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Directions: Read the following questions carefully and choose the right answer:
1
In the above figure, if area of triangle ABC is 64 sq. units, then find the area of triangle PQR, where D, E and F are mid points of sides of ΔABC and P, Q and R are midpoints of sides of ΔDEF.

» Explain it
A
Given that,

D, E and F are midpoints of BC, CA and AB and P, Q and R are midpoints of EF, FD and DE

we know that,

Area of ΔABC = 4 ΔDEF

But area of ABC = 64 sq. cm.

4 ΔDEF = 64 ⇒ ΔDEF =  64  = 16 sq. units
4

And area ΔDEF = 4 ΔPQR
⇒ 4 ΔPQR = 16 =  16  = 4 sq. units.
4

Hence, option A is correct.

2
Two medians PS and RT of ΔPQR intersect at G at right angles. If PS = 9 cm and RT = 6 cm, then the length of RS in cm is
» Explain it
C
PS = 9 cm



⇒ GS =  1  × 9 = 3 cm
3

RT = 6 cm
⇒  RG =  2  × 6 = 4 cm
3

∴   RS =  32 + 42  = 9 + 16  = 5 cm

Hence, option C is correct.

3
In the adjoining figure, if BC = a, AC = b, AB = c and ∠CAB = 120° , then the correct relation is :

» Explain it
C
Since ∠A is an obtuse angle in ∆ ABC, so

BC2 = AB2 + AC2 + 2 AB . AD

= AB2+ AC2 + 2 AB 1  AC
2

[∵   AD = AC cos 60° =  1  AC ]
2

= AB2 + AC2 + AB . AC

   a2 = b2 + c2 + bc.

Hence, option C is correct.
 
4
The length of side AB and side BC of a scalene triangle ABC are 12 cm and 8 cm respectively. The value of angle C is 59°. Find the length of side AC.
» Explain it
C


Given, AB = 12 cm, BC = 8 cm,

∠C = 59°

Let ∠A = Θ

∴   ∠B = 180° – (59° + Θ) = 121° – Θ

Now, let us see the choices. If AC = 12 cm, triangle would not be scalene. Hence, option A is ruled out. If AC = 10 cm, AB will become the largest side and ∠C the largest angle. But ∠C = 59°. Hence option B is ruled out. So, AC is either 14 cm or 16 cm. In any case, ∠B will be the largest angle and ∠A (say Θ) the smallest:

Also, ∠B = 180° – (59° + Θ) = 121° – Θ

By sine formula,

8 cm  =  12 cm  =  AC
sin Θ sin 59° sin (121° – Θ)

Thus,  8 cm  ≈  12 cm
sin Θ sin 60°

or, sin Θ ≈  8 cm × sin 60°  =  2  × 
3
 =  1  = 0.577
12 cm 3 2 3

∴   cos Θ = 
1 –  1
3
 = 
2
3

∴  sin (121° – Θ) ≈ sin (120° – Θ) = sin 120° cosΘ – cos120° sin Θ
=  
3
 × 
2
3
 –  –1  ×  1  =  1  +  1  = 0.996
2 2 3 2
2 3

Now,   AC  ≈  8 cm
sin (121° – Θ) sin Θ

or, AC =  8 cm sin(120° – Θ)  =  8 × 0.996  = 13.809 ≈ 14 cm
sin Θ 0.577

Hence, option C is correct.

5
The cordinates of the incentre of the triangle whose sides are 3x – 4y = 0, 5x + 12y = 0 and y – 15 = 0, are
» Explain it
B
We have




3x – 4y ≡ 0     ...(i)

5x + 12y ≡ 0     ....(ii)

y – 15 ≡ 0    ...(iii)

From (i) and (ii), A = (0, 0)

From (i) and (iii), B = (20, 15)

From (ii) and (iii), C = (–36, 15)

BC =   (20 + 36)2 + (15 – 15)2  = 56

AB =  (20)2 + (15)2  = 25

AC =  362 + 152  = 39

Let (α, β) be the incentre co-ordinates of ΔABC
α =  56 × 0 + 39 × 20 + 25(–36)
56 + 39 + 25

=   –900 + 780  =  –120  = –1
120 120

β =   56 × 0 + 39 × 15 + 25(15)  = 8
56 + 39 + 25

∴   (α, β) (–1, 8)

Hence, option B is correct.