Question

14. In ∆ABC, AD I BC, AB = 7.5 cm, AC = 10 cm, AD=
6 cm. Find BC. Show that ∆BAC = 90°
7.5 cm
10 cm
6 cm
.
​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

The value of $BC=12.5\ cm$ and $\angle BAC=90\ degree$ in $\triangle ABC$.

Step-by-step explanation:

Given,

In $\triangle ABC$ where $AD\perp BC,AB=7.5\ cm,AC=10\ cm\ and\ AD=6\ cm$

From figure,

$\angle ADB=\angle ADC=90\ degree$

In $\triangle ABD$,

$AB^{2}=AD^{2}+BD^{2}$

⇒$AB^{2}-AD^{2}=BD^{2}$

⇒$(7.5)^{2}-6^{2}=BD^{2}$

⇒$BD=\sqrt{56.25-36}$

⇒$BD=\sqrt{20.25}$

⇒$BD=4.5\ cm$

In $\triangle ACD$,

$CD=\sqrt{AC^{2}-AD^{2} }$

⇒$CD=\sqrt{10^{2}-6^{2} }$

⇒$DC=8\ cm$

∴ $BC=BD+CD=4.5+8=12.5\ cm$

From $\triangle ABC$,

$BC=\sqrt{AB^{2}+AC^{2} }$

⇒$BC=\sqrt{7.5^{2}+10^{2} }$

⇒$BC=12.5\ cm$

So, The value of $BC=12.5\ cm$ and $\angle BAC=90\ degree$ in $\triangle ABC$.