Challenging Geometry E Maths Question — St Andrew’s Sec 3 Maths Tuition Question

Question:

ABCD is a rectangle. M and N are points on AB and DC respectively. MC and BN meet at X. M is the midpoint of AB.

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(a) Prove that \Delta CXN and \Delta MXB are similar.

(b) Given that area of \triangle CXN: area of \triangle MXB=9:4, find the ratio of,

(i) DN: NC

(ii) area of rectangle ABCD: area of \triangle XBC. (Challenging)

[Answer Key] (b) (i) 1:3

(ii) 20:3

Suggested Solutions:

(a)
\angle MXB=\angle NXC (vert. opp. angles)

\angle MBX = \angle XNC (alt. angles)

\angle BMX = \angle XCN (alt. angles)

Therefore, \Delta CXN and \Delta MXB are similar (AAA).

(b) (i) \displaystyle\frac{NC}{BM}=\sqrt{\frac{9}{4}}=\frac{3}{2}

Let BM=2u and NC=3u

Then DC=2\times 2u=4u

So DN=4u-3u=u

Thus, DN:NC=1u:3u=1:3

(ii)

We now have a shorter solution, thanks to a visitor to our site! (see comments below)

From part (a), since \Delta CXN and \Delta MXB are similar, we have MX:XC=2:3

This means  that MC:XC=5:3

Thus \triangle MBC:\triangle XBC=5:3 (the two triangles share a common height)

Now, note that \displaystyle\frac{\text{area of }ABCD}{\triangle MBC}=\frac{BC\times AB}{0.5 \times BC \times MB}=\frac{AB}{0.5MB}=\frac{2MB}{0.5MB}=4

Hence area of ABCD=4\times\triangle MBC

We conclude that area of rectangle ABCD: area of \triangle XBC=4(5):3=20:3

Here is a longer solution, for those who are interested:

Let area of \triangle XBC =S

Let area of \triangle MXB=4u

Let area of \triangle CXN=9u

We have \displaystyle\frac{S+9u}{S+4u}=\frac{3}{2} since \triangle NCB and \triangle CMB have the same base BC and their heights have ratio 3:2.

Cross-multiplying, we get 2S+18u=3S+12u

So \boxed{S=6u}

\displaystyle\frac{\triangle BCN}{\triangle BDC}=\frac{3}{4} since \triangle BCN and \triangle BDC have the same base BC and their heights have ratio 3:4.

Hence,

\begin{array}{rcl}    \triangle BDC &=& \frac{4}{3} \triangle BCN\\    &=& \frac{4}{3} (9u+6u)\\    &=& 20u    \end{array}

Thus, area of ABCD=2 \triangle BDC=40u

area of rectangle ABCD: area of \triangle XBC=40:6=20:3

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3 Responses to Challenging Geometry E Maths Question — St Andrew’s Sec 3 Maths Tuition Question

  1. do4fun says:

    alternative methode to b(ii) :

    \triangle CXN and \triangle MXB are similar, so MX / CX = 2 / 3, implies that MC / CX = 5 / 3. Thus \triangle MBC : \triangle XBC = 5 / 3

    since Area of rectangle ABCD = 4 * \triangle MBC (M is midpoint of AB), so
    area of rectangle ABCD: area of \triangle XBC = (4*5) : 3 = 20 : 3

    Like

  2. tomcircle says:

    It is easier to see
    Rectangular ABCD is 4x MBC
    if we draw a dotted line down from midpoint M to divide the rectangula into 2 halves.

    Like

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