If there is one prayer that you should

- Samuel Dominic Chukwuemeka
**pray/sing** every day and every hour, it is the
LORD's prayer (Our FATHER in Heaven prayer)

It is the **most powerful prayer**.
A **pure heart**, a **clean mind**, and a **clear conscience** is necessary for it.

For in GOD we live, and move, and have our being.

- Acts 17:28

The Joy of a Teacher is the Success of his Students.

- Samuel Chukwuemeka

I greet you this day,

__First:__ read the notes.

__Second:__ view the videos.

__Third:__ solve the questions/solved examples.

__Fourth:__ check your solutions with my **thoroughly-explained solutions** on:
Right Triangles and
Triangles

__Fifth:__ check your answers with the
Calculators as applicable.

Comments, ideas, areas of improvement, questions, and constructive criticisms are welcome.

You may contact me.

If you are my student, please do not contact me here. Contact me via the school's system.

Thank you for visiting.

**Samuel Dominic Chukwuemeka** (SamDom For Peace) B.Eng., A.A.T, M.Ed., M.S

Students will:

(1.) Discuss triangles.

(2.) Classify triangles.

(3.) Discuss the theorems on triangles.

(4.) Discuss the laws on triangles.

(5.) Determine the mensuration of triangles.

(6.) Discuss congruent triangles.

(7.) Discuss similar triangles.

(8.) Determine the trigonometric ratios of the angles of a right triangle.

(9.) Solve applied problems on triangles.

A **Triangle** is a polygon with three sides and three angles.

**Triangles can be classified by:**

(I.) **Sides** (Side Classification)

(II).) **Angles** (Angle Classification)

(1.) **Scalene Triangle** is a triangle with **three unequal sides.**

This also implies that it has **three unequal angles.**

(2.)

This also implies that it has

In an Isosceles Triangle:

The

The

The

The

$ \underline{\triangle ABC} \\[3ex] \angle ABC = \angle ACB = 55^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle ABC \\[3ex] \angle ABC + \angle ACB + \angle BAC = 180^\circ...sum\;\;of\;\;\angle s\;\;of\;\;\triangle ABC \\[3ex] \implies \\[3ex] 55 + 55 + a = 180 \\[3ex] 110 + a = 180 \\[3ex] a = 180 - 110 \\[3ex] a = 70^\circ $

(3.) **Equilateral Triangle** is a triangle with **three equal sides.**

This also implies that it has **three equal angles (equiangular)**

**Equi** means **equal**

**lateral** means **sides**

**angular** means **angles**

So, we have: **Equilateral (equal sides)** and **Equiangular (equal angles)**

*
Teacher: What is the angle in an equilateral triangle?
*

Let the angle in an equilateral triangle = $x$

$
x + x + x = 180 ... sum\:\: of\:\: \angle s\:\: in\:\:a\:\: \triangle \\[3ex]
3x = 180 \\[3ex]
x = \dfrac{180}{3} \\[5ex]
x = 60^\circ \\[3ex]
$
Each of the angles in an equilateral triangle is $60^\circ$

(1.) **Acute Triangle** is a triangle with **three acute angles.**

*
Is the equilateral triangle an acute angle? What are your reason(s)?
*

The Equilateral Triangle is also an Acute Triangle because each of the three angles, $60^\circ$ in
an equilateral triangle is an acute angle.

(2.) **Right Triangle** is a triangle with **one right angle (an angle of $90^\circ$) and two acute angles.**

*
What may we call a $\boldsymbol{45^\circ - 45^\circ - 90^\circ}$?
*

A Right Isosceles Triangle

Right Triangle - because it has a right angle

Isosceles Triangle because it has two equal angles, each of which is $45^\circ$

(3.) **Obtuse Triangle** is a triangle with **one obtuse angle and two acute angles.**

This means that one of the angles in an Obtuse Triangle is greater than $90^\circ$ but less than $180^\circ$.

