Quadrilateral practice problems

 

Question 1 Solution:
ACQD is ||gram coz AD || CQ and AD = CQ.
So [DPQ] = [APC] from congruence.
[APC] = [BPC] coz same base(PC) and height from opposite side.
So [DPQ] = [BPC].

Question 2 Solution:
[LZY] = [XYZ] + [TXZ] + [LTX]
[MZYX] = [XYZ] + [TXZ] + [MTZ]

So we need to prove that [LTX] = [MTZ].
LMZX is a trapezium so triangles between parallel sides are similar and other 2 have equal area which can be proven using side ratios. Which gives [LTX] = [MTZ].

I am not sure what was the use of LX = XY = YN here?

Question 3 Solution:
[ADF] = [BDF] = 3 coz same base (DF) and height.
[FCE] = [ADF]  = 3due to AAS congruence. (AD = CE, F is vertical angle, angle DAF = angle CEF).
[FCB] = [FCE] = 3 coz same base(BC = EC) and height (from F to BE).
Diagonal BD splits ||gram ABCD in 2 equal areas, prove by congruence.
So [BDF] + [BFC] = 1/2[ABCD] = 3 + 3
So [ABCD] = 12.

Question 4 Solution :
Can be easily proven that [RDF] = [RAS] = [BSP] = [PFC] = k/8
where k  = [ABCD].
Hint: ASR and ABD are similar triangles with side ratio 1/2.
So [PFRS] = k - k/2 = k/2.

Question 5 Solution link.
First prove that [ABOQ] = [ORSC].
Then [OQB] = [OCR].
Then add [OBR] to both sides to get:
[BRQ] = [BRC].
Now these 2 triangles have same based and same area so they lie between same parallels. Hence proved that QC || BR.

Question 6 Solution:
[EYCB] = [BXFC] since both are ||grams with same base(BC) and between same parallels.
It also follows that BC = XF and BC = EY.
So XF = EY => EX = YF.
So [BEX] = [CYF] since same base and between same parallels.
By the same logic [AEX] = [AYF].
So [AEX] + [BEX] = [AYF] + [CYF]
=> [ABE] = [ACF].

Question 8 Solution:
Let AP = x, PB = 2x.
PD = y, DR = 2y
[PBQR] = PB*PR = 2x*(3y) = 6xy
[ABCD] = base*height = AB*PD = 3x*y = 3xy
So [ABCD]/[PBQR] = 1/2

Question 9: Answer: 35 cm^2.
Apply area by Sine formula(abSinC/2) on all triangles except DEF. You will easily figure it out.

Question 10: [CEF] = 1/2*(1/3*AB)*(height=CB) So answer is 1/6.

Q11: Solution:
By Ceva's theorem 3 cevians are concurrent iff BD/CD * CE/AE * AF/BF = 1 where D,E,F are footprints of cevians on BC,AC,AB.

Let the altitudes be cevians here. It can be shown that ACF ~ ABE by AA similarity. So AC/AB = AF/AE.
Re applying it 2 more times gives us CB/CA = CE/CD, BA/BC = BD/BF.
Multiply LHS/RHS of all 3 equations to get the desired result.

Question 12: Answer: 30.
Solution:
[ABD] = 120, [BDC] = 240 by area/ratio lemma.
[ABE] = [ADE] = 60
[EDC] = [EBC] = 120
BF/CF = [ABE]/[AEC] = 60/(60+120) = 1/3
=> [EFB]/[EFC] = 1/3
But [EBC] = 120 so [EFB] = 120/4 = 30

Q13: Answer: 5/3
Solution: We will apply Menelaus Theorem twice.
To get OD/OE we have to construct a triangle where a transversal intersects DE at O.
Let that triangle be ADE and let P be the intersection point of AD and CF.
Then AP/DP * DO/OE * EC/AC = 1

=> AP/DP * DO/OE = AC/EC = 5/3.
Now we need to figure out AP/DP and we will have our answer.

So let's construct triangle ABD. And again the same transversal CF.
Now, AF/BF * BC/DC * DP/AP = 1
=> 2/5 * 5/2 = AP/DP = 1

So DO/OE = 5/3
H.P.

Q14. Answer: 5
There are multiple ways to solve this.
First approach which I did gave me 2 answers 5, sqrt(33).
Then I tried 2 other approaches which gave me 5. So I was able to confirm that 5 is the right answer.
When it's sqrt(33), it means that angle CBA(CBX) is an obtuse angle which doesn't fit well with rest of the problem.

First approach:
Calculate [ABC] by heron's formula which gives 6.sqrt(6).
Now calculate [BCX] by heron's formula assuming CX = x and make it equal to 2.sqrt(6).
This gives a degree-4 equation in x.
Assume t = x^2 and solve the quadratic. You will get t = 25,33.
Hence x = 5,sqrt(33).

To verify using other approaches, here is the second approach.
By co-ordinate geometry.
A = (0,0). B = (6,0) X = (4,0) C = (x,y)
AC^2 = x^2 + y^2 = 49
BC^2 = (x-6)^2 + y^2 = 25
Here you will get (x,y) = (5,2.sqrt(6)) which will give CX = 5.

Similarly, if you compute Cos(B) in triangle ABC you will get 1/5.
Then you compute it in triangle BCX.
If you put CX = sqrt(33) you will get CosB = -1/5 and if you put CX = 5 you will get CosB = 1/5. So this confirms that CX = 5 is the right answer.