In this page 10th grade geometry solution11 we are going to see solutions of some practice questions.

(10) The government plans to develop a new industrial zone in an unused portion of land in a city.

Solution:

By considering the lines AD and BC,the angles

AEB = DEC (vertically opposite angles)

EAB = EDC (alternate angles)

By using AA similarity criterion ∆ EAB ~ ∆ EDC

(AB/DC) = (EF/EG)

EF = (AB/DC) x EG

= (3/1) x 1.4

= 4.2 km

Area of new industrial zone = Area of ∆ EAB

= (1/2) x AB x EF

= (1/2) x 3 x 4.2

= 6.3 km²

So the area of new industrial zone is 6.3 km²

(11) A boy is designing a diamond shaped kite, as shown in the figure where AE = 16 cm, EC = 81 cm. He wants to use a straight cross bar BD. How long should it be?

Solution:

So, EA/ED = ED/EC

ED² = EA x EC

= 16 x 81

= √16 x 81

= 4 x 9

= 36

ABD is the isosceles right triangle and the side AE is perpendicular to the side BD

BE = ED

So BD = 2 ED

= 2 (36)

= 72 cm

(12) A student wants to determine the height of flag pole. He placed a small mirror on the ground so that he can see reflection of the top of the flagpole. The distance of the mirror from him is 0.5 m and the distance of the flagpole from the mirror is 3 m .If t he eyes are 1.5 m above the ground level, then find the height of the flagpole (The foot of student, mirror and the foot of flagpole lie along the a straight line)

Let “C” be point of reflection

In triangles ABC and EDC

∠ ABC = ∠ EDC = 90 degree

∠ BCA = ∠ DCE

By using AA criterion ∆ ABC ~ ∆ EDC

AB/ED = BC/DC

1.5/ED = 0.5/3

4.5 = 0.5 ED

ED = 4.5/0.5

= 9 m

Therefore height of flag pole = 9 m

(13) A roof has a cross section as shown in the diagram

(i) Identify the similar triangles

(ii) Find the height h of the roof.

Corresponding triangles from the given picture

(i) ∆ WZY ~ ∆ YZX

(ii) ∆ WYX ~ ∆ YZX

(iii) WZY ~ ∆ WYZ

From the(ii)

WY/YZ = XY/XZ

h/8 = 6/10

h = (6/10) x 8

h =48/10

h = 4.8 m