Question

In AABC, AB = 15 cm, AC = 20 cm and ZBAC
= 90°. Find the
(i) area of AABC
(ii) length of AD if AD 1 BC
(ii) area of AABD​

Answer 1

(i) area of ∆ ABC is 150 cm²

(ii) length of AD if AD perpendicular to BC is 12 cm.

(ii) area of ∆ABD is 54 cm².

Step-by-step explanation:

It is given that,

In ΔABC, AB = 15 cm, AC = 20 cm and ∠BAC = 90°

(i) Finding the area of ΔABC

In Δ ABC, we have

Base, AB = 15 cm

Height, AC = 20 cm

∴ Area of ΔABC = 1/2 * base * height = 1/2 * 15 * 20 = 150 cm²

(ii) Finding the length of AD

Considering ΔABC and applying Pythagoras theorem, we get

BC = √[AC² + AB²] = √[20² + 15²] = √625 = 25 cm

We know that, if a perpendicular is drawn from the vertex containing a right angle of a right triangle to the hypotenuse then the triangle on each side of the perpendicular are similar to each other and to the whole triangle.

i.e., AD ⊥ BC

∴ Δ ACD ~ ΔABD or ΔACD ~ Δ CBA or ΔABD ~ ΔCBA

Taking  ΔABD ~ ΔCBA, we have

$\frac{AB}{BC} = \frac{AD}{AC}$ ..... [Corresponding sides of similar triangles are proportional to each other]

⇒ $\frac{15}{25} = \frac{AD}{20}$

⇒ AD = [15 * 20] / 25

⇒ AD = 12 cm

(iii) Area of ΔABD

Considering ΔABD and applying Pythagoras theorem, we get

BD = √[AB² - AD²] = √[15² - 12²] = √81 = 9 cm

In ΔABD, we have

Base, BD = 9 cm

Height, AD = 12 cm

∴ Area of ΔABD = 1/2 * base * height = 1/2 * 12 * 9 = 54 cm²

------------------------------------------------------------------------------------------------

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In the given figure, ∠ABC = 90° and BD ⊥ AC. If BD = 8 cm and AD = 4 cm, find CD.

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Answer 2

Answer:

Hey mates

Here is your answer

(i) area of ∆ ABC is 150 cm²

(ii) length of AD if AD perpendicular to BC is 12 cm.

(ii) area of ∆ABD is 54 cm².

Step-by-step explanation:

It is given that,

In ΔABC, AB = 15 cm, AC = 20 cm and ∠BAC = 90°

(i) Finding the area of ΔABC

In Δ ABC, we have

Base, AB = 15 cm

Height, AC = 20 cm

∴ Area of ΔABC = 1/2 * base * height = 1/2 * 15 * 20 = 150 cm²

(ii) Finding the length of AD

Considering ΔABC and applying Pythagoras theorem, we get

BC = √[AC² + AB²] = √[20² + 15²] = √625 = 25 cm

We know that, if a perpendicular is drawn from the vertex containing a right angle of a right triangle to the hypotenuse then the triangle on each side of the perpendicular are similar to each other and to the whole triangle.

i.e., AD ⊥ BC

∴ Δ ACD ~ ΔABD or ΔACD ~ Δ CBA or ΔABD ~ ΔCBA

Taking ΔABD ~ ΔCBA, we have

\frac{AB}{BC} = \frac{AD}{AC}

BC

AB

=

AC

AD

..... [Corresponding sides of similar triangles are proportional to each other]

⇒ \frac{15}{25} = \frac{AD}{20}

25

15

=

20

AD

⇒ AD = [15 * 20] / 25

⇒ AD = 12 cm

(iii) Area of ΔABD

Considering ΔABD and applying Pythagoras theorem, we get

BD = √[AB² - AD²] = √[15² - 12²] = √81 = 9 cm

In ΔABD, we have

Base, BD = 9 cm

Height, AD = 12 cm

∴ Area of ΔABD = 1/2 * base * height = 1/2 * 12 * 9 = 54 cm²

------------------------------------------------------------------------------------------------

Also View:

In the given figure, ∠ABC = 90° and BD ⊥ AC. If BD = 8 cm and AD = 4 cm, find CD.

https://brainly.in/question/6073623

In a triangle ABC, angle BAC = 90 degree and AD is perpendicular to BC . Then which one is true and why?

https://brainly.in/question/1358303

In the given figure, ∆ABC is right angled at C and DE ⊥ AB. Prove that ∆ABC ∼ ∆ADE and hence find the length of AE and DE.

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