10th Grade Geometry Solution6

In this page 10th grade geometry solution6 we are going to see solutions of some practice questions.

(1) Find the unknown values in each of the following figures. All lengths are given in centimeters (all measures are not in scale)

Solution:

From triangles ABC and triangle ADE

 ∠ABC = ∠ADE (corresponding angles)

∠A = ∠A (common angle)

So ∆ ABC ~ ∆ ADE

(AC/AE) = (BC/DE)

[x/(x + 8)] = (8/24)

[x/(x + 8)] = (1/3)

3 x = 1 (x + 8)

3 x = x + 8

3 x – x = 8

2 x = 8

  x = 8/2

x = 4 cm.

also ∆ EAG and ∆ ECF are congruent  triangles

so, (EC/EA) = (CF/AG)

AG = (EA x CF)/EC

EA = EC + CA

       = 8 + 4

      = 12 cm

y = (6 x 12)/8

 y = 72/8

    = 9 cm

The value of x is 4 cm

Value of y is 9 cm


(ii) In ∆ HBC, the sides FG and BC are parallel

∆ HFG ~ ∆ HBC

(HF/HB) = (FG/BC)

(4/10)=(x/9)

x = (4 x 10)/4

x = 3.6 cm

In triangle ∆ FBD and ∆FHG the sides BD and GH are parallel,

∠FBD = ∠FHG (alternate angles)

∠BFD = ∠HFG (vertically opposite angels)

By using AA similarity criterion ∆ FBD ~ ∆ FHG

(FG/FD) = (FH/FB)

[x/(y + 3)] = (4/6)

3.6/(y + 3) = (2/3)

3.6(3) = 2 (y + 3)

10.8 = 2 y + 6 

2 y = 10.8 – 6

2 y = 4.8

   y = 4..8/2

   y = 2.4 cm

In triangles ∆ AEG and ∆ ABC, the sides EG and BC are parallel,

∠A = ∠A (common angle)

∠ AEG = ∠ ABC (corresponding angles)

By using AA similarity criterion ∆ AEG ~ ∆ ABC

(AE/AB) = (EG/BC)

[z/(z + 5 )] = (x + y)/9

[z/(z + 5 )] = 6/9

9 z = 6 (z + 5)

9 z = 6 z + 30

9 z – 6 z = 30

3 z = 30

  z = 30/3

   z = 10 cm

So length of the side FG = 3.6 cm

Length of the side BF = 2.4 cm

Length of the side AE = 10 cm

10th grade geometry solution6 10th grade geometry solution6

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