S.chand publication New Learning Composite mathematics solution of class 7 Chapter 10 Properties of Triangles Exercise 10D

 Exercise 10D

Question 1

Complete the table for each triangle by inserting the details from the figures and verify that Pythagoras theorem is true for each triangle or not. a denotes the side opposite ∠A, b denotes side opposite ∠B and hypotenuse C denotes side opposite to ∠C

	c2	a2	b2	a2+b2
(a)	25	9	16	25
(b)	100	36	64	100
(c)	400	256	144	400
(d)	225	81	144	225

Question 2

Find the length of the hypotenuse in each triangle All measurements are in centimeters

(a) x²=(8)²+(15)²

x²=64+225

x²=289

x=√289

∴x=17

(b) x²=(12)²+(9)²

x²=144+81

x²=225

x=√225

∴x=15

(c) x²=(21)²+(20)²

x²=441+400

x²=841

x=√841

∴x=29

(d) x²=(7)²+(24)²

x²=49+576

x²=625

x=√625

∴x=25

Question 3

The length of the hypotenuse and one side are given. Find the length of the other side for each right triangle. (All lengths are in cm)

(a) y²=(15)²-(9)²

y²=225-81

y=√144

∴y=12

(b) x²=(10)²-(8)²

x²=100-64

x²=36

x=√36

∴x=6

(c) a²=(25)²-(7)²

a²=625-49

a²=576

a=√576

∴a=24

(d) b²=(26)²-(24)²

b²=676-576

b²=100

b=√100

∴b=10

Question 4

The rectangular floor of a room is 12 m long and 5 m wide. What is the length of a diagonal?

Fig to be added

∴Diagonal (AC)$=\sqrt{(12)^2+(5)^2}$

=√144+25=√169

=13

Ans : The length of a diagonal is 13cm

Question 5

A ladder 10 m long just reaches the top of a wall 8 m high. How far is the foot of the ladder from the base of wall?

Sol :

Fig to be added

∴AB$=\sqrt{(10)^2-(8)^2}$

=√100-64

=√36

=6

Ans : The foot of the ladder from the base of the wall is 6m

Question 6

What is the length of the ramp?

Sol :

Fig to be added

∴The length of the ramp$=\sqrt{(12)^2+(5)^2}$

=√144+25

=√169

=13

Question 7

ABC is a right angled triangle with ∠ACB = 90°, BC = 12 cm, AB = 15 cm and CD = 5 cm. Find the lengths of BD and AD.

Sol :

Fig to be added

From ΔABC, 

AB=15 cm

BC=12 cm

∴AC$=\sqrt{(AB)^2-(BC)^2}$

$=\sqrt{(15)^2-(12)^2}$

=√225-144

=√81

=9

∴AD=(9-5) cm

=4 cm

From ΔBCD, 

BC=12 cm

CD=5 cm

∴BD$=\sqrt{(BC)^2+(CD)^2}$

$=\sqrt{(12)^2+(5)^2}$

=√144+25

=√169

=13

Question 8

A 18 m pole is supported by 2 guy wires attached to a point $\frac{2}{3}$ of the way up the pole and to point on the ground 5 m from the base of the pole. What is the length of each guy wire?

Sol :

Fig to be added

The wire attached at $=\left(18 \times \frac{2}{3}\right)$ m

=12 m

∴The length of the each guy wire$=\sqrt{(12)^2+(5)^2}$

=√144+25

=√169

=13

Ans : The length of each guy wire is 13m

Question 9

A ship sails due east 30 km and then due south for 16 km. How far, in a direct line, is it from the starting point?

Sol :

Fig to be added

∴AB$=\sqrt{(30)^2+(16)^2}$

=√900+256

=√1156

=34

Question 10

A ship sails 60 km due north, then 20 km due west and finally 39 km due south. How far is it from its starting point?

Sol :

Fig to be added

AB=60 km and CD=39 km

∴AE=(60-39)=21 km

∴CB=DE=20 km

∴AD$=\sqrt{(AE)^2+(DE)^2}$

$=\sqrt{(21)^2+(20)^2}$

=√441+400

=√841

=29 km

Question 11

Sudhir and Deepak start from point A to walk over a rectangular park. Sudhir walks from A to D while Deepak walks directly from A to C. How much more does Deepak walk than Sudhir?

Sol :

Fig to be added

AD=BC=70 m

CD=BA=24 m

∴AC$=\sqrt{(70)^2+(24)^2}$

=√4900+576

=√5476

=74

∴Deepak walks=(70+24)=94 m

∴Deepak walks more than Sudhir=94-74=20 m