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Maths 0580 IGCSE W04 QP4 Q10 Part D

tanmay_7905

Answer to the question 0580 W04 qp 4 Part D

Reply 1

wasabiandyogurt

Original post by tanmay_7905

Answer to the question 0580 W04 qp 4 Part D

Which variant is it?

Reply 2

tanmay_7905

Original post by wasabiandyogurt

Which variant is it?

In 2004 there were no variants

Reply 3

wasabiandyogurt

Sorry for that, but I think I solved the question.

From the previous part (c) of question,

Area of kite = Area of rhombus = 120 cm^2

p = diagonal; q = diagonal

Area of rhombus => (pq/2) = 120;

pq = 240

p = 240/q ——(1)

From the previous part (b) of question,

For 1 angle of rhombus, 180 - 2x

Therefore,

4(180 - 2x) = 360

180 -2x = 90

90 = 2x

x = 45º

From the Diagram :

WhatsApp Image 2020-12-09 at 12.50.50 PM.jpeg

WhatsApp Image 2020-12-09 at 12.50.50 PM.jpeg

tan 22.5 = (p/2) / (q/2) ——(2)

Substitute (1) into (2):

tan 22.5 = 240 / q^2

q = 24.0

Thus, substitute q into (1);

p = 10

Using Pythagoras theorem:

[ (p/2) ^2 ] + [ (q/2) ^2 ] = [ (side of quadrilateral p) ^ 2 ]

(12 ^2) + (5 ^2) = 144 + 25 = 169 = 13 cm