## Miscellaneous Exercise 1

**1. Choose the correct alternative answer for each of the following questions.**

**(1) In ▯PQRS, m∠P = m∠R = 108°, m∠Q = m∠S = 72°. State which pair of sides of those given below is parallel.**

**(A) Side PQ and side QR **

**(B) side PQ and side SR**

**(C) side SR and side SP **

**(D) side PS and side PQ**

**Ans:** Option (B) – side PQ and side SR

**Solution:**

In □ PQRS,

m∠P + m∠S

= 108°+ 72

= 180°

Since, interior angles are supplementary

∴ side PQ || side SR

**(2) Read the following statements and choose the correct alternative from those given below them.**

**(i) Diagonals of a rectangle are perpendicular bisectors of each other.**

**(ii) Diagonals of a rhombus are perpendicular bisectors of each other.**

**(iii) Diagonals of a parallelogram are perpendicular bisectors of each other.**

**(iv) Diagonals of a kite bisect each other.**

**(A) Statement (ii) and (iii) are true **

**(B) Only statement (ii) is true**

**(C) Statements (ii) and (iv) are true **

**(D) Statements (i), (iii) and (iv) are true.**

**Ans:** Option (B) – Only statement (b) is true

**(3) If 19³ = 6859, find \(\sqrt[3]{0.006859}\)**

**(A) 1.9 **

**(B) 19 **

**(C) 0.019 **

**(D) 0.19**

**Ans:** Option (D) – 0.19

**Solution:**

\(\sqrt[3]{0.006859}\)

= \(\sqrt[3]{\large \frac {6859}{1000000}}\)

= \(\large \frac {\sqrt[3]{6859}}{\sqrt[3]{1000000}}\)

= \(\large \frac {\sqrt[3]{19³}}{\sqrt[3]{100³}}\)

= \(\large \frac {19}{100}\)

= 0.19

**2. Find the cube roots of the following numbers.**

**(1) 5832 **

**Solution:**

5832

= 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

= (2 × 3 × 3) × (2 × 3 × 3) × (2 × 3 × 3)

= (2 × 3 × 3)³

= (18)³

∴ \(\sqrt[3]{5832}\) = 18

**(2) 4096 **

**Solution:**

4096

= (4 × 4) × (4 × 4) × (4 × 4)

= (4 × 4)

= 16³

∴ \(\sqrt[3]{4096}\) = 16

**3. m α n, n = 15 when m = 25. Hence**

**(1) Find m when n = 87 **

**(2) Find n when m = 155**

**Given:**

m α n

∴ m = kn …(i) *[k is the constant of variation]*

When m = 25, n = 15

∴ Substituting, m = 25 and n = 15 in (i), we get,

m = kn

∴ 25 = k × 15

∴ k = \(\large \frac {25}{215}\)

∴ k = \(\large \frac {5}{3}\)

Substituting k = \(\large \frac {5}{3}\) in (i), we get,

m = kn

∴ m = \(\large \frac {5}{3}\)n …(ii)

**(1) When n = 87, m = ?**

Substituting n = 87 in (ii), we get

m = \(\large \frac {5}{3}\) n

m = \(\large \frac {5}{3}\) × 87

m = 5 × 29

m = 145

**(2) When m = 155, n = ?**

Substituting m = 155 in (ii), we get

m = \(\large \frac {5}{3}\) n

∴ 155 = \(\large \frac {5}{3}\) n

∴ 155 × \(\large \frac {3}{5}\) = n

∴ n = 31 × 3

∴ n = 93

**4. y varies inversely with x. If y = 30 when x = 12, find **

**(1) y when x = 15 **

**(2) x when y = 18**

**Solution:**

y α \(\large \frac {1}{x}\)

∴ y = k × \(\large \frac {1}{x}\) *[where, k is the constant of variation]*

∴ y × x = k …(i)

When x = 12, y = 30

∴ Substituting, x = 12 and y = 30 in (i), we get,

y × x = k

∴ 30 × 12 = k

∴ k = 360

Substituting, k = 360 in (i), we get

y × x = k

∴ y × x = 360 …(ii)

**(1) When x = 15, y = ?**

∴ Substituting x = 15 in (ii), we get

y × x = 360

∴ y × 15 = 360

∴ y = \(\large \frac {360}{15}\)

∴ y = 24

**(2) When y = 18, x = ?**

∴ Substituting y = 18 in (ii), we get

y × x = 360

∴18 × x = 360

∴ x = \(\large \frac {360}{18}\)

