AP State Syllabus AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.4 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 7th Lesson Triangles Exercise 7.1

Question 1.

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

Solution:

Let a ΔABC be right angled at ∠B.

Then ∠A + ∠C = 90°

(i.e.,) ∠A and ∠C are both acute.

Now, ∠A < ∠B ⇒ BC < AC

Also ∠C < ∠B ⇒ AB < AC

∴ AC, the hypotenuse is the longest side.

Question 2.

In the given figure, sides AB and AC of ΔABC are extended “to points P and Q respectively. Also ∠PBC < ∠QCB. Show that AC > AB.

Solution:

From the figure,

∠PBC = ∠A + ∠ACB

∠QCB = ∠A + ∠ABC

Given that ∠PBC < ∠QCB

⇒∠A + ∠ACB < ∠A + ∠ABC

⇒ ∠ACB < ∠ABC

⇒ AB < AC

⇒ AC > AB

Hence proved.

Question 3.

In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Solution:

Given that ∠B < ∠A; ∠C < ∠D

∠B < ∠A ⇒ AO < OB [in ΔAOB] ……………… (1)

∠C < ∠D ⇒ OD < OC [in ΔCOD]…… (2)

Adding (1) & (2)

AO + OD < OB + OC

AD < BC

Hence proved.

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:

Given that AB and CD are the smallest and longest sides of quadrilateral ABCD.

From the figure,

In ΔBCD

∠1 > ∠2 [∵ DC > BC] ………………(1)

In ΔBDA

∠4 > ∠3 [∵ AD > AB] ………….(2)

Adding (1) & (2)

∠1 + ∠4 > ∠2 + ∠3

∠B > ∠D

Similarly,

In ΔABC, ∠6 < ∠7 [ ∵AB < BC] ……………….(3)

In ΔACD

∠5 < ∠8 …………. (4)

Adding (3) & (4)

∠6 + ∠5 < ∠7 + ∠8

∠C < ∠A ⇒ ∠A > ∠C

Hence proved.

Question 5.

In the given figure, PR > PQ and PS bisects ∠QPR. Prove that

∠PSR > ∠PSQ.

Solution:

Given that PR > PQ;

∠QPS =∠RPS

PR> PQ

∠Q > ∠R

Now ∠Q +∠QPS > ∠R + ∠RPS

⇒ 180° – (∠Q + ∠QPS) < 180° – (∠R + ∠RPS)

⇒ ∠PSQ < ∠PSR ⇒ ∠PSR > ∠PSQ

Hence proved.

Question 6.

If two sides of a triangle measure 4 cm and 6 cm find all possible measurements (positive integers) of the third side. How many distinct triangles can be obtained ?

Solution:

Given that two sides of a triangle are 4 cm and 6 cm.

∴ The measure of third side > Differ-ence between other two sides.

third side > 6 – 4

third side > 2

Also the measure of third side < sum of other two sides

third side <6 + 4 < 10

∴ 2 < third side <10

∴ The measure of third side may be 3 cm, 4 cm, 5 cm, 6 cm, 7 cm, 8 cm, 9 cm

∴ Seven distinct triangles can be obtained.

Question 7.

Try to construct a triangle with 5 cm, 8 cm and 1 cm. Is it possible or not ? Why ? Give your justification.

Solution:

As the sum (6 cm) of two sides 5 cm and 1 cm is less than third side. It is not possible to construct a triangle with the given measures.