(4.) **Oblique Triangle** is a triangle that **does not have a right angle.**

This means that ** Acute Triangles** and

Based on the

(a.) Three acute angles OR

(b.) Two acute angles and an obtuse angle

*
Demonstrate each theorem. Explain it well.
Use several examples to enhance understanding.
*

As applicable:

Let:

$ hyp = hypotenuse \\[3ex] angles = \alpha,\;\; \beta, \;\;and\;\; 90 \\[3ex] sides = short,\;\; middle,\;\;and\;\; hyp \\[3ex] OR \\[3ex] sides = leg,\;\;leg,\;\;and\;\;hyp \\[3ex] hyp = long \\[3ex] sides = short,\;\; middle,\;\; long \\[3ex] $ The

It is the

Let:

$ angles = \alpha,\;\; \beta, \;\;and\;\; \theta\;\; OR \\[3ex] angles = A,\;\; B, \;\;and\;\; C \\[3ex] sides = a,\;\; b, \;\;and\;\; c \\[3ex] 2\;\; equal\;\; sides = side1,\;\; side2 = side \\[3ex] 3\;\; equal\;\; sides = side1,\;\; side2,\;\; side3 = side \\[3ex] unequal\;\; sides = short,\;\; middle,\;\; long \\[3ex] $

(1.)

The sum of the angles of a triangle is $180^\circ$

(2.) $\boldsymbol{30^\circ - 60^\circ - 90^\circ}$

In a $\boldsymbol{30^\circ - 60^\circ - 90^\circ}$

$hyp = 2 * short$

(3.) $\boldsymbol{30^\circ - 60^\circ - 90^\circ}$

In a $\boldsymbol{30^\circ - 60^\circ - 90^\circ}$

$middle = \sqrt{3} * short$

(4.) $\boldsymbol{45^\circ - 45^\circ - 90^\circ}$

In a $\boldsymbol{45^\circ - 45^\circ - 90^\circ}$

A $\boldsymbol{45^\circ - 45^\circ - 90^\circ}$ is a right isosceles triangle.

An isosceles triangle has two equal sides.

Therefore, a $\boldsymbol{45^\circ - 45^\circ - 90^\circ}$ has two equal sides.

$ hyp = \sqrt{2} * side1 \\[3ex] hyp = \sqrt{2} * side 2 \\[5ex] $

Applies only to right triangles.

Classroom Activity

Take that chair and mark as your starting point.

Walk

Then, turn around at an angle of 90°

In other words,

Walk

Then, turn towards me to face me.

If I asked you to come to me, how many stpes would you walk to reach me?

Ask your students and note their responses.

Ask them to give reasons for their answers.

Well, it is 5 steps.

That is correct.

But, how did you get 5 steps?

I'll just walk towards you 😊

Welcome to the

It states that:

In a

$hyp^2 = short^2 + middle^2$

OR

In a

$hyp^2 = leg^2 + leg^2$

$ A.\;\; \sqrt{22} \\[3ex] B.\;\; \sqrt{77} \\[3ex] C.\;\; \sqrt{85} \\[3ex] D.\;\; 5.5 \\[3ex] E.\;\; 11 \\[3ex] $

$ hyp^2 = leg^2 + leg^2 ...Pythagorean\:\: Theorem \\[3ex] leg = 9 \\[3ex] leg = 2 \\[3ex] hyp^2 = 9^2 + 2^2 \\[3ex] hyp^2 = 81 + 4 = 85 \\[3ex] hyp = \sqrt{85} \\[3ex] $

*
Check for prior knowledege: Ask students to discuss any buidling/shape/object/item/scenario/concept where they have
seen a right triangle.
Recall that the Pythagorean Theorem is only applicable to right triangles.
*

(1.)

Relate to the construction of staircases, flower beds, and roofs among others.

Here is an example: Question (24.) on Right Triangles

(2.)

Relate to bearings and distances.

Here is an example: Question (2.) on Bearings and Distances

(3.)

(

Examples of Pythagorean Triples are:

$ \underline{Example\;\;1} \\[3ex] 3,\;\;4,\;\;5 \\[3ex] Because: \\[3ex] 3^2 + 4^2 = 5^2 \\[3ex] 9 + 16 = 25 \\[5ex] \underline{Example\;\;2} \\[3ex] 6,\;\;8,\;\;10 \\[3ex] Because: \\[3ex] 6^2 + 8^2 = 10^2 \\[3ex] 36 + 64 = 100 \\[5ex] $ Constructing a triangle whose sides form a Pythagorean Triple is an accurate way of creating a right triangle.

This is beneficial in marking out accurate rectangles of land boundaries, in the sense that two right triangles make a rectangle where the diagonal of the rectangle is the hypotenuse.

(1.) Ask students to explain the relationship between the first two examples.

In other words, given Example 1; how did we derive Example 2?

(2.) Based on only the two examples, ask students to derive/list more Pythagorean Triples.

(3.) Extra Time/Activity: Teach students to construct right traingles based on Pythagorean Triples.

