Question

3. Find the value of a so that PQRS is isosceles trapezium.
(v) 4
Q
P
(a2 + 27)
(2a3 - 54)
S
R
CD is a parallelogram. AO = 15 units: DR
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Answer 1

Answer:

This implies that

x2+2ax=4x−4a−13

or

x2+2ax−4x+4a+13=0

or

x2+(2a−4)x+(4a+13)=0

Since the equation has just one solution instead of the usual two distinct solutions, then the two solutions must be same i.e. discriminant = 0.

Hence we get that

(2a−4)2=4⋅1⋅(4a+13)

or

4a2−16a+16=16a+52

or

4a2−32a−36=0

or

a2−8a−9=0

or

(a−9)(a+1)=0

So the values of a are −1 and 9.

Answer 2

Answer:

Given,

PQRS is an isosceles trapezium

PS = RQ ( Given)

= ∠P+∠S = 180° (sum of adjacent angles are supplementary)

= 2x + 3x = 180°

= 5x = 180°

= x =

∵∠P =∠Q ∵ PS=RQ

= 2x = y

= 2×36 = y

= y = 72°

Sana po makatulong..