In equilateral △ ABC, AD ⊥ BC and BC = x cm. Find, in terms of x, the length of AD.

In △ ABD and △ ACD, ⇒ ∠ADB = ∠ADC (Both equal to 90°) ⇒ AD = AD (Common side) ⇒ AB = AC (Since, ABC is an equilateral triangle) ∴ △ ABD ≅ △ ACD (By S.A.S. axiom) We know that, Corresponding parts of congruent triangle are equal. ∴ BD = CD = cm. In right-angled triangle ABD, By pythagoras theorem, ⇒ (Hypotenuse) 2 = (Perpendicular) 2 + (Base) 2 ⇒ AB 2 = AD 2 + BD 2 ⇒ x 2 = AD 2 + ⇒ AD 2 = x 2 - ⇒ AD 2 = ⇒ AD 2 = ⇒ AD 2 = ⇒ AD = cm. Hence, AD = cm.

ABC is an isosceles triangle right-angled at C. Then 2AC² is equal to :
BC²
AC²
AC² - BC²
AB²

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In the figure, given below, AD ⊥ BC. Prove that :
c² = a² + b² - 2ax.

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ABC is a triangle, right-angled at B. M is a point on BC. Prove that :
AM² + BC² = AC² + BM².

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M and N are the mid-points of the sides QR and PQ respectively of a △ PQR, right-angled at Q. Prove that :
(i) PM² + RN² = 5 MN²
(ii) 4 PM² = 4 PQ² + QR²
(iii) 4 RN² = PQ² + 4 QR²
(iv) 4 (PM² + RN²) = 5 PR²

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Mathematics

In equilateral △ ABC, AD ⊥ BC and BC = x cm. Find, in terms of x, the length of AD.

Pythagoras Theorem

Related Questions

ABC is an isosceles triangle right-angled at C. Then 2AC² is equal to :
BC²
AC²
AC² - BC²
AB²

In the figure, given below, AD ⊥ BC. Prove that :
c² = a² + b² - 2ax.

ABC is a triangle, right-angled at B. M is a point on BC. Prove that :
AM² + BC² = AC² + BM².

M and N are the mid-points of the sides QR and PQ respectively of a △ PQR, right-angled at Q. Prove that :
(i) PM² + RN² = 5 MN²
(ii) 4 PM² = 4 PQ² + QR²
(iii) 4 RN² = PQ² + 4 QR²
(iv) 4 (PM² + RN²) = 5 PR²

Mathematics

In equilateral △ ABC, AD ⊥ BC and BC = x cm. Find, in terms of x, the length of AD.

Pythagoras Theorem

Related Questions

ABC is an isosceles triangle right-angled at C. Then 2AC2 is equal to :BC2AC2AC2 - BC2AB2

In the figure, given below, AD ⊥ BC. Prove that :c2 = a2 + b2 - 2ax.

ABC is a triangle, right-angled at B. M is a point on BC. Prove that :AM2 + BC2 = AC2 + BM2.

M and N are the mid-points of the sides QR and PQ respectively of a △ PQR, right-angled at Q. Prove that :(i) PM2 + RN2 = 5 MN2(ii) 4 PM2 = 4 PQ2 + QR2(iii) 4 RN2 = PQ2 + 4 QR2(iv) 4 (PM2 + RN2) = 5 PR2

ABC is an isosceles triangle right-angled at C. Then 2AC² is equal to :
BC²
AC²
AC² - BC²
AB²

In the figure, given below, AD ⊥ BC. Prove that :
c² = a² + b² - 2ax.

ABC is a triangle, right-angled at B. M is a point on BC. Prove that :
AM² + BC² = AC² + BM².

M and N are the mid-points of the sides QR and PQ respectively of a △ PQR, right-angled at Q. Prove that :
(i) PM² + RN² = 5 MN²
(ii) 4 PM² = 4 PQ² + QR²
(iii) 4 RN² = PQ² + 4 QR²
(iv) 4 (PM² + RN²) = 5 PR²