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In this page 10th grade geometry solution7 we are going to see solutions of some practice questions.

(iii) EFCD is
a parallelogram, So EF = 7 cm, DE = CF = 6 cm

Solution:

By
considering triangles ∆ AEF and ∆ ABC,

∠ AEF = ∠ ABC (Corresponding angles)

∠ A = ∠ A
(common angle)

By using AA
similarity criterion ∆ AEF ~ ∆ ABC

(AF/AC) =
(EF/BC)

[x/(x + 6)] =
(7/12)

12 x = 7 (x +
6)

12 x = 7 x +
42

12 x- 7 x =
42

5 x = 42

x = 42/5

x = 8.4 cm

Considering
the triangles ∆ BDG and ∆ BCF

(BD/BC) =
(DG/CF)

DG = (BD/BC) x CF

y = (5/12) x 6

y = 30/12

= 2.5 cm

(2) The image of
the man of height 1.8 m, is length 1.5 cm on the film of a camera. If the film
is 3 cm from the lens of the camera, how far is the man from the camera?

Solution:

Here AB
represents height of man

CD represents
height of reflection

Let “L” be
the position of lens

LM represents
distance between man and lens

LN represents
distance between lens and tape.

Also the
sides AB is parallel to the side CD, AB = 1.8 m. CD = 1.5 m, LN = 3 cm

From
triangles ∆ LAB and ∆ LCD, we get

∠ LAB = ∠ LCD (alternate angles)

∠ BLA = ∠ DLC (vertically opposite angles)

By using AA
similarity criterion ∆ LAB ~ ∆ LCD

(AB/CD) =
(LM/LN)

(180/1.5) =
(LM/3)

LM = (180 x 3)/1.5

LM = 360 cm

100 cm = 1 m

LM = 360/100

LM = 3.6 m

So the
distance between man and camera is 3.6 m

(3) A girl of
height 12 cm is walking away from the base of a lamp post at a speed of 0.6
m/sec . If the lamp is 3.6 m above the ground level, then find the length of
her shadow after 4 seconds.

Solution:

Let AB be the
height of lamp. CD be the height of girl and CE be the length of shadow of the
girl.

AB = 3.6
m CD = 120 cm

= 1.2 m

The girl is
walking at the rate of 0.6 m/sec. Now we are going to find the distance traveled by the girl

AC = 4 x 0.6

= 2.4 m

From the
triangles ECD and EAB the sides CD is parallel to AB.