मराठी

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

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

उत्तर १

In equilateral Δ ABC, AD ⊥ BC.
Therefore, BC = x cm.

Area of equilateral ΔABC = `sqrt3/4 xx "side"^2  = 1/2 xx "base" xx "height"`

= `sqrt3/4 xx x^2 = 1/2 xx x xx "AD"`

AD = `sqrt3/2 x`

shaalaa.com

उत्तर २

In △ADC and △ADB,

AD = AD    ...(Common)

∠ADB = ∠ADC ...(Each 90°)

AB = AC  ....(Given, ABC is an equilateral triangle)

Thus, △ADC ≅ △ADB

BD = DC = `1/2`​BC ...(By cpct)

Hence, BD = `1/2`​​x

In △ADB,

∠D = 90°

△ADB is right angle triangle,

by Pythagoras theorem,

AB2 = BD2 + AD2

`x^2 = (1/2​x)^2 + "AD"^2`

`"AD"^2 = x^2 − (x^2)/4`

`"AD"^2 = 3/4​x^2`

`"AD" = (sqrt(3))/2x`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 13: Pythagoras Theorem [Proof and Simple Applications with Converse] - Exercise 13 (B) [पृष्ठ १६३]

APPEARS IN

सेलिना Concise Mathematics [English] Class 9 ICSE
पाठ 13 Pythagoras Theorem [Proof and Simple Applications with Converse]
Exercise 13 (B) | Q 2 | पृष्ठ १६३

संबंधित प्रश्‍न

The diagonal of a rectangular field is 16 metres more than the shorter side. If the longer side is 14 metres more than the shorter side, then find the lengths of the sides of the field.


ABC is a right triangle right-angled at C. Let BC = a, CA = b, AB = c and let p be the length of perpendicular from C on AB, prove that

(i) cp = ab

`(ii) 1/p^2=1/a^2+1/b^2`


ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2 


In the following figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that

(i) OA2 + OB2 + OC2 − OD2 − OE2 − OF2 = AF2 + BD2 + CE2

(ii) AF2 + BD2 + CE= AE2 + CD2 + BF2


The given figure shows a quadrilateral ABCD in which AD = 13 cm, DC = 12 cm, BC = 3 cm and ∠ABD = ∠BCD = 90o. Calculate the length of AB.


If P and Q are the points on side CA and CB respectively of ΔABC, right angled at C, prove that (AQ2 + BP2 ) = (AB2 + PQ2)


Show that the triangle ABC is a right-angled triangle; if: AB = 9 cm, BC = 40 cm and AC = 41 cm


In the figure below, find the value of 'x'.


Calculate the area of a right-angled triangle whose hypotenuse is 65cm and one side is 16cm.


From a point O in the interior of aΔABC, perpendicular OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that: AF2 + BD2 + CE2 = AE2 + CD2 + BF2


In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AC2 = AD2 + BC x DE + `(1)/(4)"BC"^2`


Find the unknown side in the following triangles


In triangle ABC, line I, is a perpendicular bisector of BC.
If BC = 12 cm, SM = 8 cm, find CS


Rithika buys an LED TV which has a 25 inches screen. If its height is 7 inches, how wide is the screen? Her TV cabinet is 20 inches wide. Will the TV fit into the cabinet? Give reason


Sides AB and BE of a right triangle, right-angled at B are of lengths 16 cm and 8 cm respectively. The length of the side of largest square FDGB that can be inscribed in the triangle ABE is ______.


In figure, PQR is a right triangle right angled at Q and QS ⊥ PR. If PQ = 6 cm and PS = 4 cm, find QS, RS and QR.


In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:

(i) `"AC"^2 = "AD"^2 + "BC"."DM" + (("BC")/2)^2`

(ii) `"AB"^2 = "AD"^2 - "BC"."DM" + (("BC")/2)^2`

(iii) `"AC"^2 + "AB"^2 = 2"AD"^2 + 1/2"BC"^2`


Lengths of sides of a triangle are 3 cm, 4 cm and 5 cm. The triangle is ______.


Points A and B are on the opposite edges of a pond as shown in the following figure. To find the distance between the two points, the surveyor makes a right-angled triangle as shown. Find the distance AB.


The foot of a ladder is 6 m away from its wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its top reach?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×