Question

The base BC of an isosceles triangle is 28cm and AB=AC = 50cm and AD ⊥ BC, find the length of AD and area of ΔABC

Answer 1

Answer:

answer is 662 square cm

Step-by-step explanation:

The perpendicular from vertex divides the base into two equal parts.so BD=DC=14cm.now AD^2+BD^2=AB^2. SO from this we get AD=48cm

now put in the formula of area i.e.base *height*1/2

Answer 2

⇒ Given :

In ΔABC :

BC = 28cm

AB = AC = 50cm

AD ⊥ BC

⇒ To find :

Length of AD

Area of ΔABC

⇒ Solution :

In ΔADC:

∠ADC = 90°            (Right angled)

AD² = AB² - BD²       (Pythagoras Theorem)

AD² = 50² - 14²

AD² = 2500 - 196

AD² = 2304

AD = √2304

AD = 48

★ We took BD length as 14cm , because BC length is 28cm ★

✶ BD = BC/2 ✶

⇒ Area of ΔABC :

½ × b × h

½ × 28 × 48

672

★ Area of ΔABC = 672cm² ★