Question

Plz explain the questions along with the answers.......

Answer 1

Question # 35

Answer:

K = 2

Step-by-step explanation:

Given:-

• ABCD is a parallelogram

• ∠DAB = 60°

• AP and BP bisect ∠DAB and ∠ABC respectively

To find:-

• Value of K when $\frac{CD}{CP} =K$

Solution:-

Since ABCD is a parallelogram, we have,

∠DAB+∠ABC=180°   [Adjacent angles of a parallelogram are supplementary]

→60° + ∠ABC = 180°

→∠ABC = 180° - 60°

→∠ABC = 120°

Now,

AP and BP bisect ∠DAB and ∠ABC respectively

→ ∠PAD = ∠PAB = ∠DAB / 2 and ∠PBC = ∠ABP = ∠ABC / 2

→∠PAD = ∠PAB = 60° / 2 and ∠PBC = ∠ABP = 120° / 2

→∠PAD = ∠PAB = 30° and ∠PBC = ∠ABP = 60°

Since ABCD is a parallelogram, we have

AB ║ CD

→∠APD = ∠PAB = 30°,                                               [Alternate interior angles]

and ∠BPC = ∠ABP = 60°                                          [Alternate interior angles]

But, ∠PAD = ∠PAB = 30° and ∠PBC = ∠ABP = 60°

→∠APD = ∠PAD = 30° and ∠BPC = ∠PBC = 60°

In ΔAPD, we have

∠APD = ∠PAD = 30°

→PD = AD

In ΔBPC, we have

∠BPC = ∠PBC = 60°,

and also

∠BCP = ∠DAB = 60°                                 [Opposite angles of a parallelogram]

→ΔBPC is equilateral

→BC = PC = BP

Now,

PD = AD and BC = PC = BP

but, AD = BC                                                [Opposite sides of a parallelogram]

→PD = AD = BC = PC = BP

→PD = PC

Now,

PD + PC = CD

→2PC = CD

→$\frac{CD}{PC} = 2$

but,  $\frac{CD}{CP} =K$

→ K = 2

Question # 31

Answer:

0

Step-by-step explanation:

Given:-

• $\frac{x}{(b-c)(b + c - 2a)} = \frac{y}{(c-a)(c+a-2b)} =\frac{z}{(a-b)(a+b-2c)}$

To find:-

• Value of x + y + z

Solution:-

$\frac{x}{(b-c)(b + c - 2a)} = \frac{y}{(c-a)(c+a-2b)} =\frac{z}{(a-b)(a+b-2c)}$

→ y = $\frac{x(c-a)(c+a-2b)}{(b-c)(b+c-2a)}$ , and

z = $\frac{x(a-b)(a+b-2c)}{(b-c)(b+c-2a)}$

x + y + z

= x + $\frac{x(c-a)(c+a-2b)}{(b-c)(b+c-2a)}$ + $\frac{x(a-b)(a+b-2c)}{(b-c)(b+c-2a)}$

= $\frac{x(b-c)(b+c-2a)}{(b-c)(b+c-2a)}$  + $\frac{x(c-a)(c+a-2b)}{(b-c)(b+c-2a)}$ + $\frac{x(a-b)(a+b-2c)}{(b-c)(b+c-2a)}$

= $\frac{x(b-c)(b+c-2a)+x(c-a)(c+a-2b) + x(a-b)(a+b-2c)}{(b-c)(b+c-2a)}$

=$\frac{x[(b-c)(b+c-2a)+(c-a)(c+a-2b) + (a-b)(a+b-2c)]}{(b-c)(b+c-2a)}$

= x[b² + bc - 2ab - bc - c² + 2ac + c² + ac  - 2bc - ac - a² + 2ab + a² + ab - 2ac - ab - b² + 2bc] / (b - c)(b + c - 2a)

= x[0] / (b - c)(b + c - 2a)

= 0

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