(6.) **Converse of the Pythagorean Theorem (Right Triangle Theorem):**

If the square of the long side(hypotenuse) is the
sum of the squares of the other two sides, then the triangle is a right triangle.

$long^2 = short^2 + middle^2$

(7.) **Acute Triangle Theorem:**

If the square of the long side is less than the sum of the squares of the other two sides,
then the triangle is an acute triangle.

(8.)

If the square of the long side is greater than the sum of the squares of the other two sides, then the triangle is an obtuse triangle.

(Note: An obtuse triangle has $1$ angle whose measure is greater than $90^\circ$ and less than $180^\circ$.)

$ F.\:\: \{4, 4, 5\} \\[3ex] G.\:\: \{5, 12, 13\} \\[3ex] H.\:\: \{6, 8, 10\} \\[3ex] J.\:\: \{7, 10, 12\} \\[3ex] K.\:\: \{8, 11, 16\} \\[3ex] $

Let us find the squares of the sides of the triangle.

$ \underline{Test} \\[3ex] F.\:\: \{4, 4, 5\} \\[3ex] 4^2 = 16 \\[3ex] 4^2 = 16 \\[3ex] 5^2 = 25 \\[3ex] 25 \lt (16 + 16) \\[3ex] 25 \lt 32 ...Acute\:\: Triangle ...NO \\[3ex] G.\:\: \{5, 12, 13\} \\[3ex] 5^2 = 25 \\[3ex] 12^2 = 144 \\[3ex] 13^2 = 169 \\[3ex] 169 = 25 + 144 \\[3ex] 169 = 169 ...Right\:\: Triangle ...NO \\[3ex] H.\:\: \{6, 8, 10\} \\[3ex] 6^2 = 36 \\[3ex] 8^2 = 64 \\[3ex] 10^2 = 100 \\[3ex] 100 = 36 + 64 \\[3ex] 100 = 100 ...Right\:\: Triangle ...NO \\[3ex] J.\:\: \{7, 10, 12\} \\[3ex] 7^2 = 49 \\[3ex] 10^2 = 100 \\[3ex] 12^2 = 144 \\[3ex] 144 \lt (49 + 100) \\[3ex] 144 \lt 149 ...Acute\:\: Triangle ...NO \\[3ex] K.\:\: \{8, 11, 16\} \\[3ex] 8^2 = 64 \\[3ex] 11^2 = 121 \\[3ex] 16^2 = 256 \\[3ex] 256 \gt (64 + 121) \\[3ex] 256 \gt 185 ...Obtuse\:\: Triangle ...YES $

(9.)

This is the theorem that determines if you can form a triangle using any three lengths.

The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

$ hyp + short \gt middle \\[3ex] hyp + middle \gt short \\[3ex] short + middle \gt hyp \\[3ex] $

$ long + short \gt middle \\[3ex] long + middle \gt short \\[3ex] short + middle \gt long \\[3ex] side1 + side2 \gt short \\[3ex] side1 + side2 \gt long \\[3ex] side1 + side2 \gt side3 \\[5ex] $

Which of the following CANNOT be the length of the third side, in feet?

$ F.\;\; 1 \\[3ex] G.\;\; 2 \\[3ex] H.\;\; 3 \\[3ex] J.\;\; 4 \\[3ex] K.\;\; 5 \\[3ex] $

$ \underline{Triangle\;\;Inequality\;\;Theorem} \\[3ex] Option\;\:F \\[3ex] 1 + 2.5 \lt 4 \\[3ex] 3.5 \lt 4 ...No,\;\;it\;\;cannot...Correct\;\;Option \\[3ex] Option\;\; G \\[3ex] 2 + 2.5 \gt 4 \\[3ex] 4.5 \gt 4...Yes,\;\;it\;\;can \\[3ex] Option\;\; H \\[3ex] 3 + 2.5 \gt 4 \\[3ex] 5.5 \gt 4...Yes,\;\;it\;\;can \\[3ex] Option\;\; J \\[3ex] 4 + 2.5 \gt 4 \\[3ex] 6.5 \gt 4...Yes,\;\;it\;\;can \\[3ex] Option\;\; H \\[3ex] 5 + 2.5 \gt 4 \\[3ex] 7.5 \gt 4...Yes,\;\;it\;\;can $

(10.) If any two side lengths of a triangle are unequal; the angles of the triangle are also unequal,
and the measure of an angle is opposite the length of the side facing that angle.

