Theorem (PageIndex{1})

In any ( riangle ABC), we have

(measuredangle ABC + measuredangle BCA + measuredangle CAB equiv pi.)

**Proof**First note that if ( riangle ABC) is degenerate, then the equality follows from Corollary 2.4.1. Further we assume that ( riangle ABC) is nondegenerate.

Let (X) be the reflection of (C) across the the midpoint (M) of ([AB]). By Proposition 7.2.1 (measuredangle BAX = measuredangle ABC). Note that ((AX)) is a reflection of ((CB)) across (M); therefore by Theorem 7.2.1, ((AX) parallel (CB)).

Since ([BM]) and ([MX]) do not intersect ((CA)), the points (B, M), and (X) lie on the same side of ((CA)). Applying the transversal property for the transversal ((CA)) to ((AX)) and ((CB)), we get that

[measuredangle BCA + measuredangle CAX equiv pi.]

Since (measuredangle BAX = measuredangle ABC), we have

(measuredangle CAX equiv measuredangle CAB + measuredangle ABC)

The latter identity and 7.4.1 imply the theorem.

Exercise (PageIndex{1})

Let ( riangle ABC) be a nondegenerate triangle. Assume there is a point (D in [BC]) such that

(measuredangle BAD equiv measuredangle DAC, BA = AD = DC.)

Find the angles of ( riangle ABC).

**Hint**Apply twice Theorem 4.3.1 and twice Theorem (PageIndex{1}).

Exercise (PageIndex{2})

Show that

(|measuredangle ABC| + |measuredangle BCA| + |measuredangle CAB| = pi)

for any ( riangle ABC).

**Hint**If ( riangle ABC) is degenerate, then one of the angle measures is (pi) and the other two are 0. Hence the result.

Assume ( riangle ABC) is nondegenerate. Set (alpha = measuredangle CAB), (eta = measuredangle ABC), and (gamma = measuredangle BCA).

By Theorem 3.3.1, we may assume that (0 < alpha, eta, gamma < pi). Therefore,

[0 > alpha + eta + gamma < 3 cdot pi.]

By Theorem (Pageindex{1}),

[alpha + eta +gamma equiv pi.]

From 7.4.2 and 7.4.3 the result follows.

Exercise (PageIndex{3})

Let ( riangle ABC) be an isoseles nondegenerate triangle with the base ([AC]). Suppose (D) is a reflection of (A) across (B). Show that (angle ACD) is right.

**Hint**Apply twice Theorem 4.3.1 and twice Theorem (PageIndex{1}).

Exercise (PageIndex{4})

Let ( riangle ABC) be an isosceles nondegenerate triangel with base ([AC]). Assume that a circle is passing thru (A), centered at a point on ([AB]), and tangent to ((BC)) at the point (X). Show that (measuredangle CAX = pm dfrac{pi}{4}).

- Hint
Suppose that (O) denotes the center of the circle.

Note that ( riangle AOX) is isosceles and (angle OXC) is right. Applying Theorem (PageIndex{1}) and Theorem 4.3.1 and simplifying, we should get (4 cdot measuredangle CAX equiv pi).

Show that (angle CAX) has to be acute. It follows then that (measuredangle CAX = pm dfrac{pi}{4}).

Exercise (PageIndex{5})

Show that for any quadrangle (ABCD), we have

(measuredangle ABC + measuredangle BCD + measuredangle CDA + measuredangle DAB equiv 0).

**Hint**Apply Theorem (PageIndex{1}) to ( riangle ABC) and ( riangle BDA).

## The Geometry of Triangles

**Definitions and formulas for the area of a triangle, the sum of the angles of a triangle, the Pythagorean theorem, Pythagorean triples and special triangles (the 30-60-90 triangle and the 45-45-90 triangle) Just scroll down or click on what you want and I'll scroll down for you! **

examples of triangles | area of a triangle | sum of the anglesof a triangle |

The Pythagorean Theorem | special Pythagoreantriples | special triangles45-45-90 & 36-60-90 |

**This formula is for right triangles only! The sides, a and b, of a right triangle are called the legs, and the side that is opposite to the right (90 degree) angle, c, is called the hypotenuse. This formula will help you find the length of either a, b or c, if you are given the lengths of the other two.**

**The reason that they are so special is that they are whole numbers -- none of those weird decimals, fractions or radicals!**

Ex 7.4 Class 9 Maths Question 1.

