Recent content by rushil_p

What about this: As ΔAPB is congruent with ΔBQC, AP = QB ΔANB is congruent with ΔBMC by AAS rule since ∠ANB = ∠BMC = 90°, ∠MCB = ∠NBA, AB = BC Hence, AN = BM, the third side of the congruent triangles shown. As AN = 6 cm, BM = 6 cm As ΔBMQ and ΔBNA are similar (scale factor 2), BN = BM*2 = 12...

Thank you! I think something like this would suffice: Let AX be the line parallel to MN, such that X lies on the extension of CQ. As we know ∠NMQ = 90°, ∠AXM =90° as they are interior angles between parallel lines. Since ∠MNA and ∠XAN are 90° as well, ANMX is a square, and AN = NM. Therefore MN...

Homework Statement As shown in the diagram below, the shape consists of a square and a circle with centre Q. Given that QM = 3 cm, prove that MN = 6 cm. Known data: -- triangles APB and BQC are congruent -- angle BMC = 90 -- triangles BMQ and BNA are similar and right-angled 2. Homework...