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