The measure of the smallest angle is opposite the shortest side length.

The measure of the greatest angle is opposite the longest side length.

The measure of the middle angle is opposite the middle side length.

In other words, the measure of an angle is proportional to the length of the side; side length is proportional to angle measure.

Small side faces small angle, middle side faces middle angle, big side faces big angle

**Right Triangle**

$
Let\:\: \alpha \lt \beta \lt 90 \\[3ex]
\alpha \:\:is\:\: the\:\: angle\:\: opposite\:\: short \\[3ex]
\beta \:\:is\:\: the\:\: angle\:\: opposite\:\: middle \\[3ex]
90 \:\:is\:\: the\:\: angle\:\: opposite\:\: hyp \\[3ex]
$
**Other Triangles**

$
Let\:\: \alpha \lt \beta \lt \theta \\[3ex]
\alpha \:\:is\:\: the\:\: angle\:\: opposite\:\: short \\[3ex]
\beta \:\:is\:\: the\:\: angle\:\: opposite\:\: middle \\[3ex]
\theta \:\:is\:\: the\:\: angle\:\: opposite\:\: long \\[5ex]
$
(11.) **Exterior Angle of a Triangle Theorem:**

The exterior angle of a triangle is the sum of the two interior opposite angles.

(12.) **Isosceles Base Angles Theorem:**

The base angles of an isosceles triangle are equal.

(13.) **Converse of the Isosceles Base Angles Theorem:**

If the base angles of a triangle are equal, then the triangle is an isosceles triangle.

(14.) **Perpendicular Bisector of the Base of an Isosceles Triangle Theorem:**

It states that if a line bisects the vertex angle of an isosceles triangle, then the line is also the perpendicular
bisector of the base (the line also bisects the base of that isosceles triangle at right angles).

It states that the perpendicular height drawn from the apex of an isosceles triangle to the base: (a.) bisects the base

(b.) bisects the apex angle.

What is the value of tan P?

$ \angle ABC = \angle ACB ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \underline{height\;\;drawn\;\;from\;\;the\;\;apex\;\;of\;\;an\;\;isosceles\;\;\triangle\;\;bisects\;\;the\;\;base} \\[3ex] \angle BAD = \angle CAD \\[3ex] BD = DC = 10\;cm \\[3ex] \tan P = \dfrac{30}{10} ...SOHCAHTOA \\[5ex] \tan P = 3 $

(16.) **Sine Law:**

Applies to all triangles.

It states that the ratio of the side length of any triangle to the sine of the angle measure opposite that side is the
same for the three sides of the triangle.

$
\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} \\[5ex]
$
OR

The ratio of the angle measure of any triangle to the side length opposite that angle is the same for the
three angles of the triangle.

$
\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c} \\[5ex]
$
__Sine Law is used for:__

(a.) $ASA$ when given two angles and a "common" side ("common" side is the side length between the two angles) OR

(b.) $SAA$ when given a side and two angles (side is not common) OR

(c.) $SSA$ when given two sides and a "non-included" angle. (A "non-included" angle is the angle that
is not in-between the two sides).

This is the Ambiguous Case.

*
Student: What do you mean by the "Ambiguous Case"?
Teacher: The $SSA$ is an Ambiguous Case because this case may result in no triangle, one triangle, or
one right triangle, or two triangles. The number of solutions depends on those two side lengths and the "non-included" angle.
Show students the various triangles that could be formed based on the two side lengths and the "non-included" angle.
*

Given: two side lengths: $a$ and $b$ and a "non-included" angle, $A$

If:

$a \lt b\sin A$; no triangle is formed

$a = b\sin A$; one right triangle is formed

$a \gt b\sin A$ and $a \gt b$; one triangle (acute triangle or right triangle or obtuse triangle) is formed

$a \gt b\sin A$ and $a \lt b$; two oblique triangles (one acute triangle and one obtuse triangle) are formed

Two side lengths: $a$ and $b$ and a "non-included" angle, $B$; compare $b$ and $a\sin B$

Two side lengths: $a$ and $c$ and a "non-included" angle, $A$; compare $a$ and $c\sin A$

Two side lengths: $a$ and $c$ and a "non-included" angle, $C$; compare $c$ and $a\sin C$

Two side lengths: $b$ and $c$ and a "non-included" angle, $B$; compare $b$ and $c\sin B$

Two side lengths: $b$ and $c$ and a "non-included" angle, $C$; compare $c$ and $b\sin C$

(17.) **Cosine Law:**

Applies to all triangles.

