NCERT Solutions for Class 9 Maths Chapter 7: Triangles - Exercise 7.4

Exercise 7.4

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

Ans:

Consider a right angled triangle $\text{ABC}$ right angled at $\text{B}$.

By Angle sum property of the triangle in $\vartriangle ABC,$

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

$\Rightarrow \angle A+90{}^\circ +\angle C=180{}^\circ $

$\Rightarrow \angle A+\angle C=90{}^\circ $

So the other two angles $\angle A$ and $\angle C$ must be acute angles (i.e., less than $90{}^\circ $)

Hence $\angle B$ is the largest angle in $\vartriangle ABC$ 

So $\angle B>\angle A$ and $\angle B>\angle C$

$AC>BC$ and $AC>AB$

In any triangle, the side opposite to the larger angle is longer.

Therefore, $\text{AC}$ is the largest side in $\vartriangle ABC$. $\text{AC}$ is the hypotenuse of $\vartriangle ABC$. 

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

2. In the given figure sides $\text{AB}$ and $\text{AC}$ of $\vartriangle ABC$ are extended to points $\text{P}$ and $\text{Q}$ respectively. Also, $\angle PBC<\angle QCB$. Show that $AC>AB$.

Ans: To show that $AC>AB$

Since $\angle ABC,\angle PBC$ forms the linear pair of angles,

$\angle ABC+ \angle PBC=180{}^\circ $

$\Rightarrow \angle ABC=180{}^\circ -\angle PBC$     ……(1)

Since $\angle ACB,\angle QCB$ forms the linear pair of angles,

$\angle ACB+ \angle QCB=180{}^\circ $

$\Rightarrow \angle ACB=180{}^\circ -\angle QCB$       ……(2)

Given that, $\angle PBC < \angle QCB$,

$180{}^\circ -\angle PBC > 180{}^\circ -\angle QCB$

From equation (1) and (2),

$\angle ABC>\angle ACB$

Since the side opposite to the larger angle is larger,

$AC > AB$

Hence, it has been proved that $AC>AB$

3. In the given figure, $\angle B<\angle A$ and $\angle C<\angle D$ . Show that $AD < BC$.

Ans: To show that $AD < BC$. 

Consider $\vartriangle \text{AOB}$,

Given that, $\angle B<\angle A$

Since the side opposite to smaller angle is smaller,

$\text{AO < BO}$     ……(1)

Consider $\vartriangle COD$,

Given that, $\angle C < \angle D$

Since the side opposite to smaller angle is smaller,

$OD < OC$     ……(2)

On adding equation (1) and (2),

$AO+OD < BO+OC$

$AD < BC$

Hence it has been proved that $AD < BC$

4. $\text{AB}$ and $\text{CD}$ are respectively the smallest and longest sides of a quadrilateral $\text{ABCD}$(see the given figure). Show that $\angle A>\angle C$ and $\angle B>\angle D$ .

Ans: To show that $\angle A > \angle C$ and $\angle B > \angle D$

Join $\text{AC}$.

Consider $\vartriangle ABC$,

Since $\text{AB}$ is the smallest side of the quadrilateral $\text{ABCD}$ ,

$\Rightarrow AB < BC$

Since the angle opposite to smaller side is smaller,

$\Rightarrow \angle 2 < \angle 1$     ……(1)

Consider $\vartriangle ADC$,

Since $\text{CD}$ is the largest side of the quadrilateral $\text{ABCD}$  ,

$\Rightarrow AD < CD$

Since the angle opposite to smaller side is smaller,

$\Rightarrow \angle 4 < \angle 3$     ……(2)

Adding equation (1) and (2),

$\Rightarrow \angle 2+\angle 4 < \angle 1+\angle 3$

$\Rightarrow \angle C < \angle A$

Let us join $\text{BD}$.

Consider $\vartriangle ABD$,

Since $\text{AB}$ is the smallest side of the quadrilateral $\text{ABCD}$  ,

$\Rightarrow AB < AD$

Since the angle opposite to the smaller side is smaller,

$\Rightarrow \angle 8 < \angle 5$     ……(3)

Consider $\vartriangle BDC$,

Since $\text{CD}$ is the largest side of the quadrilateral $\text{ABCD}$ ,

$\Rightarrow BC < CD$

Since the angle opposite to smaller side is smaller,

$\Rightarrow \angle 7 < \angle 6$     ……(4)

Adding equation (3) and (4),

$\Rightarrow \angle 8+\angle 7 < \angle 5+\angle 6$

$\Rightarrow \angle D < \angle B$

Hence it has been proved that $\angle A > \angle C$ and $\angle B > \angle D$ .

