11x², -100x, -11yx, 20x²y, -6x², y, 2xy, 3x
(b) 10pg, 7p, 8q, -p²q², -7qp, – 100q, -23, 12q²p², – 5p², 41, 2405p, 78qp, 13p²g, qp²,701p²
Solution:
(a) -xy² and 2xy²; -4yx² and 20x²y; 8x², -11x² and -6x²; 7y and y, -100x and 3x; -11yx and 2xy

(b) 10pq, -7qp and 78qp; 7p and 2405p; 8q and -100y; -p²q² and 12q²p², -23 and 41; -5p² and 701 p²; 13p2²q and qp²

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 7 Congruence of Triangles Ex 7.1

Question 1.
Complete the following statements:
(a) Two line segments are congruent if ___ .
(b) Among two congruent angles, one has a measure of 70°; the measure of the other angle is ___.
(c) When we write ∠A = ∠B, we actually mean ___.
Solution:
(a) They have the same length.
(b) 70°
(c) m∠A = m∠B

Question 2.
Give any two real-life examples for congruent shapes.
Solution:
(i) Sheets of same letter pad.
(ii) Biscuits in the same packet.

Question 3.
If ∆ABC ≅ ∆FED under the correspondence ABC ⟷ FED, write all the corresponding congruent parts of the triangles.
Solution:
If ∆ABC ≅ ∆FED, then the corresponding angles and sides will be equal to each other.

Question 4.
If ∆DEF ⟷ ∆BCA, write the part(s) of ∆BCA that correspond to
(i) ∠E
(ii) \(\overline{E F}\)
(iii) ∠F
(iv) \(\overline{D F}\)
Solution:
(i) ∠C
(ii) \(\overline{C A}\)
(iii) ∠A
(iv) \(\overline{B A}\)

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.5

Question 1.
PQR is a triangle, right-angled at P. If PQ= 10 cm and PR = 24 cm, find QR.
Solution:
By applying Pythagoras theorem in ∆PQR,
(PQ)² + (PR)² = (QR)²
⇒ (10)² + (24)² = (QR)2²
⇒ (QR)² = 100 + 576
= 676 = (26)²
∴ QR = 26 cm

Question 2.
ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.
Solution:
By applying Pythagoras theorem in AABC,
(AC)² + (BC)² = (AB)²
⇒ (BC)² = (AB)² – (AC)²
= (25)² – (7)²
= 625 – 49 = 576
= (24)²
∴ 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.

Solution:
By applying Pythagoras theorem,
(15)² = (12)² + a²
⇒ 225 = 144 + a²
⇒ a² = 225 – 144 = 81 = 9²
⇒ 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.
(ii) 2 cm, 2 cm, 5 cm.
(iii) 1.5 cm, 2 cm, 2.5 cm.
In the case of right-angled triangles, identify the right angles.
Solution:
(i) 2.5 cm, 6.5 cm, 6 cm
∴ (2.5)² = 6.25, (6.5)² = 42.25 and (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 measure 6.5 cm.

(ii) 2 cm, 2 cm, 5 cm
∴ (2)² = 4, (2)² = 4 and (5)² = 25
Here, (2)² + (2)² ≠ (5)²
The square of the length of one side is not equal to the sum of the squares of the lengths of the remaining two sides. Hence, these sides are not of a right-angled triangle.

(iii) 1.5 cm, 2 cm, 2.5 cm
∴ (1.5)² = 2.25, (2)² = 4 and (2.5)² = 6.25
Here, 2.25 + 4 = 6.25
⇒ (1.5)² + (2)² = (2.5)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 measure 2.5 cm.

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.
Solution:
In the drawn 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 C theorem in ∆ABC,
AC² = BC² + AB²
= (5)²+ (12)² BL
= 25 + 144 = 169
= (13)²
∴ AC = 13 m
Thus, original height of the tree = AC + CB = 13m + 5m = 18m

Question 6.
Angles Q and R of a ∆PQR are 25° and 65°. Write which of the following is true:
(i) PQ² + QR² = RP²
(ii) PQ² + RP² = QR²
(iii) RP² + QR² = PQ²

Solution:
The sum of the measures of all interior angles of a triangle is 180°.
∠P + ∠Q + ∠R = 180°
∠P + 25°+ 65° = 180°
∠P + 90° = 180°
∠P = 180° – 90° = 90°
Therefore, ∆PQR is right-angled at point P. Hence, (PR)² + (PQ)² = (QR)²
∴ (ii) is true.

