Solution for (A):

Given AP: 8, 10, 12,… with 60 terms

First term (a) = 8, Common difference (d) = 2

60th term: \(t_{60} = a + 59d = 8 + 59(2) = 126\)

51st term: \(t_{51} = a + 50d = 8 + 50(2) = 108\)

Sum of last 10 terms (51st to 60th):

\(S = \frac{n}{2}[first\ term + last\ term] \) \(= \frac{10}{2}[108 + 126] = 5 × 234 = 1170\)

Answer: 1170

Solution for (B):

Given AP: 6, 13, 20,…, 230

First term (a) = 6, Common difference (d) = 7

Let number of terms = n

Last term: \(a + (n-1)d = 230\)

\(6 + (n-1)7 = 230\)

\((n-1)7 = 224\)

\(n-1 = 32\)

\(n = 33\) (odd number of terms)

Middle term position = \(\frac{n+1}{2} = \frac{34}{2} = 17\)

17th term: \(t_{17} = a + 16d = 6 + 16(7) = 6 + 112 = 118\)

Answer: 118

Step-by-Step Solution:

Given: \(\sin(A + B) = 1\)

We know: \(\sin 90^\circ = 1\)

Therefore: \(A + B = 90^\circ\) …(1)

Also given: \(\cos(A – B) = \frac{\sqrt{3}}{2}\)

We know: \(\cos 30^\circ = \frac{\sqrt{3}}{2}\)

Therefore: \(A – B = 30^\circ\) …(2)

Adding equations (1) and (2):

\((A + B) + (A – B) = 90^\circ + 30^\circ\)

\(2A = 120^\circ\)

\(A = 60^\circ\)

Substituting in equation (1):

\(60^\circ + B = 90^\circ\)

\(B = 30^\circ\)

Answer: A = 60°, B = 30°

Proof:

Step 1: Since \(\triangle ABC \sim \triangle DEF\), we have:

\(\frac{AB}{DE} = \frac{BC}{EF}\) …(1)

Step 2: AP is median ⇒ P is midpoint of BC ⇒ \(BP = \frac{1}{2}BC\)

DQ is median ⇒ Q is midpoint of EF ⇒ \(EQ = \frac{1}{2}EF\)

Step 3: From (1): \(\frac{AB}{DE} = \frac{2BP}{2EQ} = \frac{BP}{EQ}\) …(2)

Step 4: Consider ΔABP and ΔDEQ:

From (2): \(\frac{AB}{DE} = \frac{BP}{EQ}\)

Also, \(\angle B = \angle E\) (corresponding angles of similar triangles)

Step 5: By SAS similarity criterion:

\(\triangle ABP \sim \triangle DEQ\)

Step 6: Corresponding sides of similar triangles are proportional:

\(\frac{AB}{DE} = \frac{AP}{DQ}\) (By CPST – Corresponding Parts of Similar Triangles)

Solution for (A):

Each animal can graze a sector of circle with radius 14m

The three sectors together form a semi-circle because:

Sum of angles at vertices of triangle = 180°

Area grazed = Area of semi-circle with radius 14m

Area = \(\frac{1}{2} \times \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 14^2\)

= \(\frac{1}{2} \times \frac{22}{7} \times 196\)

= \(\frac{1}{2} \times 22 \times 28 = 308 \text{ m}^2\)

Answer: 308 m²

Solution for (B):

Radius (r) = 5 cm

Angle subtended at center = 90°

Step 1: Area of minor segment = Area of sector – Area of triangle

= \(\frac{90^\circ}{360^\circ} \times \pi r^2 – \frac{1}{2} \times r^2\)

= \(\frac{1}{4} \times \pi \times 25 – \frac{1}{2} \times 25\)

= \(\frac{25\pi}{4} – \frac{25}{2} \text{ cm}^2\)

Step 2: Area of major segment = Area of circle – Area of minor segment

= \(\pi r^2 – \left(\frac{25\pi}{4} – \frac{25}{2}\right)\)

= \(25\pi – \frac{25\pi}{4} + \frac{25}{2}\)

= \(\frac{100\pi – 25\pi}{4} + \frac{25}{2}\)

= \(\frac{75\pi}{4} + \frac{25}{2} \text{ cm}^2\)

Answer: \(\left(\frac{75\pi}{4} + \frac{25}{2}\right) \text{ cm}^2\)

Solution:

Given: Circle inscribed in ΔABC, radius (r) = 4 cm

BD = 10 cm, DC = 8 cm

Area of ΔABC = 90 cm²

Step 1: Let tangents from vertices:

BD = BE = 10 cm (tangents from B)

CD = CF = 8 cm (tangents from C)

Let AE = AF = x (tangents from A)

Step 2: Area of ΔABC = Area(ΔAOB) + Area(ΔBOC) + Area(ΔAOC)

= \(\frac{1}{2} \times r \times AB + \frac{1}{2} \times r \times BC + \frac{1}{2} \times r \times AC\)

= \(\frac{r}{2}(AB + BC + AC)\)

Step 3: Sides: AB = x + 10, BC = 10 + 8 = 18, AC = x + 8

Perimeter = (x + 10) + 18 + (x + 8) = 2x + 36

Step 4: Using area formula:

90 = \(\frac{4}{2}(2x + 36) = 2(2x + 36)\)

90 = 4x + 72

4x = 18

x = 4.5 cm

Step 5: AB = x + 10 = 4.5 + 10 = 14.5 cm

AC = x + 8 = 4.5 + 8 = 12.5 cm

Answer: AB = 14.5 cm, AC = 12.5 cm

Solution for Visually Impaired:

Right triangle ABC, right angled at B

AB = 24 cm, BC = 7 cm

Step 1: Find AC using Pythagoras theorem:

\(AC^2 = AB^2 + BC^2\)  \(= 24^2 + 7^2\) \(= 576 + 49 = 625\)

AC = 25 cm

Step 2: Area of ΔABC = \(\frac{1}{2} \times AB \times BC = \frac{1}{2} \times 24 \times 7 = 84 \text{ cm}^2\) …(1)

Step 3: Let radius of inscribed circle = r

Area of ΔABC = Area(ΔAOB) + Area(ΔBOC) + Area(ΔAOC)

= \(\frac{1}{2} \times r \times AB + \frac{1}{2} \times r \times BC + \frac{1}{2} \times r \times AC\)

= \(\frac{r}{2}(AB + BC + AC)\) \(= \frac{r}{2}(24 + 7 + 25)\) \(= \frac{r}{2} \times 56 = 28r\) …(2)

Step 4: Equate (1) and (2):

84 = 28r

r = 3 cm

Answer: Radius = 3 cm