Question

The length of the midline of a trapezoid equals 4 cm and the base angles are 40° and 50°. The length of the bases if the distance of their mid-points equals 1 cm is equal to 

(a) 5 cm, 3 cm 

(b) 4 cm, 3 cm 

(c) 7 cm, 4 cm 

(d) 6 cm, 5 cm

Answer 1

(a) 5 cm, 3 cm

Let ABCD be the given trapezoid, whose base angles ∠A and ∠B equal 40° and 50°. Also mid-line EF = 4 cm.

Extend AD and BC to meet at P. 

Now in ΔAPB, 

∠APB = 180° – (40° + 50°) = 90° 

Also, ΔABP ~ ΔDCP

⇒ \(\frac{AB}{DC}\) = \(\frac{AP}{DP}\) = \(\frac{BP}{CP}\)             ....(i)

Let Q and R be the mid-points of DC and AB respectively.

∴ DC = 2DQ and AB = 2AR 

∴ From (i)

 \(\frac{AB}{DC}\) = \(\frac{AP}{DP}\) = \(\frac{BP}{CP}\) ⇒ \(\frac{2\,AR}{2\,DQ}\) = \(\frac{AP}{DP}\) = \(\frac{BP}{CP}\)

⇒  \(\frac{AR}{DQ}\) = \(\frac{AP}{DP}\) = \(\frac{BP}{CP}\)

⇒ ∆PDQ and ∆PAR are similar ⇒ P, Q, R are collinear. Now, as ∆APB and ∆DPC are right angled ∆s 

AR = RB = PR and DQ = QC = PQ.

[If D is the mid-pt. of hyp. AC, then BD = AD = CD]

∴ \(\frac{AB}{2}\) - \(\frac{DC}{2}\) = AR – DQ = PR – PQ = QR = 1

⇒ AB – CD = 2                           ....(i)

Also, \(\frac{AB+CD}{2}\) = 4  ⇒ AB + CD = 8              ...(ii)

Solving (i) and (ii) simultaneously, we get 

AB = 5 and CD = 3.