- ABCis an isosceles right angled triangle and BD - AC - DC -5 - Find x-A- 5B- 20C- 10D- 5 -2E- 10 -2

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Question

$Δ A B C$ is an isosceles right angled triangle and BD $⊥$ AC, DC = 5. Find x.

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Solution

The correct option is D
5

$Δ$ BDC $≅$ $Δ$ ADB (1 right angle and two equal sides) AD = CD = 5 $∴$ From $Δ$ ADB -- BD $^{2}$ = AB $^{2}$ - AD $^{2}$ $\Rightarrow$ BD $^{2}$ = $\frac{(5 + 5)^{2}}{2}$ - 5 $^{2}$ $\Rightarrow$ BD = x = 5.

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ΔABCis an isosceles right angled triangle and BD ⊥ AC, DC = 5. Find x.

The correct option is D 5 ΔBDC ≅ ΔADB (1 right angle and two equal sides) AD = CD = 5 ∴From ΔADB -- BD2 = AB2- AD2 ⇒ BD2= (5+5)22 - 52 ⇒ BD = x = 5.

$Δ A B C$ is an isosceles right angled triangle and BD $⊥$ AC, DC = 5. Find x.

The correct option is D
5

$Δ$ BDC $≅$ $Δ$ ADB (1 right angle and two equal sides) AD = CD = 5 $∴$ From $Δ$ ADB -- BD $^{2}$ = AB $^{2}$ - AD $^{2}$ $\Rightarrow$ BD $^{2}$ = $\frac{(5 + 5)^{2}}{2}$ - 5 $^{2}$ $\Rightarrow$ BD = x = 5.