5. In the given figure, $PR>PQ$ and $\text{PS}$ bisects $\angle QPR$ . Prove that $\angle PSR>\angle PSQ$.

Ans: To prove that $\angle \text{PSR}\angle \text{PSQ}$ 

It is given that,

$PR > PQ$

Since the angle opposite to larger side is larger,

$\Rightarrow \angle PQR > \angle PRQ$                 ……(1)

Since $\text{PS}$ is the bisector of $\angle QPR$ ,

$\Rightarrow \angle QPS = \angle RPS$                     ……(2)

$\angle \text{PSR}$ is the exterior angle of $\vartriangle PQS$

$\Rightarrow \angle PSR = \angle PQR+\angle QPS$       ……(3)

$\angle \text{PSQ}$ is the exterior angle of $\vartriangle \text{PRS}$

$\Rightarrow \angle PSQ = \angle PRQ+\angle RPS$        ……(4)

Adding equations (1) and (2),

$\Rightarrow \angle PQR + \angle QPS > \angle PRQ+\angle RPS$

Substituting the values of equations (3) and (4),

$\Rightarrow \angle PSR > \angle PSQ$

Hence it has been proved that $\angle PSR > \angle PSQ$ 

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

Ans: Let us take a line $\text{l}$ and from point $\text{P}$ (i.e., not on line $\text{l}$ ), draw two line segments $PN$ and $\text{PM}$. Let $PN$ be perpendicular to $l$ and $\text{PM}$ is drawn at some other angle.

In $\vartriangle PNM$,

$\angle N=90{}^\circ $

By angle sum property of the triangle,

$\angle P+\angle N+\angle M=180{}^\circ$

$\Rightarrow \angle P+\angle M=90{}^\circ$

Now, clearly $\text{M}$ is an acute angle.

$\Rightarrow \angle M < \angle N$

Since the side opposite to the smaller angle is smaller,

$\Rightarrow PN < PM$

Similarly, by drawing different line segments from $\text{P}$ to $\text{L}$, it can be proven that $\text{PN}$ is smaller in comparison to them.

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

An Overview: NCERT Maths Class 9 Chapter 7 Exercise 7.4

Our curriculum gives a special focus on NCERT Solutions for Class 9 Maths ch 7 ex 7.4 that have a comprehensive detail for exam preparations. There are different properties of the triangle explained in NCERT Solutions for Class 9 Maths ex 7.4  which let the students calculate the parameters like sides, angles, triangles, perimeter, etc. The chapter is a bit tricky and a student might find it difficult, but the Class 9th Maths Chapter 7 Exercise 7.4 explains all the rules precisely. All the questions can be practised and revised properly for the examinations. The study material will act as a self-assessment tool for the students.

How NCERT Solutions for Class 9 Maths Exercise 7.4 Help to Make Exam Preparations?

Maths Class 9 Chapter 7 Exercise 7.4 includes various important questions that need deep-rooted knowledge of the triangle. It has concepts like triangle properties, establishing of the triangle perimeters, rule of similarity and congruence. Although, these types of problems demand analytic skills from the students. Thus, NCERT Class 9 Maths Chapter 7 Exercise 7.4 can serve as the best guide to the students for reference. The ex 7.4 Class 9 Maths solutions are essential from the viewpoint of examination and students should become adept with the practice. Consequently, candidates can make use of Class 9 Maths NCERT Solutions Chapter 7 Exercise 7.4 to get an organised approach and short cut techniques to solve the problems.

List out Some of the Properties of Triangles Maths Class 9 Chapter 7 Exercise 7.4?

The properties of triangles mentioned in NCERT Solutions for Class 9 Maths ex 7.4 are given below:

• The summation of each angle of a given triangle is 180`.

• The summation of the length of both the sides of a triangle is greater than the third side.

• The exterior angle of the triangle is always equal to the summation of two angles of triangles on the opposite side. 

• The area of a triangle is given by ½ x Base x Height.

• The perimeter of a triangle is the summation of its three sides.

Did You Know?

The Class 9th Maths Chapter 7 Exercise 7.4 teaches about the rule of congruence and similarity. The rule of congruence states that if two triangles are exact replica of each other and are similar in size, these are congruent. If two triangles have all sides in proportion and have congruent angles then both are considered similar.