Question

ABC is a right-angled triangle with ZABC = 90°. D is any point on AB and DE is perpendicular tobAC. Prove that:
(i) AADE – AACB.
(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm, find DE and AD.
iii) Find, area of AADE: area of quadrilateral BCED.
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Answer 1

Answer:

(1). In triangle ADE and triangle ACB

angle A - angle A ( common in both triangles)

angle B - angle E ( each 90 degree )

so angle C - angle D ( by angle sum property )

by AAA property triangle ADE is similar to triangle ACB

Answer 2

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Correct Question:-

ABC is a right angled triangle with ∠ABC=90°.D is any point on AB and DE is perpendicular to AC. Prove that:-

• (i) △ADE∼△ACB.
• (ii) If AC=13cm,BC=5cm and AE=4cm. Find DE and AD.
• (iii) Find, area of △ADE : area of quadrilateral BCED.

$\qquad\underline{\pink{\sf Solution}}$

From the question it is given that,

• ∠ABC=90°

• AB and DE is perpendicular to AC

(i) Consider the △ADE and △ACB,

• ∠A=∠A … [common angle for both triangle]
• ∠B=∠E … [both angles are equal to 90°]

Therefore,

• △ADE∼△ACB

(ii) from (i) we proved that,

• △ADE∼△ACB

So,

$\bold{AE/AB\:=\:AD/AC\:=DE/BC}$

[equation (i)]

• Consider the △ABC, is a right angle triangle
• From Pythagoras theorem, we have..

From Pythagoras theorem, we have

• AC² = AB² + BC²

• 13³ = AB² + 5²

• 169 = AB² + 25

• AB² = 169 −25
• AB² =144
• AB = $\sqrt{144}$
• $\bold{AB \:=\: 12cm}$

Consider the equation (i),

• $\bold{AE/AB\:=\:AD/AC\:=\:DE/BC}$
• $\bold{Take, \:AE/AB\:=\:AD/AC}$
• $\bold{4/12\:=\:AD/13}$
• $\bold{1/3\:=\:AD/13}$
• $\bold{(1×13)/3\:=\:AD}$

AD = 4.33cm

Now, take AE/AB=DE/BC

• $\bold{4/12\:=\:DE/5}$
• $\bold{1/3\:=\:DE/5}$
• $\bold{DE\:=\:(5×1)/3}$
• $\bold{DE\:=\:5/3}$

DE = 1.67cm

(iii) Now, we have to find area of △ADE : area of quadrilateral BCED,

$\qquad\underline{\red{\sf We\:Know\: That}}$

• Area of △ADE = $\dfrac{1}{2}$× AE × DE

• = $\dfrac{1}{2}$× 4 × (5/3)

• =10/3cm²

Then, area of quadrilateral BCED= area of △ABC− area of △ADE

• $\dfrac{1}{2}$×BC × AB - 10\3
• $\dfrac{1}{2}$× 5 × 12 - 10\3

• = 1 × 5 × 6 −10/3
• =30−10/3
• =(90−10)/3
• =80/3cm²

So, the ratio of area of △ADE : area of quadrilateral BCED=(10/3)/(80/3)

• = (10/3)×(3/80)
• = (10×3)/(3×80)
• = (1×1)/(1×8)
• = 1/8

Therefore, area of △ADE : area of quadrilateral BCED is 1:8.

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