 Important for :
1
Two light rods AB = a + b, CD = a – b symmetrically lying on a horizontal AB. There are kept intact by two strings AC and BD. The perpendicular distance between rods in a. The length of AC is given by
» Explain it
D
Since, they are symmetrically on horizontal plane. ∴    AC = BD

∴    AE = BF = x

Now, AB = (a – b) + 2x

i.e. a + b = a – b + 2x ⇒ 2b = 2x   ⇒   b = x

Now in ΔACE,

x2 + a2 = AC2

 AC2 = b2 + a2   ⇒   AC = b2 + a2

Hence, option D is correct.

2
If PQRS be a rectangle such PQ = 3 QR. Then, what is ∠PRS equal to?
» Explain it
C
In rectangle PQRS, PQ || RS

∴   ∠RPQ = ∠ PRS (vertically opposite angles)

...(i)

Now in ΔPQR,

 tan ∠QPR = RQ PQ

⇒  tan ∠QPR =  QR
 3 QR

⇒ ∠QPR = 30°

∴   ∠PRS = 30°

[From the equation (i)]

Hence, option C is correct.

3
In a trapezium, the two non-parallel sides are equal in length, each being of 5 cm. The parallel sides are at a distance of 3 cm apart. If the smaller side of the parallel sides is of length 2 cm, then the sum of the diagonals of the trapezium is
» Explain it
B
In ΔBCF, By the pythagoras theorem,

BF= BC2 – CF2

(BF)2  = (5)2 – (3)2  ⇒  BF = 4 cm

AB = 2 + 4 + 4 = 10 cm

Now, in ΔACF,

AC2 = CF2 + FA2 ⇒ AC2 = 32 + 62

AC = 45 cm

Similarly, BD = 45 cm

∴   Sum of diagonal = 2 × 45 = 2 × 3 5 = 6 5 cm.

Hence, option B is correct.

4
The area of a rectangle lies between 40 cm2 and 45 cm2. If one of the sides is 5 cm, then its diagonal lies between
» Explain it
B
Area of rectangle lies between 40 cm2 and 45 cm2

Now, one side = 5 cm

Since, area can't be less than 40 cm2

 ∴    Other side can't be less than = 40 = 8 cm 5

Since, area can't be greater than 45 cm2.

 ∴    Other side can't be greater than = 45 = 9 cm 5

 ∴    Minimum value of diagonal = 82 + 52 = 89 = 9.43 cm

 ∴    Maximum value of diagonal = 92 + 52 = 106 = 10.3 cm

So, diagonal lies between 9 cm and 11 cm.

Hence, option B is correct.

5
Let ABCD be a parallelogram. Let P, Q, R and S be the mid-points of sides AB, BC, CD and DA respectively. Consider the following statements.

I. Area of triangle APS < Area of triangle DSR, if BD < AC.
II. Area of triangle ABC = 4 (Area of triangle BPQ).

Select the correct answer using the codes given below.
» Explain it
B
Area of ΔAPS = Area of ΔDSR ∵   AS = SD and AP = DR

∴   ar (ΔABC) = 4 ar (ΔBPQ).

Hence, option B is correct.