Quiz 6

xy = 4
xy could be 2 *2 or 4 * 1
221
222 These are the possible triangles.
223
441

22x will be a triangle if x is 1, 2 or 3 (trial and error).
44x is a triangle only if x is 1.
1. 221 is acute. 1² + 2² > 2²
2. 222 is equilateral. So acute.
3. 223 is obtuse. 2² + 2² < 3²
4. 144 is acute. 1² + 4² > 4²

The question is \\\" x, y, z are integer that are side of an obtuse-angled triangle. If xy = 4, find z.\\\"
And so, the third side has to be 3

Hence, the answer is 3.

Two possibilities: Two equal sides could add up to 12 or sum of 2 unequal sides = 12.
i.e. Sum of 2 equal sides = 12
Sum of 2 unequal sides = 12

=> If sum of two equal sides were 12, sides of the triangle should be 6, 6, x.
What are the values x can take?
x could range from 1 to 11.
11 integer values exist.

Now 2 unequal sides adding to 12. This could be 1 + 11, 2 + 10, 3 + 9, 4 + 8 or 5 + 7.
How many isosceles triangles are possible with the above combinations?
Isosceles triangles with the above combination:
1, 11, 11    ☑       1, 1, 11     ☒
2, 10, 10    ☑       2, 2, 10     ☒
3, 9, 9        ☑       3, 3, 9        ☒
4, 8, 8        ☑       4, 4, 8        ☒
5, 7, 7        ☑       5, 5, 7        ☑

6 possibilities. Triplets such as (1, 1, 11), (2, 2, 10), etc are eliminated as sum of the two smaller values is less than the largest value. These cannot form a triangle.

The question is \" How many isosceles triangles with integer sides are possible such that sum of two of the side is 12?\"
11 + 6 = 17 possibilities totally.

Hence, the answer is 17.

Side of a triangle are. 6, 10, x.
Area = 1/2 * 6 * 10 * sin∠BAC.
Area is maximum, when ∠BAC = 90◦
x = √(100+36) = √136
There is a more algebraic method using hero’s formula. Try that also.

The question is \"Sides of a triangle are 6, 10 and x for what value of x is the area of the △ the maximum?\"

Hence, the answer is None of these.

Let us say common tangent PQ touches the two circle at R, center of smaller circle is I.
Now, PQ is parallel to BC. AR is perpendicular to PQ. Triangle PQR is also an equilateral triangle and AORID is a straight line. (Try to establish each of these observations. Just to maintain the rigour.)

AD = √32√32a
RD = a√3a√3
AR = √32√32a - a√3a√3 = 3a−2a2√33a−2a2√3 = a2√3a2√3
AR = 1313 AD
Radius of smaller circle = 1313 radius of larger circle
Radius of smaller circle = 1313 * a2√3a2√3 = a6√3a6√3
Area of smaller circle = πr²
π (a6√3a6√3)² = πa2108πa2108
Area of triangle = √34√34a²
Ratio = πa2108πa2108 : √34√34a² = π:27√3

The question is \" What is the ratio of the area of the smaller circle to that of the equilateral triangle?\"

Hence, the answer is π:27√3.

Choice C is the correct answer.

 

Let us assume a ≤ b ≤ c.

a = 1, Possible triangle 1, 7, 7
a = 2, possible triangle 2, 6, 7
a = 3, possible triangles 3, 6, 6 and 3, 5, 7
a = 4, possible triangles 4, 4, 7 and 4, 5, 6
a = 5, possible triangle is 5, 5, 5

There are totally 7 triangles possible.

The question is \"Perimeter of a △ with integer sides is equal to 15. How many such triangles are possible?\"

Hence, the answer is 7.

APQ is an equilateral △. As PQ is parallel to BC.
Let side of the square be ‘a’
AP = a = AQ
△QRC has angles 30 – 60 – 90.

QRQCQRQC = √32√32
QC = QR * 2√32√3
AC = AQ + QC = a + 2a√32a√3
Area of equilateral △ = √34√34 AC²
= √34√34 (a + 2a√32a√3)²
= √34√34 (√3a+2a√3√3a+2a√3)²
= √34√34 * (1313) a²(4 + 3 + 4√3)
= 7+4√3127+4√312 * √3a²
= 12+7√31212+7√312 * a²
Ratio of area of square to that of equilateral △ is = a² : 12+7√31212+7√312 * a²
= 12 : 12 + 7√3

The question is \"What is the ratio of the area of the square to that of the equilateral triangle?\"

Hence, the answer is 12 : 12 + 7√3.

xz = 12
x,z can be 1, 12 or 2, 6 or 3, 4
Possible triangles
1 - 12 - 12
2 - 6 - 5; 2 - 6 - 6; 2 - 6 - 7
3 - 4 - 2; 3 - 4 - 3; 3 - 4 - 4; 3 - 4 - 5; 3 - 4 - 6.

The question is \"△ABC has integer sides x, y, z such that xz = 12. How many such triangles are possible?\"

Hence, the answer is 9 triangles.

Choice C is the correct answer.

Join OE and OD.
Internal angle of regular pentagon = 108^°
∠EAB = ∠EDC = 108^°
∠OAB = 60^°
∠EAO = 48^°
AO = OB = AB as the triangle is equilateral.
AB = AE as this is a regular pentagon.

Triangle AEO is isosceles as AO = EA.
∠AEO = ∠AOE = x (say)

In triangle AEO,
∠OAE + 2x = 180^°
48^° + 2x = 180^°
2x = 132^°, or x = 66^°

The question is \" What is ∠OEA?\"
∠OEA = 66^°

Hence, the answer is 66^°

Assume a ≤ b ≤ c,
we have a² + b² > c².
This implies the triangle with sides a, b, c has to be acute-angled.

The question is \"△ has sides a², b² and c². Then the triangle with sides a, b, c has to be:\"

Hence, the answer is Acute-angled .

r = 2
R = 7 (Half of hypotenuse)
Hypotenuse = 14
r = (a + b - h)/2
2 = (a + b - 14)/2
a + b - 14 = 4
a + b = 18
a² + b² = 14 ²
a² + (18 - a)² = 14 ²
a² + 324 + a² - 36a = 196
2a² - 36a + 128 = 0
a² - 18a + 64 = 0
Now the 2 roots to this equation will effectively be a, 18 – a. Product of the roots = 64.
Area = 1/2 * product of roots (How? Come on, you can figure this out.)
= 32 sq. cms

The question is \"What is the area of the triangle?\"

Hence, the answer is 32 sq. cms

Choice A is the correct answer.