Show that in a right angled triangle, the hypotenuse is the longest side.

Solution:

Ex 7.4 Class 9 Maths Question 2.

In given figure sides AB and AC of Δ ABC are extended to points P and Q respectively. Also, ∠ PBC ∠ QCB. Show that

AC > AB.

Solution:

Ex 7.4 Class 9 Maths Question 3.

In figure ∠ B < ∠ A and ∠ C< ∠ D.show that AD<BC

Solution:

Ex 7.4 Class 9 Maths Question 4.

AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see fig.). Show that ∠ A > ∠ C and ∠ B > ∠ D.

Solution:

Given: ABCD is a quadrilateral. AB is the shortest side and CD is the longest side.

To Prove : ∠ A> ∠ C and ∠ B > ∠ D.

Construction: Join A and C, also B and D.

Proof: In Δ ABC, AB is the smallest side.

BC > AB

∠ 1 > ∠ 2 …(i)

[∵ angle opposite to longer side greater]

Ex 7.4 Class 9 Maths Question 5.

In the figure, PR > PQ and PS bisect ∠ QPR. Prove that ∠ PSR > ∠ PSQ.

Solution:

In Δ PQR, we have

Ex 7.4 Class 9 Maths Question 6.

Show that of all line segments drawn from a given point, not on it, the perpendicular line segment is the shortest.

Solution:

Given: x is a line and A is a point not lying on x. AB ⊥ x, C is any point on x other than B.

To prove : AB < AC

Proof: In ΔABC, ∠ B is the right angle.

∴ ∠ C is an acute angle.

∴ ∠ B > ∠ C

=> AC >AB

(∵ the Side opposite to greater angle is longer)

=>AB< AC

Hence, the perpendicular line segment is the shortest.

We hope the NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4, drop a comment below and we will get back to you at the earliest.

## NCERT Solutions for Class 9 Maths Chapter 7

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.1, Exercise 7.2, Exercise 7.3, Exercise 7.4 and Exercise 7.5 in English Medium as well as Hindi Medium updated for new academic session 2021-2022. According to Utter Pradesh Board (Prayagraj) students of class 9 will use NCERT Books for 9th Maths as course books. So, UP Board Solutions for Class 9 Maths Tribhuj ki Prashnavali 7.1, Prashnavali 7.2, Prashnavali 7.3, Prashnavali 7.4 and Prashnavali 7.5 in Hindi Medium PDF format are given to free download for academic session 2021-22. NCERT Solutions in Hindi Medium and English Medium or View in Video Format, for the students studying NCERT Books 2021-22, are based on the CBSE Curriculum 2021-2022.

Questions and solutions of all the five exercise of 9th Maths Chapter 7 are given below. Steps for solving questions are described properly by subject experts. Easy and simplified solutions are given, so that student can understand easily.

You can classify a triangle either by side length or internal angle.

Types of triangles by length of sides.

Type of Triangle by Lengths of SidesDescription

An isosceles triangle has two sides of equal length, and one side that is either longer or shorter than the equal sides. Angle has no bearing on this triangle type.

**Equilateral**

All sides and angles are equal in length and degree.

All sides and angles are of different lengths and degrees.

Types of triangles by angle.

Type of Triangle by Internal AngleDescription

**Right (right angled)**

Each of the three angles measure less than 90 degrees.

One angle is greater than 90 degrees.

Triangles classified by side and angles.

- The golden triangle is uniquely identified as the only triangle to have its three angles in the ratio 1 : 2 : 2 (36°, 72°, 72°). [3]

- Golden triangles can be found in the spikes of regularpentagrams.
- Golden triangles can also be found in a regular decagon, an equiangular and equilateral ten-sided polygon, by connecting any two adjacent vertices to the center. This is because: 180(10−2)/10 = 144° is the interior angle, and bisecting it through the vertex to the center: 144/2 = 72°. [1]
- Also, golden triangles are found in the nets of several stellations of dodecahedrons and icosahedrons.

