RS Aggarwal 2019,2020 solution class 7 chapter 15 Property of Triangles Exercise 15C

Exercise 15C

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Question 1:

Is it possible to draw a triangle, the lengths of whose sides are given below?

(i) 1 cm, 1 cm, 1 cm
(ii) 2 cm, 3 cm, 4 cm
(iii) 7 cm, 8 cm, 15 cm
(iv) 3.4 cm, 2.1 cm, 5.3 cm
(v) 6 cm, 7 cm, 14 cm

Answer 1:

(i) Consider numbers 1, 1 and 1.
Clearly, 1 + 1 $>$1
             1 + 1 $>$1
             1 + 1 $>$1

Thus, the sum of any two sides is greater than the third side.
Hence, it is possible to draw a triangle having  sides 1 cm, 1 cm and 1 cm.

(ii)
Clearly, 2 + 3 $>$4
             3 + 4 $>$2
             2+ 4 $>$3
Thus, the sum of any two sides is greater than the third side.
Hence, it is possible to a draw triangle having sides 2 cm, 3 cm and 4 cm.

(iii)
Clearly, 7 + 8 = 15

Thus, the sum of these two numbers is not greater than the third number.
Hence, it is not possible to draw a triangle having sides 7 cm, 8 cm and 15 cm.

(iv) Consider the numbers 3.4, 2.1 and 5.3.

Clearly: 3.4 + 2.1 $>$5.3
             5.3 + 2.1 $>$ 3.4
             5.3 + 3.4 $>$ 2.1

Thus, the sum of any two sides is greater than the third side.
Hence, it is possible to draw a triangle having sides 3.4 cm, 2.1 cm and 5.3 cm.

(v) Consider the numbers 6, 7 and 14.
Clearly, 6+7 $\ngtr$ 14

Thus, the sum of these two numbers is not greater than the third number.
Hence, it is not possible to draw a triangle having sides 6 cm, 7 cm and 14 cm.

Question 2:

Two sides of a triangle are 5 cm and 9 cm long. What can be the length of its third side?

Answer 2:

Let the length of the third side be x cm.

Sum of any two sides of a triangle is greater than the third side.

∴ 5 + 9 $>$x
$\Rightarrow x<14$

Hence, the length of the third side must be less than 14 cm.

Question 3:

If P is a point in the interior of ∆ABC then fill in the blanks with > or < or =.

(i) PA + PB ...... AB
(ii) PB + PC ...... BC
(iii) AC ...... PA + PC

Answer 3:

(i) $>$
(ii) $>$
(iii) $<$

The reason for the above three is that the sum of any two sides of a triangle is greater than the third side.

Question 4:

AM is a median of ∆ABC. Prove that (AB + BC + CA) > 2AM.

Answer 4:

Sum of any two sides of a triangle is greater than the third side.

In $△$AMB:
AB + BM >AM........(i)

In $△$AMC:
AC + CM >AM.........(ii)

Adding the above two equation:
AB + BM + AC + CM >AM + AM
AB + BC + AC > 2AM

Hence, proved.

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Question 5:

In the given figure. P is a point on the side BC of ∆ABC. Proce that (AB + BC + CA) > 2AP.

Figure

Answer 5:

Sum of any two sides of a triangle is greater than the third side.

 $$\begin{gathered} In △APB: \\ AB + BP >AP \\ In △APC: \\ AC + PC >AP \\ Adding the correspondong sides: \\ AB + BP + AC + PC> AP + AP \\ AB + AC+ BC >2AP \end{gathered}$$

Hence, proved.

Question 6:

ABCD is quadrilateral.
Prove that (AB + BC + CD + DA) > (AC + BD)

Figure

Answer 6:

Sum of any two sides of a triangle is greater than the third side.

In $△$ABC:
AB + BC > AC

In $△$ADC:
CD + DA > AC

Adding the above two:

AB + BC + CD + DA  > 2 AC               ... (i)

In $△$ADB:
AD + AB > BD

In $△$BDC:
CD + BC > BD

Adding the above two:

AB + BC + CD + DA  >2 BD             ... (ii)

Adding equation (i) and (ii):

AB + BC + CD + DA+AB + BC + CD + DA> 2(AC+BD)
=> 2(AB + BC + CD + DA)>2(AC+BD)
=> AB + BC + CD + DA > AC+BD

Question 7:

If O is a point in the exterior of ∆ABC, show that 2(OA + OB + OC) > (AB + BC + CA).

Figure

Answer 7:

We know that the sum of any two sides of a triangle is greater than the third side.

In $△$AOB:
OA + OB > AB...........(1)

In $△$BOC:
OB + OC > BC........................... (2)

In $△$AOC:
OA + OC > CA.............................(3)

Adding (1), (2) and (3):

OA + OB + OB + OC  + OA + OC >  AB + BC + CA

2( OA + OB  + OC) >  AB +BC + CA

Hence, proved.