Skakil draws an equilateral triangle PQR. I draw three perpendiculars from appoint inside of that equilateral triangle on three sides, of which lengths are 10 cm, 12 cm and 8 cm. Let us write by calculating the area of triangle.
Answers
Solution :
ΔPQR is an equilateral triangle
let side be 'a' and o is the point of ΔPQR
let,
AO = 10 cm
OB = 12 cm
OE = 8 cm
Area of ΔPQR = ΔOQP + ΔOQR + ΔOPR
=> √3/2 a² = ½ x PQ x OA + ½ x QR x OC + ½ x PR x OB
=> √3/2 a² = ½ x a x 10 + ½ x a x 12 + ½ x a x 8
=> √3/2 a² = ½ (10a + 12a + 8a)
=> √3/2 a² = ½ x 30a
=> √3/2 a² = 15a
=> a = 15 x 4 x a/√3
=> a = 60/√3
area of ΔPQR
=> √3/4 a²
=> √3/2 x (60/√3)²
=> √3/2 x 3600/ √3 x √3
=> 900/√3
=> 519.6 cm²
★ QUESTION :
Skakil draws an Equilateral Triangle PQR. I draw 3 perpendiculars from appoint inside of that equilateral triangle on three sides, of which lengths are 10 cm , 12 cm and 8 cm. Let us write by calculating the Area of Triangle.
GIVEN :
- Here, It's given that – ∆ PQR is an Equilateral Triangle.
- The length are given as 10cm, 12cm and 8 cm respectively.
TO FIND :
- Area of the triangle (∆PQR) = ?
STEP-BY-STEP EXPLAINATION :
Let us consider here angles of the triangle, as follows :
⭐ ∠ AO = 10cm
⭐ ∠ OE = 8cm
⭐ ∠ OB = 12cm
━━━━━━━━━━━━━━━━━━━━━━━━━
Now, let us find here the area of triangle :
let us apply here is the formula for finding area of equilateral triangle (∆PQR)
➠ ∆PQR = ∆OQP + ∆OQR + ∆OPR
substituting the values as per given formula :
➠ √3/2 a^2 = 1/2 × PQ × OA + 1/2 × QR × OC + 1/2 × PR × OB
➠ √3/2 a^2 = 1/2 × a × 10 + 1/2 × a × 12 + 1/2 × a × 8
➠ √3/2 a^2 = 1/2 × 30a
➠ √3/2 a^2 = 15a
➠ a = 15 × 4 × a /√3
➠ a = 60 / √3
_____________________
➠ √3/4 a^2
➠ √3/2 × ( 60 / √3 ^2)
➠ √3/2 × 3,600 / √3 × √3
➠ √ 900/√3 (when solved) = 519.6 cm^2
━━━━━━━━━━━━━━━━━━━━━━━━━
Therefore, area of the triangle (∆PQR) = 519.6cm²