Question 7.
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
Solution:
Let ABCD be the given rectangle with length AB = 40 cm and diagonal AC = 41 cm.
In ∆ABC,
(AC)² = (AB)² + (BC)²
⇒ (4)² = (40)²+ (BQ)²
⇒ (BC)² = 1681 – 1600 = 81 = 9²
∴ BC = 9 cm

∴ Perimeter of rectangle = 2(40 + 9)
= 2(49) = 98 cm.

Question 8.
The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
Solution:
Let ABCD be the given rhombus with diagonals AC = 16 cm and BD = 30 cm.
We know that diagonals of a rhombus bisect each other at right angle.
Let diagonals AC and BD bisects at O.

Hence, perimeter of rhombus = 4 × (17) = 68 cm

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 13 Exponents and Powers Ex 13.3

Question 1.
Write the following numbers in the expanded forms:
279404, 3006194, 2806196, 120719, 20068
Solution:
279404 = 2 × 10⁵ + 7 × 10⁴ + 9 × 10³ + 4 × 10² + 0 × 10¹ + 4 × 10⁰
3006194 = 3 × 10⁶ + 0 × 10⁵ + 0 × 10⁴ + 6 × 10³ + 1 × 10² + 9 × 10¹ + 4 × 10⁰
2806196 = 2 × 10⁶ + 8 × 10⁵ + 0 × 10⁴ + 6 × 10³ + 1 × 10² + 9 × 10¹ + 6 × 10⁰
120719 = 1 × 10⁵ + 2 × 10⁴ + 0 × 10³ + 7 × 10² + 1 × 10¹ + 9 × 10⁰
20068 = 2 × 10⁴ + 0 × 10³ + 0 × 10² + 6 × 10¹ + 8 × 10⁰

Question 2.
Find the number from each of the following expanded forms:
(a) 8 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 10¹ + 5 × 10⁰
(b) 4 × 10⁵ + 5 × 10³ + 3 × 10² + 2 × 10⁰
(c) 3 × 10⁴ + 7 × 10² + 5 × 10⁰
(d) 9 × 10⁵ + 2 × 10² + 3 × 10¹
Solution:
(a) 8 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 10¹ + 5 × 10⁰ = 86045
(b) 4 × 10⁵ + 5 × 10³ + 3 × 10² + 2 × 10⁰ = 405302
(c) 3 × 10⁴ + 7 × 10² + 5 × 10⁰ = 30705
(d) 9 × 10⁵ + 2 × 10² + 3 × 10¹ = 900230

Question 3.
Express the following numbers in standard form:
(i) 5,00,00,000
(ii) 70,00,000
(iii) 3,18,65,00,000
(iv) 3,90,878
(v) 39087.8
(vi) 3908.78
Solution:
(i) 50000000 = 5 × 10⁷
(ii) 7000000 = 7 × 10⁶
(iii) 3186500000 = 3.1865 × 10⁹
(iv) 390878 = 3.90878 × 10⁵
(v) 39087.8 = 3.90878 × 10⁴
(vi) 3908.78 = 3.90878 × 1³

Question 4.
Express the number appearing in the following statements in standard form.
(a) The distance between Earth and Moon is 384.0. 000 m.
(b) Speed of light in vacuum is 300,000,000 m/s
(c) Diameter of the Earth is 1,27,56,000 m.
(d) Diameter of the Sun is 1,400,000,000 m.
(e) In a gala×y there are on an average 100,000,000,000 stars.
(f) The universe is estimated to be about 12,0,000,000 years old.
(g) The distance of the Sun from the centre of the Milky Way Gala×y is estimated to be 300,000,000,000,000,000,000 m.
(h) 60, 230, 000, 000, 000, 000, 000, 000 molecules are contained in a drop of water weighing 1.8 gm.
(i) The earth has 1, 353, 000,000 cubic km of sea water.
(j) The population of India was about 1,027,0,000 in March, 2001.
Solution:
(a) 3.84 × 10⁸ m
(b) 3 × 10⁸ m/s
(c) 1.2756 × 10⁷ m
(d) 1.4 × 10⁹ m
(e) 1 × 10¹¹ stars
(f) 1.2 × 10¹¹ years
(g) 3 × 10²⁰ m
(h) 6.023 × 10²² molecules
(i) 1.353 × 10⁹ cubic km
(j) 1.027 × 10⁹

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 6 The Triangles and Its Properties Ex 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
Solution:
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) cm = 5 cm
However, 5 cm = 5 cm
Hence, this triangle is not possible.

