Xavier Aptitude Test 2015 Paper

Question : 68 of 113

Marks: +1, -0

The parallel sides of a trapezoid ABCD are in the ratio of 4:5.ABCD is divided into an isosceles triangle ABP and aparallelogram PBCD (as shown below). ABCD has a perimeterequal to 1120 meters and PBCD has a perimeter equal to1000 meters. Find Sin ABC,given 2 ∠DAB = ∠BCD.

4/5
16/25
5/6
24/25
A single solution not possible.

Validate

Solution:

(a) Produce CB to E such that AE A CE.Draw BFAAD.

Given BC : AD = 4 : 5
If BC = 4x, then AD = 5x
Hence PD = 4x and AP = x
Now,∠ BCD = ∠BPD = ∠BAP + ∠ABP = 2∠DAB
Therefore ∠BAP = ∠ABP ⇒ AP = BP = CD.
Let AB = y
Now, perimeter of trapezoid ABCD
= y + 4x + x + 5x = 1120
⇒ 10x + y = 1120
Perimeter of parallelogram = x + 4x + x + 4x = 1000
⇒ x = 100 m therefore y = 120 m
Now, area of 'ABP
=

√{160× 40× 60× 60}

= 80 × 60 = 4800m

^{2}

Also, area of ΔABP
=

1/2

× AP × BF =

1/2

× 100 × BF = 50 × BF
∴ 50 × BF = 4800 ⇒ BF = 96m = AE
sin∠ ABC = sin (180° – ∠ABE = sin ∠ABE
[ sin (180° – θ) = sinθ]
=

{AE}/{AB}

{96}/{120}

{4}/{5}

Go to Question:

Prev Question

Next Question