Maths 4 Marks Questions

Let two chords AB and CD are intersecting each other at point O.

In $\triangle\text{AOB}$ and $\triangle\text{COD},$

OA = OC (Given)

OB = OD (Given)

$\angle\text{AOB}=\angle\text{COD}$ (Vertically opposite angles)

$\triangle\text{AOB}\cong\triangle\text{COD}$ (SAS congruence rule)

AB = CD (By CPCT)

Similarly, it can be proved that $\triangle\text{AOD}\cong\triangle\text{COB}$

$\therefore\text{AD = CB}$ (By CPCT)

Since in quadrilateral ACBD, opposite sides are equal in length, ACBD is a parallelogram.

We know that opposite angles of a parallelogram are equal.

$\therefore\angle\text{A}=\angle\text{C}$

However, $\angle\text{A}+\angle\text{C}=180^\circ$ (ABCD is a cyclic quadrilateral)

$\Rightarrow\angle\text{A}+\angle\text{A}=180^\circ$

$\Rightarrow2\angle\text{A}=180^\circ$

$\Rightarrow\angle\text{A}=90^\circ$

As ACBD is a parallelogram and one of its interior angles is 90°, therefore, it is a rectangle.

$\angle\text{A}$ is the angle subtended by chord BD. And as $\angle\text{A}=90^\circ$ therefore, BD should be the diameter of the circle. Similarly, AC is the diameter of the circle.

Given,

AB is equal to the radius of the circle.

In $\triangle\text{OAB},$

OA = OB = AB = radius of the circle.

Thus, $\triangle\text{OAB}$ is an equilateral triangle.

$\angle\text{AOC}=60^\circ$

also,

$\angle\text{ACE}=\frac{1}{2}\angle\text{AOB}=\frac{1}{2}\times60^\circ=30^\circ$

ACBD is a cyclic quadrilateral,

$\angle\text{ACB}+\angle\text{ADB}=180^\circ$ (Opposite angles of cyclic quadrilateral)

$\Rightarrow\angle\text{ADB}=180^\circ-30^\circ=150^\circ$

Thus, angle subtend by the chord at a point on the minor arc and also at a point on the major arc are 150° and 30° respectively.

Let two circles having their centers as O and O' intersect each other at point A and B respectively. Let us join OO'.

In $\triangle\text{AO}{\text{O}'}$ and $\triangle\text{BO}{\text{O}'},$

OA = OB (Radius of circle 1)

O'A = O'B (Radius of circle 2)

OO' = OO' (Common)

$\triangle\text{AO}{\text{O}'}\cong\triangle\text{BO}{\text{O}'}$ (By SSS congruence rule)

$\triangle\text{OA}{\text{O}'}=\triangle\text{OB}{\text{O}'}$ (By CPCT)

Therefore, line of centers of two intersecting circles subtends equal angles at the two points of intersection.

Let A, B and C represents the positions of Ankur, Syed and David respectively. All three boys at equal distances thus ABC is an equilateral triangle.
$\text{AD}\perp\text{BC}$ is drawn. Now, AD is median of $\triangle\text{ABC}$ and it passes through the centre O.
Also, O is the centroid of the $\triangle\text{ABC}.$ OA is the radius of the triangle.

$\text{OA}=\frac{2}{3}\text{AD}$

Let the side of a triangle a meters then $\text{BD}=\frac{\text{a}}{2}\text{m}.$
Applying Pythagoras theorem in $\triangle\text{ABD}.$

$\text{AB}^2=\text{BD}^2+\text{AD}^2$

$\Rightarrow\text{AD}^2=\text{AB}^2-\text{BD}^2$

$\Rightarrow\text{AD}^2=\text{a}^2-\Big(\frac{\text{a}}{2}\Big)^2$

$\Rightarrow\text{AD}^2=\frac{3\text{a}^2}{4}$

$\Rightarrow\text{AD}=\sqrt{\frac{3\text{a}}{2}}$

$\text{OA}=\frac{2}{3}\text{AD}\Rightarrow20\text{m}=\frac{2}{3}\times\sqrt{\frac{3\text{a}}{2}}$

$\Rightarrow\text{a}=20\sqrt{3}\text{m}$

Length of the string is $20\sqrt{3}\text{m}.$

Join chords AP and DQ.

For chord AP,

$\angle\text{PBA}=\angle\text{ACP}$ (Angles in the same segment) ...(1)

For chord DQ,

$\angle\text{DBQ}=\angle\text{QCD}$ (Angles in the same segment) ...(2)

ABD and PBQ are line segments intersecting at B.

