∆LMN ~ ∆PQR, 9 × A (∆PQR ) = 16 × A (∆LMN). If QR = 20 then Find MN.
Practice Set 1.4 | Q 4 | Page 25
∆LMN ~ ∆PQR, 9 × A (∆PQR ) = 16 × A (∆LMN). If QR = 20 then Find MN.
Solution
Given:
∆LMN ~ ∆PQR
9 × A (∆PQR ) = 16 × A (∆LMN)
Consider, 9 × A (∆PQR ) = 16 × A (∆LMN)
[frac{Aleft( ∆ LMN right)}{Aleft( ∆ PQR right)} = frac{9}{16}]
[ Rightarrow frac{{MN}^2}{{QR}^2} = frac{3^2}{4^2}]
[ Rightarrow frac{MN}{QR} = frac{3}{4}]
[Rightarrow MN = frac{3}{4} times QR]
[ Rightarrow MN = frac{3}{4} times 20 left[ because QR = 20 right]]
[ Rightarrow MN = 15]
[]
Answer:-
Given: ∆LMN ~ ∆PQR 9 × A (∆PQR ) = 16 × A (∆LMN) QR = 20
According to theorem of areas of similar triangles “When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides”.
Hence, A(∆ LMN) / A(∆ PQR) = (MN)^2 / (QR)^2
Given, 9 × A (∆PQR ) = 16 × A (∆LMN)
9 / 16 = A(∆ LMN) / A(∆ PQR)
Substituting the value, we get
(MN)^2 / (QR)^2 = 9 / 16
(MN / QR) = 3 / 4
Now, QR = 20
Substituting the value, we get
(MN / 20) = 3 / 4
MN = (3/4) * 20
MN = 15
Hence, the value of MN is 15.
Chapter 1. Similarity- Practice Set 1.4 – Page 25
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