Just tell me how to start the last two question and how to solve qn 18 part (b)
Hello,
For 18.b), I haven't found any quicker way to solve it...
You know the area of triangle ADC.
You know the ratio of the areas of triangles CFD and ADF.
From these two points, you can get the area of CFD.
Then you can easily find the area of triangle DCE.
Thereafter, area of triangle CFE=area of triangle DCE-area of triangle CFD.
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For 19.a)
Note that $\displaystyle \frac{24}{24+12}=\frac{20}{20+10}$, that is $\displaystyle \frac{CR}{CD}=\frac{CQ}{CB}$
So you can use the similarity of triangles CRQ and CBD.
Now you just have to find BC, because with the similarity of the two triangles, the length of RQ will follow.
To find BC, consider the triangles APS and ABD. (look at the ratios of the lengths)
I don't have time to do the third one :/
Now I have some time
When you're given parallel lines and lengths, the first thing that should come in mind is "similar triangles" !
Why ? Because there are equal angles formed by the parallel lines.
a) See what you can say about triangles AED and ABC (comparing the angles / or saying that all sides are parallel to each other)
Then use the properties of similar triangles concerning the lengths of the sides.
Same thing for triangles AEF and ACD.
Tell me what values for x,y,z you find.
b) (iii) is immediate. with the similar triangles AED and ABC, you can get the area of ABC.
Then substract the area of AED to the area of ABC.
(ii) is immediate if you have the area of AEF. Because AEF and ACD are similar, you can have the area of ACD. Then substract the area of AED to the area of ACD to get the area of CDE.
(i) still thinking on how to get this one ><