$\therefore\angle\text{PBA}=\angle\text{DBQ}$ (Vertically opposite angles) ...(3)

From equations (1), (2), and (3), we obtain

$\angle\text{ACP}=\angle\text{QCD}$

1. AD || BC.
2. EB = EC.

1. Since EA = ED

Then, $\angle\text{EAD}=\angle\text{EDA}\dots(1)$ [Oppo. angles to equal sides]

Since, ABCD is a cyclic quadrilateral

Then, $\angle\text{ABC}+\angle\text{ADC}=180^\circ$

But $\angle\text{ABC}+\angle\text{EBC}=180^\circ$ [Linear pair of angles]

Then, $\angle\text{ADC}=\angle\text{EBC}\dots(2)$

Compare equations (1) and (2)

$\angle\text{EAD}=\angle\text{EBC}\dots(3)$

Since, corresponding angles are equal

Then, BC || AD

2. From equation (3)

$\angle\text{EAD}=\angle\text{EBC}\dots(3)$

Similarly $\angle\text{EDA}=\angle\text{ECB}\dots(4)$

Compare equations (1)(3) and (4)

$\angle\text{EBC}=\angle\text{ECB}$

$\Rightarrow \text{EB}=\text{EC}$ [Opposite angles to equal sides]

It is given that BE is the bisector of $\angle\text{B}.$

$\therefore\angle\text{ABE}=\frac{\angle\text{B}}{2}$

However, $\angle\text{ADE}=\angle\text{ABE}$ (Angles in the same segment for chord AE)

$\Rightarrow\angle\text{ADE}=\frac{\angle\text{B}}{2}$

Similarly, $\angle\text{ACF}=\angle\text{ADF}=\frac{\angle\text{C}}{2}$ (Angle in the same segment for chord AF)

$\angle\text{D}=\angle\text{ADE}+\angle\text{ADF}$

$=\frac{\angle\text{B}}{2}+\frac{\angle\text{C}}{2}$

$=\frac{1}{2}(\angle\text{B}+\angle\text{C})$

$=\frac{1}{2}(180^\circ-\angle\text{A})$

$=90^\circ-\frac{1}{2}\angle\text{A}$

Similarly, it can be proved that

$\angle\text{E}=90^\circ-\frac{1}{2}\angle\text{B}$

$\angle\text{F}=90^\circ-\frac{1}{2}\angle\text{C}$

Given: $\angle\text{ACB}$ is an angle in minor segment.

To prove: $\angle\text{ACB}<90^\circ$

Proof: By degree measure theorem

$\angle\text{AOB}=2\angle\text{ACB}$

And $\angle\text{AOB}<180^\circ$

Then, $2\angle\text{ACB}<180^\circ$

$\Rightarrow\angle\text{ACB}<\frac{180^\circ}{2}$

$\Rightarrow\angle\text{ACB}<90^\circ$

We have, $\angle\text{B}=70^\circ$

Since, ABCD is a cyclic quadrilateral

Then, $​​\angle\text{B}+\angle\text{D}=180^\circ$

$\Rightarrow70^\circ+\angle\text{D}=180^\circ$

$\Rightarrow\angle\text{D}=180^\circ-70^\circ=110^\circ$

Since, AB || DC

Then, $\angle\text{B}+\angle\text{C}=180^\circ$ [Co-interior angles]

$\Rightarrow70^\circ+\angle\text{C}=180^\circ$

$\Rightarrow\angle\text{C}=180^\circ-70^\circ=110^\circ$

Now, $\angle\text{A} +\angle\text{C}=180^\circ$ [Opposite angles of cyclic quad.]

$\Rightarrow\angle\text{A}+110^\circ=180^\circ$

$\Rightarrow\angle\text{A}=180^\circ-110^\circ=70^\circ$

By degree measure theorem

$\angle\text{AOB}=2\angle\text{APB}$

$\Rightarrow\angle\text{APB}=2\times50^\circ=100^\circ$ since OA = OB [Radius of circle]

Then $\angle\text{OAB}=\angle\text{OBA}$ [Angles opposite toequalsides]

Let $\angle\text{OAB}=\text{x}$

In $\triangle\text{OAB},$ by angle sum property

$\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^\circ$

$\Rightarrow\text{x}+\text{x}+100^\circ=180^\circ$

$\Rightarrow\text{2x}=180^\circ-100^\circ$

$\Rightarrow\text{2x}=80^\circ$

$\Rightarrow\text{x}=40^\circ$

$\angle\text{OAB}=\angle\text{OBA}=40^\circ$

$\Rightarrow\text{AC}=\sqrt{20}$

$\frac{\text{OA}}{\text{OD}}=\frac{2}{1}$

$\Rightarrow\frac{\text{4OM}}{\text{OD}}=\frac{2}{1}$

$\Rightarrow\text{OD}=20\text{m}$

$\text{AC}^2 = \text{AD}^2 + \text{DC}^2$

$\text{AC}^2=60^2+\big(\text{AC}^2\big)^2$

$\text{AC}^2=3600+\frac{\text{AC}^2}{4}$

$\Rightarrow\frac{3}{4}\text{AC}^2=3600$

$\Rightarrow\text{AC}^2=4800$

$\Rightarrow\text{AC}=40\sqrt{3}\text{m}$