$Δ A B C$ is right-angled at A and $A D ⊥ B C$ . If BC = 13 cm and AC = 5 cm, find the ratio of the areas of $Δ A B C a n d Δ A D C$ .

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Solution

In ΔBAC and ΔADC, $∠ B A C = ∠ A D C = 90^{o}$ (AD ⊥ BC)

∠ACB = ∠DCA (common)

So, Δ BAC ~ ΔADC

$\frac{a r (Δ A B C)}{a r (Δ A D C)} = \frac{a r (Δ B A C)}{a r (Δ A D C)} = \frac{B C^{2}}{A C^{2}}$

$= \frac{13^{2}}{5^{2}} = \frac{169}{25}$

Therefore, the ratio of the areas of ΔABC and ΔADC = 169 : 25

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Question 7

$Δ$ ABC is right angled at A, AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12, find the area of $Δ$ ABC. Also, find the length of AD.

Δ A B C

is a right-angle at

A

A D

is perpendicular to

B C

A B = 5 c m

B C = 13 c m

and

A C = 12 c m

. Find the area of

Δ A B C

. Also , find the length of

A D

Δ A B C

is right angle at A(figure) . AD is perpendicular to BC. if

A B = 5 c m, B C = 13 c m, A C = 12 c m,

Find the area of

Δ A B C

. Also find the length of AD.

ΔABC is right angled at A (see the given figure). AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, Find the area of ΔABC. Also find the length of AD.

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ΔABC is right-angled at A and AD⊥BC. If BC = 13 cm and AC = 5 cm, find the ratio of the areas of ΔABC and ΔADC.

In ΔBAC and ΔADC, ∠BAC=∠ADC=90o (AD ⊥ BC) ∠ACB = ∠DCA (common) So, Δ BAC ~ ΔADC ar(ΔABC)ar(ΔADC)=ar(ΔBAC)ar(ΔADC)=BC2AC2 =13252=16925 Therefore, the ratio of the areas of ΔABC and ΔADC = 169 : 25

$Δ A B C$ is right-angled at A and $A D ⊥ B C$ . If BC = 13 cm and AC = 5 cm, find the ratio of the areas of $Δ A B C a n d Δ A D C$ .