37. In the Rangoli competition, the students are asked to prepare a Rangoli in triangular shape. Dimensions of APQR are 28 cm, 26 cm and 18 cm. The winner team has decorated Rangoli with the flower petals in the ALMN as shown in the figure, where L, Mand N are the midpoints of sides PQ, PR and QR respectively. Find the area of ALMN. (110 = 3.16 [5]
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Answer:
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Solution :-
Side of ∆PQR are 28 cm, 26 cm and 18 cm .
So,
→ Semi - perimeter of ∆PQR = s = (18 + 26 + 28)/2 = 36
now, using heron's formula to find the area of ∆PQR,
→ Area (∆PQR) = √[s * (s - a) * (s - b) * (s - c)]
→ Area (∆PQR) = √[36 * (36 - 28) * (36 - 26) * (36 - 18)]
→ Area (∆PQR) = √[36 * 8 * 10 * 18]
→ Area (∆PQR) = √[2 * 8 * 18 * 18 * 10]
→ Area (∆PQR) = 4 * 18√10
→ Area (∆PQR) = 72 * 3.16 { given that, √10 = 3.16 }
→ Area (∆PQR) = 227.52 cm²
therefore,
→ ∆LMN area = (1/4) * Area (∆PQR) { since , ∆ is formed by joining the mid points. }
→ ∆LMN area = (1/4) * 227.52
→ ∆LMN area = 56.88 cm² (Ans.)
Hence, the area of ∆LMN is equal to 56.88 cm² .
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