Geometry practice problems

Solutions:

Q1: sqrt(56)/4

Q2: 3

Q3: 25/2

Q4: sqrt(3) - 1

Q5: 2

Q6: 14

Q7: 208

A1B1C1D1 is a parallelogram by applying midpoint theorem multiple times. 

A1B1 || AC and B1C1 || BD. Also A1B1  = AC/2 and B1C1 = BD/2.

Again by MP theorem diagonals of A1B1C1D1 are 8 and 12 and also perpendicular to each other.

If diagonals of a parallelogram are perpendicular to each other, it's a rhombus.

So A1B1C1D1 is a rhombus. With all sides equal.

A1B1 = B1C1 so AC = BD so it's enough to find one diagonal.

In the rhombus A1B1C1D1, taken one quarter formed by its diagonals and find one side.

A1B1 = sqrt(16 + 36) = sqrt(52)

AC = 2sqrt(52) = sqrt(208)

AC.BD = 208

Q8: 36

[BXCY] = 13k

[ABC] = 18k = 6AC => AC = 3k

[AXY] = 18k - 13k = 5k

AXY ~= BXY by SAS

So [BXY] = 5k
[BCY] = 18k - 5k - 5k = 8k = 6.CY => CY = 4k/3

AY = AC - CY = 5k/3 = BY by congruence earlier.

In triangle BCY, BY^2 = 12^2 + CY^2= 144+(4k/3)^2 = (5k/3)^2

=> 144 = k^2 => k = 12 => AC = 12.3 = 36

Q9: 49

Sides are 8,24,17

Q10: 45

Apply Area = 1/2(bcSinA) on both triangles.

Q11: 26

Prove that 5(XY^2) = 4(XN^2 + YM^2)

Q12: 25

Pick the smaller side from both and put them together. You can't get lesser than that.

Q13: 84

c^2 - b^2 = 144 = (c+b)(c-b)

Now do various factors of 144 and equate them to c+b and c-b.
Biggest perimeter happens with 37,35,12.

 

Q14: 16sqrt(3)
Note that AFB is right angle triangle and FE = AE = EB.

AMB is isosceles and AM = MB. So opposite angles are equal. So angle MAB = angle MBA = x.

In triangle BNE, angle NEB = 90 - x, so angle NEA = 90 + x.

In triangle AEF, AE = EF so x = 90 - 2x, hence x = 30 deg.

AC = BC/sin(30) = 16sqrt(3) = DB.

Solution link

Q15: 0

Make 3 inequalities for triangle sides. You will get (a-1)^2 < b^2 < (a+1)^2.
But a,b are positive so a-1 < b < a+1.
So b = a, since both are integers. And don't forget to check the third inequality.

Q16: Answer: 73

Total possible triples 9C3 = 84. Remove invalid from this(11).

Q17: Answer: 48

Find area of equilateral triangle and equate to sum of parts.

Q18: Answer: 30
Solution link

Q19: Answer: 13
Solution link

Hint: 2 opposite triangles are similar. Other 2 have equal area. Use sin(x) formula for area. After doing multiplications you will get 3 possible products. Handle each one by one.

Q20: Answer 24

Solution link

Q21: [ABC] = 9 [G1G2G3]

Assume ABC is equilateral. If the problem is valid for any acute angle triangle it will be valid for an equilateral triangle also.

Let H be the orthocenter. In an equilateral triangle, H,G,I,C all are same so this is centroid as well.
Let AD be the median on BC.

Let 'a' be a side of ABC.

So AH = 2/3(a.sqrt(3)/2) = a/sqrt(3) and HD = a/(2.sqrt(3))

HG1 = HD.2/3 = a/(3.sqrt(3)) = HG2 = HG3
By symmetry, G1G2 = G1G3 = G2G3 hence G1G2G3 is equilateral as well.

In the triangle G1G2G3, H is again ortho/circum/incenter/centroid.

Let G1G2 = x. 

HG1 = x/(sqrt(3)) = a/(3.sqrt(3)) => x = a/3
=> [ABC]/[G1G2G3] = 9