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

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(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:

∆ EAD ~ ∆ EDC

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)