**Triangles**

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

**Solution:**

Let us consider ∆ABC such that ∠B = 90°

∴ ∠A + ∠B + ∠C = 180°

⇒ ∠A + 90°-+ ∠C = 180°

⇒ ∠A + ∠C = 90°

⇒∠A + ∠C = ∠B

∴ ∠B > ∠A and ∠B > ∠C

⇒ Side opposite to ∠B is longer than the side opposite to ∠A

i.e., AC > BC.

Similarly, AC > AB.

Therefore, we get AC is the longest side. But AC is the hypotenuse of the triangle. Thus, the hypotenuse is the longest side.

**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

The side opposite to ∠ABC > the side opposite to ∠ACB

⇒ AC > AB.

**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 have

OB + OC > OA + OD

⇒ BC > AD

**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 ∆ is greater]

Adding (1) and (2), we get

∠BAC + ∠CAD > ∠BCA + ∠ACD

⇒ ∠A > ∠C

Similarly, by joining BD, we have ∠B > ∠D.

**Question 5.In figure, PR > PQ and PS bisect ∠QPR. Prove that ∠PSR >∠PSQ.**

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

∴ ∠QPS = ∠RPS

and PR > PQ [Given]

⇒ ∠PQS > ∠PRS [Angle opposite to longer side of A is greater]

⇒ ∠PQS + ∠QPS > ∠PRS + ∠RPS …(1) [∵∠QPS = ∠RPS]

∵ Exterior ∠PSR = [∠PQS + ∠QPS]

and exterior ∠PSQ = [∠PRS + ∠RPS]

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

Now, from (1), we have

∠PSR = ∠PSQ.

**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°

Since, ∠M + ∠N+ ∠P = 180°

[Sum of angles of a triangle is 180°]

∵ ∠M = 90° [PM ⊥ l]

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 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.