It states that the square of a side of a triangle is the difference between the sum of the squares of the other two sides
and twice the product of the two sides and the included angle.

$
a^2 = b^2 + c^2 - 2bc \cos A \\[3ex]
\cos A = \dfrac{b^2 + c^2 - a^2}{2bc} \\[5ex]
\rightarrow A = \cos^{-1} \left(\dfrac{b^2 + c^2 - a^2}{2bc}\right) \\[5ex]
b^2 = a^2 + c^2 - 2ac \cos B \\[3ex]
\cos B = \dfrac{a^2 + c^2 - b^2}{2ac} \\[5ex]
\rightarrow B = \cos^{-1} \left(\dfrac{a^2 + c^2 - b^2}{2ac}\right) \\[5ex]
c^2 = a^2 + b^2 - 2ab \cos C \\[3ex]
\cos C = \dfrac{a^2 + b^2 - c^2}{2ab} \\[5ex]
\rightarrow C = \cos^{-1} \left(\dfrac{a^2 + b^2 - c^2}{2ab}\right) \\[5ex]
$
__Cosine Law is used for:__

(a.) $SAS$ when given two sides and an "included" angle ("included" angle is the angle between the two sides) OR

(b.) $SSS$ when given three sides

(18.)

Applies to all triangles with "inserted" parallel lines as applicable.

It states that if a line is parallel to one side of a triangle and intersects the other two sides of the triangle, then it divides those two sides proportionally.

Let us review examples:

Triangle in which a line is parallel to a side of the triangle.

$ BC\; || \;DE \\[3ex] \dfrac{AB}{BD} = \dfrac{AC}{CE} \\[5ex] $

$ KJ\; || \;GH \\[3ex] \dfrac{MK}{KG} = \dfrac{MJ}{JH} \\[5ex] $

$ QR\; || \;PT \\[3ex] \dfrac{NQ}{QP} = \dfrac{NR}{RT} \\[5ex] $ Triangle in which two lines are parallel to a side of the triangle.

In this case, we have to consider/separate each set of parallel lines relative to the side of the triangle (each line that is parallel to the side of the triangle)

$ Side\;\;of\;\;the\;\;triangle\;\;in\;\;question = AG \\[3ex] CE \;||\; AG \\[3ex] and \\[3ex] BF \;||\; AG \\[3ex] Consider\;\;parallel\;\;lines\;\; CE \;\;and\;\; AG \dfrac{DC}{CA} = \dfrac{DE}{EG} \\[5ex] Consider\;\;parallel\;\;lines\;\; BF \;\;and\;\; AG \dfrac{DB}{BA} = \dfrac{DF}{FG} \\[5ex] $

What is the length of $\overline{AC}$, in feet?

$ A.\;\; 13 \\[3ex] B.\;\; 26 \\[3ex] C.\;\; 29 \\[3ex] D.\;\; 42 \\[3ex] E.\;\; 48 \\[3ex] $

$ Consider\;\;parallel\;\;lines:\;\; \overline{DE} \;\;and\;\; \overline{BC} \\[3ex] \dfrac{\overline{AD}}{\overline{DB}} = \dfrac{\overline{AE}}{\overline{EC}} \\[5ex] \dfrac{8}{7 + 6} = \dfrac{16}{\overline{EC}} \\[5ex] \dfrac{8}{13} = \dfrac{16}{\overline{EC}} \\[5ex] 8 * \overline{EC} = 13(16) \\[3ex] \overline{EC} = \dfrac{13(16)}{8} \\[5ex] \overline{EC} = 13(2) \\[3ex] \overline{EC} = 26 \\[3ex] \overline{AC} = \overline{AE} + \overline{EC}...figure\;\;shown \\[3ex] \overline{AC} = 16 + 26 \\[3ex] \overline{AC} = 42\;units $

(19.)

Applies to all similar figures including similar triangles.

It states that for two similar figures, the ratio of the perimeters is the same as the scale factor; and the ratio of the areas is the ratio of the square of the scale factor.

This implies that:

$ if: \\[3ex] Scale\;\;Factor = \dfrac{c}{d} \\[5ex] then: \\[3ex] Perimeter\;\;Ratio = \dfrac{c}{d} \\[5ex] and: \\[3ex] Area\;\;Ratio = \left(\dfrac{c}{d}\right)^2 = \dfrac{c^2}{d^2} \\[5ex] $

How many inches long is the shortest side of a similar triangle that has a perimeter of 60 inches?

