ABC is a right angled triangle with - ABC- 90 o- D is any point on AB and DE is perpendicular to AC- Prove that -i ADE- ACB-ii If AC-13 cm- BC-5 cm and AE-4 cm- Find DE and AD-iii Find- area of ADE - area of quadrilateral BCED-

To Prove : #160; ADE ∼ ACB Proof :(i) In ADE and ACB #160; #160; (1) ∠ A=∠ A #160; #160; #160; #160; #160; #160; ...

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Standard X

Mathematics

Relation between Areas and Sides of Similar Triangles

ABC is a righ...

Question

$A B C$ is a right angled triangle with $∠ A B C = 90^{o}$ . $D$ is any point on $A B$ and $D E$ is perpendicular to $A C$ . Prove that :
(i) $△ A D E \sim △ A C B$ .
(ii) If $A C = 13 c m$ , $B C = 5 c m$ and $A E = 4 c m$ . Find $D E$ and $A D$ .
(iii) Find, area of $△ A D E :$ area of quadrilateral $B C E D$ .

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Solution

To Prove : $△ A D E \sim △ A C B$
Proof :
(i) In $△ A D E$ and $△ A C B$
(1) $∠ A = ∠ A$ [common]
(2) $∠ A E D = ∠ A B C = 90^{o}$ (given)
$∴$ $△ A D E \sim △ A C B$ [AA axiom]
(ii) ${(A C)}^{2} = {(A B)}^{2} + {(B C)}^{2}$
$169 = {(A B)}^{2} + 25$
$A B = 12 c m$
$∵$ $△ A D E \sim △ A C B$
$∴$ $\frac{D E}{B C} = \frac{A D}{A C} = \frac{A E}{A B}$
$∴$ $\frac{D E}{B C} = \frac{A E}{A B}$
$\frac{D E}{5} = \frac{4}{12}$
$D E = \frac{20}{12} = \frac{5}{3} = 1 \frac{2}{3} c m$
Now, $\frac{A D}{A C} = \frac{A E}{A B}$
$\frac{A D}{13} = \frac{4}{12}$
$A D = \frac{13 \times 4}{12} = \frac{13}{3} = 4 \frac{1}{3} c m$ .
(iii) $\frac{A r . o f (△ A B C)}{A r . o f (△ A D E)} = \frac{{A B}^{2}}{{A E}^{2}} = \frac{144}{16} = \frac{9}{1}$
$\frac{A r . o f (△ A D E) + A r . o f (B C E D)}{A r . o f (△ A D E)} = 9$
$1 + \frac{A r . o f (B C E D)}{A r . o f (△ A D E)} = 9$
$\frac{A r . o f (△ A D E)}{A r . o f (B C E D)} = \frac{1}{8}$

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ABC is a right angled triangle with $∠$ ABC = $90^{\circ}$ . D is any point on AB and DE is perpendicular to AC. Prove that : (i) $Δ$ ADE~ $Δ$ ACB. (ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD. (iii) Find, area of $Δ$ ADE : area of quadrilateral BCED.

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ABC is a right angled triangle with ∠ABC=90o. 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.