HSBC Geometry Questions and Answers Quiz-1

We have given that,

Radius of circle is $3.$

In this question, we have to find the equation of the circle at point $\left(3,0\right)$

.

Then,

The equation of the circle is

${(x-h)^}^2+{(y-k)^}^2={\mathrm{r^}}^2$

Here, we have

$\left(h,k\right)=\left(3,0\right)$

and $r=3$

So, put these value in above equation

$\begin{array}{cc}{(x-3)^}^2+{(y-0)^}^2& ={(3)^}^2\end{array}$

$\begin{array}{cc}{(x-3)^}^2+{\mathrm{y^}}^2& =9\end{array}$

Hence, this is our required equation.

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

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.

11 + 6 = 17 possibilities totally.

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.

Hence, the answer is None of these.

In-radius of equilateral triangle of side a = a2√3
Diameter of larger circle = a2√3

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

a² = π:27√3

Hence, the answer is π:27√3.

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.

QRQC

* a²
= 12 : 12 + 7√3

Hence, the answer is 12 : 12 + 7√3.

Perimeter = 12.
Let the side of the rhombus be “a”, then 4a = 12 => a = 3.
One angle = 120°.
Adjacent angles of a rhombus are supplementary. Therefore, the other angle = 60^°.
Diagonals of a rhombus bisect each other, therefore, ∠DAC = 60^° ∠DCA =60^° ∠BAC =60^°
Therefore, Triangle DAC and BAC are equilateral triangles.
Therefore, Area of Rhombus = 2 * Area of the Equilateral Triangle
= 2 * (√3/4) * a² = (√3/2) * 9 = 9 * (√3/2).

Hence, the answer is 9 * (√3/2).

Given :
Radius of the circle = 25 cm
Center is O
Length of Chord AB = 14 cm
Therefore, △ AOB will be an isosceles triangle with 2 sides OA and OB as 25 cm and side AB as 14 cm.
For an isosceles triangle with the equal sides as 'a' and base as 'b' then

area = b4

* (4(25)² - 14²)^1/2 = 168cm²

Hence, the answer is 168 cm².

When 2 chords AB and CD intersect at P then AP * PB = CP * PD
Hence 4 * 6 = 3 * PD
Thus, PD = 8
Now AB = AP + PB = 10
And CD = CP + PD
Thus, CD = 11
Consider the circle with center O.
Drop a perpendicular from O to chord AB and CD.
This will bisect the chords at X and Y i.e AX=XB and CY = YD.
Here AX = AP + PX i.e 5 = 4 + PX
PX = 1
Similarly, since CD = 11, PY+CP+YD = 11,
=> PY = 11-3-5.5 = 2.5.
PY = OX and PX = OY.
So, PXOY will from a rectangle as seen in the figure.
Now consider the triangle BOX, it is a right triangle where OB is the radius.
XB = 5, OX = 2.5
Then OB = ( OX² + XB² )^1/2 OB = 31.25^1/2 Thus radius = 31.25^1/2 The radius can also be found out using the triangle YOD.

Hence, the answer is 31.25^1/2.

Side of the rhombus 's' = 6 cm
Angle between sides 'a' = exterior angle of an octagon
Therefore a = 45 degrees
Area of rhombus = sin a * s²
Area = (1/√2) * 36
Area = 18√2 cm²

Hence, the answer is 18√2 cm².