$ F.\;\; 9 \\[3ex] G.\;\; 11 \\[3ex] H.\;\; 20 \\[3ex] J.\;\; 24 \\[3ex] K.\;\; 27 \\[3ex] $

$ Perimeter\;\;of\;\;1st\;\;\triangle = 3 + 8 + 9 = 20\;inches \\[3ex] Perimeter\;\;of\;\;similar\;\;\triangle = 60\;inches \\[3ex] Shortest\;\;side\;\;of\;\;1st\;\;\triangle = 3\;inches \\[3ex] Shortest\;\;side\;\;of\;\;similar\;\;\triangle = p\;inches \\[3ex] \dfrac{20}{60} = \dfrac{3}{p}...Perimeter\;\;Ratio = Scale\;\;Factor \\[5ex] \dfrac{1}{3} = \dfrac{3}{p} \\[5ex] 1 * p = 3 * 3 \\[3ex] p = 9\;inches $

*
Demonstrate each theorem. Explain it well.
Use several examples to enhance understanding.
*

Applies to similar triangles.

It states that two triangles are similar if two angles of one triangle are congruent to two angles of the other triangle.

PRT is a triangle with points Q and S on PR and RT respectively such that SQ is parallel to TP.

(a) Show that triangles RTP and RSQ are similar.

(b) Calculate the size of SQ

(c) Given that the area of triangle RTP is 35 $cm^2$, calculate:

(i) the area of triangle RSQ

(ii) the area of quadrilateral QSTP

$ (a) \\[3ex] \angle RQS = \angle RPT...corresponding\;\;\angle s\;\;are\;\;equal \\[3ex] \angle RSQ = \angle RTP ...corresponding\;\;\angle s\;\;are\;\;equal \\[3ex] \angle QRS = \angle PRT...common\;\;\angle\;\;R \\[3ex] Any\;\;of\;\;the\;\;two\;\;can\;\;be\;\;used\;\;for\;\;the\;\;AA\sim\;\;Postulate \\[3ex] \therefore \triangle RTP \sim \triangle RSQ...AA\sim\;\;Postulate \\[3ex] (b) \\[3ex] \dfrac{|SQ|}{7.8} = \dfrac{7.2}{7.2 + 6} \\[5ex] \dfrac{|SQ|}{7.8} = \dfrac{7.2}{13.2} \\[5ex] |SQ| = \dfrac{7.8(7.2)}{13.2} \\[5ex] |SQ| = \dfrac{56.16}{13.2} \\[5ex] |SQ| = 4.254545455 \\[3ex] |SQ| \approx 4.3\;cm \\[3ex] (c) \\[3ex] (i) \\[3ex] Scale\;\;Factor = \dfrac{|RQ|}{|RP|} \\[5ex] Area\;\;Ratio = \dfrac{|RQ|^2}{|RP|^2} ...Area\;\;Ratio-Scale\;\;Factor\;\;Theorem \\[5ex] Also: \\[3ex] Area\;\;Ratio = \dfrac{Area\;\;of\;\;\triangle RSQ}{Area\;\;of\;\;\triangle RTP} \\[5ex] \implies \\[3ex] \dfrac{Area\;\;of\;\;\triangle RSQ}{Area\;\;of\;\;\triangle RTP} = \dfrac{|RQ|^2}{|RP|^2} \\[5ex] \dfrac{Area\;\;of\;\;\triangle RSQ}{35} = \dfrac{(7.2)^2}{(7.2 + 6)^2} \\[5ex] \dfrac{Area\;\;of\;\;\triangle RSQ}{35} = \dfrac{(7.2)^2}{(13.2)^2} \\[5ex] \dfrac{Area\;\;of\;\;\triangle RSQ}{35} = \dfrac{51.84}{174.24} \\[5ex] Area\;\;of\;\;\triangle RSQ \\[3ex] = \dfrac{35 * 51.84}{174.24} \\[5ex] = \dfrac{1814.4}{174.24} \\[5ex] = 10.41322314 \\[3ex] \approx 10.4\;cm^2 \\[3ex] (ii) \\[3ex] Area\;\;of\;\;Quadrilateral\;QSTP = Area\;\;of\;\;\triangle RTP - Area\;\;of\;\;\triangle RSQ ...diagram \\[3ex] = 35 - 10.41322314 \\[3ex] = 24.58677686 \\[3ex] \approx 24.6\;cm^2 $

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