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The triangle AEC is right-angled at E, and there is a point B on the line EC, BD is the altitude of triangle ABC. In the triangle AC = 25 cm, BC = 7 cm and AE = 15 cm. Determine the area of triangle ABC and the length of DB.

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120

c m^{2}

and 6 cm

140

c m^{2}

and 8 cm

150

c m^{2}

and 5 cm

150

c m^{2}

and 4.2 cm

answer is D.

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Detailed Solution

Here, we are given that a

Δ A E C

which is a right triangle at E,
BD is the altitude of a triangle △ABC, i.e., BD⊥AC.

A C = 25

cm, BC=7 cm and AE

= 15

Now, we will find the base EC of

Δ A E C

using Pythagoras Theorem as follows,

A E 2 + E C 2 = A C 2 225 + E C 2 = 625 E C 2 = 400 E C = 20 cm .

Now, as per the diagram, in

△ A E C

E B + B C = E C

So, substituting the dimensions off EC and BC in the above equation,

E B = E C − B C E B = 20 − 7 E B = 13 cm

Now, using the formula for the area of a triangle,

Area △ A B C = 12 × 15 × 20 = 150 cm 2

EB=EC-BC

∴ EB = 20 - 7

= 13 cm

Now, the area of

△ ABC is,

= \frac{1}{2} \times 15 \times 20

= 150 c m^{2}

Now, in the

△ A E C

and

△ B D C

,we have,

∠ A E B = ∠ B D C = 90 °

∠ C = ∠ C

(Common for both the triangles)

△ A E C ≈ △ B D C

Therefore, we have,

D B B C = E A A C D B 7 = 1525 D B = 4.2 cm .

So, the area of

△ A B C

is obtained as 150

cm 2

and the length of DB is obtained as 4.2 cm.
Hence, the correct option is (4).

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