(ii) Given that, the sides of the triangle are
3 cm, 6 cm, 7 cm.
Here, (3 + 6) cm = 9 cm > 7 cm
(6 + 7) cm = 13 cm > 3 cm
(3 + 7) cm = 10 cm > 6 cm
Hence, this triangle is possible.

(iii) Given that, the sides of the triangle are
6 cm, 3 cm, 2 cm.
Here, (6 + 3) cm = 9 cm > 2 cm
However, (3 + 2) cm = 5 cm < 6 cm Hence, this triangle is not possible.

Question 2.
Take any point 0 in the interior of a triangle PQR. Is

(i) OP + OQ > PQ?
(ii) OQ + OR > QR?
(iii) OR + OP > RP?
Solution:
If O is a point in the interior of a given triangle, then three triangles ∆OPQ, ∆OQR and ∆ORP 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 OQR is a triangle with sides OR, OQ and QR.

∴ OQ + OR> QR

(ii) Yes, as OQR is a triangle sides OR, OQ and QR.
∴ OQ + OR > QR

(iii) Yes, as A ORP 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 > 2AM?
(Consider the sides of ∆ABM and ∆AMC.)

Solution:
In a triangle, the sum of the lengths of either two sides is always greater than the third side.
In ∆ABM,
AB + BM > AM …(i)
Similarly, in ∆ACM,
AC + CM > AM …(ii)
Adding (i) and (ii), we obtain
AB + BM + MC +AC > AM + AM
AB + BC + AC > 2AM
Yes, the given expression is true.

Question 4.
ABCD is a quadrilateral.

Is AB + BC + CD + DA > AC + BD ?
Solution:
In a triangle, the sum of the lengths of either two sides is always greater than the third side.
In ∆ABC; AB + BC > CA …(i)
In ∆BCD; BC + CD > DB …(ii)
In ∆CDA; CD + DA > AC …(iii)
In ∆DAB; DA + AB > DB …(iv)
Adding (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.

Question 5.
ABCD is a quadrilateral.

Is AB + BC + CD + DA < 2(AC + BD)?
Solution:
In a triangle, the sum of the lengths of either two sides is always greater than the third side.
Let diagonals AC and BD of the quadrilateral ABCD intersects at O.

In ∆OAB, OA + OB > AB … (i)
In ∆OBC, OB + OC > BC … (ii)
In ∆OCD, OC + CD > CD … (iii)
In ∆ODA, OD + OA > DA … (iv)
OA + OB + OB + OC + OC + OD + OD + OA > AB + BC + CD + DA
⇒ 20A + 20B + 20C + 20D > AB + BC + CD + DA
⇒ 2(OA + OC) + 2(OB + OD) > AB + BC + CD + DA
⇒ 2(AC) + 2(BD) > AB + BC + CD + DA
⇒ 2(AC + BD) > AB + BC + CD + DA
Yes, the given expression is true.

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?
Solution:
In a triangle, the Sum of the lengths of either two side is always greater than the third side.
Lengths of two sides of a triangle are 12 cm and 15 cm.
Let the third side be x cm.
∴ x + 12 > 15 ⇒ x > 3
x + 15 > 12 ⇒ x > – 3 but side length never be negative.
and 12 + 15 > x ⇒ 27 > x
Hence, third side can measure between 3 and 27.

