Question

what is the area of an equilateral triangle whose inscribed circle has radius 2cm?

Answer 1

Step-by-step explanation:

• Here, OC=r, BC=l2, AB=l. Since ABC∼BOC, taking ratios, we get AC=l24r.
• Here, OC=r, BC=l2, AB=l. Since ABC∼BOC, taking ratios, we get AC=l24r.By the Pythagorean theorem, AB2=AC2+BC2,
• Here, OC=r, BC=l2, AB=l. Since ABC∼BOC, taking ratios, we get AC=l24r.By the Pythagorean theorem, AB2=AC2+BC2,Therefore,
• Here, OC=r, BC=l2, AB=l. Since ABC∼BOC, taking ratios, we get AC=l24r.By the Pythagorean theorem, AB2=AC2+BC2,Therefore,l=l24+l416r2−−−−−−−−−√
• Here, OC=r, BC=l2, AB=l. Since ABC∼BOC, taking ratios, we get AC=l24r.By the Pythagorean theorem, AB2=AC2+BC2,Therefore,l=l24+l416r2−−−−−−−−−√Simplifying, we get l=r12−−√
• Here, OC=r, BC=l2, AB=l. Since ABC∼BOC, taking ratios, we get AC=l24r.By the Pythagorean theorem, AB2=AC2+BC2,Therefore,l=l24+l416r2−−−−−−−−−√Simplifying, we get l=r12−−√The area would be 3√4l2, which would be
• Here, OC=r, BC=l2, AB=l. Since ABC∼BOC, taking ratios, we get AC=l24r.By the Pythagorean theorem, AB2=AC2+BC2,Therefore,l=l24+l416r2−−−−−−−−−√Simplifying, we get l=r12−−√The area would be 3√4l2, which would be33–√r2

Answer 2

Answer:

$inscribed \: circle \: has \: radius \: 2cm$

$by \\ nisha$