### Logarithmic spiral Edit

The golden triangle is used to form some points of a logarithmic spiral. By bisecting one of the base angles, a new point is created that in turn, makes another golden triangle. [4] The bisection process can be continued indefinitely, creating an infinite number of golden triangles. A logarithmic spiral can be drawn through the vertices. This spiral is also known as an equiangular spiral, a term coined by René Descartes. "If a straight line is drawn from the pole to any point on the curve, it cuts the curve at precisely the same angle," hence *equiangular*. [5]

Closely related to the golden triangle is the golden gnomon, which is the isosceles triangle in which the ratio of the equal side lengths to the base length is the reciprocal 1 φ

"The golden triangle has a ratio of base length to side length equal to the golden section φ, whereas the golden gnomon has the ratio of side length to base length equal to the golden section φ." [6]

### Angles Edit

(The distances AX and CX are both *a*′ = *a* = φ , and the distance AC is *b*′ = φ², as seen in the figure.)

## Strategies for Solving a Triangle Question

Because there are so many different kinds of triangle problems, it is difficult to break down one exact path for problem solving them.

That said, your greatest assets and strategies when solving triangle problems will be to:

**1) Write down your formulas**

Because you are not given any formulas, you must keep them in your head and in your heart. The good news is that more you practice, the better you’ll be at rattling off triangle areas or side lengths of 30-60-90 triangles or anything else you’ll need.

But if you feel like you’ll forget your formulas as you go through your test, take a few seconds and write them down before you start solving your questions. Once you do, they will be there indelibly for you to work from for the rest of the math section, and you won’t have to worry about forgetting them.

**2) Use your formulas (and take your short-cuts)**

Once you’re sure that you’ve remembered your formulas, using them is the absolute most crucial step for any triangle problem. And, considering that most of your formulas essentially act as short-cuts (why bother solving with the Pythagorean theorem when you know that the legs of a 30-60-90 triangle are $x, x√3, 2x$?), you will save yourself a great deal of time and energy when you can keep your formulas on hand and in order.

**3) When working with multi-shapes, break it into small steps**

Remember that dealing with a multi-shape triangle problem is like working with dominos. Each successive piece of information makes way for finding the next piece of information.

Don’t get intimidated that you don’t have enough information or that there are too many shapes or lines to deal with. You will always have enough data to go on--just focus on finding one shape and one piece of information at a time, and the dominos will fall into place.

**4) Draw it out**

Draw your own diagrams if you are given none. Draw *on top* of your diagrams when you *are* given pictures. Write in your givens and all the measurements you find along the way to your missing variable (or variables), mark congruent lines and angles.

The more you can clarify your diagrams, the less likely you’ll be to make careless errors in misplacing or confusing your numbers and equalities.

## A more general Pythagorean Triple Calculator

- The
**Total count**of all Pythagorean triangles with a hypotenuse in the range 20-30 is 6 - If we
**Show all**of them they are:

*Since there are infinite number of PTs with any given side difference - see above on Hypotenuse and Longest leg are consecutive and The two legs are consecutive - these options are marked with the &infin sign and an extra input box will appear for difference searches to limit the search to a given maximum side length.*

**Sizes**reports the sizes (of the side|perimeter|area|inradius requested) in the given range, so that if a side|perimeter|area|inradius is found in more than one triple, it is reported once for each separate triple.

**Show all**lists all the triples found but if you want just one example use

**Show one**.

The

**results**are printed in the Results box, triples being given with their area, perimeter and inradius. Select and copy from this area to use the output as text or in other applications.

### A General Pythagorean Calculator

## Angle bisector of a triangle - Angle bisector theorem

Angle bisector theorem states that:

*An angle bisector of a triangle angle divides the opposite side into two segments that are proportional to the other two triangle sides.*

*The ratio of the BD length to the DC length is equal to the ratio of the length of side AB to the length of side AC:*

## 7.4: Angles of triangles - Mathematics

**For setting triangles:** Use a large square and cut __twice__ on the diagonal--the long side of each of these triangles is on the straight of grain. You will cut 4 triangles from each square and need to do some simple math to determine how many setting triangles you need for your particular quilt.

To determine the size of the square, use this simple formula:

**Finished Block Size x 1.41 = Finished Diagonal + 1.25 =Size of Square to Cut**

*(round off to the nearest 1/8")*

**For corner triangles:** Use a large square and cut __once__ on the diagonal--the short sides of each of these triangles is on the straight of grain. You will cut two triangles from each square--for the 4 corners of your quilt you will need to cut two squares the same size and cut into 4 triangles.

To determine the size of the square, use this simple formula:

**Finished Block Size x 1.41 = Finished Diagonal divided by 2 + .875" = Size of Square to Cut**