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 13 Exponents and Powers Ex 13.2

Question 1.
Using laws of exponents, simplify and write the answer in exponential form:
(i) 3² × 3⁴ × 3⁸
(ii) 6¹⁵ ÷ 6¹⁰
(iii) a³ × a²
(iv) 7^x × 7²
(v) (5²)³ 4 5³
(vi) 2⁵ × 5⁵
(vii) a⁴ × b⁴
(viii) (3⁴)³
(ix) (2²⁰ ÷ 2¹⁵) × 2³
(x) (8^t ÷ 8²)
Solution:
(i) 3² × 3⁴ × 3⁸ = (3)^{2 + 4 + 8} = 3¹⁴ [∵ a^m × aⁿ = a^m+n]
(ii) 6¹⁵ ÷ 6¹⁰ = (6)^{15 – 10} = 6⁵ [∵ a^m ÷ aⁿ = a^{m – n}]
(iii) a³ × a² = a^{3 + 2} = a⁵[∵ a^m × aⁿ = a^m+n]
(iv) 7^x × 7² = 7^{x + 2}[∵ a^m × aⁿ = a^m+n]
(v) (5²)³ ÷ 5³ = 5^{2 × 3} ÷ 5³ [∵ (a^m)ⁿ = a^mn]
= 5⁶ ÷ 5³ = 5^{(6 – 3)} [∵ a^m ÷ aⁿ = a^{m – n}]
= 5³
(vi) 2⁵ × 5⁵ = (2 × 5)⁵      [∵ a^m × b^m = (a × b)^m]
= 10⁵
(vii) a⁴ × b⁴ = (ab)⁴
(viii) (3⁴)³ = 3^{4 × 3} = 3¹²       [∵ (a^m)ⁿ = a^mn]
(ix) (2²⁰ ÷ 2¹⁵) × 2³ = (2^{20 – 15}) × 2³     [∵ a^m ÷ aⁿ = a^{m – n}]
= 2⁵ × 2³ = 2^{5 + 3} [∵ a^m × aⁿ = a^m+n]
(x) 8^t ÷ 8² = 8^{(t – 2)}    [∵ a^m ÷ aⁿ = a^{m – n}]

Question 2.
Simplify and express each of the following in exponential form:

Solution:

(ii) [(5²)³ × 5⁴] ÷ 5⁷
= [5^{2 × 3} × 5⁴] ÷ 5⁷ [∵ (a^m)ⁿ = a^mn]
= [5⁶ × 5⁴] ÷ 5⁷ [∵ a^m × aⁿ = a^{m + n}]
= [5^{6 + 4}] ÷ 5⁷
= 5¹⁰ ÷ 5⁷
= 5^{10 – 7} [∵ a^m ÷ aⁿ = a^{m – n}]
= 5³

(iii) 25⁴ ÷ 5³ = (5 × 5)⁴ ÷ 5³
= (5²)⁴ ÷ 5³ = 5^{2 × 4} ÷ 5³ [∵ (a^m)ⁿ = a^mn]
= 5⁸ ÷ 5³ = 5^{8 – 3} [∵ a^m ÷ aⁿ = a^{m – n}]
= 5⁵

(vi) 2⁰ + 3⁰ + 4⁰ = 1 + 1 + 1 = 3     [∵ a⁰ = 1]
(vii) 2⁰ × 3⁰ × 4⁰ = 1 × 1 × 1 = 1
(viii) (3⁰ + 2⁰) × 5⁰ = (1 + 1) × 1 = 2

Question 3.
Say true or false and justify your answer:
(i) 10 × 10¹¹ = 100¹¹
(ii) 2³ > 5²
(iii) 2³ × 3² = 6⁵
(iv) 3⁰ = (1000)⁰
Solution:
(i) L.H.S = 10 × 10¹¹
= 10^{1 + 11} = 10¹²
R.H.S = 100¹¹ = (10 × 10)¹¹ = (10⁰)¹¹ = 10^{2 × 11} = 10²²
⇒ L.H.S. ≠ R.H.S.
Hence, the given statement is false.

(ii) 2³ > 5²
L.H.S. = 2³ = 2 × 2 × 2 = 8
R.H.S. = 5³ = 5 × 5 = 25
⇒ L.H.S ≠ R.H.S
Hence, the given statement is false.

(iii) 2³ × 3² = 6⁵
L.H.S. = 2³ × 3² = 2 × 2 × 2 × 3 × 3 = 72
R.H.S. = 6⁵ = 6 × 6 × 6 × 6 × 6 = 7776
⇒ L.H.S. ≠ R.H.S.
Hence, the given statement is false.