∴ x = 20

**5. Draw a line l. Draw a line parallel to line l at a distance of 3.5 cm.**

**Solution:**

**6. Fill in the blanks in the following statement. **

**The number \((256)^{\large \frac {5}{7}}\) is ____ th root of ____ th power of ____.**

**Ans:** The number \((256)^{\large \frac {5}{7}}\) is **7** th root of **5** th power of **256**.

**7. Expand. **

**(1) (5x – 7) (5x – 9) **

**Solution:**

(5x – 7) (5x – 9)

= (5x)² + (– 7 – 9) 5x + (– 7) × (– 9) …*[∵ (x + a) (x + b) = x² + (a + b)x + ab]*

= 25x² + (– 16) × 5x + 63

= 25x² – 80x + 63

**(2) (2x – 3y)³ **

**Solution**:

Here,

a = 2x and b = 3y

∴ (2x – 3y)³

= (2x)³ – 3 (2x)² (3y) + 3 (2x) (3y)² – (3y)³ …*[∵ (a – b)³ = a³ – 3a²b + 3ab² – b³]*

= 8x³ – 3 (4x²) (3y) + 3 (2x) (9y²) – 27y³

= 8x³ – 36x²y + 54xy² – 27p³

**(3) \(\large (\)a + \(\large \frac {1}{2})^{3}\)**

**Solution:**

Here,

A = a

B = \(\large \frac {1}{2}\)

∴ \(\large (\)a + \(\large \frac {1}{2})^{3}\)

= (a)³ + 3(a)²\(\large (\frac {1}{2})\) + 3(a)\(\large (\frac {1}{2})^{2}\) + \(\large (\frac {1}{2})^{3}\) …*[(A + B)³ = A³ + 3A²B + 3AB² + B³]*

= a³ + \(\large (\frac {3a²}{2})\) + 3(a)\(\large (\frac {1}{4})\) + \(\large \frac {1}{8}\)

= a³ + \(\large \frac {3a²}{2}\) + \(\large \frac {3a}{4}\) + \(\large \frac {1}{8}\)

**8. Draw an obtuse angled triangle. Draw all of its medians and show their point of concurrence.**

**Solution:**

P, Q and R are the mid-points of sides TV, SV and ST respectively.

Hence, Seg SP, Seg TQ and VR are the medians. They are concurrent. Point of concurrence is G.

**9. Draw ∆ABC such that l (BC) = 5.5 cm, m∠ABC = 90°, l (AB) = 4 cm. Show the orthocentre of the triangle.**

**Solution:**

Vertex B is the orthocentre of ∆ABC.

**10. Identify the variation and solve. It takes 5 hours to travel from one town to the other if the speed of the bus is 48 km/hr. If the speed of the bus is reduced by 8 km/hr, how much time will it take for the same travel ?**

**Solution:**

Let us assume the speed of the vehicle to be s and time taken to travel be t.

s α \(\large \frac {1}{t}\)

∴ st = k …[where k is the constant of variation]

∴ k = s × t

∴ k = 48 × 5

∴ k = 240

Now ,

The value of t when s = (48 – 8) = 40 km/hr is,

s × t = 240

∴ 40 × t = 240

∴ t = \(\large \frac {240}{6}\)

∴ t = 6

**Ans:** Time taken to travel the same distance at the speed 40 km/hr is 6 hours.

**11. Seg AD and seg BE are medians of ∆ABC and point G is the centroid. If l(AG) = 5 cm, find l(GD). If l(GE) = 2 cm, find l(BE).**

**Solution:**

Centroid G divides the median in the ratio 2 : 1

**(i)** Point G is the centroid and seg AD is the median.

\(\large \frac {AG}{GD}\) = \(\large \frac {2}{1}\)

∴ \(\large \frac {5}{GD}\) = \(\large \frac {2}{1}\)

∴ 5 = 2 × l(GD)

∴ l(GD) = \(\large \frac {5}{2}\)

∴ l(GD) = 2.5 cm

**(ii)** Point G is the centroid and seg BE is the median

\(\large \frac {BG}{GE}\) = \(\large \frac {2}{1}\)

∴ \(\large \frac {BG}{2}\) = \(\large \frac {2}{1}\)

∴ l(BG) = 2 × 2

∴ l(BG) = 4

Now,

l(BE) = l(BG) + l(GE)

