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MATHMAA

19)

Given : $$\Delta ABC$$  ~ $$\Delta DEF$$
ar($$\Delta ABC$$) = ar($$\Delta DEF$$)

RTP   : $$\Delta ABC$$ $$\cong$$ $$\Delta DEF$$

Proof :

$$\large \frac{ar(\Delta ABC}{ar \Delta DEF} = \frac{AB²}{DE²}$$

1    = $$\large \frac{AB²}{DE²}$$

AB²= DE²

AB = DE

Similarly, we can prove BC = EF, AC = DF.

In $$\Delta ABC$$  and  $$\Delta DEF$$ are congruent by SSS.

AB      =      DE
BC      =      EF
CA      =      FD

Hence proved.

(OR)

Given : AB $$\perp$$BC, DE$$\perp$$AC, GF$$\perp$$BC.

RTP   : $$\triangle AED$$~$$\triangle GCf$$

Proof : In $$\triangle ABC$$  and $$\triangle AED$$
$$\angle ABC$$   =   $$\angle DEA$$ = 90
$$\angle BAC$$   =   $$\angle EAD$$

Therefore, $$\triangle ABC$$  ~ $$\triangle AED$$ by AA similarity
criterion.

Similarly we can prove that $$\triangle ABC$$ ~ $$\triangle GFC$$ by
AA similarity criterion.

$$\triangle AED$$ ~ $$\triangle ABC$$ ~ $$\triangle GFC$$

So, $$\triangle AED$$ ~ $$\triangle GFC$$

Hence proved.

20)

Given  : ABCD is a rhombus, AC and BD are diagonals and E is the meeting
point of these diagonals.

RTP    : AC² + BD² = 4AB²

Proof  : In triangle AEB,
AB² = AE² + BE²

AB² = $$\large \frac{AC²}{4} + \frac{BD²}{4}$$

AB² = $$\large \frac{AC² + BD²}{4}$$

4AB² = AC² + BD²
Hence Proved.

21)

LHS

= $$\large \frac{cos(90-A) sin(90-A)}{tan(90-A)}$$

= $$\large \frac{(sinA cosA)}{\frac{cosA}{sinA}}$$

= $$\large \frac{(sinA cosA)sinA}{cosA}$$

= sin²A

= RHS

Hence Proved.

22)

cosΘ = $$\large \frac{adj}{hyp}$$ = $$\large \frac{3}{5} adj = 3 , hyp = 5 opp = √(hyp² - adj²) = √(5² - 3²) = √(25 - 9) = √16 = 4. cosecΘ = \(\large \frac{hyp}{opp}$$

= $$\large \frac{5}{4} cosΘ + cosecΘ = \(\large \frac{3}{5} + \frac{5}{4}$$

= $$\large \frac{12+25}{20}$$

= $$\large \frac{35}{20}$$

= $$\large \frac{7}{4}$$

(OR)

= cos²20° + cos²70° + sin48° sec42° + cos40° cosec50°

= cos²20° + cos²(90-20)° + sin(90-42)° sec42° + cos(90-50)° cosec50°

= 1 - sin²20° + sin²20° + cos42° sec42° + sin50° cosec50°

= 1+1+1

= 3

23)

mode = l + $$\large \left [ \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right]$$ h

Here, l = 20, $$f_1 = 22, f_0 = 12, f_1 = 8$$, h = 10

mode = 20 + $$\large \left [ \frac{22 - 12}{2(22) - 12 - 8} \right]$$ 10

= 20 + $$\large \frac{100}{22 }$$

= $$\large \frac{220 + 50}{11}$$

= 24.5.

Therefore mode is 24.5.