(iv) 3⁰ = (1000)⁰
L.H.S. = 3⁰ = 1
R.H.S. = (1000)⁰ = 1
⇒ L.H.S. = R.H.S.
Hence, the given statement is true.

Question 4.
Express each of the following as a product of prime factors only in exponential form:
(i) 108 × 192
(ii) 270
(iii) 729 × 64
(iv) 768
Solution:
(i) 108 × 192
= (2 × 2 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2 × 3)
= (2² × 3³) × (2⁶ × 3)
= 2²⁺⁶ × 3³⁺¹ = 2⁸ × 3⁴
(ii) 270 = 2 × 3 × 3 × 3 × 5 = 2 × 3³ × 5
(iii) 729 × 64 = (3 × 3 × 3 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2) = 3⁶ × 2⁶
(iv) 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2⁸ × 3

Question 5.
Simplify:

Solution:

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.3

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


Solution:
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°
⇒ 2x = 180° – 50° = 130°

(v) x + x + x = 180°
⇒ 3x = 180°

(vi) x + 2x + 90° = 180°
⇒ 3x = 180° – 90° = 90° = 90°

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

Solution:
(i) y + 120° = 180° (Linear pair)
⇒ y = 180° – 120° = 60°
Now, x + y + 50° = 180°
(Angle sum property)
⇒ x + 60° + 50° = 180° x = 180° – 60° – 50° = 70°

(ii) y = 80° (Vertically opposite angles)
Now, y + x + 50° = 180°
(Angle sum property)
⇒ 80° + x + 50° = 180°
⇒ x = 180° – 80°- 50° = 50°

(iii) y + 50° + 60° = 180°
(Angle sum property)
⇒ y = 180° – 60° – 50° = 70°
Now, x + y = 180° (Linear pair)
⇒ x = 180° – y = 180° – 70° = 110°

(iv) x = 60° (Vertically opposite angles)
Now, 30° + x + y = 180°
(Angle sum property)
⇒ 30° + 60° + y = 180° y = 180° – 30° – 60° = 90°

(v) y = 90° (Vertically opposite angles)
Now, x + x + y = 180°
(Angle sum property)
⇒ 2x + 90° = 180°
⇒ 2x = 180° – 90° = 90°

(vi) y = x (Vertically opposite angles)
a = x (Vertically opposite angles)
b = x (Vertically opposite angles)
Now, a + b + y = 180°
(Angle sum property)
⇒ x + x + x= 180°
⇒ 3x = 180°


Hence, y = x = 60°

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.2

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


Solution:
(i) x = 50° + 70° (Exterior angle property) = 120°
(ii) x = 65° + 45° (Exterior angle property)
= 110°
(iii) x = 40° + 30° (Exterior angle property)
= 70°
(iv) x = 60° + 60° (Exterior angle property)
= 120°
(v) x = 50° + 50° (Exterior angle property)
= 100°
(vi) x = 30° + 60° (Exterior angle property)
= 90°

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


Solution:
(i) x + 50° = 115°
(Exterior angle property)
⇒ x = 115° – 50° = 65°
(ii) 70° + x = 100° (Exterior angle property) ⇒ x = 100° – 70° = 30°
(iii) x + 90° = 125° (Exterior angle property) ⇒ x = 125° – 90° = 35°
(iv) x + 60° = 120° (Exterior angle property) ⇒ x = 120° – 60° = 60°
(v) x + 30° = 80° (Exterior angle property) ⇒ x = 80° – 30° = 50°
(vi) x + 35° = 75° (Exterior angle property) ⇒ x = 75° – 35° = 40°

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.1

Question 1.
In ∆PQR, D is the mid-point of \(\overline{Q R}\).
\(\overline{P M}\) is ___.
PD is ___.
Is QM = MR?

Solution:
(i) PM is an altitude
(ii) PD is a median
(iii) No, QM ≠ MR.

Question 2.
Draw rough sketches for the following:
(a) In ∆ABC, BE is a median.
(b) In ∆PQR, PQ and PR are altitudes of the triangle.
(c) In ∆XYZ, YL is an altitude in the exterior of the triangle.
Solution:

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

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

MP Board Class 7th Maths Solutions