∴ l(BE) = 4 + 2

∴ l(BE) = 6 cm

**12. Convert the following rational numbers into decimal form. **

(1) \(\large \frac {8}{13}\)

**Solution:**

∴ \(\large \frac {8}{13}\) = \(0.\overline{615384}\)

(2) \(\large \frac {11}{7}\)

**Solution:**

∴ \(\large \frac {11}{7}\) = \(1.\overline{571428}\)

(3) \(\large \frac {5}{16}\)

**Solution:**

∴ \(\large \frac {5}{16}\) = 0.3125

(4) \(\large \frac {7}{9}\)

**Solution:**

∴ \(\large \frac {7}{9}\) = \(0.\dot{7}\)

**13. Factorise. **

**(1) 2y² – 11y + 5 **

**Solution:**

2y² – 11y + 5

= 2y² – 10y – y + 5

= 2y(y – 5) – 1(y – 5)

= (y – 5)(2y – 1)

**∴ 2y² – 11y + 5 = (y – 5)(2y – 1)**

**(2) x² – 2x – 80 **

**Solution:**

x² – 2x – 80

= x² – 10x + 8x – 80

= x (x – 10) + 8 (x – 10)

= (x – 10)(x + 8)

**∴ x² – 2x – 80 = (x – 10)(x + 8) **

**(3) 3x² – 4x + 1**

**Solution:**

3x² – 4x + 1

= 3x² – 3x – x + 1

= 3x(x – 1) – 1(x – 1)

= (x – 1) (3x – 1)

**∴ 3x² – 4x + 1 = (x – 1) (3x – 1)**

**14. The marked price of a T. V. Set is ₹ 50000. The shopkeeper sold it at 15% discount. Find the price of it for the customer. **

**Solution:**

Marked price = Rs 50,000,

Discount = 15%

Let the discount percent be x

∴ x = 15%

We know that,

Discount = \(\large \frac {Marked\, Price\,×\,x}{100}\)

∴ Discount = \(\large \frac {50000\,×\,15}{100}\)

∴ Discount = 500 × 15

∴ Discount = ₹ 7500

Now,

Selling price = Marked price – Discount

∴ Selling price = 50000 – 7500

∴ Selling price = Rs 42500

**Ans:** The price of the T.V. set for the customer is Rs 42,500.

**15. Rajabhau sold his flat to Vasantrao for ₹ 88,00,000 through an agent. The agent charged 2 % commission for both of them. Find how much commission the agent got. **

**Solution:**

Selling price of the flat = Rs 88,00,000

Rate of commission = 2%

∴ Commission

= 2% of selling price

= \(\large \frac {2}{100}\) × 88,00,000

= 2 × 88,000

= Rs 1,76,000

∴ Total commission

= Commission from Rajabhau + Commission from Vasantrao

= 1,76,000 + 1,76,000

= Rs 3,52,000

**Ans:** The agent got a commission of Rs 3,52,000.

**16. Draw a parallelogram ABCD. such that l(DC) = 5.5 cm, m∠D = 45°, l(AD) = 4 cm.**

**Solution:**

Opposite sides of a parallelogram are congruent.

∴ l(AB) = l(DC) = 5.5 cm and l(AD) = l(BC) = 4 cm

Opposite angles of a parallelogram are congruent.

∴ m∠D = m∠B = 45⁰

Adjacent angles are of a parallelogram supplementary

∴ m∠A + m∠D = 180⁰

∴ m∠A = 180⁰ – m∠D

∴ m∠A = 180 – 45⁰

∴ m∠C = m∠A = 135⁰

**Rough Figure**

**This is the required construction.**

**17. In the figure, line l || line m and line p || line q. Find the measures of ∠a , ∠b, ∠c and ∠d.**

**Solution:**

line l || line m and line p is a transversal.

∴ m∠a = 78° …(i) *[Corresponding angles]*

line p || line q and line m is a transversal.

∴ m∠d = m∠a …*[Corresponding angles]*

∴ m∠d = 78° …(ii) *[From (i)]*

m∠b = m∠d …*[Vertically opposite angles]*

∴ m∠b = 78° …*[From (ii)]*

line l || line m and line q is a transversal.

∴ m∠c + m∠d = 180° …*[Interior angles]*

∴ m∠c + 78° = 180° …*[From (ii)]*

∴ m∠c =180° – 78°

∴ m∠c = 102°

**Ans:** The measure of m∠a is 78°, m∠b is 78°, m∠c is 102° and m∠d is 78°