**NCERT Solutions for Class 9
Maths Chapter 7 Triangles Ex 7.4**

NCERT Solutions for Class
9 Maths Chapter 7 Triangles Ex 7.4 are the part of NCERT Solutions for Class 9 Maths.
Here you can find the NCERT Solutions for Chapter 7 Triangles Ex 7.4.

**NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1****NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2****NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3****NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4****NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.5**

**Ex 7.4 Class 9 Maths**** ****Question 1.****
**Show that in a right angled
triangle, the hypotenuse is the longest side.

**Solution:****
**Let us consider a ∆ABC in which ∠B = 90°.

∴ ∠A + ∠B + ∠C = 180°

⇒ ∠A + 90° + ∠C = 180°

⇒ ∠A + ∠C = 90°

⇒ ∠A + ∠C = ∠B

∴ ∠B > ∠A and ∠B > ∠C

Therefore, AC > BC.

Similarly, AC > AB.

Hence, AC is the longest side of the ∆ABC. But AC is the hypotenuse of the triangle. Thus, the hypotenuse is the longest side.

**Ex 7.4 Class 9 Maths**** ****Question 2.****
**In figure, sides AB and AC of ∆ABC
are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

**Solution:**

∠ABC + ∠PBC = 180° [Linear pair]

and ∠ACB + ∠QCB = 180° [Linear pair]

But ∠PBC < ∠QCB [Given]

⇒ 180° – ∠PBC > 180° – ∠QCB

⇒ ∠ABC > ∠ACB

⇒ AC > AB [The side opposite to ∠ABC > the side opposite to ∠ACB]

**Ex 7.4 Class 9 Maths**** ****Question 3.****
**In figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

**Solution:**

Since
∠A > ∠B [Given]

∴ OB > OA …(1)
[Side opposite to greater angle is
longer]

Similarly, OC > OD …(2)

Adding (1) and (2), we get

OB + OC > OA + OD

⇒ BC > AD

Or, AD < BC

**Ex 7.4 Class 9 Maths**** ****Question 4.****
**AB and CD are respectively the
smallest and longest sides of a quadrilateral ABCD (see figure). Show that ∠A > ∠C and ∠B > ∠D.

**Solution:**

Let us join AC.

Now, in ∆ABC, AB < BC [∵ AB is the smallest side of the quadrilateral ABCD]

⇒ BC > AB

⇒ ∠BAC > ∠BCA …(1) [Angle opposite to longer side of a ∆ is greater]

Again, in ∆ACD, CD > AD [CD is the longest side of
the quadrilateral ABCD]

⇒ ∠CAD > ∠ACD …(2) [Angle opposite to longer side of a ∆
is greater]

Adding eqs. (1) and (2), we get

∠BAC + ∠CAD > ∠BCA + ∠ACD

⇒ ∠A > ∠C

Similarly, by joining BD, we can prove that ∠B > ∠D.

**Ex 7.4 Class 9 Maths**** ****Question 5.
**In figure, PR > PQ and PS bisect ∠QPR. Prove that ∠PSR > ∠PSQ.

**Solution:****
**In ∆PQR, PS bisects ∠QPR [Given]

∴ ∠QPS = ∠RPS …(1)

and PR > PQ [Given]

⇒ ∠PQS > ∠PRS …(2) [Angle opposite to longer side of a ∆ is greater]

Adding eqs. (1) and (2), we get

∠PQS + ∠QPS > ∠PRS + ∠RPS …(3)

∵ Exterior ∠PSR = [∠PQS + ∠QPS]

and exterior ∠PSQ = [∠PRS + ∠RPS]

[An exterior angle is equal to the sum of interior
opposite angles]

Now, from (3), we have

∠PSR > ∠PSQ.

**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:****
**Let us consider the ∆PMN such that ∠M = 90°

[Sum of angles of a triangle is 180°]

∵ ∠M = 90° [PM ⊥ MN]

So, ∠N + ∠P = ∠M

⇒ ∠N < ∠M

⇒ PM < PN …(1)

Similarly, PM < PN

_{1}…(2)

and PM < PN

_{2}…(3)

From (1), (2) and (3), we have PM is the smallest line segment drawn from a given point P on the line l.

Thus,
the perpendicular line segment is the shortest line segment drawn on a line
from a point not on it.

**NCERT Solutions for Maths Class 10**